Inter 1st Year

Students get through AP Inter 1st Year Chemistry Important Questions 13th Lesson Organic Chemistry-Some Basic Principles and Techniques which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 13th Lesson Organic Chemistry-Some Basic Principles and Techniques

Very Short Answer Questions

Question 1.
Write the reagents required for conversation of benzene to methyl benzene.
Answer:

Therefore, the required reagents are methyl chloride and anhydrous AlCl₃.
This reaction is known as Friedel Craft’s alkylation.

Question 2.
How is nitrobenzene prepared? [AP 15]
Answer:
Benzene on heating with the mixture of cone. HNO₃ & cone. H₂SO₄ at 60°C forms Nitro benzene.

Question 3.
Write the conformations of ethane.
Answer:
Conformational isomers are due to rotation along C-C single bond in alkanes.

Conformations of ethane:
Free rotation about C-C single bond in ethane gives an infinite number of confotmers. Out of these. Eclipsed and Staggered conformations are most significant.

Question 4.
How do you prepare ethyl chloride from ethylene?
Answer:
Ethylene combines with hydrogen chloride to form ethyl chloride.
CH₂ = CH₂2 + HCl → CH₃ – CH₂Cl

Question 5.
Write HJPAC names of: [AP, TS 16, 18][Mar’ 13]

Answer:
a) 1 – Pentene
b) 2 -Pentanone, 3- Pentanone, 1,3-Butadiene
c) 3-nitro benzene carbaldehyde, 4-nitro benzene carbaldehyde
d) 4,4 Dimethyl pentanoic acid

Question 6.
Write the structures of Trichloroethanoic acid, Neopentane, p-nitrobenzaldehyde. [TS 22]
Answer:

Question 7.
Discuss Lassaigne’s test.
Answer:
Lassaigne’s test is used to detect nitrogen, sulphur, halogens and phosphorous in a given organic compound.

In Lassaigne’s test, the organic compound is mixed with a sodium metal and fused strongly.

If Nitrogen is present in the organic compound, it reacts with sodium & carbon resulting sodium cyanide.

If sulphur is present in the organic compound, it reacts with sodium resulting sodium sulphide.

It halogens are present in the organic compound, they result sodium halides.

Question 8.
Explain the principle of Chromatography. [Imp.Q]
Answer:
Chromatography is an important technique extensively used to separate mixtures into their components. Here, purification is based on adsorption (or) partition principle.

The mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent or a mixture of solvents is allowed to move slowly over the stationary phase.

The components of the mixture get gradually separated from one another.

Question 9.
Explain why an organic liquid vaporizes at a temperature below its boiling point in its steam distillation?
Answer:
In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid (P₁) and that due to water (P₂) becomes equal to the atmospheric pressure (P), i.e.,P = P₁ + P₂. Hence even though P₁ is lower than P, the organic liquid vaporizes at lower temperature than its boiling point.

Question 10.
Explain the following (a) Crystallisation (h) Distillation
Answer:
a) Crystallisation :
This technique is used for the purification of solid organic compounds.

This method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.

The impure compound is dissovled in a solvent, in which it is partially soluble at room temperature and appreciably soluble at higher temperature.

The solution is heated to get a saturated solution further the solution is cooled to get pure organic compound crystals by filtration.

On repeating this process finally we get very pure compound.

b) Distillation :
Distillation is the process of converting a liquid into vapour on heating and then cooling the vapour back to liquid state.

This method is used to separate

1. Volatile liquids from non-volatile impurities and
2. The liquids having a sufficient difference in their boiling points (>40°C),

Liquids having different boiling points vapourises at different temperatures.

These vapours are cooled and the liquids formed are collected separately.

Short Answer Questions

Question 1.
Write the corresponding equations for the following reactions and name the products A, B, C. [IPE’ 14][TS 15,16][AP I6]

Answer:

Question 2.
Name the products A, B, C formed in the following reactions, (live equations for the reactions. [Mar’09][TS 15]

Answer:

Question 3.
How does acetylene react with: (a) Bromine (b) Hydrogen? Write tile balanced equations for the above reactions. Name the products. [AP 16]
Answer:
a) Acetylene reacts with bromine in the presence of CCl₄ and forms an addition compound 1, 1,2, 2-Tetra bromoethane.

b) Hydrogen reacts with acetylene in the presence of Ni or Pt catalyst and forms ethane.

Question 4.
What is substitution reaction? Explain any three substitution reactions of Benzene. [May’ 10]
Answer:
Substitution reaction :
The reactions in which an atom or group present in a compound is replaced by another atom or group are known as substitution reactions.

The characterisitc reaction of benzene is Electrophilic substitution reaction.

Substitution reactions of benzene :
a) Benzene reacts with bromine or chlorine in the presence of AlCl₃ and gives corresponding halo benzene, ie., Halogenation.

b) Benzene on heating with nitration mixture, below 60°C gives Nitrobenzene, ie., Nitration.

c) Benzene reacts with methyl chloride in the presence of anhydrous AlCl₃ and gives Toluene. This reaction is known as Friedel crafts alkylation.

Question 5.
What is Dehydrohalogenation? Give the corresponding equation for the formation of alkene from alkyl halide.
Answer:
Dehydrohalogenation:
Removal of one hydrogen atom and one halogen atom from adjacent carbon atoms of a compound is called dehydrohalogenation.

Alkene from alkyl halide:
Ethyl Bromide on heating with ale. KOH undergoes dehydrohalogenation and gives ethylene.

Question 6.
Which type of compounds react with Ozone? Explain with one example. [IPE’ 14]
Answer:
Unsaturated hydrocarbons usally react with ozone.

Ozone reacts with unsaturated hydrocarbons to form addition compounds called ozonides.

On hydrolysis the ozonides give carbonyl compounds. This process is called ozonolysis. Ozonolysis is used for the location of the double bond or triple bond in unsaturated compounds like alkene, alkyne and benzene.
Ex: Ethylene undergoes addition reaction with ozone and forms ethylene ozonide.
It on hydrolysis in the presence of Zn dust gives formaldehyde and H₂O₂.

Question 7.
Give two examples each for position and functional isomerism. [TS 22][AP,TS 16]
Answer:
Position Isomerism is due to difference in the position of a substituent or functional group.

Functional isomerism is due to difference in the functional group in the molecules.

Question 8.
Explain the mechanism of halogenations of methane.
Answer:
Halogenation:
Alkanes exhibit free radical substitution reactions.
Ex: Halogenation of methane: In the presence of sunlight or U.V.light, methane reacts with chlorine and undergoes substitution with chlorine atoms successively forming mono, di, tri and tetra chloro methane respectively.

Mechanism:
Halogenation of alkanes takes place by free radical mechanism. It involves three steps.
(i) Chain initiation:
In the presence of heat or light, chlorine molecule undergoes homolysis to form chlorine free radicals.

(ii) Chain propagation:
Chlorine free radical attacks methane molecule, breaking one of the C-H bonds and generating methyl free radical with the formation of HCl.

The methyl free radical thus obtained attacks second chlorine molecule to form CH3C/
with the liberation of another chlorine free radical.

Steps (a) and (b) repeat several times making the reaction a chain reaction. Many other propagation steps are possible and may occur, there by forming other chlorine substituted products.

(iii) Chain termination:
When two free radicals directly combine, the chain reaction terminates

Question 9.
How is ethylene prepared from ethyl alcohol? Write the reaction.
Answer:
Methods of preparation of Ethylene:
1) Dehydration of ethyl alcohol Ethyl alcohol on heating with con. H₂SO₄ at 170°C gives ethylene.

2) Dehydrohalogenation of ethyl halides:
Ethyl halides on heating with alcoholic NaOH or KOH gives ethylene.

Question 10.
Explain the reactions of acetylene with (a) Na in NH₃ (b) chromic acid:
Write the equations and name the products.
Answer:
(a) Action with Na in NH₃:
Acetylene combines with sodium in liquid NH₃ and gives mono sodium and disodium acetylides.

Question 11.
Explain crystallization and sublimation phenomena which are used in the puritlcation of organic compounds:
Answer:
a) Crystallisation:
This technique is used for the purification of solid organic compounds.

This method is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.

The impure compound is dissovled in a solvent, in which it is partially soluble at room temperature and appreciably soluble at higher temperature.

The solution is heated to get a saturated solution further the solution is cooled to get pure organic compound crystals by filtration.

On repeating this process finally we get very pure compound.

b) Sublimation:
Sublimation is a purification method of solids.Some solid substances on heating directly pass into vapour state without melting. Those vapours on cooling form directly solid without condensing to liquid. This phenomenon is called sublimation.
Solid ⇌ Vapour
The compound to be purified is taken in a beaker covered with a watch glass and heated. The compound sublimes and solidifies on the lower surface of the watch glass. Impurities remain in the beaker. The compound is separated by scratching the watch glass.

Question 12.
Describe solvent extraction method to purify a compound.
Answer:
Solvent extraction:
It is used to separate an organic compound present in aqueous medium by using a solvent which is immisible with water and in which organic substance is more soluble. The aqueous solution of impure organic compound is shaken with a suitable solvent.

Organic compound goes into the organic solvent which is immisible with water.

The organic layer is separated and distilled to remove the liquid solvent.
The compound remains in distillation flask.

Question 13.
Explain the estimation of phosphorous and sulphur in the given organic compounds.
Answer:
Estimation of sulphur:
A known mass of the organic compound is heated with sodium peroxide in a Carius tube. Then ‘S’ is oxidised to H₂SO₄. The acid is precipitated as BaSO₄ by adding excess of BaCl₂ aq.solution. The ppt. is filtered, washed, dried and weighed.

Calculations:
Mass of organic compound = a g
Mass of barium sulphate = b g
Molecular weight of BaSO₄ = 23
1 mole of BaSO₄ or 233 g of BasO₄ contains 32 g of sulphar.

Estimation of phosphorous:
A known mass of organic compound is heated with fuming nitric acid in a Carius tube. Then phosphorus is oxidised to phosphoric acid. The acid is precipitated as ammonium phosphomolybdate by adding ammonia and ammonium molybdate solutions.

Calculations:
Mass of organic compound = a g
Mass of barium sulphate = b g
Molecular weight of ammonium phosphomolybdate = 1877
1877 g of ammonium phosphomolybdate contains 31 g of phosphorus

Question 14.
Explain addition of HBr to Propene with the ionic mechanism.
Answer:
Electrophilic addition of HBr to CH₃-CH=CH₂ molecule:
Addition of HBr to unsymmetrical alkene like propene generally follows Markownikoffs rule. According to this rule, the negative part of the reagent adds to the double bonded carbons bearing less number of hydrogens.

Mechanism:
1) HBr ionises to give H⁺ and Br (HBr → H⁺ + Br^–)
2) Electrophile (H) attacks the double bond to form carbonium ion.

3) Nucleophile (Br^–) attacks the carbonium ion to form 2-bromo propane as major product.

Question 15.
What is the product formed when sodium propionoate is heated with soda lime?
Answer:
When sodium propionoate is heated with soda lime, Ethane gas is obtained

This reaction is known as Decarboxylation. In these reactions the number of carbon atoms in the product is less than that of reactants ie., one carbon is less.

Long Answer Questions

Question 1.
Explain the classification of hydrocarbons.
Answer:
Hydrocarbons are classified as open chain(acyclic/aliphatic) hydrocarbons and closed chain hydrocarbons. Aliphatic hydrocarbons are again classified as open chain hydrocarbons and closed-chain hydrocarbons i.e., Cyclo hydrocarbons.

Both open chain and closed chain hydrocarbons are again classified as hydrocarbons containing
i) C-C (alkanes and cycloalkanes)
ii) > C = C < (alkenes and cycloalkenes)
iii) – C≡C – (alkynes and cycloalkynes)

Question 2.
Write IUPAC names of the following compounds: [AP 26]

Answer:
(a) 1, 3-butadiene
(b) Pent-l-en-3-yne
(c) 2-methy 1-2-butene
(d) 4-pheny 1-1 -butene
(e) 4- Ethyl Deca-l,5,8-triene

Question 3.
a) Describe the methods of preparation of Ethane. [AP 19]
b) Explain the chemical properties of Ethane with equations. [TS 22]
Answer:
a) Preparation of ETHANE (C₂H₆) :
1) Decarboxylation:
Ethane is prepared by heating sodium propionate with sodalime (Sodalime is a mixture of NaOH & CaO)

2) Wurtz reaction :
Ethane is prepared by heating methyl iodide with sodium metal in the presence of dry ether. [AP 17][TS 15]

3) Kolbe’s electrolysis:
Ethane is obtained by the electrolysis of aqueous potassium acetate (or aqueous sodium acetate ). [TS 15]

4) Sabatier – Senderen’s reduction reaction :
On a large scale, ethane is prepared by the catalytic hydrogenation of ethylene.

5) Reduction of ethyl iodide :
Ethane is prepared by the reduction of ethyl iodide in the presence of Zn and HCl.

b) Chemical Properties of Ethane:
1) Action with Chlorine (Chlorination):
Ethane reacts with chlorine in the presence of sunlight or UV light and gives ethyl chloride.

2) Action with Nitric acid(Nitration):
Ethane reacts with nitric acid vapours at 400°C and gives nitroethane.

3) Action with Oxygen (Oxidation):
Ethane reacts with oxygen on combustion forms CO₂ and H₂O

Question 4.
Write the structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated.
(a) C₄H₈ (one double bond)
(b) C₅H₈ (one triple bond)
(c) C₅H₁₂ (no multiple nonds)
Answer:

Question 5.
Write chemical equations for combustion reaction of the following hydrocarbons,
a) Butane b) Pentene c) Hexyne
Answer:

Question 6.
Addition HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Answer:
Addition of Hydrogen halides:
Addition of HX to a double bond generally follows Markownik off rule.

Statement of Markownikoff rule:
The rule states that addition of hydrogen halides to unsymmetrical alkenes takes place in such a manner that the positive part of the reagent attach itself to that carbon atom which has more number of hydrogen atoms.

Explanation:
It follows electrophilic addition mechanism.

Since 2° carbocation is more stable than l° carbocation, the major product formed is 2-Bromopropane.
Note: Markownikoff s rule is only applicable for unsymmetrical alkenes.

Anti Markovnikov’s rule (or) Kharasch effect:
In the presence of peroxide (R- O-O-R) the addition of hydrogen halides (HBr) to an unsymmetrical alkene like propene takesplace in such a way that the negative part of the reagent attach itself to that carbon atom which has more number of hydrogen atoms.

Stability of free radicals follows as 3°>2°>1°
Therefore, 1-bromopropane is major product.

Uses of Ethylene:

1. It is used in the manufacture of ethyl alcohol, polythene etc.,
2. In the preparation of ethylene glycol as antifreeze
3. In the preparation of mustard gas.
   

Question 7.
Describe two methods of preparation of ethylene. Give equation for the reactions of ethylene with the following. [TS 18][AP 16,19]
a) Ozone
b) llypohalous acid
c) Cold and dil.alk. KMnO₄
d) Heated with O₂ at high pressure
Answer:
a) preparation of Ethylene (C₂H₆):
1) Dehydration:
Ethylene is prepared by heating ethyl alcohol with cone, sulphuric acid at 170°C.

2) Dehydrohalogenation:
Ethylene is prepared by heating ethyl chloride with alcoholic potassium hydroxide.

Reactions of Ethylene:
a) Action with Ozone (Ozonolysis):
Ethylene reacts with ozone to form unstable ozonide. This undergoes hydrolysis in presence of Zinc and gives Formaldehyde.

b) Action with hypohaloos acid:
Ethylene reacts with hypochlorous acid and gives Ethylene chlorohydrin.

c) Action with cold dil. alk. KMnO₄:
Ethylene reacts with cold dil .alk.KMnO₄ (Baeyer’s reagent) at 273K and gives Ethylene glycol.

In this reaction, Baeyer’s reagent loses its prime colour. This test is used to detect carbon- carbon double bond or triple bond and is known as Baeyer’s test.

d) Action with O₂ (Polymerisation):
Ethylene when heated with 02 at high pressure gives Polythene. [AP 17]

Question 8.
How does ethylene react with the following reagents? Give the chemical equations and names of the products formed in the reactions. [TS 16]
a) Hydrogen.halide
b) Hydrogen
c) Bromine
d) Water
e) Oxygen in presence of Ag at 200°C
f) H₂SO₄
Answer:
a) Ethylene undergoes addition with hydrogen halides forming ethyl halides.

b) Ethylene reacts with hydrogen in presence of Pt, Pd or Ni catalyst to form ethane.

c) Ethylene reacts with bromine in presence of CCl₄ to form 1, 2- dibromoethane.

d) When ethylene gas is passed through dil.H₂SO₄, it undergoes addition with water forming ethyl alcohol.

e) Ethylene on heating in oxygen in presence of Ag at 200°C give ethylene oxide.

f) Ethylene reacts with H₂SO₄ to form ethylene hydrogen sulphate
CH₂ = CH₂ + H₂SO₄ → CH₃.CH₂.HSO₄

Question 9.
An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pcntan3-onc. Write the reaction, structure of the products and alkene-A. Give the IUPAO name of alkene A.
Answer:

Question 10.
An alkene ‘A’ contains three C-C, eight C-H bonds and once C=C bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44. Write IUPAC name of ‘A’.
Answer:
The aldehyde with molar mass 44 is CH₃CHO(ethanal).

Question 11.
Give two methods of preparation of acetylene. How does it react with water and ozone? [AP 20][TS 18,20]
Answer:
Preparation of Acetylene (C₂H₂):
1) Dehydro halogenation :
Acetylene is prepared by heating dibromoethane with alcoholic KOH. This undergoes dehydrohalogenation and gives acetylene.

2) From iodoform:
Acetylene is prepared by heating Iodoform with silver powder.

Properties of Acetylene:
1) Action with water:
Acetylene gas when passed through dil.H₂SO₄ below 60°C in the presence of H₂SO₄, undergoes addition reaction with water and forms acetaldehyde. [ AP 17]

2) Action with ozone:
Acetylene reacts with ozone to form acetylene ozonide. This on hydrolysis in the presence of Zn forms glyoxal.

Question 12.
How does acetylene react with the following reagents? Give the corresponding equations and name the products formed in the reactions. [TS 16,18]
a) Acetic acid
b) Water
c) Hydrogen
d) Halogens
e) Hydrogen halide
f) Ammonial AgNO₃ and Cu₂Cl₂.
Answer:
a) Action ith CH₃COOH:
Acetylene reacts with acetic acid in presence of Hg⁺² ions as catalyst, forming at first vinyl acetate and finally ethylidene diacetate.

b) Action with water:
Acetylene gas when passed through dil.H₂SO₄ below 60°C in the presence of H₂SO₄, undergoes addition reaction with water and forms acetaldehyde.

c) Action with Hydrogen:
Acetylene reacts with H₂ in presence of Ni or Pt catalyst and undergoes addition reaction forming ethylene and ethane.

d) Action with Halogens:
Acetylene reacts with halogens in presence of CCl₄ and undergoes addition reaction giving finally 1, 1,2, 2 -tetrachloroethane.

e) Action with Hydrogen halides:
Acetylene undergoes addition with hydrogen halides and gives finally 1,1 -dichloroethane. [TS 20]

f) Action with ammoniacal AgNO₃ solution:
when acetylene gas is passed through ammoniacal AgNO₃ solution, a white ppt. of silver acetylide is formed.

g) Action with ammonical Cu₂Cl₂ solution:
When acetylene gas is passed through ammoniacal Cu₂Cl₂ solution, a red ppt.of cuprous acetylide is formed.

Question 13.
Describe any two methods of preparation of benzene with corresponding equations. Benzene does not behave like an alkenes, why? How do we get methyl benzene from benzene? [AP 19]
Answer:
Preparation of Benzene:
Laboratory Method:
Sodium benzoate on distillation with soda lime gives benzene.

A mixture of NaOEl and CaO is called soda lime.

2) Polymersiation of acetylene:
Acetylene when passed through red hot Cu, polymerises and gives benzene.

Reason for not behaving as an alkenc:
Benzene has a number of resonance structures.

Because of resonance, benzene has more stability . The resonance energy of benzene is also very high. It also indicates more stability to the benzene molecule. Hence, stability of a molecule is due to saturation. Hence, from these two points it is evident that benzene exhibits more the properties of a saturated compound (alkane) rather than the properties of an unsaturated compound (alkene).

Methyl benzene from benzene :
In the presence of anhydrous AlCl₃, benzene reacts with methyl chloride and forms methyl benzene (Toluene).

Question 14.
How do we get benzene from acetylene? Give the corresponding equation. Explain the halogenation, alkylation, acylation, nitration and sulphonation of benzene. [AP 15][TS 19][AP 18]
Answer:
Preparation of benzene from acetylene:
Acetylene when passed through red hot Cu, polymerises and gives benzene.

1) Halogenation :
Benzene reacts with chlorine in the presence of Anhy. AlCl₃ and forms. [AP 18][IPE’ 14]

2) Friedel Craft’s alkylation :
Benzene reacts with methyl chloride in the presence of anhy. AlCl₃ and forms methyl benzene. [AP 18,22][IPE 13,14]

3) Friedel – Craft’s acylation:
Benzene reacts with acetyl chloride in the presence of anhy. AlCl₃ and forms acetophenone.

4) Nitration :
Benzene reacts with Nitric acid in the presence of cone. H₂SO₄ at 60°C and forms nitro benzene. [May’13][AP 17,22]

5) Sulphonation:
Benzene reacts with filming sulphuric acid and gives benzene sulphonic acid.

Question 15.
Explain the differences between structural isomers and stereo isomers.
Answer:

Structural Isomers	Stereo Isomers
1) The compounds having same moiecuiar formula but differ in their structural arrangements of atoms or group of atoms	1) The compounds having same moiecuiar formula but differ in their spatial arrangement of atoms or group of atoms.
2) Spatial arrangement is not considered.	2) Structure of the compound is not changed.
3) Strucutral isomers possess different physical properties. Ex: With increase in number of branches in alkanes with same molecular fomrula, boiling points are decreased.	3) Stereo isomers may or may not differ in their physical properties. Ex: a)Diastereomers and Cis-trans isomers differ in their boiling points and melting points. b) Enantiomers have same physical properties like melting & boiling points.
4) Chain isomers, positional isomers, functional isomers, metamers, keto-enol tautomers and ring isomers are called structural isomers.	4) Configurational and conformational isomers are called stereoisomers.

Question 16.
What is the difference between conformation and configuration in open chain molecules.
Answer:
Stereo isomers are two types known as configurational and conformational isomers.

Conformation isomers:
These are the stereo isomers that are easily converted into one another by the rotation around ‘C-C’ (Sigma ) bonds. These are in dynamic equilibrium with one another. This type of isomerism is seen in alkanes. Ex: n-butane.

Configuration Isomers:
They cannot be interconverted into one another with out making and breaking of new bonds. Configurational stereo isomers possess certain types of rigidity in the molecules which are discrete, stable.

Question 17.
What do you understand about geometrical isomerism? Explain the geometrical isomers of 2-butene. Draw the cis and trans isomers of (i) CHCl=CHCl (ii) C₂H₅CCH₃=CCH₃C₂H₅ [AP 22]
Answer:
Geometrical isomerism arises due to restricted rotation of atoms or groups along C=C bond.
Ex: Geometrical isomers of 2-butene are as follows.

The isomer in which same groups or atoms are on the same side of the C=C is called cis-isomer while the isomer having same groups or atoms on the opposite sides of the C=C is called trans¬isomer. In cis-2-Butene the two methyl groups are on one side while the two hydrogen atoms are on the other side. In trans-2-butene two methyl groups and two hydrogen atoms are present opposite to the double bond. This isomerism is due to restricted rotation around C=C bond.

Question 18.
Explain the method of writing E – Z configurations for geometrical isomers taking CHCl = CFBr as an example.
Answer:
E – Z configurations for geometrical isomers – Steps :
i) Arrange the atoms/groups attached to each doubly bonded carbon in the order of their atomic numbers.

ii) Choose the atom/group of higher priority on each doubly bonded carbon. If the atoms/groups of higher priority on each carbon are on the same side of the molecule, the letter ‘Z’ is used to denote the configuration of such isomer. When the atoms/groups of higher priority on each carbon are on the opposite sides of the molecule, the letter ‘E’ is prefixed before the name to indicate configuration.
Ex : CHCl = CFBr
Among H and Cl, ₁₇Cl gets more priority than ₁H.
Among F and Br, ₃₅Br gets more priority than ₉F.

Question 19.
If an alkene contains on carbons at double bond Cl.Br-CH₂-CH₂-OH and -CH(CH₃)₂. Write the E and Z configurations of it.
Answer:

Question 20.
Write a note on : (a) Distillation (b) Fractional distillation
(c) Distillation under reduced pressure (d) Steam distillation
Answer:
a) Distillation :

1. This process is useful for the purification of liquids contaminated with nonvolatile impurities.
2. The impure liquid is boiled in a distillation flask and the vapours are condensed and collected in a receiver.
3. This method can also be used to separate liquids if only their boiling points differ by above 40°C.

b) Fractional distillation :

1. It is used to separate a liquid mixture which contains the components that have boiling point difference less than 40°C.
2. Let a mixture of A and B is taken in a distillation flask which is fitted with a fractionating column at its mouth.
3. The upper end of the column is connected to the water condenser.
4. When the mixture is heated, the vapours of both A and B pass through the fractionating column.
5. While moving through column, the vapours of liquid which has high boiling point (sayA) condenses and come back to the flask.
6. The vapours of other liquid (say B) which has low boiling point still in vapour state and move out of column as pure vapours of B to the condenser.
7. Vapours of B condense and get collected in the receiver.

c) Distillation under reduced pressure :

1. This method is useful to purity liquids that have very high boiling points and those which decompose at or below their boiling points.
2. If the external pressure is reduced the liquid boils at lower temperature than its normal boiling point without decomposition.
3. At reduced pressure the boiling point of a liquid is also reduced, hence its decomposition is avoided
   Ex: Glycerol and H₂O₂ are purified by this method.

d) Steam distillation :

1. This method is used to purify the liquids which are insoluble in water, possess high vapour pressure and steam volatile.
2. In this method steam is passed into the hot liquid mixture taken in the distillation flask.
3. The liquid with low boiling point comes out along the steam by leaving, behind the liquid which has comparatively high boiling point.
4. They are passed through the condenser and condensed and finally collected in the receiver.
5. The water layer and the organic liquid layer are separated using a separating funnel. Aniline can be purified by this method.

Question 21.
Write a brief note on Chromatography.
Answer:
Chromatography is developed as a method of separating components of a mixture based on adsorption, generally between the stationary phase and a mobile phase.

Chromatography involves the three steps given below :
a) Adsorption and retention of a mixture of substances on the stationary phase and separation of adsorbed substances by the mobile phase to different distances on the stationary phase.
b) Recovery of the substances separated by a continuous flow of the mobile phase (known as elution)
c) Qualitative and quantative analysis of the eluted substances.

Classification :
Two general chromatography techniques are
i) adsorption chromatography
ii) partition chromatography.

Question 22.
Explain the following:
(a) Column chromatography
(b) Thin layer chromatography
(c) Partition chromatography
Answer:
a) Column chromatography :

1. In the column chromatography ,the components of a mixture are separated by a column of adsorbent packed in a glass tube.
2. The column is fitted with a stopcock at its lower end.
3. The mixture to be adsorbed on the adsorbent is placed at the top of the starionary phase.
4. A suitable eluant, either a single solvent or a mixture of solvents is allowed to flow down the column slowly.
5. Depending on the degree to which the compounds are adsorbed the components are separated.
6. The most readily absorbed substances are retained near the top and other come down accordingly to various distances.

b) Thin layer chromatography (TLC):

1. This also involves adsorption differences. Here the adsorbent, say silica gel or alumina is coated over a glass plate of suitable size in thin layer.
2. The plate is called TLC plate or chromoplate.
3. The solution of the mixture to be separated is applied as a small spot at about 2 cm from the bottom of the plate.
4. The plate is then kept in a closed jar containing the eluant,
5. As the eluant rises up the plate, the components of the mixture move up along with the eluant to various distances depending on their degree of adsorption.
6. The relative adsorption of a component of the mixture is expressed in terms of its Retardation Factor (R_f) value.
   

Partition chromatography :

1. This principle is similar to thin layer chromatography, except that a strip of paper acts as adsorbent.
2. It is based on continuous differential separation of components of a mixture between the stationary phase and the mobile phase.
3. In paper chromatography, for example, a special paper called chromatography paper contains water trapped in it which acts as the stationary phase.
4. The chromatography paper spotted with the solution of the mixture at the base is suspended in a suitable solvent or a mixture of solvents.
5. The solvent rises up the paper by capillary action and moves over the spot.
6. The paper selectively retains different components as per their differing partition in mobile and stationaiy phase.
7. The paper strip so developed is known as chromatogram.
8. The spots of the separated coloured compounds are detected and for colourless compounds other methods like spraying suitable reagent are used.

Question 23.
Explain the estimation of nitrogen of an organic compound by
a) Dumas method
b) KjeldahPs method
Answer:
a) Estimation of nitrogen by Duma’s method:
In this method a known weight of organic compound is heated strongly with coarse cupric oxide. Then carbon and hydrogen get oxidised to CO₂ and H₂O vapour respectively. Nitrogen is converted to N₂ gas. Some nitrogen converted into its oxides, gets reduced by copper gauze to nitrogen. The liberated gases are passed over a solution of KOH. Then CO₂ gets absorbed. Nitrogen is collected over KOH solution and its volume is found out.

Calculations:
Suppose ‘a’ g of nitrogen compound gives V₁ ml of N₂ gas at room temperature T K. If atmospheric pressure is P and aqueous tension at TK is P₀, then the pressure of N₂ gas at TK is P₁ = (P – P₀).

b) Estimation of N₂ by Kjeldahl’s method:
In this method the organic compound is heated with conc. H₂SO₄ in presence of a small amount of CuSO₄. Then N₂ is quantitatively converted into ammonium sulphate. The contents of the flask are transferred into another flask and heated with excess of NaOH solution to liberate ammonia gas. The gas so liberated is passed and absorbed in a known vol. of known cone. H₂SO₄(excess). Now the excess of acid remained after the neutralisation by NH₃ is titrated against a standard solution of alkali. From this, the amount of H₂SO₄ used to neutralise NH₃ is calculated. From this the mass of ammonia formed is calculated and from that the % of N₂ gas is calculated.
Organic compound + H₂SO₄ → (NH₄)₂SO₄
(NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O + 2NH₃
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Calculation:

a = Weight of organic compound = a g
M = Molarity of H₂SO₄ solution
V = Volume of H₂SO₄ solution
V₁ = Volume of H₂SO₄ solution consumed by NaOH.

Question 24.
Explain Inductive effect with a suitable example.
Answer:
Inductive effect :
The permanent displacement of sigma (a) bonded electrons towards the more electronegative atoms (or) group is known as Inductive effect.

Illustration :
Consider the molecue CH₃-CH₂-CH₂-Cl. There is a ‘σ’ covalent bond between carbon atom and chlorine atom. The electron pair between them is not equally shared. The more electronegative chlorine atom tends to attract the sharped pair, more towards itself. Due to this, the electron density tends to be greater, nearer chlorine atom than carbon atom.

This is generally represented as . But, carbon atom bonded to chlorine atom is itself attached to other carbon atoms. Therefore, the effect can be transmitted further.

Typcs of Inductive effect:
-I effect:
Electron withdrawing groups are said to have negative inductive effect. An order of -I effect
-NO₂ > CHO > CO > COOH > F > Cl > Br > I > OH > OR > NH₂ > C₆H₅ > H.

+ I effect :
Electron donating groups are said to have positive inductive effect.
An order of+1 effect: -C(CH₃)₃ >-CH(CH₃)₂ > -CH₂CH₃ > -CH₃

Question 25.
Write a on Mesomeric effect.
Answer:
Mesomeric effect :
The permanent transfer of ‘π’ electrons from multiple bonds to adjacent atom (or) bond (or) transfer of lone pair of electrons from an atom to adjacent bond is known as mesomeric effect.

Salient features of the mesomeric effect :

1. Mesomeric effect is a permanent effect and it operates in the ground state of the molecule.
2. Lone pairs and π electrons are involved in this effect and this operate through eonjugative mechanism of electron displacement.
3. This effect influences the physical properties, reaction rates etc.

Types of mesomeric effect :
+M effect:
Groups which tend to increase the electron density of the rest of the molecule are said to have +M effect. Such groups tend to posses lone pairs of electrons.
Ex: H₂N-C=C- Here – NH₂ group has + M effect.

-M effect:
Groups that decrease the electron density of the rest of the molecule are said to have -M effect. Unsaturated groups having polar character have -M effect. Ex: -C=C-C=0 Here C=0 group decreases the electron density of the remaining molecule.Hence it has -M effect.

Question 26.
Describe Resonance effect with one example.
Answer:
Resonance effect:
This is the polarity produced in a molecule by the interactions of two π bonds or between a π bond and a lone pair of electrons present on adjacent atoms. This effect i s tran sm itted through the chain.

If the transfer of electron is away from the atom or substituent group attached to the conjugated system, then the molecule gets some of its positions high electron density as in aniline and it is given +R.

If the shift of electrons are towards the atom or substituent group it is (-R) as in nitrobenzene and it is given -R
Groups showing (+R) effect : X, -OH, -OR, -NH₂, -NHR, – NR₂, -NHCOR
Groups showing (-R) effect : – COOH, CHO, > C = O, -CN, – NO₂.

Question 27.
Explain how many types of organic reactions are possible.
Answer:
Organic reactions are mainly classified into four types known as
i) Addition reactions
ii) Substitution reaction
iii) Elimination reactions
iv) Molecular rearrangements.

i) Addition reactions :
In these reactions the reagent and the substrate combine together to give a single product.

Depending on the reagent added in the slow rate determining step, addition reactions are agian classified as a) Electrophilic addition reactions b) Nucleophilic addition reactions c) Free radical addition reactions.

ii) Substitution reactions :
In these reactions an atom or a group of the substrate species is replaced by another atom or group. These are again classified as a) electrophilic substitution b) Nucleophilic substitution and c) Free radical substitution reactions on the basis of the reagent involved in the rate determining step.
Ex: OH^–_(aq) + R – X → HO – R + X^–_(aq)

iii) Elimination reactions :
In these reactions two or more atoms or groups of an organic substrate are removed to form multiple
CH₃CH₂Br + KOH_(alc) → CH₂ = CH₂ + KBr + H₂O

iv) Molecular rearrangements :
Flere one organic species (generally less stable) rearranges to other species (generally more stable). For example, Fries rearrangement.

Question 28.
Write the possible conformations of ethane and epxlain which is more stable. [Mar’ 09]
Answer:
Conformational isomers are obtained due to rotation about C-C single bond in alkanes.

Conformations of ethane:
Free rotation about C-C single bond in ethane gives an infinite number of conformers. Out of these, Eclipsed conformation and Staggered conformation are most significant. All the remaining conformers of ethane are called Skew conformations.

1) Staggered confirmation :
The C-H bond on one carbon are arranged such that they bisects the angle between two C-H bonds on adjascent carbon.

2) Eclipsed confirmation :
Each O H bond on one carbon are aligned with C-H bonds on adjacent (C-H bond) carbon.

Stagged conformer is most stable than the eclipsed. Because in Eclipsed conformer of ethane hydrogen atoms are very close to each other. So, repulsions are more and stability is less. The difference in the energy content of Staggered and Eclipsed Conformations is 12.5kJmol^-1.

Question 29.
Explain aromatic electrophilic substitution reactions of benzene.
Answer:
Electrophilic substitution reaction of benzene proceeds in three steps.

Step I :
Generation of electrophile: The reagent undergoes heterolysis to form electrophile.
E – A → E^⊕ + A^Ꮎ

Step 2:
The electrophile (E⁺) takes two electrons of the six-electron π system, to form a σ bond with a carbon atom of the benzene ring. Then that carbon atom becomes sp³ hybridised. As a result carbocation is formed which is stabilised by resonance.

Step 3:
A proton is removed from the carbon atom of the σ complex that bears the E (electrophile). Then that carbon atom changes to sp² state again. Then the benzene derivative with fully delocalised six π electrons is formed.

(The proton H+ combines with the anion obtained from the molecule E – A → H⁺ + A^– → HA)

Question 30.
Explain electrophilic addition reactions of ethylene with mechanism.
Answer:
1) Addition of halogens:
Ethylene combines with halogens at room temperature in an inert solvent medium to give a vicinal dihalide.
Ex: Bromination.

Mechanism: It involves 2 steps.
Step(1):
The π-electrons of the double bond attack the halogen molecule.Then a loose π complex (cyclic bromonium ion), with a 3- membered ring is formed.

Step(2) :
The negative bromide ion attacks the cyclie bromonium ion from the backside.
This is called ‘trans’ addition of halogens.

2) Addition of hydrogen halide:
Ethylene reacts with hydrogen halide (HX) to give ethyl halide.

Mechanism:
This mechanism is quite similar to that of the halogen addition, involving 2 steps

Question 31.
With the help of mechanism explain free radical halogenations of alkanes.
Answer:
Chlorination of ethane takes place in three steps known as (1) Chain Initiation (2) Chain Propagation and (3) Chain Termination.

i) Chain Initiation:
in this step Cl₂ molecule absorbs energy and splits into Cl free radicals. (C-C bonds and C-H bonds of ethane being relatively stronger, they do not break at this stage.)

ii) Chain Propagation:
Chlorine free radicals react with ethane molecule.

(a) and (b) repeat several times making the reaction a chain reaction and these steps are called propagation steps. In these steps the main products are formed.

In addition to (a) and (b) other reactions to replace other hydrogen atoms of ethane also take place.

iii) Chain Termination:
When free radicals directly combine the chain reaction stops down.

Question 32.
Explain Markownikoff’s rule and Kharash effect(Anti-markownikoff).
Answer:
Addition of Hydrogen halides:
Addition of HX to a double bond generally follows Markownikoff s rule.

Statement of Markownikoff s rule:
The rule states that addition of hydrogen halides to unsymmetrical alkenes takes place in such a manner that the positive part of the reagnet attach itself to that carbon atom which has more number of hydrogen atoms.

Explanation:
It follows electrophilic addition mechanism.

Since 2° carbocation is more stable than l° carbocation, the major product formed is 2-Bromo propane.

Note: Markownikoff s rule is only applicable for unsymmetrical alkenes.

Anti Markownikoffs rule or peroxide effect or Kharasch effect:
In the presence of peroxide (R-O-O-R) the addition of HBr to unsymmetrical alkene like propene takes place in such a way that the negative part of the reagent attach itself to that carbon atom which has more number of hydrogen atoms.


Stability of free radicals follows as 30>2°>1°. Therefore, 1-bromopropane is major prodcut

Uses of Ethylene : Ethylene is usded

1. In the manufacture of ethylalcohol, polythene etc.
2. In the preparation of ethylene glycol an antifreeze.
3. In the preparation of mustard gas.

Question 33.
How would you convert the following compounds into benzene? a) Chlorobenzene b) Toluene c) p-nitro toluene
Answer:
a) Reduction of chlorobenzene with Ni-Al alloy andNaOH gives benzene.
C₆H₅Cl + 2[H] → C₆H₆ + HCl

b) Toluene is first oxidised to benzonic acid with dil.HNO₃ and alkaline or acidic KMnO₄. Then decarboxylation of benzoic acid gives benzene.

This on dizotization forms diazonium salt which on treating with KCN/Cu followed by hydrolysis gives p-methyl benzoic acid. The methyl group is also oxidised to acid group with dil. HNO₃. This on decarboxylation gives benzene.

Question 34.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate with one example?
Answer:
In Wurtz reaction when alkyl halide is treated with sodium metal, an alkane having double the number carbon atoms present in alkyl halide will be formed. So we always get an alkane with even number of carbon atoms. If an alkyl halide with even number of carbon atoms and another alkyl halide with odd number of carbon atoms are used in Wurtz reaction a mixture of hydrocarbons are produced. So Wurtz reaction is not preferable for the preparation of alkanes with odd number of carbon atoms.

This can be illustrated with the following examples.

Question 35.
Write the equations involved in the detection of Nitrogen, Halogens and sulphur in organic compounds.
Answer:
Lassaigne’s Test:
In Lassaigane’s test, the organic compound is mixed with sodium metal and fused strongly. The fused mass is extracted with water by plunging the red hot ignition tube in distilled water and the contents are boiled for 5 minutes and filtered. The filtrate is called Sodium fusion extract or Lassaigne’s extract.

If nitrogen is present in the organic compound, it reacts with sodium and carbon forming sodium cyanide.
Na + C + N → NaCN

If sulphur is present in the organic compound, it reacts with sodium forming sodium
2Na + S → Na₂S

If halogens are present in the organic compound, it reacts with sodium fonning sodium halides.
Na + X → NaX [X= Cl, Br, I]

Test for Nitrogen:
A little of the Lassaigne’s extract is made alkaline with NaOH solution and then freshly prepared FeSO₄ solution is added. To it 2-3 drops of FeCl₃ solution is added.
A prussian blue colouration is observed.

Test for Halogens:
A little of the Lassaigne’s extract is acidified with nitric acid and treated with AgNO₃ solution.

If white ppt. soluble in NH₄OH is formed, the halide is Cl^–. Hence, chlorine is present.

If pale yellow ppt. partially soluble in NH₄OH is formed, the halide is Br^–. Hence, bromine is present.

If yellow ppt. insoluble in NH₄OH is formed, the halide is I^–. Hence, iodine is present.

Test for Sulphur:
To a little of the Lassaigne’s extract freshly prepared sodium nitroprusside solution is added. Deep purple colouration is observed.

Question 36.
Explain how Carbon and Hydrogen are quantitatively determined in an organic compound.
Answer:
Liebig’s method for the quantitative estimation of carbon and hydrogen:
A known weight of the organic compound is taken and completely burnt in excess of air and copper(II) oxide. Then carbon oxidises to CO₂ and hydrogen oxidises to H₂O. The CO^– and H₂O so obtained are passed through already weighed U tubes containing anhydrous CaCl₂ and caustic potash respectively. The increased weights of these two tubes give the weights of H₂O and CO₂ formed.

Suppose that ‘a’ g of organic compound on combustion gives ‘b’ g of water vapourand ‘c’ g of CO₂.

Question 37.
Describe Dumas and Kjeldahl’s method for the estimation of Nitrogen.
Answer:

a) Estimation of nitrogen by Duma’s method:
In this method a known weight of organic compound is heated strongly with coarse cupric oxide. Then carbon and hydrogen get oxidised to CO₂ and H₂O vapour respectively. Nitrogen is converted to N₂ gas. Some nitrogen converted into its oxides, gets reduced by copper gauze to nitrogen. The liberated gases are passed over a solution of KOH. Then CO₂ gets absorbed. Nitrogen is collected over KOH solution and its volume is found out.

Calculations:
Suppose ‘a’ g of nitrogen compound gives V₁ ml of N₂ gas at room temperature T K. If atmospheric pressure is P and aqueous tension at TK is P₀, then the pressure of N₂ gas at TK is P₁ = (P – P₀).

b) Estimation of N₂ by Kjeldahl’s method:
In this method the organic compound is heated with conc. H₂SO₄ in presence of a small amount of CuSO₄. Then N₂ is quantitatively converted into ammonium sulphate. The contents of the flask are transferred into another flask and heated with excess of NaOH solution to liberate ammonia gas. The gas so liberated is passed and absorbed in a known vol. of known cone. H₂SO₄(excess). Now the excess of acid remained after the neutralisation by NH₃ is titrated against a standard solution of alkali. From this, the amount of H₂SO₄ used to neutralise NH₃ is calculated. From this the mass of ammonia formed is calculated and from that the % of N₂ gas is calculated.
Organic compound + H₂SO₄ → (NH₄)₂SO₄
(NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O + 2NH₃
2NH₃ + H₂SO₄ → (NH₄)₂SO₄

Calculation:

a = Weight of organic compound = a g
M = Molarity of H₂SO₄ solution
V = Volume of H₂SO₄ solution
V₁ = Volume of H₂SO₄ solution consumed by NaOH.

Question 38.
How do you determine Sulphur, Phosphorous and Oxygen are determined quantitatively in an organic compound?
Answer:
Estimation of Sulphur:
A known mass of the organic compound is heated with sodium peroxide in a Carius tube. Then ‘S’ is oxidised to H₂SO₄. The acid is precipitated as BaSO₄ by adding excess of BaCl₂ aq. solution. The ppt. is filtered, washed, dried and weighed.

Mass of organic compound = a g
Mass of barium sulphate = b g
Molecular weight of BaSO₄ = 233
l mole of BaSO₄ or 233 g of BaSO₄ contains 32 g of sulphur.

Estimation of phosphorous:
A known mass of organic compound is heated with fuming nitric acid in a Carius tube. Then phosphorus is oxidised to phosphoric acid. The acid is precipitated as ammonium phosphomolybdate by adding ammonia and ammonium molybdate solutions.

Calculation:
Mass of organic compound = a g.
Mass of ammonium phosphomolybdate = b g.
Molecular weight of ammonium phosphomolybdate = 1877.
1877 g of ammonium phosphomolybdate contains 31 g of phosphorus

Estimation of Oxygen:
A known weight of organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red hot coke to convert all oxygen in those oxides to CO. Then the mixture is passed through hot iodine pentoxide to convert to CO into CO₂ and iodine liberates.

Calculation:
Mass of organic compound = a g
Mass of carbon dioxide = b g
44 g of CO₂ contains 16 g of oxygen.

Question 39.
Explain Carius method for the determination of Halogens quantitatively in an organic compound.
Answer:
Halogens can be estimated by Carius method.

In this method a known weight of the organic compound is heated with fuming nitric acid in presence of AgNO₃ in a hard glass tube. Carbon and hydrogen are oxidised to CO₂ & H₂O. Halogens are converted into silver halides. The silver halide is filtered off, washed, dried and weighed.

Calculation:
Mass of the organic compound = ‘a’ g.
Mass of the silver halide formed (AgX) = ‘b’g.
I mole of AgX contains l mole of X

(Molecular Masses: AgCl = 143.5(108 + 35.5); AgBr = 188 (108 + 80); Agl = 235 (108 + 127)

Question 40.
What is Carcinogenicity? Explain with two examples?
Answer:
Benzene and several polynuclear hydrocarbons are toxic and are said to be carcinogenic (Cancer producing).

Most of these are formed due to incomplete combustion of organic substances like tobacco, coal, petroleum etc.

They undergo various chemical changes in human body and finally damage DNA to cause cancer.

Multiple Choice Questions

Question 1.
Which of the following is the correct SUPAC name?
1) 3-Ethyl-4,4-dimethylheptane
2) 4,4-Dimethyl-3-ethylheptane
3) 5-Ethyl-4,4-dimethylheptane
4) 4,4-Bis(methyl)-3-ethylheptane
Answer:
2) 4,4-Dimethyl-3-ethylheptane

Question 2.
The IUPAC name for

1) 1-hydroxypentane-1,4-dione
2) 1,4-dioxopentanol
3) 1-carboxybutan-3-one
4) 4-oxopentanoic acid
Answer:
4) 4-oxopentanoic acid

Question 3.
The IUPAC name for is

1) 1 -Chloro-2-nitro-4-methylbenzene
2) 1 -Chloro-4-methyl-2-nitrobenzene
3) 2-Chloro-1 -nitro-5-methy lbenzene
4) m-Nitro-p-chlorotoluene
Answer:
2) 1 -Chloro-4-methyl-2-nitrobenzene

Question 4.
The number of sigma (σ) and pi(π) bonds in pent-2-en-4-yne is
1) 13 σ bonds and no π bond
2) 10 σ bonds and no 3π bonds
3) 8 σ bonds and no 5π bonds
4) 11 σ bonds and no 2π bonds
Answer:
2) 10 σ bonds and no 3π bonds

Question 5.
In which of the following, functional group isomerism is not possible?
1) Alcohols
2) Aldehydes
3) Alkyl halides
4) Cyanides
Answer:
3) Alkyl halides

Question 6.
Dihedral angle of least stable conformer of ethane is
1) 0°
2) 120°
3) 180°
4) 60°
Answer:
1) 0°

Question 7.
Arrange the following in decreasing order of their boiling points.
(A) 77-butane
(B) 2-methylbutane
(C) 77-pentane
(D) 2,2-dimethylpropane
1) A > B > C > D
2) B > C > D > A
3) D > C > B > A
4) C > B > D > A
Answer:
4) C > B > D > A

Question 8.
The compound which shows metamerism is
1) C₄H₁₀O
2) C₅H₁₂
3) C₃H₈O
4) C₃H₆O
Answer:
1) C₄H₁₀O

Question 9.
During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?
1) column chromatography
2) solvent extraction
3) distillation
4) thin layer chromatography
Answer:
4) thin layer chromatography

Question 10.
The principle involved in paper chromatography is
1) Adsorption
2) Partition
3) Solubility
4) Volatility
Answer:
2) Partition

Question 11.
Paper chromatography is an example of
1) adsorption of chromatography
2) partition of chromatography
3) thin layer of chromatography
4) column chromatography
Answer:
2) partition of chromatography

Question 12.
The increasing order of reduction of alkyl halides with zinc and dilute HCl is
1) R-Cl < R-I < R-Br
2) R-Cl < R-Br < R-I
3) R-I < R-Br < R-Cl
4) R-Br < R-I < R-Cl
Answer:
2) R-Cl < R-Br < R-I

Question 13.
Arrange the following hydrogen halides in order of their decreasing reactivity with propene.
1) HCl > HBr > HI
2) HBr > HI > HCl
3) HI > HBr > HCl
4) HCl > HI > HBr
Answer:
3) HI > HBr > HCl

Question 14.
Arrange the halogens F₂, Cl₂, Br₂, I₂, in order of their increasing reactivity with alkanes.
1) I₂ < Br₂ < Cl₂ < F₂
2) Br₂ < Cl₂ < F₂ < I₂
3) F₂ < Cl₂ < Br₂ < I₂
4) Br₂ < I₂ < Cl₂ < F₂
Answer:
1) I₂ < Br₂ < Cl₂ < F₂

Question 15.
Arrange the following carbanions in order of their decreasing stability.
(A) H₃C – C ≡ C^–
(B) H – C ≡ C^–
(C) H₃C-CH^–₂
1) A > B > C
2) B > A > C
3) C > B > A
4) C > A > B
Answer:
2) B > A > C

Question 16.
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?
1) -I effect of -CH₃ groups
2) +R effect of -CH₃ groups
3) -R effect of -CH₃ groups
4) Hyperconjugation
Answer:
4) Hyperconjugation

Question 17.
Which of the following is correct with respect to -I effect of the substituents? (R=alkyl)
1) -NH₂ < -OR < -F
2) – NR₂ < OR < F
3) – NH₂ > -OR > -F
4) – NR₂ > -OR > -F
Answer:
1) -NH₂ < -OR < -F

Question 18.
In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?

Answer:
1

Question 19.
Which of the following alkane cannot be made in good yield by Wurtz reaction?
1) n-Hexane
2) 2,3-Dimethylbutane
3) n-Heptane
4) n-Butane
Answer:
3) n-Heptane

Question 20.

Consider the above reaction and identify the missing reagent/chemical
1) DIBAL-H
2) B₂H₆
3) Red Phoshporous
4) CaO
Answer:
4) CaO

Students get through AP Inter 1st Year Chemistry Important Questions 12th Lesson Environmental Chemistry which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 12th Lesson Environmental Chemistry

Very Short Answer Questions

Question 1.
Define the terms atmosphere, biosphere.
Answer:
Atmosphere:
The protective blanket of gases present around the earth is called atmosphere.

Biosphere:
The part of the earth in which all living organisms exist is called biosphere.

Question 2.
Explain the terms Lithosphere, Hydrosphere.
Answer:
Hydrosphere:

1. All the natural water resources like oceans, seas, rivers, lakes, glaciers, polar ice caps and ground water constitute the hydrosphere.
2. Water occupies 3/4^th of the earth’s surface. Out of this 97% is present in the form of sea water and the remaining 3% in the form of ice in polar ice caps.
3. Small percentage of water is available for drinking, agriculture and other human purposes.

Lithosphere:

1. The outer mantle of the solid earth consisting of minerals and the soil is called Lithosphere.
2. The inner layers of earth contain minerals, the deeper inner layers contain natural gas and oil.

Question 3.
Define the term Soil Pollution.
Answer:
Any factor which reduce the quality, texture and mineral content of the soil and disturbs the biological balance of living organisms in it is called soil pollution.

Question 4.
What is Chemical Oxygen Demand (COD)? [AP 18; AP,TS 15,16,17,19]
Answer:
Chemical Oxygen Demand (COD):
The amount of oxygen required to oxidise organic substances present in polluted water is chemical oxygen demand (COD).

Question 5.
What is Biochemical Oxygen Demand (BOD)? [TS 18,19,19] [AP 15, 16,18,20]
Answer:
Biochemical Oxygen Demand (BOD):
The amount of oxygen used by suitable microorganisms present in water during five days at 20°C is called B.O.D. For pure water BOD is about 1 ppm.

Question 6.
What are Troposphere and Stratosphere?
Answer:
Troposphere:
The lowest region of atmosphere in which the human beings along with other organisms live is called troposphere.

Stratosphere:
Above the troposphere, extends from 10 to 50 km above sea level is called stratosphere.

Question 7.
Name the major particulate pollutants present in Troposphere. [AP 19]
Answer:
Particulate pollutants present in troposphere are dust, mist, smoke and smog.

Question 8.
List out four Gaseous Pollutants present in the polluted air.
Answer:
Oxides of Sulphur, Nitrogen and Carbon, Ozone, Hydrocarbons etc., are gaseous pollutants present in polluted air.

Question 9.
Green house effect is caused by …..and ……..gases. [AP,TS 20][IPE’ 14][TS 18]
Answer:
Green house effect is caused by gases such as CO₂, CH₄, O₃, CFCs (Chloro Fluoro Carbons) and water vapour in the atmosphere.

Question 10.
Which oxides cause acid rain? What is its pH value? [IPE 13][TS 19]
Answer:
Oxides of Nitrogen, Sulphur and Carbon dissolved in rain water causes acid rain.
The pH value of acid rain is 5.6.

Question 11.
Name two adverse effects caused by acid rains. [AP, TS 15,17,18][AP 16]
Answer:

1. Acid rains reduce the life of buildings and historical monuments.
2. Acid rains decrease the fertility of soil by reducing the pH values of the soil.
3. Acid rains decrease quality of drinking water.
4. Acid rains decrease the productivity of fish in water.

Question 12.
What are smoke and mist?
Answer:
Smoke:
A mixture of CO₂, Water vapour with very small soot particles produced by burning and combustion of organic matter are called smoke.

Mist :
The particles produced by spray liquids, formed by condensation of vapours in air is called mist.

Question 13.
What is classical smog? and What is its Chemical Character (Oxidizing (or) reducing)?
Answer:
Classical smog occurs in cool humid climate. It is a mixture of smoke, fog and sulphur dioxide. Chemically it is a reducing mixture and so it is also called as reducing smog.

Question 14.
Name the common components of Photochemical smog. [AP 16,19]
Answer:
Common components of photochemical smog are ozone, nitric oxide, formaldehyde, acrolein and PAN (peroxy acetyl nitrate). It is also called as oxidising smog.

Question 15.
What is PAN? What effect is caused by it? [AP 19]
Answer:
PAN means Peroxy acetyl nitrate is called
PAN. It causes photochemical smog.

Question 16.
How is Ozone formed in the Stratosphere?
Answer:
The formation of O₃ in stratosphere takes place in two steps.

In the first step the Ultraviolet radiation coming from sun split the dioxygen into two oxygen atoms. In second step, the oxygen atoms react with the molecular oxygen to form ozone.

Question 17.
Give the Chemical equations involved in (he Ozone depletion by CF₂ Cl₂.
Answer:
CFC’s undergo decomposition in the presence of sunlight as given below.

Reactions:

Chain reaction continues in which ozone layer is depleted.

Question 18.
What is Ozone hole? Where was it first observed?
Answer:
The depletion of ozone layer is commonly known as ozone hole. It was first observed in Antarctica region of south pole.

Question 19.
What is the value of dissolved Oxygen in pure cooled water?
Answer:
The value of dissolved oxygen in pure cooled water is 4-6 mg/lit (or) 4-6ppm.

Question 20.
Give the possible BOD values of clean water and the polluted water.
Answer:
For pure water BOD is 1 ppm.
BOD value of clean water is less than 5 ppm.
BOD value of polluted water is 17ppm or more.
Municipal Sewage -l 00-4000 ppm.

Question 21.
Name three industrial Chemicals that pollute water.
Answer:
Poly chlorinated biphenyls, detergents and fertilizers.

Question 22.
What agrochemicals arc responsible for water pollution? [AP 19]
Answer:
Fertilizers, insecticides, herbicides, fungicides etc are responsible for water pollution.

Short Answer Questions

Question 1.
What are different segments of the earth’s environment?
Answer:
Environment is divided into four segments.
They are 1) Atmosphere
2) Hydrosphere
3) Lithosphere and
4) Biosphere.

1) Atmosphere:
The blanket of gases that surrounds the earth is called Atmosphere. It contains large proportions N₂ and O₂ and in small proportions CO₂., water vapour etc. It absorbs the harmful radiations coming from the sun and plays an important role in maintaining the heat balance on earth.

2) Hydrosphere:
All the natural water resources together constitute the Hydrosphere. It includes oceans, seas, rivers, lakes, streams, reservoirs, polar ice caps and ground water etc.

3/4th of the earth’s surface is occupied by water. Out of this 97% is present in the form of sea water and the remaining 3% is in the form of ice in polar ice caps and only very small percentage of water is available for drinking, agriculture and other human purposes.

3) Lithosphere:
1/4th of the total earth surface is in the form of land Inner layers of earth contain minerals. Deeper inner layers of earth contain natural gas and oil. All these things including hills and mountains come under this segment.

4) Biosphere:
All living organisms, trees, plants, animals, and human beings constitute Biosphere. This biosphere is interrelated to other segments of environment. Biosphere is dependent on atmosphere and hydrosphere. Polluted atmosphere stops the plants growth and can create health problems among animals and human beings.

Question 2.
Define the terms Sink, COD, BOD and TLV. [TS 15,16,18]
Answer:
Sink :
The medium which interacts with pollutants and reduces its effect is called sink.

COD (Chemical Oxygen Demand):
The amount of oxygen required to oxidise organic substances present in polluted water is called chemical oxygen demand (COD).

BOD(Biochemicai Oxygen Demand):
The amount of oxygen used by the suitable micro-organisms present in water during five days at 20°C is called B.O.D.

TLV(Threshold Limit Value):
The permissible level of a toxic pollutant without any adverse effect on a healthy industrial worker, working for 8 hours per day, in a polluted atmosphere is called TLV.

Question 3.
Name the gaseous pollutants present in the air and explain their formation.
Answer:
Substances which mix with air and affect the human beings, animals, plants and global temperature are called air pollutants.

Examples:

1. Oxides of carbon: CO and CO₂
2. Oxides of Nitrogen: N₂O and NO
3. Oxides of Sulphur: SO₂
4. Ozone
5. CFCs: Chlorofluorocarbons
6. Methane and Butane
7. Smog
8. Metals: Lead, Mercury.

Air pollution :
In cities, 80% of air pollution is due to exhausts of automobiles. At peak times in cities the level of CO will be upto 50-100ppm. If it is inhaled, it causes adverse effects.

During the combustion of fossil fuels NO, N₂O, NO₂ etc. are also released. When the level of these oxides is greater than 10 ppm, the plants cannot perform photosynthesis.

SO₂ is released into the atmosphere during the burning of sulphur and roasting of sulphide ores. SO₂ causes respiratory diseases in human beings. It bleaches the green colour of the leaf apexes in plants to yellow colour and thus prevent photosynthesis process.

Chlorofluoro carbons when percolate into stratosphere cause depletion of ozone layer. Harmful pesticides and biocides mix up with the air at their manufacturing units and pollute the air. They cause major health hazards.

Question 4.
What is Green house effect? and how is it caused? [TS 15][AP 18]
Answer:
The progressive warming up of the earth’s surface due to blanketing effect of CO₂ and water vapour in the atmosphere is called global warming (or) green house effect.

Reason :
If the concentrations of gases such as CO₂, NO, N₂O, CH₄, O₃ etc increase in atmosphere, the gases act like the glass panels of a green house (or) the window glass of a closed car. They allow the sunrays to the earth, but prevent the longer wavelength (Infrared) radiations coming from the earth. As a result, more energy is radiated back to the Earth by the gases due to their increased concentration and their blanketing effect. Hence the temperature of the earth increases.

Effects :

1. The ice caps of polar region melt. This results in the increase of the level of the sea water upto 90cm. Because of this, low lying countries will get submerged.
2. The rate of evaporation of water from the seas, river, ponds will increase. This leads to cyclones and hurricanes.
3. Agriculture sector will be affected badly.

Preventions:

1. The number of sinks (Plants) should be increased to absorb CO₂.
2. The growth of blue-green algae in sea water should be prevented.

Question 5.
Explain with chemical equation involved the formation of acid rain.
Answer:
Acid rain :
Acid rain is the rainwater containing sulphuric acid and nitric acid which are formed from the oxides of sulphur and nitrogen present in the air as pollutants and has a pH of 4 – 5.

Formation of acid rain:
Oxides of nitrogen combine with oxygen and ozone to form higher oxides of nitrogen. These oxides dissolve in water to form nitric acid.
NO + O₃ → NO₂ + O₂
NO₂ + O₃ → NO₃ + O₇
NO₂ + NO₂ → N₇O₅
N₂O₅ + H₂O → 2HNO₃

Sulphur dioxide reacts with oxygen and water to form sulphuric acid.
SO₇ + ½O₂ → SO₃
SO₃ + H₂O → H₂SO₄

These acids obtained in the atmosphere dissolve in rain water and come down to earth as rain. It is called acid rain.

Question 6.
Explain in detail the adverse effects caused by the acid rain.
Answer:

1. Acid rain is harmful for agriculture. It washes away the nutrients required for the growth of plants. So crops cannot grow in the areas where the acid rain falls.
2. Aquatic animals and plants cannot live in acid water. So they die in the lakes, ponds, rivers where acid rain falls.
3. The metal pipes which carry water corrodes in the acid water where the acid rain falls.
4. Acid rain damages buildings and other structures made with stone or metal.
   Ex: Tajmahal in India has been affected by acid rain.

Question 7.
How is photochemical smog formed? What arc its ill effects.
Answer:
Photochemical smog is produced from the action of sunlight on nitrogen oxides and hydrocarbons present in the exhaust gases of the automobiles and factories. The NO is oxidised by the atmospheric oxygen to NO₂.

NO₂ then undergoes photochemical decomposition in the presence of sunlight.

The free oxygen atom so formed reacts with atmospheric oxygen to form ozone.
O_2(g) + O_(g) → O_3(g)

Ozone then reacts rapidly with NO_(g) to form NO_2(g)
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Ozone is a toxic gas and both NO₂ and O₂ are strong oxidising agents and can react with the unburnt hydrocarbons in the polluted air to produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate (PAN). The common components of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and PAN. It is also called as oxidising smog.

Effects:

1. Photochemical smog causes eye irritation.
2. It decreases visibility and thus effects the air and road traffic.
3. It leads to cracking of rubber and extensive damage to plant life.

Question 8.
How is ozone layer depleted in the atmosphere and what are the harmful effects caused by ozone layer depletion. [TS 15,16]
Answer:
The process of destroying the existence of ozone molecules in the stratosphere is called depletion of ozone layer.

Depletion of ozone layer is due to release of certain chemical substances like CFC’s, NO, Cl₂ and gases evolved from Volcanoes.

CFC’s absorb U.V radiation and decompose to liberate chlorine free radicals which catalyse the decomposition of ozone, causing depletion of ozone layer.

Effects of ozone holes:
The ‘ozone layer’ prevents the U V rays coming from the Sun. When holes are formed in ozone layer, then U.V. rays will pass through these holes and reach the earth. These U.V. rays cause (1) Skin cancer (2) Cataract of eyes (3) Decrease moisture content of the soil (4) Decrease in the efficiency of photosynthesis in plants

Question 9.
List out the industrial wastes that cause water pollution and what are the international standards fixed for drinking water.
Answer:
Process Wastes:
These are form inorganic process wastes and organic process wastes. Inorganic process wastes are present in the effluents from chemical industries, electroplating industries, metallurgical and petroleum industries etc. These are mainly toxic but do not generally produce biological problems. Organic process wastes are from food processing industries, diaries, distilleries, paper, textile mills etc. It is very difficult to dispose organic process wastes.

Chemical Wastes :
Industries that manufacture acids, bases, detergents, explosives, dyes, insecticides, fungicides, fertilisers, silicones, plastics, resins etc., which are generally used as raw materials for further manufacturing processes, contain chemical wastes. These wastes are produced during sedimentation, flocculation, washing, filtering, evaporation, distillation, electrolysis, absorption, crystallisation, burning, centrifusing etc. Chemical wastes require biological oxidation treatment methods like thickening filters, activated sludge method. The international standards fixed for drinking water.
Fluoride concentration – 1 ppm
Lead – 50 ppb
Sulphate -<500 ppm
Nitrate – 50 ppm

Question 10.
Explain in detail the strategies adopted in Green chemistry to avoid environmental pollution.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce the pollution of the environment as far as possible are known as green chemistry.

The over exploitation of the soil using fertilisers and pesticides polluted the soil, water and air. But farming irrigation etc cannot be stopped. Then, methods to reduce the environmental pollution must be developed.

Generally in a reaction some by-products are formed. In many processes these by-products become the pollutants. Green chemistry works for not producing wasteful by-products in these processes.

Normal process:

As the reaction (2) gives no by-product it is called an environmental-friendly reaction. Even if a by-product is formed, it must be made a useful product instead of polluting the environment. Green chemistry suggests that instead of using conventional fuels and energy systems, non-non-conventional fuels and non-conventional systems must be put into practice. Because of this, pollution would be reduced.

Green chemistry is a cost effective approach that involves minimum chemical usage, minimum energy consumption and minimum wastef pollutant) generation.
Ex: In the dry cleaning of clothes, tetrachloroethane was used earlier. This compound contaminates the ground water. Therefore, this compound is replaced by liquefied CO₂ along with a suitable detergent. This would not pollute ground water much. Nowadays, H₂O₂ is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.

Long Answer Questions

Question 1.
What is environmental pollution? How many types of pollution are encountered?
Answer:
Due to increase in the population and industrialisation, the natural resources have diminished. To prepare many natural things artificially, many industries were started. For improving the yields many technologies were introduced. Along with this development, many waste products were released into environment. Thus environment got polluted. This is known as environmental pollution.

Some reasons for environmental pollution are:

1. Increase in population and decrease in natural resources.
2. Industrialisation
3. Urbanisation
4. Deforestation

The different types of pollutions are:

1. Air pollution
2. Water pollution.
3. Soil pollution
4. Sound pollution
5. Thermal pollution
6. Radiological pollution

Question 2.
Explain the following in detail.
a) Global warming b) Ozone depletion c) Acid Rain d) Eutrophication
Answer:
a) Global Warming :
Gases like CO₂, CFCs, O₃, NO and water vapour absorb I.R radiations coming to the earth and reflect them back to the earth’s surface. Due to this, the surface of the earth gets heated. This is called greenhouse effect or global warming. The gases which are responsible for this effect are called greenhouse gases.

Effect of global warming: [AP 17]

1. For a 1°C rise in temperature, the ice caps of polar regions melt and level of the sea water increases. Thereby many coastal countries get submerged.
2. Due to global warming, the rate of evaporation of water from the seas, rivers, ponds will increase. This leads to unwarranted rains, cyclones and hurricanes.
3. Due to global wanning, short of water supply for agriculture occurs. Global warming can be prevented by growing trees, forests, stopping production of CFCs etc.

b) Ozone depletion: Ozone layer is depleted by the pollutant gases such as ehlorofluorocarbons, NO₂ and HOCl.

The Chlorofluoro carbons break down in the presence of sunlight and deplete the ozone layer as follows.

The chloride radicals are continuously regenerated and cause the breakdown of ozone. One CFC molecule destroys one lakh O₃ molecules.

In winter season the Cl\(\dot{\mathrm{O}}\) reacts with NO₂ forming chlorine nitrate. This hydrolyses giving hypochlorous acid and HCl. In summer season, these breakup providing C\(\dot{\mathrm{ḷ}}\) radicals which can deplete the ozone layer.

c) Acid Rain:
Oxides of nitrogen and sulphur released from automobiles and industries enter into atmosphere and dissolves in water to form the acids HNO₃ and H₂SO₄. These acids dissolve in rain water and come down to earth as acid rain.
2SO₂ + 2H₂O + O₂ → 2H₂SO₄
4NO₂ + 2H₂O + O₂ → 4HNO₃

1. Acid rain is harmful for agriculture. It washes away the nutrients required for the growth of plants. So crops cannot grow in the areas where the acid rain falls.
2. Aquatic animals and plants cannot live in acid water. So they die in the lakes, ponds, rivers where acid rain falls.
3. The metal pipes which carry water corrodes in the acid water where the acid rain falls.
4. Acid rain damages buildings and other structures made with stone or metal.
   Ex: Tajmahal in India has been affected by acid rain.

d) Eutrophication :
When organic waste from industry and agriculture are released into ponds or lakes, then the lakes become over bounded with over nutrients and large amounts of phosphates. These support the growth of algae. Thus the lake becomes marshy, filled with sediments and ultimately becomes dry. This process is known as eutrophication.

Question 3.
Green Chemistry is to avoid environmental pollution. Explain.
Answer:
The ways of using the knowledge and the principles of chemistry and other sciences to develop methods to reduce the pollution of the environment as far as possible are known as green chemistry.

The over exploitation of the soil using fertilisers and pesticides polluted the soil, water and air. But farming irrigation etc cannot be stopped. Then, methods to reduce the environmental pollution must be developed.

Generally in a reaction some by-products are formed. In many processes these by-products become the pollutants. Green chemistry works for not producing wasteful by-products in these processes.

Normal process:

As the reaction (2) gives no by-product it is called an environmental-friendly reaction. Even if a by-product is formed, it must be made a useful product instead of polluting the environment. Green chemistry suggests that instead of using conventional fuels and energy systems, non-conventional fuels and non-conventional systems must be put into practice. Because of this, pollution would be reduced.

Green chemistry is a cost effective approach that involves minimum chemical usage, minimum energy consumption and minimum waste( pollutant) generation.
Ex: In the dry cleaning of clothes, tetrachloroethane was used earlier. This compound contaminates the ground water. Therefore, this compound is replaced by liquefied CO₂ along with a suitable detergent. This would not pollute ground water much. Nowadays, H₂O₂ is used for bleaching clothes in laundries. This gives better results and decreases the consumption of water.

Multiple Choice Questions

Question 1.
Which of the following gases is not a green house gas?
1) CO
2) O₃
3) CH₄
4) H₂O vapour
Answer:
1) CO

Question 2.
Excessive release of CO₂ into the atmosphere results in
1) global warming
2) formation of smog
3) polar vortex
4) depletion of ozone.
Answer:
1) global warming

Question 3.
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not amongst the components of photochemical smog, identify it.
1) NO₂
2) O₃
3) SO₂
4) Unsaturated hydrocarbon
Answer:
3) SO₂

Question 4.
The higher concentration of which gas in air can cause stiffness of flower buds?
1) SO₂
2) CO
3) NO₂
4) CO₂
Answer:
1) SO₂

Question 5.
Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?
1) N₂O₅
2) NO₂
3) N₂O
4) NO
Answer:
1) N₂O₅

Question 6.
Air pollution that occurs in sunlight is
1) oxidising smog
2) fog
3) reducing smog
4) acid rain.
Answer:
1) oxidising smog

Question 7.
The layer of atmosphere between 10 km and 50 km above the sea level is called
1) thermosphere
2) mesosphere
3) stratosphere
4) troposphere.
Answer:
3) stratosphere

Question 8.
The upper stratosphere consisting of the ozone layer protects us from the sun’s radiation that falls in the wavelength region of
1) 400-550 nm
2) 600-750 nm
3) 200-315 nm
4) 0.8-1.5nm
Answer:
3) 200-315 nm

Question 9.
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was
1) phosgene
2) methylisocyanate
3) methylamine
4) ammonia.
Answer:
2) methylisocyanate

Question 10.
The smog is essentially caused by the presence of
1) O₂ and O₃
2) O₂ and N₂
3) oxides of sulphur and nitrogen
4) O₃ and N₂
Answer:
3) oxides of sulphur and nitrogen

Question 11.
Which of the following is a sink for CO?
1) Microorganisms present in the soil
2) Oceans
3) Plants
4) Haemoglobin
Answer:
1) Microorganisms present in the soil

Question 12.
Which one of the following is not a common component of photochemical smog?
1) Ozone
2) Acrolein
3) Peroxyacetylnitrate
4) Chlorofluorocarbons
Answer:
4) Chlorofluorocarbons

Question 13.
Which of the following statements is not true about classical smog?
1) Its main components are produced by the action of sunlight on emissions of automobiles and factories.
2) Produced in cold and humid climate.
3) It contains compounds of reducing nature.
4) It contains smoke, fog and sulphur dioxide.
Answer:
1) Its main components are produced by the action of sunlight on emissions of automobiles and factories.

Question 14.
Which of the following is not correct about carbon monoxide?
1) It forms carboxyhaemoglobin
2) It reduces oxygen carrying ability of blood.
3) The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin
4) It is produced due to incomplete combustion
Answer:
3) The carboxyhaemoglobin (haemoglobin bound to CO) is less stable than oxyhaemoglobin

Question 15.
Which is wrong with respect to our responsibility as a human being to protect our environment?
1) Avoiding the use of floodlighted facilities
2) Setting up compost tin in gardens
3) Using plastic bags
4) Restricting the use of vehicles
Answer:
3) Using plastic bags

Question 16.
BOD stands for
1) Biochemical Oxidation Demand
2) Biological Oxygen Demand
3) Biochemical Oxygen Demand
4) Bacterial Oxidation Demand.
Answer:
3) Biochemical Oxygen Demand

Question 17.
Biochemical Oxygen Demand, (BOD) is a measure of organic material present in water. BOD value less than 5 ppm indicates a water sample to be
1) rich in dissolved oxygen.
2) poor in dissolved oxygen.
3) highly polluted.
4) not suitable for aquatic life.
Answer:
1) rich in dissolved oxygen.

Question 18.
Water samples with BOD values of 4 ppm and 18 ppm, respectively, are
1) clean and highly polluted
2) highly polluted and highly polluted
3) highly polluted and clean
4) clean and clean.
Answer:
1) clean and highly polluted

Question 19.
Addition of phosphate fertilizers to water bodies causes
1) enhanced growth of algae
2) increase in amount of dissolved oxygen in water
3) deposition of calcium phosphate
4) increase in fish population.
Answer:
1) enhanced growth of algae

Question 20.
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to
1) acid rain
2) soil pollution
3) water pollution
4) global warming
Answer:
1) acid rain

Question 21.
Identify the wrong statement in the following.
1) Acid rain is mostly because of the oxides of nitrogen and sulphur.
2) Chlorofluorocarbons are responsible for ozone layer depletion.
3) Greenhouse effect is responsible for global warming.
4) Ozone layer does not permit infrared radiation from the sun to reach the earth.
Answer:
4) Ozone layer does not permit infrared radiation from the sun to reach the earth.

Question 22.
Which of the following conditions in drinkipg water causes methemoglobinemia?
1) > 50 ppm of chloride
2) > 50 ppm of nitrate
3) > 50 ppm of lead
4) > 100 ppm of sulphate
Answer:
2) > 50 ppm of nitrate

Question 23.
The maximum prescribed concentration of copper in drinking water is
1) 0.05 ppm
2) 3 ppm
3) 5 ppm
4) 0.5 ppm
Answer:
2) 3 ppm

Question 24.
Among the following the one that is not a green house gas is
1) sulphur dioxide
2) nitrous oxide
3) methane
4) ozone
Answer:
1) sulphur dioxide

Question 25.
Which one of the following substances used in dry cleaning is a better strategy to control environmental pollution?
1) Sulphur dioxide
2) Carbon dioxide
3) Nitrogen dioxide
4) Tetrachloroethylene
Answer:
2) Carbon dioxide

Students get through AP Inter 1st Year Chemistry Important Questions 11th Lesson The p-Block Elements – Group 14 which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 11th Lesson The p-Block Elements – Group 14

Very Short Answer Questions

Question 1.
Discuss the variation of oxidation states in the group -14 elements.
Answer:

1. The common oxidation states exhibited by group-14 elements are +4 and +2.
2. Carbon exhibits negative oxidation states.
3. Heavier elements exhibit +2 oxidation state.
4. The tendency to show +2 oxidation state increases in the order Ge < Sn < Pb.
5. Pb exhibits +2 oxidation state as stable state because of inert pair effect.

Question 2.
How the following compounds behave with water? (a) BCl₃ (b) CCl₄
Answer:
a) When BCl₃ is treated with water, boric acid (H3BO3 )is formed.
BCl₃ + 3 H₂O → H₃BO₃ + 3HCl

b) When CCl₄ is treated with super heated steam in the presence of iron or copper, phosgene (COCl₂) is fonned.
CCl₄ + H₂O → COCl₂ + 2 HCl

Question 3.
Are BCl₃ and SiCl₄ electron-deficient compounds? Explain.
Answer:
The compounds in which the central atom has incomplete octet configuration are called electron deficient compounds.

BCl₃ is an electron deficient molecule because boron atom has sextet configuration (6 electrons). SiCl₄ is not an electron deficient molecule, because silicon atom has octet configuration.

Question 4.
Give the hybridisation of carbon in
a) CO₃^-2 b) diamond c) graphite d) fullerene [AP,TS 16] [TS 18]
Answer:
a) Hybridisation of ‘C in CO^-2₃ is sp²
b) Hybridisation of ‘C in diamond is sp³
c) Hybridisation of ‘C’ in graphite is sp²
d) Hybridisation of ‘C’ in fullerene is sp²

Question 5.
Why is CO poisonous? [AP, TS 16][AP 18]
Answer:
Carbon monoxide (CO) forms a stable complex with haemoglobin called carboxy haemoglobin. This prevents haemoglobin from carrying oxygen into the cells of the body and it ultimately results death.

Question 6.
What is allotropy? Give the crystalline allotropes of carbon. [AP 16,20][TS 22]
Answer:
The phenomenon of existence of an element in different physical forms having same chemical properties is called allotropy. Crystalline allotropes of carbon are
a) Diamond
b) Graphite.

Question 7.
Classify the following oxides as neutral, acidic, basic or amphoteric.
a) CO b) B₂O₃ c) SiO₂ d) CO₂ e) Al₂O₃ f) PbO₂ g) Tl₂O₃
Answer:
Neutral oxide : CO
Acidic oxides : CO₂, SiO₂, B₂O₃
Basic oxide : Tl₂O₃
Amphoteric oxides : Al₂O₃, PbO₂

Question 8.
Name any two man-made silicates. [AP 15]
Answer:
Two important man made silicates:
Glass and cement.

Question 9.
Write the outer electron configuration of group -14 elements.
Answer:
The general outer most electron configuration of group -14 elements is ns²np².
1) Carbon – [He]2s²2p²
2) Silicon – [Ne]3s²3p²
3) Germanium – [Ar]3d¹⁰4s²4p²
4) Tin – [Kr]4d¹⁰5s²5p²
5) Lead – [Xe]4f¹⁴5d¹⁰6s²6p²

Question 10.
How does Graphite function as a lubricant? [TS 15,19]
Answer:
Graphite has two -dimensional layer structure and these layers can easily slide one over the other, as they are held by weak vanderwaal forces. Hence graphite is used as a lubricant.

Question 11.
Graphite is a good conductor. Explain. [TS 19,22][AP 15,17,22]
Answer:
In Graphite, each carbon atom is in sp- hybridised state. Each carbon atom contains one electron in pure ‘p’ orbital. Due to the presence of these free electrons, graphite acts as a good conductor of electricity.

Question 12.
Explain the structure of silica.
Answer:

1. Silica is a giant molecule with 3- dimensional structure.
2. In Silica, eight membered rings are formed with alternative Silicon and Oxygen atoms.
3. Each ‘Si’ is tetrahedrally surrounded by four oxygen atoms.
4. Each Oxygen at the vertex of the tetrahedron is shared by two Silicon atoms.
5. In SiO₂, Silicon atom undergoes sp³ hybridisation.
   

Question 13.
What is Synthesis gas? [TS 15,18][AP 20]
Answer:
The mixture of CO and H₂ is known as water gas or synthesis gas or Syngas.
It is used for the synthesis of methanol and a number of hydrocarbons.

Question 14.
What is ‘producer gas’? [TS 16]
Answer:
The mixture of CO and N₂ is known as Producer gas.
It is prepared by passing air over hot coke.

Question 15.
Diamond has high melting point – explain. [AP 19,22]
Answer:
In Diamond sp³ hybridisation. It has a three dimensional network involving strong C-C bonds, which are very difficult to break. Hence diamond has high melting point.

Question 16.
Give the use of CO₂ in photosynthesis.
Answer:
The process of converting the atmospheric CO₂ into carbohydrates by green plants is known as ‘photosynthesis’.

In photosynthesis, CO₂ changes to carbohydrates such as glucose.

Question 17.
How does CO₂ increase the green house effect?
Answer:

1. Due to deforestation, decomposition of lime stone and burning of fossil fuels etc. CO₂ concentration is increased in atmosphere.
2. A blanket of CO₂ gas in the atmosphere traps and reflects the infra red radiations. So atmosphere gets heated up. This is called green house effect or global warming.

Question 18.
What are silicones? [AP 15]
Answer:
Silicones:
Silicones are organo silicon polymers in which silicon is strongly linked with oxygen and carbon.

Silicones contain R₂SiO as repeating unit. The structure of R₂SiO unit is similar to that of ketone.

Question 19.
Give the uses of silicones.
Answer:
Silicones are used

1. in the preparation of silicone rubber.
2. to prepare water proof clothes and paper.
3. to prepare grease and lubricants that are used in aeroplanes.
4. in paints and enamels.

Question 20.
What is the effect of water on tin?
Answer:
Tin decomposes steam and liberates H₂.

Question 21.
Write an account of SiCl₄.
Answer:

1. Silicon tetrachloridel (SiCl₄) is also called as silane.
2. SiCl₄ can acts as Lewis acid due to availability of 3d orbital in ‘Si’.
3. SiCl₄ undergoes hydrolysis and forms Si(OH)₄

Uses:

1. SiCl₄ and NH₃ mixture is used to produce smoke screens.
2. Ultra pure Silicon is used to make transistors.
3. SiO₄ prepared from SiCl₄ is used in epoxy paints, resins etc.

Question 22.
SiO₂ is a solid while CO₂ is a gas explain.
Answer:

1. SiO₂ is a solid because it has giant molecule structure with strong Si-0 single bonds.
2. CO₂ is gas because it is made up of descrete linear O=C=O molecules with weak vander wall forces.

Question 23.
Write the use of ZSM-5. [TS 19,20]
Answer:
ZSM -5 is one type of zeolite. It is used to convert alcohols directly into gasoline.

Question 24.
What is the use of dry ice? [AP 15]
Answer:
Dry ice (Solid CO₂) is used as a refrigerant for ice-cream and frozen food.

Question 25.
How is water gas prepared? [AP 18]
Answer:
The mixture of CO and H₂ is called water

Water gas is prepared by passing steam over red hot coke.

Question 26.
How is producer gas prepared? [TS 19]
Answer:
The mixture of CO and N₂ is call ed producer gas.

Producer gas is prepared by passing air over red hot coke.

Question 27.
C-C bond length in graphite is shorter than C-C bond length in diamond. Explain.
Answer:

1. Graphite has sp² hybridisation and C-C bond involves sp² – sp² hybridized carbon.
2. Diamond has sp³ – hybridisation and C-C bond involves sp³ – sp³ hybridized carbon.
3. If the ‘s’ character (in sp²) in a hybridised atom increases then the size of the hybridized orbital decreases. This results in more overlapping which leads to shorter bond length.

Question 28.
Diamond is used as precious stone. Explain.
Answer:
Diamond has high refractive index. Diamonds reflect & refract the incident radiations. So diamonds glitter due to total reflection. Hence, diamonds are used as precious stones in jewellery.

Question 29.
Carbon never shows co-ordination number greater than four while other members of carbon family show coordination number as high as six-explain.
Answer:
The maximum coordination number of carbon is 4 due to absence of d-orbitals in the outermost orbit. While other members of carbon family show maximum coordination number 6 due to the presence of d-orbitals in the outer most orbit.

Question 30.
Producer gas is less efficient fuel than water gas- explain.
Answer:
At Producer gas is less fuel efficient due to presence of large proportion of non combustible nitrogen in it.

Also due to less calorific value, producer gas is less fuel efficient.

Question 31.
SiF^2-₆ is known while SiCl^2-₆ is not. Explain. [AP 16,19]
Answer:
The reasons for non-existence of SiCl^2-₆:

1. Six large chloride ions cannot be accommodated around Si⁴⁺ due to small size.
   1. Interaction between lone pair of chloride ion and Si⁴⁺ is not very strong.

Short Answer Questions

Question 1.
Explain the difference in properties of diamond and graphite on the basis of their structure. [TS 17]
Answer:

Diamond	Graphite
1) Diamond is the hardest material. It is bad conductor of electricity due to absence of free electrons.	1) Graphite is soft. It is good conductor of electricity due to presence of free electrons.
2) Each carbon is bonded to 4 other carbon atoms tetrahedrally.	2) Each carbon is bonded to 3 other carbon atoms to form hexagonal rings.
3) It is a 3 dimensional polymer.	3) It is a 2 dimensional polymer.
4) C – C bond length is 1.54 Å and bond angle is 109°28¹.	4) C – C bond length is 1.42 Å and bond angle is 120°.
5) Carbon atoms are strongly held by covalent bonds.	5) The hexagonal layers of carbons are held by weak Vander Waal’s forces.
6) sp³ hybridisation.	6) sp² hybridisation.

Question 2.
Explain the following.
a) PbCl₂ reacts with Cl₂ to give PbCl₄.
b) PbCl₄ is unstable to heat.
c) Lead is not known to form Pbl₄.
Answer:
a) Pb can exist in its compounds in two
different oxidation states of+2 and +4. +2 +4

Due to inert pair effect, Pb is more stable in +2 than in +4 oxidation state.
So PbCL react with chlorine to form unstable PbCl₄.

b) Pb²⁺ is more stable than Pb⁴⁺ due to inert pair effect. So PbCl₂ is more stable than PbCl₄. Thus on heating PbCl₄ converts into stable PbCl₂ by losing Cl₂.
PbCl₄ → PbCl₂ + Cl₂

c) PbI₄ does not exist because I^– ion is a powerful reducing agent. It reduces Pb⁴⁺ to stable Pb²⁺ion in solution. So PbI₄ is not formed.

Question 3.
Explain the following:
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper
(b) SiO₂ is treated with HF.
(c) Graphite is a lubricant
(d) Diamond is an abrasive.
Answer:
(a) Silicon on heating with methylchloride at 300°C in the presence of copper catalyst gives dimethyl dichlorosaline.

(b)When Silica (SiO₂) reacts with HF, it forms silicon tetra fluoride.
SiO₂ + 4HF → SiF₄ + 2H₂O

(c) Graphite contains layer like structure and between two layers weak Vander waal forces of attractions are present. As a result, it is slippery in nature. Hence graphite is used as lubricant.

(d) Diamond contains three dimensional polymeric structure. Hence diamond has giant molecule structure.
Due to this giant molecular structure, diamond is very hard and having high melting point.
Hence diamond is abrasive in nature.

Question 4.
What do you understand by (a) Allotropy (b) inert pair effect (c) Catenation. [AP 19]
Answer:
(a) Allotropy:
The phenomenon of existence of an element in different physical forms having same chemical properties is called allotropy.
Crystalline allotropes of carbon are
(a) Diamond (b) Graphite

(b) lncrt pair effect:
The reluctance of’ns’ pair of electrons to take part in bond formation is known as inert pair effect. Kg; Lead exhibits +2 oxidation state as stable oxidation state due to inert pair effect (instead of +4 state).

(c) Catenation:
The phenomenon of self-linkage of atoms among themselves to form long chains (or) rings is called as catenation. [TS 18]
Carbon has highest catenation tendency due to its high bond energy (348 KJ/mole)

Question 5.
If the starting material for the manufacture of silicones is R SiCl₃, write the structure of the product formed.
Answer:
When trichloro silane undergoes hydrolysis followed by polymerisation it forms cross linked silicones.
RSiCl₃ + 3H₂O → RSi(OH)₃.

Condensation of the hydrolysed product gives three dimensional silicone.

Question 6.
Write a short note on zeolites.
Answer:

1. Zeolites are three dimensional alumino silicates.
2. If some silicon atoms in three dimensional network silicon dioxide are replaced by aluminium atoms, then alumino silicate is formed.
3. They acquire a negative charge which is balanced by cations like Na⁺, K^-1, Ca²⁺ ions.
4. Generally zeolites have cage like structures with cavities.

Uses:

1. Zeolites acts as ion exchangers & molecular sieves.
2. Used in softening of hard water.
3. ZSM-5 Zeolite is used to convert alcohols directly into gasoline.

Question 7.
Write a short note on Silicates.
Answer:
Silicates :
Silicates are regarded as the metal derivatives of orthosilicic acid (H₄SiO₄). Most of the building materials are nothing but silicates.
Ex: Granites, slates, bricks and cement. Ceramics and glass are also silicates.

The Si-O bonds in silicates are very strong. They do not dissolve in any of the common solvents nor do they mix with other substances readily.

The silicates can be divided into six types.
(i) Orthosilicate or Nesosilicates:
Their general formula is M₂(SiO₄)
Ex: Willemite Zn₂(SiO₄)

(ii) Pyrosilicates or sorosilicates or Disil icatcs:
These contain Si₂O₇^-6 units. Pyrosilicates are rare.
Ex: Thortvteitite Ln₂(Si₂O₇)

(iii) Chain silicates:
They contain(SiO₃)_n^2n- units
Ex: Spodumene LiAl (SiO₃)₂.

Amphiboles are one type chain silicates. Generally double chains are formed in them.

(iv) Cyclic silicates:
They are silicates having ring structures. They are formed of general formule (SiO₃)_n^2n-. Rings containing three, four, six and eight tetrahedral units are known. But rings with three and six are the most common.
Ex: Beryl Be₃Al₂(Si₆O₁₈)

(v) Sheet silicates:
When SiO₄ units share three comers the structure formed is an inifinite two dimensional sheet. The empirical formula is (SiO₃)_n^2n-. These compounds appear in layer structures. They can be cleaved.
Ex: Kaolin Al₂(OH)₄ Si₂O₅.

(vi) Frame work silicates or three dimensional silicates:
Sharing all the four corners of a Si04 tetrahedron results in three dimensional lattice of formula SiO₂.
Ex: Quartz, Tridymite, Cristobalite, Feldspar and ultramarine (Na₈[Al₆Si₆O₂₄]S₂), zeolites.

Question 8.
What are silicones? How are they obtained.
Answer:
Silicones are a group of organo silicon polymers containing (R₂Si-0) as repeating unit

Preparation:
First alkyl or aryl substituted chlorosilanes are prepared. These on hydrolysis followed by condensation, polymerisation gives polymeric silicones.

This reaction continue to give chain polymers. If water molecule is eliminated from the terminal OH groups of same chain cyclic Silicones are formed.

Question 9.
Write a short note on fuilerene.
Answer:
Fullerenes are the crystalline allotropic form of carbon with cage like structure.

Preparation:
These are formed by strong heating graphite in an electric arc, in presence of inert gases like He or Ar.

Properties of Fuilerene:

1. In fuilerene, the carbon undergoes sp² hybridisation.
2. C₆₀ molecule has a shape like soccer ball and called “Buckminster fuilerene”.
3. The spherical fullerenes are called bucky balls.
4. It contains 20 six membered rings and 12 five membered rings. A six membered ring fused with six or five membered rings. But five membered ring can only fuse with six membered rings.
5. Ball shaped fuilerene has 60 vertices and each one is occupied by one carbon atom and it is also contains both single and double bonds.
   

Question 10.
Why SiO₂ does not dissolve in water?
Answer:

1. Silica (SiO₂) is giant molecule.
2. It has a three dimensional polymeric structure.
3. Each silicon atom is of sp³ hybridisation.
4. Each silicon is linked to four oxygen atoms tetrahedrally by strong covalent bonds. So it does not dissolve in water. But at high pressures, if heated with water, it dissolves slightly.

Question 11.
Why is diamond hard? [TS 22]
Answer:

1. In diamond, every carbon is in sp hybridisation. Every carbon is bonded with four other carbon atoms tetrahedrally.
2. The structure extends in space and produce a rigid three dimensional network of carbon atoms.
3. In this structure, directional covalent bonds are present throughout the lattice.
4. To break these covalent bonds, very high energy is required. Therefore diamond is the hardest substance and thus can be used as abrasive.
   

Question 12.
What happens when the following are heated?
a) CaCO₃ b) CaCO₃ and SiO₂
c) CaCO₃ and excess of coke.
Answer:
a) CaCO₃
CaCO₃ on heating gives quick lime and liberates CO₂.

b) CaCO₃ and SiO₂
CaCO₃ on heating forms calcium silicate (CaSiO₃) with SiO₂
CaCO₃ + SiO₂ → CaSiO₃ + CO₂ ↑

c) CaCO₃ and excess of coke
CaCO₃ on heating with excess of coke forms calcium carbide(CaC₂).
CaCO₃ + 4C → CaC₂ + 3CO

Question 13.
Why does Na₂CO₃ solution turn into a suspension when saturated with CO₂ gas?
Answer:
By passing excess of CO₂ through saturated solution of sodium carbonate, sodium bicarbonate is formed. Being sparingly soluble in water, it gets precipitated.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

Question 14.
What happens when
a) CO₂ is passed through slaked lime
b) CaC₂ is heated with N₂
Answer:
a) CO₂ is passed through slaked lime:
When CO₂ is passed through slaked lime, it turns milky due to the formation of insoluble CaCO₃
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

If excess CO₂ is passed, the milkyness disappears due to the conversion of insoluble CaCO₃ to soluble Ca(HCO₃)₂.
CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂

b) CaC₂ is heated w ith N₂:
On heating CaC₂ with N₂, a mixture of calcium cyanamide and graphite is fonned.

Question 15.
Write a short note on the anomalous behaviour of carbon in the group-14.
Answer:
Anomalous behaviour of carbon is due to small size, absence of d-orbitals and high nuclear charge.

Carbon differs from the other members of its group in the following respects:

1. Carbon occurs in free state, whereas other elements of group-14 are almost not available in free state.
2. Carbon does not contain d-orbitals in its valence shell while the other elements of its group contain d-orbitals.
3. The maximum covalency of carbon is 4 whereas other elements exhibit a maximum covalency of 6.
4. Carbon has a very high catenating ability.
5. Carbon alone forms multiple bonds among themselves.
6. Carbon forms a large number of hydrides known as hydrocarbons, which are thermally very stable. The other elements form limited number of hydrides which are thermally not very stable.
7. Reducing nature of carbon is very high while the reducing nature of others is less.
8. The halogen compounds of carbon are not hydrolysed, while the other halogen compounds are readily hydrolysed.

Long Answer Questions

Question 1.
What are silicones? How are they prepared? Give one example. What are their uses?
Answer:
Silicones are organosilicon polymers, which contain -[R₂ -Si -o}_n as repeating unit. They are prepared in two stages.
1) In the first stage alkyl or aryl substituted chlororsilanes are prepared.
2) In the second stage, these are hydrolysed followed by condensation polymerisation.

Preparation:
When silicon is heated with alkyl or aryl halides, different methyl or aryl substituted

The chain length of the polymer can be controlled by adding (CH₃)₃ SiCl. This blocks the ends as shown below’.

The alkyl and aryl groups present in silicones are water repelling in nature. They are thermally more stable. They have high dielectric constant. They are resistant to oxidation and chemicals.

They are used as sealants, greases, electrical insulators. They are used in making water proof fabrics. They are bio compatible. Hence they are used in surgical and cosmetic implants.

Question 2.
Explain the structure of silica. How docs it react with a) NaOH b) HF?
Answer:

Structure of Silica:
In silica, each silicon atom is linked to four oxygen atoms by covalent bonds and they are arranged tetrahedrally around silicon atom. But each oxygen atom is surrounded by two silicon atoms only. SiO₂ has a three dimensional giant Si molecular, structure

a) Reaction with NaOH:
Silica reacts with NaOH to form sodium silicate
2Na0H + SiO₂2 → Na₂SiO₃ + H₂O

b) Reaction with HF:
Silica reacts with HF to form silicon tetrafluoride.
SiO₂ + 4HF → SiF₄ + 2H₂O

Question 3.
Write a note on the allotropy of carbon.
Answer:
Allotropy:
The existence of an element in different physical forms but possessing similar chemical properties is known as allotropy.

Carbon has many allotropic forms. They are of both crystalline and amorphous. The crystalline allotropes of carbon are diamond, graphite and fullerenes.

Diamond:
In diamond, each carbon atom undergoes sp³ hybridisation and is in bond with four other carbon atoms in a tetrahedral form. This structure extends in space and produces a rigid three dimensional network of carbon atoms. These bonds are present throughout the lattice. To break all these extended bonds, large amount of energy is required. So diamonds are very hard and have high melting point. They are used as precious stones, abrasives, cutting tools etc.

Graphite:
Graphite has layered lattice structure, in which every carbon is in sp² hybridisation and in bond with three carbon atoms. The carbon atoms are arranged in sheets containing hexagonal rings. The fourth electron is in 7t bond and delocalised on all the carbon atoms in a sheet. The distance between the layers is 340 pm. They are held together by weak vander Waals forces. So they are slippery in nature. Hence graphite can be used as solid lubricant. Due to delocalised electrons graphite act as a good electrical conductor.

Fullerene:
Fullerene is a crystalline allotrope of carbon. Fullerenes are made by heating, graphite in an electric arc, in the presence of inert gas. The sooty material formed by condensation of vapour of carbon contains mainly C60 fullerene molecules. It is knowm as Buckminster fullerene. It has soccer ball shape.

C₆₀ fullerene contains 20 six membered rings and 12 five membered rings. A six membered is fused with either six membered ring or with five membered ring. But a five membered ring is alw ays fused with six membered rings. All carbon atoms in fullerene are in sp² hybridisation and are in bond with three other carbon atoms. The fourth electron of carbon delocalises on all carbon atoms giving aromatic character to fullerene.

Question 4.
Write a short note on (a) Silicates (b) Zeolites (c) Fiilierenes
Answer:
a) Silicates :
Silicates compounds formed from the orthosilicic acid. All silicates have the basic structural unit SiO^4-₄ which has tetrahedral structure.

When discrete SiO^4-₄ unit is present in silicates they are called orthosilicates.

Different silicates are formed by joining a number of SiO^4-₄ units through the comers of SiO^4-₄ tetrahedron by sharing 1, 2, 3 or 4 oxygen atoms per silicate unit. Thus 1, 2, 3, 4 -Pyro, chain(or) ring, sheet or three dimensional silicates are formed depending on the number of oxygen atoms shared. Negative charge on the silicate is neutralised by the positive charged metal ions.

b) Zeolites :
Zeolities are three dimensional alumino silicates. If some silicon atoms in three dimensional network silicon dioxide are replaced by aluminium atoms then alumino silicate is formed. They will have some negative charge. To balance these negative charges some extra cations such as Na⁺, K¹⁺, Ca²⁺ are present in these silicates.

Hydrated zeolites are used as ion exchangers in softening of hard water.
ZSM-5 is used as a catalyst in direct conversion of alcohols into gasoline.

c) Fiilierenes:
Fullerene is a crystalline allotrope of carbon. Fullerenes are made by heating graphite in an electrict arc in the presence of inert gas. The soot formed by condensation of vapour of carbon contains mainly C₆₀ fullerene molecules. It is knowrn as Buckminster fullerenes. It has soccer ball shape.

C₆₀ fullerene contains 20 six membered rings and 12 five membered rings. A six membered is fused with either six membered ring or with five membered ring. But a five membered ring is always fused with six membered rings. All carbon atoms in fullerene are in sp² hybridisation and are in bond with three other carbon atoms. The fourth electron of carbon delocalises on all carbon atoms giving aromatic character to fullerene.

Multiple Choice Questions

Question 1.
Dry ice is
1) Solid NH₃
2) Solid SO₂
3) Solid CO₂
4) Solid N₂
Answer:
3) SoliclCO₂

Question 2.
The correct order of atomic radii in group 13 elements is
1) B < Al < In< Ga < Tl
2) B < Al < Ga < In < Tl
3) B < Ga < Al < Tl < In
4) B < Ga < Al < In < Tl
Answer:
4) B < Ga < Al < In < Tl

Question 3.
Which one of the following elements is tinahle to form MF^3-₆ ion?
1) Ga
2) Al
3) B
4) In
Answer:
3) B

Question 4.
The most commonly used reducing agent is
1) AlCl₃
2) PbCl₂
3) SnCl₄
4) SnCl₂
Answer:
4) SnCl₂

Question 5.
The element which exists in liquid state for a wide range of temperature and can he used tor measuring high temperature is
1) B
2) Al
3) Ga
4) In
Answer:
3) Ga

Question 6.
Ionisation enthalpy (∆_i H₁ kJ mol^-1) for the elements of Group 13 follows the order.
1) B > Al > Ga > In > Tl
2) B < Al < Ga < In < Tl
3) B < Al > Ga < In > Tl
4) B > Al < Ga > In < Tl Answer: 4) B > Al < Ga > In < Tl

Question 7.
Which of the following oxides is acidic in nature?
1) B₂O₃
2) Al₂O₃
3) Ga₂O₃
4) In₂O₃
Answer:
1) B₂O₃

Question 8.
Which of the following is a Lewis acid?
1) AlCl₃
2) MgCl₂
3) CaCl₂
4) BaCl₂
Answer:
1) AlCl₃

Question 9.
In the structure of diborane
1) All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane.
2) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
3) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
4) All the atoms are in the same plane.
Answer:
3) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.

Question 10.
Boric acid is an acid because its molecule
1) contains replaceable H⁺ ion
2) gives up a proton
3) accepts OH^– from water releasing proton
4) combines with proton from water molecule
Answer:
3) accepts OH^– from water releasing proton

Question 11.
BF₃ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are
1) sp² and 8
2) sp³ and 4
3) sp³ and 5
4) sp² and 6
Answer:
4) sp² and 6

Question 12.
Quartz is extensively used as a piezoelectric material, as it contains ______
1) Pb
2) Si
3) Ti
4) Sn
Answer:
2) Si

Question 13.
Catenation i.e., linking of similar atoms depends on size and electronic contlguration of atoms. The tendency of catenation in Group 14 elements follows the order:
1) C > Si > Ge > Sn
2) C > > Si > Ge ≈ Sn
3) Si > C > Sn > Ge
4) Ge > Sn > Si > C
Answer:
2) C > >Si > Ge ≈ Sn

Question 14.
Which of the following is incorrect statement?
1) SnF₄ is ionic in nature
2) PbF₄ is covalent in nature
3) SiCl₄ is easily hydrolysed
4) GeX₄ (X=F,Cl,Br,I) is more stable than GeX₂
Answer:
2) PbF₄ is covalent in nature

Question 15.
Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixture?
1) group 2
2) groups 2, 13 and 14
3) groups 2 and 13
4) groups 2 and 14
Answer:
2) groups 2, 13 and 14

Question 16.
Which of the following species is not stable?
1) [SiCl₆]^2-
2) [SiF₆]^2-
3) [GeCl₆]^2-
4) [Sn(OH)₆]^2-
Answer:
1) [SiCl₆]^2-

Question 17.
Identify the correct statements from the following:
A) CO₂(g) is used as refrigerant for icecream and frozen food
B) The structure of C₆₀ contains twelve six carbon rings and twenty five carbon rings.
C) ZSM-5 a type of zeolite, is used to convert alcohols into gasoline
D) CO is colourless and odourless gas
1) (A), (B) and (C) only
2) (A) and (C) only
3) (B) and (C) only
4) (C) and (D) only
Answer:
4) (C) and (D) only

Question 18.
Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding
1) MeSiCl₃
2) Me₂SiCl₂
3) Me₃SiCl
4) Me₄Si
Answer:
3) Me₃SiCl

Students get through AP Inter 1st Year Chemistry Important Questions 10th Lesson The p-Block Elements – Group 13 which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 10th Lesson The p-Block Elements – Group 13

Very Short Answer Questions

Question 1.
Discuss the pattern of variation in the oxidation states of Boron to Thallium.
Answer:
All the elements of III A group exhibit a common oxidation state of +3. Except boron and aluminium all other elements show +1 oxidation state. This +1 state becomes more and more stable, as we go down the group.

Question 2.
(low do you explain higher stability of BCI3 over TlCl₃?
Answer:
Due to the poor shielding effect of the s-electrons of the valency shell by inner electrons, inert pair effect is maximum in Tl. But such effect is minimum in B. Hence BCl₃ is more stable when compared to TlCl₃.

Question 3.
Why does BF₃ behave as a Lewis acid? [AP 22]
Answer:
BF₃ is an electron deficient compound because it contains sextet configuration (6 electrons) in the valency shell of Boron. So it can accept a pair of electrons to get octet.
Hence BF₃ behaves as Lewis acid.

Question 4.
Is boric acid a protic acid? Explain.
Answer:
Boric acid is a weak mono basic acid. In Boric acid, planar BO₃ units are joined by hydrogen bonds. But it acts as a Lewis acid by accepting electrons from a hydroxyl ion. Hence it is not a protic acid.

Question 5.
What happens when boric acid is heated?
Answer:
When Boric acid is heated above 370K it forms meta boric acid. This on further heating forms Boric oxide.

Question 6.
Describe the shapes of BF₃ and BH^–₄ Assign the hybridization of boron in these species.
Answer:
Shape of BF₃ molecule is Trigonal planar. Hybridisation of ‘B’ in BF₃ is sp²

Shape of BH^–₄ molecule is Tetrahedral.
Hybridisation of ‘B’ in BH^–₄ is sp³.

Question 7.
Explain why atomic radius of Ga is less than that of ‘Al’. [AP 22]
Answer:
In Gallium, penultimate shell contains 10-d electrons. Due to this 10-d electrons, shielding effect becomes poor on outer most electrons. As a result, the electrons in gallium experience greater force of attraction by the nucleus than in Al. Hence atomic size of Ga(135pm) is slightly less than that of Al( 143pm).

Question 8.
Explain inert pair effect. [TS 17]
Answer:
The reluctance of ‘ns²’ electron pairs to take part in bond formation is called inert pair effect.
Ex: In Group -13, Tl exhibits + l stable oxidation state instead of +3 oxidation state due to inert pair effect.

Question 9.
Write balanced equations for

Answer:

Question 10.
Why is boric acid polymeric?
Answer:
H₃BO₃ (Boric acid) has layer like structure. In this structure planar BO₃ units are joined by hydrogen bonds Hence it forms a polymeric structure.

Question 11.
What is the hybridization of B in diborane and borazine?
Answer:
a) In diborane (B₂H₆), B undergoes sp³ hybridisation.
b) In borazine (B₃N₃H₆), B undergoes sp² hybridisation.

Question 12.
Write the electronic configuration of group-13 elements.
Answer:
The general electronic configuration of group-13 elements is ns²np¹

Question 13.
Give the formula of borazine. What is its common name? [TS 17]
Answer:
The formula of borazine is B₃N₃H₆.

It’s common name is “Inorganic benzene”. Because its structure resembles the structure of benzene ie., Cyclohexagon.

Question 14.
Give the formulae of
a) Borax b) Colemanite
Answer:
a) Borax : Na₂B₄O₇.10H₂O
b) Colemanite: Ca₂B₆O₁₁.5H₂O

Question 15.
Give two uses of aluminium.
Answer:
Aluminium is used :

1. for making electrical cables, trays, picture frames.
2. as a deoxidiser in metallurgy.
3. for making alloys.
4. as a reducing agent in thermite welding

Question 16.
What happens when a) LiAlH₄ and BCl₃ mixture in dry ether is warmed and b) Borax is heated with H₂SO₁₁?
Answer:
a) When LiAlH₄ and BCl₃ mixture is warmed in dry ether, diborane is formed
4BCl₃ + 3LiAIH₄ → 2B₂H₆ + 3LiCl + 3AlCl₃

b) When Borax is heated with H₂SO₄ then boric acid is formed.
Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃

Question 17.
Sketch the structure of Boric acid.
Answer:

Question 18.
Write the structure of AlCl₃ as dimer.
Answer:

Question 19.
Metal borides (having ¹⁰B) are used as protective shield -Why?
Answer:
Boron -10 (¹⁰B) has high ability to absorb neutrons. Hence metal borides (having ¹⁰B) are used as protective shields and control rods in nuclear industry.

Short Answer Questions

Question 1.
Write reactions to justify amphoteric nature of aluminium.
Answer:
Aluminium dissolves in both acids and alkalies. Hence it is amphoteric.

Reaction with Acid: Aluminium dissolves in dilute HCl and liberates dihydrogen.
2Al + 6HCl → 2AlCl₃ + 3H₂

Reaction with Base:
Aluminium also dissolves in aqueous alkali and liberates dihydrogen.
2Al + 2NaOH + 6H₂O → 2Na + [Al(OH)₄]^– + 3H₂

These reaction indicate the amphoteric nature of aluminium.

Question 2.
What are electron deficient compounds? Is BCl₃ an electron deficient species? Explain. [TS 22]
Answer:
The compounds in which the central atom has incomplete octet of electrons in its outermost orbit are called electron deficient compounds.

BCl₃ is an electron deficient compound.
In BCl₃ molecule, the central atom boron has only six electrons in its valency shell. Therefore BCl₃ molecule accepts an electron pair to achieve stable electronic configuration and thus behave as Lewis acid.

Eg: BCl₃ molecule easily accepts a lone pair of electrons from ammonia to form

Question 3.
Suggest reason why the B-F bond lengths in BF₃ (130pm) and BF^–₄ (143 pm) differ.
Answer:
BF3 is a planar molecule in which B is sp² hybridised. It has an empty 2p-orbital. F-atom has three lone pairs of electrons in the 2p- obritals. Because of its small size and strong inter electronic repulsions, pπ – pπ back bonding or back donation occurs in which a lone pair is transfered from F to B.

As a result of this back bonding, B-F bond acquires some double bond character. In contrast, in [BF^–₄] ion, B is sp-’ hybridised and hence does not have an empty p-orbital available to accept the electrons donated by the F atom. Consequently, in B-F is a purely single bond. Since double bonds are shorter than single bonds, the B-F bond length in BF₃ is shorter (130 pm) than B-F bond length (143pm) in [BF^–₄]

Question 4.
B-Cl bond has a bond moment. Explain why BCl₃ molecule has zero dipole moment.
Answer:
B-Cl bond has a certain dipole moment because it has polar nature. But BCl₃ has zero dipole moment since the molecule is symmetrical (planar triangular) in which net dipole moment (μ) is zero.

Question 5.
Explain the structure of Boric acid.
Answer:
Boric acid contains planar BO₃ units which are linked together through hydrogen bonding. Hydrogen atoms act as bridge between two oxygen atoms of different BO₃ units. This results in a trigonal planar structure. Boron atom in boric acid is sp² hybridized.

The dotted lines represent hydrogen bonds.

Question 6.
What happens when
(a) Borax is heated strongly
(b) Boric acid is added to water
(c) Aluminium is heated with dilute NaOH
(d) BF₃ is treated with ammonia
(e) Hydrated alumina is treated with aq. NaOH solution.
Answer:
(a) On heating, borax loses water molecules and swells into a white, opaque mass of anhydrous sodium tetraborate.
On further heating, it turns into a transparent liquid, which modifies into glass like material known as borax bead.

(b) Boric acid is slightly soluble in cold water and is readily soluble in hot water.
B(OH)₃ + 2H₂O → [B(OH)₄]^– + H₃O⁺

(c) When Aluminium is dissolved in dilute NaOH it liberates hydrogen gas.
2Al + 2NaOH + 6H₂O → 2Na⁺[Al(OH₄)]^– + 3H₂ ↑

(d) When BCl₃ is treated with NH₃ it forms Ammonium boron trifluoride.
BCl₃ + NH₃ → [BCl₃←NH₃] → [BCl₃.NH₃]

(e) When Hydrated Aluminium is treated with aq.NaOH solution it forms Sodium meta aluminate.

Question 7.
Give reasons
a) Cone HNO₃ can be transported in aluminium container.
b) A mixture of dil. NaOH and aluminium pieces is used to open drain.
c) Aluminium alloys are used to make aircraft body.
d) Aluminium utensils should not be kept in water overnight.
e) Aluminium wire is used to make transmission cables.
Answer:
a) Cone. HNO₃ initially reacts with aluminium to form aluminium oxide (Al₂O₃). This forms a protective coating inside the container. The metal becomes passive and it does not react with the acid any more. Therefore, the acid can be safely stored in aluminium container.

b) NaOH reacts with Al to evolve dihydrogen gas, whose high pressure is enough to open clogged drains.

c) Alloys of aluminium like duralumin are light, tough and resistant to corrosion. Hence they are used in making air craft bodies.

d) Aluminium reacts with water and dissolved O₂ to form a thin film of aluminium oxide.
2Al + O₂ + H₂O → Al₂O₃ + H₂

e) Al metal is not affected by air, moisture. And also due to its good conductivity, it is used to make transmission cables.

Question 8.
Explain why the electronegativity of Ga, In and Tl will not vary very much.
Answer:
The d-electrons present in Ga, In and Tl cannot shield the valence electrons from the nuclear attractions. Hence their valence electrons are attracted strongly by the nucleus. Hence, the electro negativity of Ga, In and Tl will not vary much.

Question 9.
Explain Borax bead test with a suitable examples. [AP 18, 20][TS 16, 19, 20]
Answer:
Borax bead test:
This test is used for the identification of basic radicals in qualitative analysis.

On heating, borax loses water molecules and swells into a white, opaque mass of anhydrous sodium tetraborate.

On further heating, it turns into a transparent liquid, which modifies into glass like material known as borax bead.

The metaborates of many transition elements have characteristic colours and therefore, borax bead test can be used to identify them in the laboratory.
Ex: When borax is heated in a bunsen burner flame with cobalt oxide on a loop of platinum wire, a blue coloured Co(BO₂)₂ bead is fonned.

Question 10.
Explain the structure of diborane. [AP,TS 15,16,17,18,19,19]
Answer:
I) Structure of diborane:
The molecular formula of diborane is B₂H₆. Electron diffraction studies have shown that diborane contains two coplanar BH₂ groups.
Its structure can be represented as follows:

II) Structure of diborane:

1. Diborane contains 2 coplanar BH₂ groups.
2. The four hydrogen atoms located at the ends as shown in the above figure are known as terminal hydrogen atoms (H_t).
3. The middle positioned 2 hydrogen atoms are called bridge hydrogens (H_b).
4. These two bridge hydrogens lie in a plane perpendicular to the plane of the BH₂ groups.
5. One of the bridge hydrogens lies above the plane and the other lies below the plane.
6. In diborane, Boron undergoes sp³ hybridisation to form 4 sp³ hybrid orbitals.
7. Out of 4 orbitals, 3 orbitals contain one electron each and the fourth orbital is vacant.
8. The 2 sp³ hybrid orbitals of each Boron atom fonns 2 sigma bonds with 2 H atoms.
9. The bridge between two Boron atoms is formed due to overlap of vacant sp³ orbital of one Boron, ‘s’ orbital of hydrogen and sp³ orbital of another Boron containing one electron.
10. Hence this hydrogen bridge is considered as ‘three centered two electron bond’ (or) Banana bond (or) Tau bond.
    

Question 11.
Explain the reactions of aluminium with acids.
Answer:
1) Aluminium reacts with dilute or cone. HCl and liberates H₂.
2Al + 6HCl → 2AlCl₃ + 3H₂

2) Al reacts slowly with dil. H₂SO₄ in cold condition but reacts fast in hot condition.
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

3) Al reacts with cone. H₂SO₄ and liberates SO₂.
2Al + 6H₂SO₄ → Al₂(SO₄)₂ + 6H₂O + 3SO₂

4) Al reacts with very dil. HNO₃ and gives NH₄NO₃.
8Al + 30HNO₃ → 8Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O

5) Concentrated nitric acid renders the aluminium passive.

Question 12.
Write a short note on the anomalous behaviour of boron in the group-13.
Answer:
Anomalous of behaviour of boron is due to
a) Its small size.
b) The difference in the penultimate shell configuration.
c) High first ionization potential.
d) Absence of d-orbitals.

Anomalous properties of Boron:

1. Boron is a non-metal. ‘Al’ is amphoteric. Ga, In and Tl are metals.
2. B₂O₃ is an acidic oxide. The trioxides of others are either amphoteric or basic in nature.
3. B(OH)₃ is an acid while the hydroxides of other elements are either amphoteric or basic.
4. Boron always forms covalent compounds, while the others form ionic compounds.
5. Boron has a maximum covalency of 4 only. But others exhibit a maximum covalency of 6.
6. Boron shows diagonal relation ship with silicon. Such a relationship is not shown by other elements.
7. Boron does not displace hydrogen from acids while others displace hydrogen from acids under suitable conditions.
8. Boron can form stable hydrides while the others cannot form stable hydrides.
9. Simple borates and silicates can polymerise readily forming polyacids while others do not form such polymers.

Question 13.
Aluminium reacts with dil.HNO₃ but not with cone. HNO₃ – explain.
Answer:
Dilute HNO_3- reacts with aluminium slowly and forms aluminium nitrate and ammonium nitrate.
8Al + 30HNO₃ → 8Al(NO₃)₃ + 3NH₄NO₃ + 9H₂O
Aluminium does not react with cone. HNO₃.

Reasons:
Aluminium is passive towards cone. HNO₃ due to the formation of thin film of Al₂O₃ layer on the surface.
Because of this passivity between Al and cone. HNO₃, cone. HNO₃ is transported in Aluminium containers.

Question 14.
Give two methods of preparation of diborane.
Answer:
1) Diborane can be prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃

2) In the laboratory, it is conveniently prepared by the oxidation of sodium borohydride with iodine.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

3) On large scale, it is prepared by reaction of BF3 with sodium hydride.

Question 15.
How does diborane react with [AP 19]
a) H₂O
b) CO
c) N(CH₃)₃
d) NH₃
Answer:
a) Diborane reacts with water to give Boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃(aq) + 6H₂(g)

b) Diborane reacts with carbon monoxide to give Boran Carbonyl.
B₂H₆ + 2CO → 2BH₃.CO

c) Diborane reacts with trimethyl ammine to give an adduct.
B₂H₆ + 2N(CH₃)₃ → 2BH₃N(CH₃)₃

d) Dibroane reacts with ammonia it forms an addition product with asymmetric cleavage which on heating forms borazine which is known as inorganic benzene.

Question 16.
Al₂O₃ is amphoteric -explain with suitable reactions.
Answer:
Al₂O₃ reacts with both acids and bases. While reacting with acids it behaves like base. In the reaction with bases it behaves like acid. So it is amphoteric.

Reaction with acid:
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Reaction with base:
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Question 17.

Answer:
When borax is heated with conc.H₂SO₄ boric acid is formed.
A mixture of ethyl alcohol with boric acid burns with green edged flame due to the formation of ethylborate.

Long Answer Questions

Question 1.
How are borax and boric acid prepared? Explain the action of heat on them.
Answer:
Borax (Sodium tetraborate-Na₂B₄O₇) is an important compound of boron.

I) Preparation of Borax:
1) From tincal:
Borax is obtained from tincal by boiling it with water. The solution is filtered to remove in soluble impurities of sand, clay etc. The solution is concentrated till the crystals of borax separate out.

2) From Colemanite: Colemanite ore (Ca₂B₆O₁₁). When boiled with sodium carbonate solution, borax is formed along with sodium metaborate.

3) From boric acid:
Small quantities of borax are obtained by neutralizing boric acid solution with soda ash.
4H₃BO₃ + Na₂CO₃ → Na₂B₄O₇ + 6H₂O + CO₂
On cooling, crystals of Na₂B₄O₇.10H₂O separate out.

II) Preparation of Orthoboric acid(H₃BO₃):
1) Orthoboric acid is usually prepared by acidifying an aqueous solution of borax.
Na₂B₄O₇ + 2HCl + 5H₂O → 2NaCl + 4B(OH)₃

2) Orthoboric acid is also prepared by the hydrolysis of boron compounds[halides, hydrides etc]
BCl₃ + 3H₂O → H₃BO₃ + 3HCl
BN + 3H₂O → H₃BO₃ + NH₃↑

III) Action of heat on Borax:
1) On heating borax swells into a white, opaque mass of anhydrous sodium tetraborate.
When it is fused, borax glass is obtained. This contains metaborate and B₂O₃.

2) When borax is heated with cone. H₂SO₄ boric acid is formed.
Na₂B₄O₇ + H₂SO₄ + 5H₂O → Na₂SO₄ + 4H₃BO₃

IV) Action of heat on boric acid:
Boric acid on heating first gives metaboric acid at low temperature but at red hot condition forms boron trioxide.

Question 2.
How is diborane (B₂H₆)prepared? Explain its structure.
Answer:
I) Structure of diborane:
The molecular formula of diborane is B₂H₆. Electron diffraction studies have shown that diborane contains two coplanar BH₂ groups.
Its structure can be represented as follows:

II) Structure of diborane:

1. Diborane contains 2 coplanar BH₂ groups.
2. The four hydrogen atoms located at the ends as shown in the above figure are known as terminal hydrogen atoms (H_t).
3. The middle positioned 2 hydrogen atoms are called bridge hydrogens (H_b).
4. These two bridge hydrogens lie in a plane perpendicular to the plane of the BH₂ groups.
5. One of the bridge hydrogens lies above the plane and the other lies below the plane.
6. In diborane, Boron undergoes sp³ hybridisation to form 4 sp³ hybrid orbitals.
7. Out of 4 orbitals, 3 orbitals contain one electron each and the fourth orbital is vacant.
8. The 2 sp³ hybrid orbitals of each Boron atom fonns 2 sigma bonds with 2 H atoms.
9. The bridge between two Boron atoms is formed due to overlap of vacant sp³ orbital of one Boron, ‘s’ orbital of hydrogen and sp³ orbital of another Boron containing one electron.
10. Hence this hydrogen bridge is considered as ‘three centered two electron bond’ (or) Banana bond (or) Tau bond.
    

1) Diborane can be prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃

2) In the laboratory, it is conveniently prepared by the oxidation of sodium borohydride with iodine.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

3) On large scale, it is prepared by reaction of BF3 with sodium hydride.

Question 3.
Write any two methods of preparation of diborane. [AP 18]
How does it react with a) Carbon monoxide and b) Ammonia?
Answer:
1) Diborane can be prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃

2) In the laboratory, it is conveniently prepared by the oxidation of sodium borohydride with iodine.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

3) On large scale, it is prepared by reaction of BF3 with sodium hydride.

4) Diborane reacts with water to give Boric acid.
B₂H₆(g) + 6H₂O(l) → 2B(OH)₃(aq) + 6H₂(g)

5) Diborane reacts with carbon monoxide to give Boran Carbonyl.
B₂H₆ + 2CO → 2BH₃.CO

6) Diborane reacts with trimethyl ammine to give an adduct.
B₂H₆ + 2N(CH₃)₃ → 2BH₃N(CH₃)₃

7) Dibroane reacts with ammonia it forms an addition product with asymmetric cleavage which on heating forms borazine which is known as inorganic benzene.

Students get through AP Inter 1st Year Chemistry Important Questions 9th Lesson The s-Block Elements which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 9th Lesson The s-Block Elements

Very Short Answer Questions

Question 1.
Give reasons for the diagonal relationship observed in the periodic table.
Answer:
The diagonal relationship is due to nearly equal
a) Ionic or atomic sizes.
b) Electronegative values.
c) Polarising powers.

Question 2.
Write completely’ the electronic configuration of K and Rb.
Answer:
The E.C of ‘K’ with atomic number Z = 19 is
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

The E.C of ‘Rb’ with atomic number Z = 37 is
1s² 2s² 2p6 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹

Question 3.
Lithium salts are mostly hydrated. Why? [TS 15, 22]
Answer:
Lithium is the smallest in size among the alkali metals. Hence, Li⁺ ion can polarize water molecules more easily than other alkali metals.

As a result, Li⁺ has maximum degree of hydration and lithium salts are mostly hydrated.
Ex: LiCl.2H₂O

Question 4.
Which of the alkali metals shows abnormal density? What is the order of the variation of density among the IA group elements? [AP 18]
Answer:
a) In the alkali metals, Potassium(K) shows abnormal density due to
i) abnormal increase in its atomic size.
ii) Presence of vacant 3d orbitals
iii) Empty space in its crystal lattice.

b) Order of density of IA group elements is Li < Na > K < Rb < Cs.

Question 5.
Lithium reacts with water less vigorously than sodium. Give your reasons. [TS 18]
Answer:
Lithium reacts with water less vigorously than Sodium, because Lithium has small size and very high hydration energy than Sodium.

Question 6.
Lithium iodide is the most covalent among the alkali metal halides. Give the reasons.
Answer:
Reasons:

1. Li⁺ ion has very small size and I^– ion has large size.
2. The polarising power of lithium ion is high.
3. Li⁺ ion has high tendency to distort electron cloud around the iodide ion.
4. So Lithium iodide is the most covalent among the alkali metal halides.

Question 7.
In what respects, lithium hydrogen carbonate differs from other alkali metal hydrogen carbonates?
Answer:
Except Lithium hydrogen carbonate, other alkali metal hydrogen carbonates are solids. Lithium does not exist in solid state due to its less electropositive nature.

Except Lithium hydrogen carbonate, other alkali metal hydrogen carbonates are decomposed on heating.

Question 8.
Write the complete electronic configurations of any two alkaline earth metals.
Answer:
The E.C. of Be with atomic number Z = 4 is 1s² 2s²
The E.C. of Mg with atomic number Z = 12 is 1s² 2s² 2p⁶ 3s²

Question 9.
Tell about the variation of melting points and boiling points among the alkaline earth metals.
Answer:
Due to low I.P. values and different crystalline structures, the variation of melting points and boiling points among alkaline earth metals is irregular.

Variation of M.P:
Be>MgSr>Ba>Ra (Be>Ca>Sr>Ba>Ra>Mg)

Variation of B.P:
Be>MgSrRa (Be>Ba>Ca>Sr>Mg)

Question 10.
What are the characteristic colours imparted by the IIA elements? [TS 22]
Answer:
Be and Mg do not produce any colour in bunsen flame. But other metals produce the following colours.
Calcium – Brick red
Strontium – Crimson red
Barium – Apple green

Question 11.
What happens when magnesium metal is burnt in air? [TS 15, 18, 19]
Answer:
When Magnesium is burnt in air, it bums with dazzling brilliance and gives MgO and Mg₃N₂.
2Mg + O₂ → 2MgO
3Mg + N₂ → Mg₃N₂

Question 12.
Lithium carbonate is not so stable to heat as the other alkali metal carbonates. Explain.
Answer:
Lithium ion being very small in size and polarises the bigger carbonate ion and distort its electron cloud. So it dissociates forming stable Li₂O.
Li₂CO₃ → Li₂O + CO₂

Question 13.
Write a balanced equation for the formation of ammoniated group IIA metal ions from the metals in liquid ammonia.
Answer:
Alkaline earth metals dissolve in liquid ammonia and give deep blue black solutions . forming ammoniated ions.
M + (x + 2y)NH₃ → MM(NH₃)_x]⁺² + 2[e(NH₃)_y]^–

Question 14.
The fluorides of alkaline earth metals are relatively less soluble than their respective chlorides in water. Why?
Answer:
The fluorides of alkaline earth metals are relatively less soluble in water than the chlorides due to their high lattice energies.

Question 15.
What happens when hydrated Mg(NO₃)₂ is heated? Give the balanced equation. [AP 22]
Answer:
When hydrated Mg(NO₃)₂ is heated it decomposes and gives MgO.
2Mg(NO₃)₂ → 2MgO + 4NO₂ + O₂

Question 16.
Why does the solubility of alkaline earth metal hydroxides in water increases down the group? [AP 20]
Answer:
The lattice enthalpy decreases much more than the hydration enthalpy, with increasing ionic size down the group. So the solubility increases as we go down the group.

Question 17.
Why does the solubility of alkaline earth metal carbonates and sulphates in water decrease down the group?
Answer:
Down the group, with increase atomic size the lattice enthalpies and hydration enthalpies of carbonates and sulphates decrease. But the decrease in hydration enthalpies is rapid than lattice enthalpies. So the solubilities of carbonates and sulphates decrease down the group.

Question 18.
Write the average compositiqn of Portland cement.
Answer:
Average composition of portland cement:
CaO : 50%-60%
SiO₂ : 20% – 25%’ :
Al₂O₃ : 5 % -10%
MgO : 2% – 3%
SO₃ : 1% – 2%
Fe₂O₃ : 1% – 2%

Question 19.
Why is gypsum added to cement? [TS 15, 19]
Answer:
Gypsum (Calcium sulphate dihydrate) is added to cement to slow down the setting time. Hence hardness of cement increases.

Question 20.
Why are alkali metals not found in the free state in nature? [Mar 13][ AP 17]
Answer:
Alkali metals are highly reactive. Hence they do not found in the free state in nature.

Reason :
All alkali metals have one valance electron, ns¹. The loosly held s-electron in the outermost valence shell of these elements makes them the most electropositive metals. They readily lose electron to give monovalent M⁺ ions.

Question 21.
Potassium carbonate cannot be prepared by Solvay process. Why? [AP 19]
Answer:
The Solvay process cannot be extended for the manufacture of K₂CO₃, because KHCO₃ is more soluble in water unlike NaHCO₃. So it cannot be isolated.

Question 22.
Describe the important uses of Caustic Soda (or) sodium hydroxide. [TS 18, 19][AP 15, 16, 18]
Answer:
Sodium hydroxide (NaOH) is used

1. in the manufacture of Soaps, paper industries.
2. in ‘petroleum refining’.
3. in ‘textile finishing’.
4. as a ‘laboratory reagent’.

Question 23.
Describe the important uses of sodium carbonate. [AP 20]
Answer:
Sodium Carbonate (Na₂CO₃) is used

1. to remove hardness of water.
2. in the preparation of glass, caustic soda.
3. in laundries as washing soda.
4. in paper, paints and petroleum industries.

Question 24.
Describe the important uses of quick lime [AP 19]
Answer:
Quick lime(CaO) is used:

1. as a primary material for manufacturing cement.
2. in the manufacture of Na₂CO₃ from caustic soda.
3. in the purification of sugar and dye stuffs.

Question 25.
Draw the structures of i) BeCl₂ (vapour) and ii) BeCl₂ (solid).
Answer:
a) Structure of BeCl₂ (vapour) :
BeCl₂ exists as a dimer in vapour state.

b) Structure of BeCl₂ (solid):
BeCl₂ exists as a polymer in solid state.

Question 26.
Describe the importance of Plaster of Paris.
Answer:
Plaster of paris is used in

1. Surgical bandages for bone fracture
2. making white chalks.
3. making casts for statues, roofs, toys etc.

Question 27.
Which of the alkaline earth metal carbonates is thermally the most stable? Why?
Answer:
BaCO₃ is thermally most stable among Alkaline earth metal carbonates

Reason:
The bigger cation Ba²⁺ having less polarising power, cannot distort the carbonate ion. So its stability is more.

Question 28.
Write balanced equations for the reactions between
(i) Na₂O₂ and water (ii) K₂O and water.
Answer:
i) Na₂O₂ + 2H₂O → 2NaOH + H₂O₂
(ii) K₂O + H₂O → 2KOH

Short Answer Questions

Question 1.
Alkali metals and their salts impart characteristic colours to an oxidizing flame. Explain the rea$onizing flame. Explain the reason.
Answer:
Alkali metals and their salts impart characteristic colours to oxidising flame. Reason: The heat from the flame excites the outer most electron to a higher energy level. When excited electrons come back to the ground state, they emit the absorbed energy in the form of light in visible region.
Lithium – Crimson red
Sodium – Yellow
Potassium – Lilac or pale violet
Rubidium – Red violet
Caesium – Blue violet

Question 2.
What makes caesium and potassium useful as electrodes in photoelectric cells?
Answer:
Potassium(K) and Caesium(Cs) have low- ionisation energy and can easily lose electrons.

Alkali metals when irradiated with light, the light energy absorbed is just sufficient to make an atom lose electron.

This makes caesium and potassium useful as electrodes in photo electric cells.

Question 3.
Write a short note on the reactivity of alkali metals towards air.
Answer:
Reactivity of alkali metals towards air: The alkali metals tarnish in dry air due to formation of their oxides, which in turn react with moisture to form hydroxides. They bum vigorously in oxygen and form oxides. Lithium forms only monoxide.
4Li + O₂ → 2Li₂O

Sodium forms monoxide in limited supply of air but forms peroxide in excess of oxygen.
4Na + O₂ → 2Na₂O
2Na + O₂ → Na₂O₂
Other alkali metals form superoxides.
M + O₂ (excess) → MO₂ (M = K, Rb, Cs)

Question 4.
Give any two uses for each of the following metals, i) Lithium ii) Sodium
Answer:
i) Lithium is used

1. in making electrochemical cells.
2. in thermo nuclear reactions.
3. in the preparation of alloys.
   Ex: White metal is an alloy of lithium and lead used in making bearings for motor engines. The alloy of lithium-magnesium is used to make armour plates.

ii) Uses of Sodium ;

1. Sodium-lead alloy is used in making tetraethyl lead, an anti knocking agent in petrol.
2. Liquid sodium metal is used as coolant in nuclear reactors.

Question 5.
Give an account of properties of Washing soda?
Answer:
Properties of Washing soda:

1. Na₂CO₃ is a white crystalline solid.
2. Na₂CO₃ exists as a decahydrate ,(NaCO₃.10H₂O).
   This is called washing soda
3. Na₂CO₃ is readily soluble in water
4. On heating it loses water of crystallization and froms monohydrate.
   
5. Above 373K, the monohydrate becomes completely anhydrous and changes to a white powder called soda ash.
   
6. Carbonate part of sodium carbonate gets hydrolysed by water to fomi an alkaline solution.
   CO^2-₄ + H₂O → HCO^–₃ + OH^–

Question 6.
Mention r some uses of Sodium carbonate.
Answer:
Sodium earbonate(Washing soda) is used

1. in water softening, cleaning and laundries.
2. in the manufacture of glass, water glass, caustic soda, paper dyes.
3. in paper, paints and textile industries. ,
4. as a reagent in the laboratory in qualitative and quantitative analysis.

Question 7.
How do you obtain pure sodium chloride from a crude sample?
Answer:

1. Crude sodium chloride, contains sodiumsulphate, calcium sulphate, calcium chloride and magnesium chloride as impurities.
2. First the crude sodium chloride is dissolved in minimum amount of water and filtered to remove insoluble impurities.
3. Now hydrogen chloride gas is passed into the saturated solution of sodium chloride.
4. Then pure sodium chloride crystallises out.

Question 8.
What do you know about Castner-Kellner process? Write the principle involved in it.
Answer:

1. Castner-Kellner process is a commercial method used for the preparation of sodium hydroxide.
2. In this process sodium hydroxide is prepared by the electrolysis of sodium chloride,
3. Brine solutiont (NaCl solution) is electrolysed using a mercury cathode and a carbon anode.
4. Sodium metal is formed at cathode and it combine with mercury to form sodium amalgam.
5. Chlorine gas is evolved at anode.
6. The amalgam is treated with water to form sodium hydroxide.
   

This amalgam is treated with water to give sodmmhydroxide and hydrogen gas.
2(Na – Hg) + 2H₂O → 2NaOH + 2Hg + H₂
This process is also called as mercury cathode process.

Question 9.
Write a few’ applications of caustic soda. [May’ 13][AP 15]
Answer:
Caustic Soda (NaOH) is used

1. in the manufacture of Soaps, paper- industries.
2. in ‘petroleumrefining’.
3. in ‘textile finishing’.
4. as a ‘laboratory reagent’.

Question 10.
Give an account of biological importance of Na⁺ and K⁺ ions. [AP 18][TS 17]
Answer:
Sodium(Na):

1. Sodium ions are found primarily in the blood plasma.
2. They are also found in the interstitial fluids surrounding cells.
3. Sodium ions help in the transmission of nerve signals.
4. They help in regulating the flow of water across the cell membranes.
5. They also help in transporting sugars and aminoacids into the cells.

Potassium (K):

1. Potassium ions are found in the highest quantity within the cell fluids.
2. K⁺ ions help in activating many enzymes.
3. They also participate in oxidising glucose to produce ATP.
4. They also help in transmitting nerve signals.

Question 11.
Mention the important uses of Mg metal. [TS 19]
Answer:
Uses of Magnesium:

1. Mg forms alloys with Al, Zn, Mn and Sn.
2. Mg is used in flash powders and bulbs, incendiary bombs and signals.
3. A suspension of Mg(OH)₂ in water is known as Milk of magnesia. It is used as antacid and laxative in medicine.
4. MgCO₃ is an ingredient of toothpaste.

Question 12.
Show that Be(OH)₂ is amphoteric in nature.
Answer:
Beryllium hydroxide is amphoteric in nature as it reacts with both acids and bases.

1) Reaction with Acid:
Be(OH)₂ + 2HCl → BeCl₂ + 2H₂O
The reaction of Be(OH)₂ with hydrochloric acid shows its basic nature.

2) Reaction with Base:
Be(OH)₂ + 2NaOH → Na₂BeO₂ + 2H₂O
The reaction with sodium hydroxide shows its acidic nature.
So Be(OH)₂ is amphoteric.

Question 13.
Write a note on the anomalous behaviour of Beryllium.
Answer:
Anomalous characters of Beryllium:

1. Compounds of Be are predominantly covalent.
2. Be is not easily affected by air and does not decompose water at ordinary temperature.
3. Be is an amphoteric metal. It dissolves in alkali solutions forming beryllates.
4. BeSO₄ is soluble in water whereas the sulphates of Ca, Sr and Ba are insoluble.
5. Be and its salts do not respond to flame test while Ca,Sr and Ba give characteristic flame colours.
6. Be forms many complexes, while heavier elements do not form complexes easily.
7. Be has a maximum covalency of 4, while others have a maximum covalency of 6.

Question 14.
Be shows diagonal relationship with Al. Discuss.
Answer:
Beryllium shows a diagonal relationship with aluminium in the following respects.

1. Both Be and Al have the same EN value (1.50)
2. Both the compounds ofBe and Alundergo hydrolysis.
   BeCl₂ + 2H₂O → Be (OH)2 + 2HCl
   AlCl₃ + 3H₂O → Al (OH)₃ + 3HCl
3. Be and Al are rendered passive by cone HNO₃.
4. Both Be and Al form complexes.
5. Both Be and Al are amphoteric metals. They dissolve in alkali and form beryllates and aluminates respectively.
   Be + 2 NaOH → Na₂BeO₂ + H₂
   2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂
6. The carbides of Be and Al liberate CH₄ gas on treatment with water.
   Be₂C + 4 H₂O → 2Be(OH)₂ + CH₄
   Al₄Cl₃ + 12 H₂O → 4Al(OH)₃ + 3CH₄

Question 15.
What is Plaster of Paris? Write a short note on it. [AP 15, 16, 17, 18, 19, 22][TS 16, 17]
Answer:
1) Plaster of Paris:
Hemih.ydrate of calcium sulphate (CaSO₄ \(\frac{1}{2}\) H₂O) is called plaster of paris.

2) It is obtained when gypsum (CaSO₄. 2H₂O) is heated to 393K .

3) Above 393K, it loses water molecules to form anhydrous calcium sulphate (CaSO₄). This is known as ‘dead burnt plaster’.

4) It has remarkable property of setting with water on mixing with an adequate quantity of water. It forms a plastic mass that gets into a hard solid in 5 to 15 minutes.
5) Plaster of paris is used
(i) in making casts for statues, toys etc,
(ii) in surgical bandages for bone fracture
(iii) in making white chalks.

Question 16.
In what ways lithium shows similarities to magnesium in its chemical behavior?
Answer:
Li shows similarity with Mg in the following respects.

1. Li & Mg have diagonal relationship.
2. Both Lithium and Magnesium are harder and lighter than other elements in the respective groups.
3. Both Lithium and Magnesium give monoxides only.
4. Lithium and Magnesium react slowly with water.
5. LiCl deliquescent ( absorbs water from atmosphere) like MgCl₂.
6. Halides of Lithium and Magnesium are soluble in ethanol
7. Both Li⁺ and Mg⁺² ions are highly hydrated.
8. The carbonates, phosphates and fluorides of both Lithium and Magnesium are sparingly soluble in water.
9. Alkyl Lithium is chemically similar to Grignard reagents(R-Mg-X) in organic synthesis.

Question 17.
When an alkali metal dissolves in liquid ammonia, the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:

1. The Alkali metals dissolve in liquid ammonia and they give deep blue solutions. They are conducting in nature.
   M + (x + y)NH₃ → MM(NH₃)_x]⁺ + [e(NH₃)_y]^–
2. The blue colour of the solution is due to the ammoniated electrons which absorb energy in the visible region of light and thus impart blue colour to the solution.
3. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide.
   M⁺ + e^– +NH₃ → MNH₂ + \(\frac{1}{2}\)H₂
4. In concentrated solution, on warming, the blue colour changes to bronze colour and becomes diamagnetic.

Question 18.
What happens when (i) Sodium metal is dropped in water? (ii) Sodium metal is heated in a free supply of air? (iii) Sodium peroxide dissolves in water?
Answer:

1. When Sodium metal is dropped in water, it reacts with water vigorously and liberates H₂ gas.
   2Na + 2H₂O → 2NaOH + H₂
2. When Sodium metal is heated in free supply of air it forms Sodium peroxide.
   
3. When Sodium peroxide is dissolved in water it forms NaOH and Hydrogen peroxide.
   Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 19.
State as to why i) An aqueous solutions of Na₂CO₃ is alkaline. ii)Alkali metals are prepared by the electrolysis of their fused chlorides?
Answer:
i) When Sodium carbonate is added to water, it undergoes anionic hydrolysis to give weak carbonic acid (H₂CO₃) and strong base sodium hydroxide (NaOH). NaOH being strong electrolyte ionises completely but H₂CO₃ does not ionise. So the solution is alkaline.
Na₂CO₃ + 2H₂O ⇌ 2Na⁺ + 2OH^– + H₂CO₃

ii) Alkali metals themselves are strong reducing agents. Any reducing agent stronger than alkali metal is no where available. Hence they cannot be extracted by chemical reduction methods.

When aqueous solutions of alkali metal salts are electrolysed, hydrogen gas will be liberated at cathode instead of alkali metal. This is because, the discharge potential of H⁺ is less than Na⁺.

Hence alkali metals are prepared by the electrolysis fused chlorides. To decrease the melting points of the alkali metal halides, they are mixed with some other compounds.

Question 20.
How would you explain the following observations?
i) BcO is almost insoluble but BeSO₄ is soluble in water?
ii) BaO is soluble but BaSO₄ is insoluble in water?
Answer:
i) BeO is almost insoluble due to its high degree of covalency.
But, BeSO₄ is soluble in water due to high hydration energy of Be⁺² ion.

ii) BaO is soluble in water due to its high ionic nature.
But. BaSO₄ is insoluble in water due to low heat of hydration of Ba⁺²

Long Answer Questions

Question 1.
Justify the inclusion of alkali metals in the same group of the periodic table with reference to the following, i) Electronic configuration ii) Reducing nature iii) Oxides and hydroxides.
Answer:
i) The electronic configurations of alkali metals are as follows.

All the alkali metals have one valence electron (ns¹). Because of this similarity in electronic configuration, they are placed in the same group and hence they resemble in their physical and chemical properties.

ii) Reducing nature:
All the Alkali metals have bigger atomic sizes and have low I.P values. So they have a tendency to lose electron and thus they act as strong reducing agents. The large hydration energy of small Li⁺ ion makes the Lithium, the strongest reducing agent. The order of reducing powers of alkali metals: Li>Cs>Rb>K>Na

iii) Oxides and Hydroxides:
Alkali metals when heated with oxygen form three types of oxides, namely oxides, peroxides and superoxides depending up on the activity of the metal. The oxides and hydroxides of all the alkali metals are strongly alkaline. Because of High electropositive character of alkali metals, their oxides and hydroxides are strongly basic.

The basic character of alkali metals hydroxides increases down the group:
LiOH<NaOH<KOH<RbOH<CsOH.

Thus there is a gradual change in the properties with the increase of their atomic number. This justifies their inclusion in the same group of the periodic table.

Question 2.
Write an essay on the differences between lithium and other alkali metals.
Answer:

1. The anomalous behaviour of Lithium is due to its exceptionally small size and high polarising power
2. Lithium is a much harder metal while other alkali metals are soft.
3. Lithium is least reactive but the strongest reducing agent among all the alkali metals.
4. When Lithium is burnt in air, it forms Li₂O and Li₃N. But remaining alkali elements form only oxides.
   4Li + O₂ → 2 Li₂O; 6Li + N₂ → 2 Li₃N
5. Li Cl is deliquescent (absorbs water form atmosphere) and crystallises as a hydrate, LiCl.2H₂O whereas other alkali metal chlorides do not form hydrates.
6. Lithium hydrogen carbonate is not obtained in the solid form while all other elements form solid hydrogen carbonates.
7. Lithium does not react with acetylene, but remaining elements react with acetylene.
8. Lithium nitrate when heated gives Lithium oxide Li₂O, whereas other alkali metal nitrates decompose to give the corresponding nitrite.
   4LiNO₃ → 2LiO + 4NO₂ + O₂
   2NaNO₃ → 2NaNO₂ + O₂
9. Lithium compounds (LiF, Li₂O) are less soluble in water, than the corresponding compounds of other alkali metals.
   Among the carbonates of IA group only Li₂CO₃ is decomposable remaining are stable.
   Li₂CO₃ → Li₂O + CO₂

Question 3.
Discuss the preparation and properties of sodium carbonate.
Answer:
1) Preparation:
SodtUm carbonate is prepared by Solvay process.

2) The required raw materials:
1) Brine solution 2) Ammonia 3)Lime Stone.

3) Principle:
Brine solution is saturated with Ammonia and C02 gas is passed through it. Then sodium bicarbonate is formed.
NH₃ + H₂O + CO₂ → NH₄HCO₃
NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl

The sodium bicarbonate thus formed on heating decomposes to give sodium carbonate.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Process:
Step 1 :
When carbondioxide is passed through a concentrated brine solution saturated with ammonia, it results in the formation of ammonium bicarbonate.
2NH₃ + H₂O + CO₂ → (NH₄)2CO₃
(NH₄)2CO₃ + H₂O + CO₂ → 2NH₄HCO₃

Step 2 :
The ammonium bicarbonate then reacts with common salt forming sodiumbicarbonate.
NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl
The solution containing crystals of NaHCO₃ is fdtered to obtain NaHCO₃.

Step 3 :
Fusion of sodium bicarbonate: NaHC03 is heated to high temperatures to convert it into Na₂CO₃.

Note: Recovery of Ammonia:
The filtrate NH₄Cl obtained in step 2, is mixed with Ca(OH)₂ and heated with steam. Then NH₃ gas is liberated which is sent back to saturation tower. In this process CaCl₂ is obtained as a by-product.
Ca(OH)₂ + 2NH₄Cl → 2NH₃ + 2H₂O + CaCl₂

The overall reaction taking place in Solvay process is
2NaCl + CaCO₃ → Na₂CO₃ + CaCl₂

Chemical properties of Na₂CO₃:
1) Na₂CO₃ is a white crystalline solid.

2) Na₂CO₃ exists as a decahydrate Na₂CO₃.10H₂O which is called washing soda.

3) Na₂CO₃ is readily soluble in water.

4) Aqueous Na₂CO₃ solution is basic in nature. So methyl orange produces yellow colour in that solution.

5) Action with acids:
When Na₂CO₃ is treated with HCl then CO₂ gas is liberated.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

6) Action with non-metals and their oxides:
When Sodium carbonate reacts with a mixture of S and S02, Hypo (Sodium thiosulphate) is formed.
Na₂CO₃ + SO₂ + S → Na₂S₂O₃ + CO₂

7) Action with CO₂ :
An aq. solution of sodium carbonate when saturated with CO₂ gives ppt. of sodium bicarbonate.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

8) Action with silica :
When Na₂CO₃ is fused with SiO₂., water glass is formed.
Na₂CO₃ + SiO₂ → Na₂SiO₃ + CO₂

Question 4.
Discuss the similarities between alkaline earth metals and gradation in the following aspects:
i) Electronic configuration ii)Hydration enthalpies iii) Nature of the oxides and hydrides,
Answer:
i) Electronic configuration of alkaline earth metals:

All the alkaline earth metals have same outer electrons (ns²). But, the atomic size increases gradually due to increase in the number of orbits.

ii) Hydration enthalpies: Alkali metal ions (M²⁺) have more charge and small size. So they have high hydration enthalpies. But with the increase in ionic size, the attraction towards water molecules decreases. So hydration enthalpies of alkaline earth metal ions decreases gradually down the group.
Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺

iii) Nature of oxides and hydroxides :
The oxides and hydroxides of alkaline earth metals are strongly alkaline in nature. BeO and Be(OH)₂ are amphoteric. The oxides of other elements are ionic and basic in nature. The oxides react with water forming hydroxides.
MO + H₂O → M(OH)₂
The solubility thermal stability and the basic character of these oxides and hydroxides increase with increasing atomic number.

Question 5.
Discuss on: i) Carbonates ii) Sulphates and iii) Nitrates of alkaline earth metals.
Answer:
i) Carbonates:
Carbonates of alkaline earth metals are insoluble in water. The solubility of these carbonates decreases down the group. Thermal stability of these carbonates increases down the group with increasing cationic size. So their decomposition temperatures increase down the group, These are prepared by the addition of sodium or ammonium carbonate solution to the solutions of soluble compounds of these metals.

ii) Sulphates :
Sulphates are white solids. Their solubility decreases down the group. Thermal stability increases down the group. BeSO₄ and MgSO₄ are soluble. The greater hydration enthalpies of Be²⁺ and Mg²⁺ ions overcome the lattice enthalpy factor and therefore their sulphates are soluble.

iii) Nitrates :
They exist as hydrated salts. These can be prepared by dissolving their carbonates in dilute nitricacid. Barrium nitrate crystallise as anhydrous salt. This is because of the decrease in the hydration enthalpies. All these nitrates decompose on heating.
2M(NO₃)₂ → 2MO + 4NO₂ + O₂, (M = Be, Mg, Ca, Sr, Ba)

Question 6.
What are the common physical and chemical features of alkali metals?
Answer:
Physical properties:

1. All the alkali metals are soft metals with low melting points and boiling points.
2. In each period the alkali metals have large atomic sizes.
3. All the alkali metals exhibit only one oxidation state +1.
4. All the alkali metals exhibit flame colours.
5. Alkali metals have low IP and have tendency to lose electrons. So they act as strong reducing agents.

Chemical properties:

1. All the alkali metals react with oxygen in air forming oxides.
   Ex: 4Li + O₂ → 2Li₂O
2. All the alkali metals react with water liberating hydrogen.
   Ex: 2M + 2H₂O → 2MOH + H₂
3. All the alkali metals react with hydrogen forming ionic hydrides.
4. All the alkali metals react with halogen forming similar halides of the type MX.
5. The oxides and hydroxides of the alkali metals are strongly alkaline.
6. All the alkali metals dissolve in ammonia forming blue coloured solution due to the presence of ammoniated electrons. These solutions are good reducing agents, good conductors of electricity and are paramagnetic. In concentrated solution, the blue colour changes to bronze colour on warming and becomes diamagnetic.

Question 7.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:

1. The general electronic configuration of alkaline earth metals is ns².
2. Atomics Size increases from top to bottom in the group due to increase in the number of orbits.
3. Density increases from top to bottom in the group but Ca is less denser than Mg.
4. Melting points and boiling points do not vary regularly.
5. Ionisation enthalpies decrease from top to bottom.
6. Hydration enthalpies decrease from top to bottom in the group.
7. All these elements react with halogens forming halides.
8. All these elements except beryllium react with hydrogen directly forming ionic hydrides.
9. These elements readily react with acids liberating hydrogen.
10. Alkaline earth metals are good reducing agents and their reduction power increases from top to bottom.
11. Reactivity towards air and water increases from top to bottom in the group. All these elements bum in air forming metal oxides and metal nitrides. Be and Mg do not react with water but other elements react with water and the reactivity increases from Ca to Ba.

Question 8.
Discuss the various reactions that occur in the Solvay process. [AP 16]
Answer:
Various reactions that occur in the Solvay process:
Reaction 1:
When carbondioxide is passed through a concentrated brine solution saturated with ammonia, it results in the formation of ammonium bicarbonate.
2NH₃ + H₂P + CO₂ → (NH₄)2CO₃
(NH₄)₂CO₃ + H₂O + CO₂ → 2NH₄HCO₃

Reaction 2:
The ammonium bicarbonate then reacts with common salt forming sodium bicarbonate.
NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl

Reaction 3:
NaHCO₃ is heated to high temperatures to convert it into Na₂CO₃.

The overall reaction taking place in Solvay process is
2NaCl + CaCO₃ → Na₂CO₃ + CaCl₂

Question 9.
Starting with sodium chloride how would you proceed to prepare
i) Sodium metal;
ii) Sodium hydroxide
iii) Sodium peroxide
iv) Sodium carbonate.
Answer:
i) Sodium nictal :
When molten sodium chloride is electrolysed sodium metal is formed. To decrease the melting point of sodium chloride it is mixed with KCl and CaCl₂.

ii) Sodium hydroxide:
Electrolysis of aqueous sodium chloride either in Nelson’s cell or in Castner-Kellner cell gives sodium hydroxide.

iii) Sodium peroxide :
First sodium metal is prepared from sodium chloride as above.
Then it is burnt in excess of O₂ to produce sodium peroxide.
2Na + O₂ → Na₂O₂

iv) Sodium carbonate :
Brine is saturated with ammonia. Then CO₂ is passed into the solution. Then Sodium bicarbonate is formed due to the following reactions.
NH₃ + H₂O + CO₂ → NH₄HCO₃
NaCl + NH₄HCO₃ → NaHCO₃ + NH₄Cl
This sodium bicarbonate on calcination gives sodium carbonate.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Question 10.
What happens when (i) Magnesium is burnt in air?
(ii) Quick lime is heated with silica?
(iii)Chlorinc reacts with slaked lime?
(iv) Calcium nitrate is strongly heated?
Answer:
i) When magnesium is burnt in air, it bums with brilliant white light to form MgO and Mg₃N₂.
2Mg + O₂ → 2MgO
3Mg + N₂ → Mg₃N₂

ii) When quick lime is heated with silica, calcium silicate is formed.
CaO + SiO₂ → CaSiO₃

iii) When Cl₂ reacts with slaked lime, bleaching powder(CaOCl₂) is formed.
Ca(OH)₂ + Cl₂ → CaOCl₂ + HCl

iv) When calcium nitrate is heated strongly, calcium oxide is formed with the liberation of NO₂ and O₂ gases.

Question 11.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids. [TS 16,20]
Answer:
Significance of Na, K in biological fluids:

1. Na⁺ ions participate in the transmission of nerve signals.
2. Na’ ions regulate the flow of water across cel! membranes.
3. Na⁺ ions responsible for transport of sugars and amino acids into cells.
4. K⁺ ions are useful in activating enzymes.
5. K⁺ ions participate in the oxidation of glucose to produce ATP.
6. K⁺ along with Na⁺ responsible for the transmission of nerve signals.

Biological importance of Mg and Ca: [TS 18]
Significance of Mg²⁺ :

1. Mg²⁺ ions are concentrated in animal cells.
2. Enzymes like ‘phosphohydrolases’ and ‘Phospho transferases’ contain Mg²⁺ ions.
   These enzymes participate in ATP reactions and release energy in the process. Mg²⁺ forms a complex with ATP.
3. Mg²⁺ is a constituent of chlorophyll, the green component of plants.

Significance of Ca²⁺:

1. About 99% of body calcium is present in bones and teeth. It plays important roles in neuro muscular function, intemeuronal transmission, cell membrane integrity and blood coagualation.
2. The calcium concentration in plasma is regulated at about 100 mg/Lit. It is maintained by two hormones, calcitonin and parathyroid hormone. Ca²⁺ ion are necessary for muscle contraction.

Question 12.
Write a few lines about cement.
Answer:
Cement :
Cement is an important building material. It is also called portland cement. It is the product obtained by combining materials rich in lime (CaO) wdth other material such as clay which contains silica (SiO₂) along with oxides of Al, Fe and Mg.

Raw material :
The raw materials used for manufacture of cement are
i) Limestone ii) Clay iii) Gypsum

When clay and lime are heated strongly together, they fuse and react to form cement clinker. This clinker is mixed with 2 to 3% by weight of gypsum to form cement. The important ingredients present in portland cement are dicalcium silicate (Ca₂SiO₄ : 26%) tricalcium silicate (Ca₃SiO₄ :51%) and tricalcium aluminate (Ca₃Al₂O₆ : 11%).

Setting of Cement:
When cement is mixed with water, the setting of cement takes place to give a hard mass. This is due to hydration of the molecules of the constituents and their rearrangement. The purpose of adding gypsum is only to slow’ down the process of setting time so that it gets sufficiently hardened.
Uses: Cement is used

1. in concrete and reinforced concrete.
2. in plastering
3. in the construction of buildings, dams and bridges.

Multiple Choice Questions

Question 1.
The formula of soda ash is
1) Na₂CO₃.10H₂O
2) Na₂CO₃.2H₂O
3) Na₂CO₃.H₂O
4) Na₂CO₃
Answer:
4) Na₂CO₃

Question 2.
Suspension of slaked lime in. water is known as
1) lime water
2) quick lime
3) milk of lime
4) aqueous solution of slaked lime
Answer:
3) milk of lime

Question 3.
By adding gypsum to cement
1) setting time of cement becomes less.
2) setting time of cement increases.
3) colour of cement becomes light.
4) shining surface is obtained.
Answer:
2) setting time of cement increases.

Question 4.
The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room temperature rises to 30°C?
1) Na
2) K
3) Rb
4) Cs
Answer:
4) Cs

Question 5.
The order of decreasing ionisation enthalpy in alkali metals is
1) Na > Li > K > Rb
2) Rb < Na < K < Li
3) Li > Na > K > Rb
4) K < Li < Na < Rb
Answer:
3) Li > Na > K > Rb

Question 6.
When sodium is dissolved in liquid ammonia, a solution of deep blue colour is obtained. The colour of the solution is due to
1) ammoniated electron
2) sodium ion
3) sodium amide
4) ammoniated sodium ion
Answer:
1) ammoniated electron

Question 7.
The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution.
1) Sublimation enthalpy
2) Ionisation enthalpy
3) Hydration enthalpy
4) Electron-gain enthalpy
Answer:
3) Hydration enthalpy

Question 8.
Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously?
1) Li
2) Na
3) K
4) Cs
Answer:
1) Li

Question 9.
Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to
1) Ionic nature of lithium fluoride
2) High lattice enthalpy
3) High hydration enthalpy for lithium ion.
4) Low ionisation enthalpy of lithium atom
Answer:
2) High lattice enthalpy

Question 10.
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
1) K
2) Rb
3) Li
4) Na
Answer:
3) Li

Question 11.
Which of the alkali metal chloride (MCl) forms its dihydrate salt (MCl.2H₂O) easily?
1) LiCl
2) CsCl
3) RbCl
4) KCl
Answer:
1) LiCl

Question 12.
HCl was passed through a solution of CaCl₂, MgCl₂ and NaCl. which of the following compounds crystallises?
1) Both MgCl₂ and CaCl₂
2) Only NaCl
3) Only MgCl₂
4) NaCl, MgCl₂ and CaCl₂
Answer:
2) Only NaCl

Question 13.
Crude sodium chloride obtained by crystallisation of brine solution does not contain
1) MgSO₄
2) Na₂SO₄
3) MgCl₂
4) CaSO₄
Answer:
1) MgSO₄

Questionn 14.
In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH4CI with Ca(OH)2. The by-product obtained in this process is
1) CaCl₂
2) NaCl
3) NaOH
4) NaHCO₃
Answer:
1) CaCl₂

Question 15.
A substance which gives brick red flame and breaks down on heating to give oxygen and a brown gas is
1) Magnesium nitrate
2) Calcium nitrate
3) Barium nitrate
4) Strontium nitrate
Answer:
2) Calcium nitrate

Question 16.
The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and w ith Na, is responsible for the transmission of nerve signals.
1) Iron
2) Copper
3) Calcium
4) Potassium
Answer:
4) Potassium

Question 17.
Which of the following elements does not form hydride by direct heating with dihydrogen?
1) Be
2) Mg
3) Sr
4) Ba
Answer:
1) Be

Question 18.
Among CaH₂, BeH₂, BaH₂ the order of ionic character is
1) BeH₂ < CaH₂ < BaH₂
2) CaH₂ < BeH₂ < BaH₂
3) BeH₂ < BaH₂ < CaH₂
4) BaH₂ < BeH₂ < CaH₂
Answer:
1) BeH₂ < CaH₂ < BaH₂

Question 19.
Which of the following is an amphoteric hydroxide?
1) Be(OH)₂
2) Sr(OH)₂
3) Ca(OH)₂
4) Mg(OH)₂
Answer:
1) Be(OH)₂

Question 20.
The structures of beryllium chloride in solid state and vapour phase are
1) chain in both
2) chain and dimer respectively
3) linear in both
4) dimer and linear, respectively.
Answer:
2) chain and dimer respectively

Question 21.
Among the following alkaline earth metal halides one which is covalent and soluble in organic solvents is
1) beryllium chloride
2) calcium chloride
3) strontium chloride
4) magnesium chloride
Answer:
1) beryllium chloride

Question 22.
Which of the follow ing statements is true about Ca(OH)₂?
1) It is used in the preparation ofbleaching powder
2) It is a light blue solid
3) It does not possess disinfectant property.
4) It is used in the manufacture of cement.
Answer:
1) It is used in the preparation ofbleaching powder

Question 23.
Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is
1) BeCl₂
2) MgCl₂
3) CaCl₂
4) SrCl₂
Answer:
1) BeCl₂

Question 24.
Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s²2s²2p³, the simplest formula for this compound is
1) Mg₂X₃
2) MgX₂
3) Mg₂X
4) Mg₃X₂
Answer:
4) Mg₃X₂

Question 25.
Metals form basic hydroxides. Which of the following metal hydroxide is the least basic?
1) Mg(OH)₂
2) Ca(OH)₂
3) Sr(OH)₂
4) Ba(OH)₂
Answer:
1) Mg(OH)₂

Question 26.
Which of the carbonates given below is unstable in air and is kept in C02 atmosphere to avoid decomposition.
1) BeCO₃
2) MgCO₃
3) CaCO₃
4) BaCO₃
Answer:
1) BeCO₃

Question 27.
Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?
1) MgCO₃
2) CaCO₃
3) SrCO₃
4) BaCO₃
Answer:
4) BaCO₃

Question 28.
Dead burnt plaster is
1) CaSO₄
2) CaSO₄½H₂O
3) CaSO₄.H₂O
4) CaSO₄.2H₂O
Answer:
1) CaSO₄

Students get through AP Inter 1st Year Chemistry Important Questions 8th Lesson Hydrogen and its Compounds which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 8th Lesson Hydrogen and its Compounds

Very Short Answer Questions

Question 1.
The three isotopes of hydrogen differ in their rates of reaction. Give the reasons.
Answer:
Due to variation in enthalpies of bond dissociation of H₂, D₂, T₂ molecules, they differ in their rates of reaction.
The reactivity order: H₂ > D₂ > T₂

Question 2.
Why is dihydrogen used in welding of high-melting metals?
Answer:
Atomic hydrogen and oxy-hydrogen torches generate very high temperatures of 4000K. At such high temperatures only, metals like platinum and quartz having high melting points will melt. So dihydrogen is used in the welding of high melting metals.

Question 3.
Describe One method of producing high purity hydrogen.
Answer:
High purity (> 99.95%) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes.

Question 4.
Explain the term’SYNGAS’?
Answer:
The mixture of CO and H₂ is usually Called as Water gas. It is also called as SYNGAS, because, it is used in the synthesis of “methanol” and some other hydrocarbons.

Question 5.
What is meant by coal gasification? Explain with relevant, balanced equation. [TS19]
Answer:
The process of producing ‘syngas’ from coal is called ‘coal gasification’.

Question 6.
Define the term hydride. How many categories of hydrides are known? Name them. [AP 22]
Answer:
Binary compounds of hydrogen formed with other elements are called hydrides.
Three categories of hydrides are known
a) Ionic or salt like hydrides.
b) Covalent or molecular hydrides.
c) Metallic hydrides.

Question 7.
The unusual property of water in condensed phase leads to its high heat Of vaporization. What is that property?
Answer:
The unusual property of water in the condensed phase is due to the presence of intermolecular hydrogen bonding between water molecules.

Question 8.
During Photosynthesis water is oxidized to O₂. Which element is reduced?
Answer:
The photosynthesis reaction is
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

The oxidation number of Carbon in CO₂ is +4. But in the Carbohydrate(C₆H₁₂O₆) the oxidation number of Carbon is 0. Thus the element Carbon is reduced.

Photosynthesis:
CO₂ reacts with H₂O in the presence of Chlorophyll and Sunlight and releases Carbohydrates in to atmosphere.

Question 9.
What do you understand by ‘auto protolysis’ of water? Give the equation to represent the autoprotolysis of water.
Answer:
Self-ionisation of water is known as auto protolysis of water.

Significance :
From the above reaction, it is clear that water has the ability to act as an acid as well as base. Thus, water behaves as an amphoteric substance.

Question 10.
Water behaves as an amphoteric substance in the Bronsted sense. How do you explain.
Answer:
Water has ability to act as an acid (proton donor) and a base(proton acceptor). So water behaves as an amphoteric substance in the Bronsted sense.
Ex 1: H₂O + HCl ⇌ H3O⁺ + Cl^–

In this reaction, H₂O accepts one proton from HCl. So H₂O acts as a base in this reaction.
Ex 2: H₂O + NH₃ ⇌ NH⁺4 + OH^–

In this reaction, H₂O donates one proton to NH₃. So H₂O acts as an acid in this reaction.

Short Answer Questions

Question 1.
The boiling points of NH₃, H₂O and HF are higher than those of hydrides of the subsequent members of the group. Give your reasons.
Due to the presence of hydrogen bonds, the boiling points of NH₃, H₂O and HF are higher than those of hydrides of the subsequent members of the group.

Reasons for forming hydrogen bonds :
High Electronegativity differences between H and N, O, F in their hydrides.

But, the electronegativities of the subsequent members of N, O, F groups are less. So their hydrides are less polar and cannot form hydrogen bonds. Hence the boiling points of the hydrides of the subsequent members of the N,O,F groups have low boiling points.

Question 2.
Discuss the position of hydrogen in the periodic table on the basis of its electronic configuration. [AP 20]
Answer:
Electronic configuration of H is 1s¹ This configuration is responsible for its dual nature. It behaves like both Alkali metals and Halogens.

Points in support of placing H in IA group:
a) Just like alkali metals, Hydrogen has one electron in its outer shell (1s¹).
b) Just like alkali metals, Hydrogen forms unipositive ion, H⁺(aq).
c) Just like alkali metals, Hydrogen shows valency 1 in its compounds.

Also, it is quite reasonable to start the periodic table with an element having the least atomic number (Z = 1).

Points in support of placing H in VIIA group: 1
a) Just like Halogens, Hydrogen is a gaseous non-metal.
b) Just like Halogens, Hydrogen forms uninegative ion, H^–.
c) Just like Halogens (F2, Cl2….), Hydrogen is a diatomic molecule H2.

Also, Hydrogen has a tendency of gaining one electron to attain stable electronic configuration of He.
Even though it resembles in many properties both with alkali and halogens, it differs from them as well. Hence, the position of hydrogen in the periodic table has become a matter of choice. It can be placed along with alkali metals in IA group (or) along with halogens in VIIA group.

Question 3.
How is the electronic configuration of hydrogen suitable for its chemical reactions?
Answer:
Atomic number of Hydrogen is 1. Hence, its electronic configuration is 1s¹. Due to this half filled orbital configuration, Hydrogen takes part in a reaction by
(i) Loss of the only one electron to form H⁺.
(ii) Gain of one electron to form Hydride H^–
Ex: Na + H → Na⁺ + H^–
(iii) Sharing electrons to form a single covalent bond.

But, Hydrogen is inert at room temperature, due to its high bond dissociation energy.

Question 4.
What happens when dihydrogen reacts with (a) Chlorine and (b) Sodium metal. Explain
Answer:
(a) Reaction of dihydrogen with chlorine:
Hydrogen reacts with chlorine and forms hydrogen chloride gas.

Here, an electron pair is shared between H and Cl leading to the formation of a covalent molecule.

(b)Reaction of dihydrogen with sodium metal:
Hydrogen reacts with highly reactive sodium and forms sodium hydride.

Here, an electron is transferred from Na to H, leading to the formation of ionic compound NaH.

Question 5.
Write a note on heavy water.
Answer:
1) Molecular formula of heavy water is D₂O (Deuterium oxide).

2) Preparation:
Heavy water is obtained by the exhaustive electrolysis of water.

3) Physical Properties:
The physical properties like molecular mass, melting point, Boiling point etc., of D20 are more than that of water.

4) Chemical Properties:
(a) Heavy water reacts with calcium carbide and forms Deutero acetylene.
CaC₂ + 2D₂O → C₂D₂ + Ca(OD)₂

(b) Heavy water reacts with sulphur trioxide and forms Deutero sulphuric acid.
SO₃ + D₂O → D₂SO₄

(c) Heavy water reacts with aluminium carbide and forms Deutero methane
Al₄C₃ + 12D₂O → 3CD₄ + 4Al(OD)₃

5) Uses of D₂O:
(a) It is used as a moderator in nuclear reactors to decrease the speed of neutrons.
(b) It is used to study the reaction mechanism in exchange reactions.
(c) It is used for the preparation of deuterium and its compounds.

Question 6.
Name the isotopes of hydrogen. What is the ratio of the masses of these isotopes? [TS 22]
Answer:
Isotopes of hydrogen:
(i) Hydrogen ₁¹H
(ii) Deuterium ₁²H or ₁²D
(iii) Tritium ₁³H or ₁³T
The ratio of their masses is 1 : 2 : 3

Question 7.
What is water-gas shift reaction? How can the production of dihydrogen be increased by this reaction?
Answer:
‘Water gas’ is produced by passing steam, over red hot coke.

To increase the amount of H2. Now water gas is mixed with steam over iron chromate catalyst. This process is called water gas shift reaction.

The CO₂ formed is removed by scrubbing with sodium arsenite solution. Thus we get pure H₂.

Question 8.
Complete and balance the following reactions:

Answer:

Question 9.
What is the nature of the hydrides formed by elements of 13 group?
Answer:

1. Group 13 elements form covalent hydrides when they react with hydrogen.
2. Boron forms a number of stable covalent hydrides with a general formula B_nH_n+4.
   These are called boranes. Ex: B₂H₆.
3. Gallium forms a dimeric hydride Ga₂H₆ (digallane) and indium forms a polymeric hydride (InH₃)_n
4. The hydrides of gallium and indium are not very stable.
5. Boron, aluminium and gallium form complex anionic hydrides such as
   NaBH₄ – Sodium borohydride
   Li AlH₄ – Lithium aluminium hydride
   LiGaH₄ – Lithium gallium hydride
   These are powerful reducing agents.
6. Thallium does not form any hydride.

Question 10.
Discuss the principle and the method of softening of hard water by synthetic, ion- exchange resins.
Answer:
I) Ion Exchange method:

1. This method is useful to remove the permanent hardness of water.
2. It is also named as permutit (or) zeolite process.
3. Permutit is the artificial zeolite i.e., sodium aluminium orthosilicate (Na₂Al₂Si₂O₈. XH₂O (or) NaAlSiO₄)
4. Permutit in short form is written as Naz.
5. When permutit is added to hard water, the following ion-exchange reactions take place.
   2Na_(s) + Ca⁺²_(aq) → CaZ_2(s) + 2Na⁺_(aq)
   2Na_(s) + Mg⁺²_(aq) → MgZ_2(s) + 2Na⁺_(aq)
6. CaZ₂ and MgZ₂ are called as exhausted permutit. These are regenerated to permutit by the treatment with brine solution (aq NaCl solution).
   CaZ_2(s) + 2Na⁺_(aq) → 2NaZ_(s) + Ca⁺²_(aq)

II) Synthetic resins method:

1. Hardwater is softened by using synthetic cation exchangers.
2. This method is more efficient than zeolite process.
3. Pure water (de-ionized) is obtained by passing water successively through a Cation exchange resin and an anion exchange resin:
4. Cation exchange resin contains acidic groups like (RSO₃H) or RCOOH. (Here R stands for Resin.)
5. Anion exchange resin contains basic groups like (RNH₃OH)
6. First hard water is passed through cation exchange resin. Then it absorbs all cations Ca²⁺ and Mg²⁺ ions present in the hard water and releases H⁺ions.
   2RCOOH + Ca²⁺ → (RCOO)₂Ca + 2H⁺
   2RCOOH + Mg²⁺ → (RCOO)₂Mg + 2H⁺
7. Now the water coining from the cation tank is passed through a tank containing anion exchange resin. This resin absorbs all anions like Cl^–, SO^2-₄ etc. from the hard water and releases OH^– ions.
   RNH⁺₃OH^– +Cl^– → RNH3Cl^– + OH^–
   2RNH⁺₃OH^– + SO^2-₄ → (RNH⁺₃)₂SO^–₄ + 2OH^–
8. The H+ and OH^– ions unite to form Pure water (de-ionized).
   H⁺ + OH^– → H₂O

Question 11.
Write a few lines on the utility of hydrogen as a fuel. [AP 17]
Answer:

1. Hydrogen has high ‘Heat of combustion’. When compared to any other fuels like methane, L.P.G etc. Hence it is used for industrial purposes.
2. Pollutants in combustion of hydrogen will be less than petrol. The only pollutants will be the oxides of nitrogen and it can be minimised by injecting small amount of water into the hydrogen cylinder.
3. Hydrogen is used as a rocket fuel.
4. Hydrogen is used in fuel cells for generating electrical energy.
5. Atomic hydrogen and oxy hydrogen torches are used for welding and cutting metals.

Question 12.
A 1% solution of H₂O₂ is provided to you. What steps do you take to prepare pure H₂O₂.
Answer:
To obtain pure H₂O₂ from the given 1 % H₂O₂ the following steps are involved.

Step I : Carcful evaporation of the solution
The given 1% H₂O₂ is carefully evaporated on a water bath under reduced pressure by distillation.
Then 20% – 30% H₂O₂ solution is obtained.

Step II : Distillation under reduced pressure
The obtained solution in step-1 is heated in a distillation flask at a reduced pressure of 15mm.
Due to this process 85% H₂O₂ is obtained.

Step III : Crystallization
The sample obtained in step-II is crystallized by freezing.
Then 100% pure hydrogen peroxide is obtained.

Question 13.
Mention any three uses of H202 in modern times. [TS 19]
Answer:

1. H₂O₂ is used as antiseptic in medicine and surgery.
2. H₂O₂ is used to bleach silk, wool, ivory, hair etc.
3. H₂O₂ is used to restore the colour of old and spoiled lead paintings.
4. H₂O₂ is used as an oxidising and reducing agent in the laboratory.
5. 90% solution of H₂O₂ is used as a fuel in submarines and rockets.

Long Answer Questions

Question 1.
Write an essay on the commercial preparation of dihydrogen. Give balanced equations.
Answer:
Commercially hydrogen is prepared in the following methods.
1) Electrolysis method :
Electrolysis of acidified or alkaline water gives hydrogen.

2) In the electrolysis of brine solution along with NaOH and Cl₂, hydrogen is obtained as ‘by product’ at cathode.
Ionisation: 2NaCl → 2Na⁺ + 2Cl^–
At anode: 2Cl^–(aq) → Cl_2(g) + 2e^– (oxidation)
At cathode: 2H₂O_(l) + 2e^– → H_2(g) + 2OH^–(aq) (reduction)

3) When coke (or) hydrocarbons react with steam at high temperature in presence of catalysts produces hydrogen.

The mixture of CO and H₂ is known as ‘Syngas’.
4) When Syngas mixture reacts with steam in the presence of iron chromate (FeCrO₄) gives hydrogen.

Question 2.
Illustrate the chemistry of dihydrogen by its reaction with
i) N₂
ii) Metal ions and metal oxides and
iii) Organic compounds How is dihydrogen used in the manufacture of chemicals?
Answer:
Chemistry of dihydrogen:
(i) Reaction with N₂ :
Hydrogen reacts with Nitrogen in the presence of iron powder as catalyst and forms ammonia.

This reaction is used in the manufacture of ammonia by Haber’s process.

ii) Reaction with metal ions, metal oxides :
Hydrogen is a good reducing agent and reduces several metal oxides and metal ions to their corresponding metals.
Pd²⁺ + H₂ → Pd + 2H⁺
Cuo + H₂ → Cu + H₂O

iii) Reaction with Organic compounds:
a) Hydrogenation of oils :
When H₂ gas is passed into vegetable oils, in the presence of Ni catalyst, at 5 atm pressure and 473K, Vanaspati is produced.

Vegetable oil +H2 Ni/473K so] fat

b) Hydrofomrylation of alkenes gives aldehydes, which further reacts with H₂ and fonns alcohols on reduction.
RCH = CH₂ + H₂ + CO → RCH₂ – CH2CHO
RCH₂CH₂CHO + H₂ → RCH₂CH₂CH₂OH

In Hie manufacture of Chemicals:
Dihydrogen is used in the preparation of so many chemicals like Ammonia, HCl, Methanol,
i) H₂ reacts with N₂ and forms NH₃.( Manufacture of Ammonia by Haber’s process)

ii) H₂ reacts with Cl₂ and gives hydrochloric acid.
H₂ + Cl₂ → 2HCl
iii) Water gas and H₂ mixture is passed on ZnO+CrO₃ catalyst at 300°C gives Methanol.

Question 3.
Explain, with suitable examples, the following: [IPE’ 14][AP I6][TS 18, 20]
i) Electron deficient ii) Electron-precise and iii) Electron-rich hydrides.
Answer:
i) Electron deficient hydrides :
These hydrides have lesser number of electrons than required for conventional Lewis structure.
Ex: B₂H₆ (Diborane)
Group 13 elements form such hydrides.

ii) Electron precise hydrides :
These hydrides have exactly the required number of electrons for conventional Lewis structures.
Ex: CH₄, SiH₄, GeH₄ Group 14 elements form such hydrides.

iii) Electron rich hydrides :
These hydrides have excess number of electrons than required for conventional Lewis structure.
Ex: NH₃, H₂O, HF.
Group 15-17 elements form such hydrides.

The excess number of electrons are present as lone pairs.
Ex: NH₃ has 1- lone pair; H₂O has 2- lone pairs; HF has 3- lone pairs.

Question 4.
Write in brief on i)ionic hydrides ii)interstitial hydrides [TS 17]
Answer:
i) Ionic hydrides:

1. Ionic hydrides are formed by the combination of H₂ with highly electropositive s-block metals.
2. These are stoichiometric compounds.
3. But hydrides such as LiH, BeH₂, MgH₂ due to polymeric structure have covalent nature.
4. Usually the ionic hydrides have crystalline structures. They are non-volatile compounds. They are non-conducting in solid state. But in molten condition, if electrolysed, they give metal at cathode and hydrogen at anode.
   Anode: 2H^– → H₂ + 2e^–
   This reaction confirms that ionic hydrides contain H^– ion.
5. Ionic hydrides react violently with water and produces H₂ gas.
   NaH + H₂O → NaOH + H₂
   So ionic hydrides are used as a source of hydrogen.

ii) Interstitial hydrides:

1. Interstitial hydrides are formed by the combination of H₂ with d – block and f – block elements.
2. The are non stoichiometric hydrides. Ex: LaH_2.87 YbH_2.55, ZrH_1.3-1.75
3. But metals of 7, 8 and 9 group elements do not form hydrides.
4. These hydrides conduct electricity and heat but not as efficiently as their parent metals.
5. Some metals like Pd and Pt absorb large amounts of hydrogen. This property is known as occlusion. This property is also used for storage of H₂ and purification of H₂.
6. Absorption of H₂ by transition metals is used in catalytic hydrogenation and reduction.

Question 5.
Explain any four of Hie chemical properties of water.
Answer:
1) Amphoteric nature:
Water can act both as acid and base.
In the Bronsted sense, it acts as an acid with NH₃ and a base with H₂S.
H₂O(l) + NH₃(aq) ⇌ OH^–(aq) + NH⁺₄(aq)
H₂O(l) + H₂S(aq) ⇌ H₃O⁺(aq) + HS^–(aq)

2) Photosynthesis :
CO₂ reacts with H₂O in the presence of Chlorophyll and Sunlight forms carbohydrates with the liberation of O₂.
6CO₂ + 6H₂O → C₆H₂O₆ + 6O₂

3) Reaction with Metals :
Water reacts with metals and liberates hydrogen.
Here water is reduced to hydrogen and metals are oxidised.
2Na + 2H₂O → 2NaOH + H₂

4) Reaction with Non Metals: H20 reacts with F₂ and it is oxidised to O₂.
2F_2(g) + 2H₂O_(l) → 4HF_(aq) + O₂

Question 6.
Explain the terms hard water and soft water. Write a note on the [AP 16, 18, 22]
I) ion-exchange method and [TS 15, 19]
II) Calgon method for the removal of hardness of water.
Answer:
Soft water :
Water which gives lather readily with soap is called soft water.

Soft water is free from dissolved salts of calcium, iron, or magnesium.

Hard water:
Water which do not give lather readily with soap is called hard water. Hardness of water is due to the presence of soluble compounds of magnesium and calcium such as MgCl₂, MgSO₄, Mg(HCO₃)₂, CaCl₂, CaSO₄, Ca(HCO₃)₂.

I) Ion Exchange method:

1. This method is useful to remove the permanent hardness of water.
2. It is also named as permutit (or) zeolite process.
3. Permutit is the artificial zeolite i.e., sodium aluminium orthosilicate (Na₂Al₂Si₂O₈. XH₂O (or) NaAlSiO₄)
4. Permutit in short form is written as Na₂z.
5. When hard water is passed through permutit, the following ion-exchange reactions take place.
   Na₂Z(s) + Ca⁺²(aq) → CaZ (s) + 2Na⁺(aq)
   Na₂Z(s) + Mg⁺²(aq) → MgZ(s) + 2Na⁺(aq)
6. CaZ and MgZ are called as exhausted permutit. These are regenerated to permutit by the treatment with Brine solution(NaCl).
   CaZ(s) + 2Na⁺(aq) → Na₂Z(s) + Ca⁺²(aq)

II) Calgon method :
Sodium hexametaphosphate is commercially called as calgon. When calgon is added to hard water it reacts with calcium and magnesium ions forming complex anions.
Na₂[Na₄(PO₃)₆] + 2Mg²⁺ → Na₂[Mg₂(PO₃)₆] + 4Na⁺
Na₂[Na₄(PO₃)₆] + 2Ca²⁺ → Na₂[Ca₂(PO₃)₆] + 4Na⁺

Due to the formation of the complex, the Mg²⁺ and Ca²⁺ ions become inactive and cannot react with soap. So the water gives good lather. This method is used only for laundry process.

Question 7.
Write the chemical reaction to justify the hydrogen peroxide can function as on oxidizing as well as reducing agent. [AP 15, 18, 19][TS 16, 18]
Answer:
Hydrogen peroxide can act as both as oxidising and reducing agent in both acid and basic medium. In H₂O₂ oxidation state of oxygen is -1. Here, it is oxidised to O₂. Hence H₂O₂ is reductant. H₂O₂ can be reduced to H₂O(or) OH^–. Here H₂O₂ is oxidant.

As oxidising agent :
H₂O₂ oxidises black lead sulphide to white lead sulphate.
(a) PbS + 4H₂O₂ → PbSO₄ + 4H₂O,
(b) 2Fe²⁺ + H₂O₂ → 2Fe³⁺ + 2OH^–
(c) Mn²⁺ + H₂O₂ → Mn⁴⁺ + 2OH^–,
(d) 2Fe²⁺ + 2H⁺ + H₂O₂ → 2Fe³⁺ + 2H₂O

As reducing agent :
H₂O₂ reduces acidified potassium permanganate to manganese sulphate.
(a) 2KMnO₄ + 3H₂SO₄ + 5H₂O₂ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
(b) HOCl + H₂O₂ → Cl^– + H₃O⁺ + O₂
(c) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂
(d) 2MnO₄ + 3H₂O₂ → 2MnO₂ + 3O₂ + 2OH^– + 2H₂O

Question 8.
Complete and balance the following chemical reactions:
i) PbS(s) + H₂O₂(aq) →
ii) MnO^–₄(aq) + H₂O₂(aq) →
iii) CaO(s) + H₂O(g) →
iv) Ca₃N₂(s) + H₂O(l) →
Classify the above into (a) hydrolysis (b) redox and (c) hydration reactions.
Answer:
i) PbS + 4H₂O₂ → PbSO₄ + 4H₂O
This reaction is a redox reaction. Here, Pbs is oxidised and H₂O₂ is reduced.

ii) 2MnO^–₄ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
This reaction is a redox reaction. Here, H₂O₂ is oxidised and MnO^–₄ is reduced.

iii) CaO(s) + H₂O(g) → Ca(OH)₂
This reaction is a hydration reaction.

iv) Ca₃N₂ + 6H₂O(g) → 3Ca(OH)₂ + 2NH₃
This reaction is a hydrolysis reaction.

Question 9.
Discuss, with relevant chemical equations, various methods of preparing hydrogen . peroxide. Which of these methods is useful to prepare D₂O₂?
Answer:
Preparation of H₂O₂ :
1) Acidified barium peroxide on removing excess water by evaporation under reduced pressure gives hydrogen peroxide.
BaO₂.8H₂O(s) + H₂SO₄ (aq) → BaSO₄ + H₂O₂ (aq) + 8H₂O_(l)

This BaSO₄ can be removed by filtration. Excess water can be removed by evaporation under reduced pressure.

2) Auto oxidation method :
Industrially H₂O₂ is prepared by the auto oxidation of 2-ethyl anthraquinols.

In this case 1% H₂O₂ is formed.

3) Electrolysis of 50% H₂SO₄ using high current density gives peroxodisulphuric acid.

4) Now deuterolysis of peroxodisulphuric acid gives D₂O₂
K₂S₂O₈ + 2D₂O → 2KDSO₄ + D₂O₂
This method is useful to prepare D₂O₂

Question 10.
In how many ways can you express the strength of H₂O₂? Calculate the strength of 15 volume solution of H₂O₂ in g/L. Express this strength in normality and molarity.
Answer:
The strength of H₂O₂ can be expressed in four ways:
(i) Volume strength
(ii) Percentage of Wt.
(iii) Molarity
(iv) Normality

(i) Volume strength: ‘x’ vol. H₂O₂ means 1 ml of that H₂O₂ gives ‘x’ ml of O₂ at STP.
Ex: 10 vol. H₂O₂, 20 vol. H₂O₂ etc.
10 vol. H₂O₂ solution means 1 ml of this solution liberates 10 ml of O₂ gas at STP.

(ii) Percentage of Wt :
‘x’ % (W/V) of H₂O₂ indicates x gm of H₂O₂ present in 100ml solution.
Ex: 10 vol. H₂O₂ = 3.036% (W/V)

(iii) Molarity of H₂O₂ :
It indicates number of moles of H₂O₂ in 1 lit solution.

Ex: 10 vol. H202 =0.893M .

(iv) Normality of H₂O₂ :
It indicates number of gram equivalents of H₂O₂ present in 1 lit solution.

Ex: 10 vol. H202=1.786N

Solution to the given problem:
We know the strength of 10 vol. H₂O₂ is equal to 3% W/V

Multiple Choice Questions

Question 1.
Radioactive elements emit α, β and γ are characterised by their half lives. The radioactive isotope of hydrogen is
1) Protium
2) Deuterium
3) Tritium
4) Hydronium
Answer:
3) Tritium

Question 2.
Tritium a radioactive isotope of hydrogen, emits which of the following particles?
1) Neutron(n)
2) Beta (β^–)
3) Alpha (α)
4) Gamma (γ)
Answer:
2) Beta (β^–)

Question 3.
Which of the following reactions is an example of use of water gas in the synthesis of other compounds?

Answer:
4

Question 4.
Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is
1) LiH > NaH > CsH > KH>RbH
2) LiH < NaH < KH < RbH < CsH
3) RbH > CsH > NaH > KH > LiH
4) NaH > CsH > RbH > LiH > KH
Answer:
2) LiH < NaH < KH < RbH < CsH

Question 5.
Elements of which of the following group(s) of periodic table do not form hydrides.
1) Groups 7, 8, 9
2) Group 13
3) Groups 15, 16, 17
4) Groupl4
Answer:
1) Groups 7, 8, 9

Question 6.
Which of the following reactions increases production of dihydrogen from synthesis gas?

Answer:
3

Question 7.
Which of the-following hydrides is electron-precise hydride?
1) B₂H₆
2) NH₃
3) H₂O
4) CH₄
Answer:
4) CH₄

Question 8.
Which of the following statements about hydrogen is incorrect?
1) Hydronium ion, H₃O⁺ exists freely in solution
2) Dihydrogen does not act as a reducing agent
3) Hydrogen has three isotopes of which tritium is the most common
4) Hydrogen never acts as cation in ionic salts.
Answer:
3 And 4

Question 9.
Only one element of _____ forms hydride.
1) group 6
2) group 7
3) group 8
4) group 9
Answer:
1) group 6

Question 10.
Which Of the following ions will cause hardness in water sample?
1) Ca²⁺
2) Na⁺
3) Cl^–
4) K⁺
Answer:
1) Ca²⁺

Question 11.
Which of the following compounds is used for water softening?
1) Ca₃(PO₄)₂
2) Na₃PO₄
3) Na₆P₆O₁₈
4) Na₂HPO₄
Answer:
3) Na₆P₆O₁₈

Question 12.
The method used to remove temporary hardness of water is
1) synthetic resins method
2) Calgon’s method
3) Clark’s method
4) ion-exchange method
Answer:
3) Clark’s method

Question 13.
Match the following and identify the correct option

(A) CO(g) + H2(g)	(i) Mg(HCO3)2 + Ca(HCO3)2
(B) Temporary	(ii) an electron deficient hardness of water hydride
(C) B2H6	(iii) Synthesis gas
(D) H2O2	(iv) Non-planar structure

Answer:
1

Question 14.
Hydrogen peroxide is obtained by the electrolysis of _________
1) water
2) sulphuric acid
3) hydrochloric acid
4) fused sodium peroxide
Answer:
2) sulphuric acid

Question 15.
When sodium peroxide is treated with dilute sulphuric acid, we get
1) sodium sulphate and water
2) sodiun sulphate and oxygen
3) sodium sulphate, hydrogen and oxygen
4) sodium sulphate and hydrogen peroxide
Answer:
4) sodium sulphate and hydrogen peroxide

Question 16.
Hydrogen peroxide is
1) an oxidising agent
2) a reducing agent
3) both an oxidising and a reducing agent
4) neither oxidising nor reducing agent
Answer:
3) both an oxidising and a reducing agent

Question 17.
Which’of the following equations depict the oxidising nature of H₂O₂?
1) 2MnO^–₄ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
2) 2Fe⁺₃2H⁺ +H₂P₂ → 2Fe²⁺ +2H₂O + O₂
3) 2I^– + 2H⁺ + H₂O₂ → I₂ + 2H₂O
4) KIO₄ + H₂O₂ . KIO₃ + H₂O + O₂
Answer:
3) 2I^– + 2H⁺ + H₂O₂ → I₂ + 2H₂O

Question 18.
Which of the following equation depicts reducing nature of H₂O₂?
1) 2[Fe(CN)₆]^4- + 2H⁺ + H₂O₂ → 2[Fe(CN)₆]^3- + 2H₂O
2) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂
3) Mn²⁺ + H₂O₂ → Mn⁴⁺ + 2OH^–
4) PbS + 4H₂O₂ → PbSO₄ + 4H₂O
Answer:
2) I₂ + H₂O₂ + 2OH^– → 2I^– + 2H₂O + O₂

Question 19.
The oxide that gives H₂O₂ on treatment with dilute H₂SO₄is
1) PbO₂
2) BaO₂.8H₂O + O₂
3) MnO₂
4) TiO₂
Answer:
2) BaO₂.8H₂O + O₂

Question 20.
Some statements about heavy water are given below:
A) Heavy water is used as a moderator in nuclear reactors
B) Heavy water is more associated than ordinary w;ater
C) Heavy water is more effective solvent than ordinary water
Which of the above statements are correct?
1) A and B
2) A, B and C
3) B and C
4) A and C
Answer:
1) A and B

Students get through AP Inter 1st Year Chemistry Important Questions 7th Lesson Chemical Equilibrium and Acids-Bases which are most likely to be asked in the exam.

AP Inter 1st Year Chemistry Important Questions 7th Lesson Chemical Equilibrium and Acids-Bases

Very Short Answer Questions

Question 1.
State Saw of chemical equilibrium.
Answer:
Law of chemical equilibrium:
In a balanced chemical equation, the ratio of the product of molar concentrations of the products raised to their respective stoichiometric coefficients to the product of the concentrations of the reactants raised to their individual stoichiometric coefficients, is a constant, at a given temperature.

Question 2.
Can equilibrium be achieved between water and its vapours in an open vessel?
Answer:
No. In an open vessel, the vapours escape into atmosphere. Hence the rate of evaporation will be more than the rate of condensation.

So equilibrium cannot be achieved.

Question 3.
Why the concentrations of pure liquids and pure solids are ignored from equilibrium constant expressions.
Answer:
The concentrations of pure solids and pure liquids are taken as unity (1). This value T does not effect the value of K_c.

Hence concentrations of pure solids and pure liquids are ignored in the expressions.

Question 4.
What is homogeneous equilibrium? Write two homogenous reactions. [IPE’ 14]AP 17,18]
Answer:
When the reactants and products are in the same physical state then the equilibrium is called homogeneous equilibrium.
Ex1: H_2(g) + I_2(g) ⇌ 2HI_(g)
Ex2: N_2(g) +3H_2(g) ⇌ 2NH_3(g)

Question 5.
What is heterogeneous equilibrium? Write two heterogeneous reactions. [AP 17,18]
Answer:
When the reactants and products are in different physical states then the equilibrium is said to be heterogeneous equilibrium.

Question 6.
Write reaction quotient, Q for each of the following reactions.

Answer:

Question 7.
Define equilibrium constant.
Answer:
Equilibrium constant:
In a balanced chemical reaction, the ratio of the product of molar concentrations of products to the product of molar concentrations of reactants, at a given temperature, is called equilibrium constant.

Question 8.
The equilibrium constant expression for

Write the balanced chemical equation corresponding to this expression.
Answer:

Question 9.
Write the relation between K_p and K_c.
Answer:
K_p = K_c (RT)^∆n
K_p = equilibrium constant in terms of partial pressures
K_c = equilibrium constant in terms of molar concentrations
R = gas constant
T = Temperature in kelvin
∆n = Difference in the number of moles gaseous products and reactants.

Question 10.
Under what conditions for a reaction K_p and K_c are numerically equal?
Answer:
If the number of moles of gaseous products = Number of moles of gaseous reactants
then n_p – n_R = 0 ⇒ ∆n = 0
Now, K_p = K_c (RT)^∆n = K_c (RT)⁰
= K_c(1) = K_c
∴ K_p = K_c, when ∆n=0.

Question 11.
Give two chemical equilibrium reactions for which K_p = K_c
Answer:

Question 12.
Give two chemical equilibrium reactions for which K_p > K_c [AP 20]
Answer:

Question 13.
Give two chemical equilibrium reactions for which K_p < K_c
Answer:

Question 14.
Write the equations for the conversion of K_c to K_p for each of the following reactions.

Answer:

Question 15.
What are the factors which influence the chemical equilibrium?
Answer:
Factors influencing chemical equilibrium:

1. Concentration of reactants and products.
2. Temperature of reaction.
3. Pressure of reaction.
4. Addition of Inert gas

Question 16.
What is the effect of pressure on a gaseous chemical equilibrium? [AP 19]
Answer:
1) If the number of moles of gaseous products is equal to the number of moles of gaseous reactants (∆n = n_P – n_R = 0) then pressure has no effect on the equilibrium.
2) If ∆n ≠ 0 then
(a) Increase of pressure, shifts the position of equilibrium in the direction of decreasing the number of moles.
Ex: N_2(g) +3H_2(g) ⇌ 2NH_3(g), ∆n < 0 (fw)
(b) Decrease of pressure, shifts the position of equilibrium in the direction of increasing the number of moles.
Ex: PCl_5(g) ⇌ PCl_3(g) +Cl_2(g), ∆n > 0(bw)

Question 17.
What is the effect of increase in concentration of reactants of a chemical reaction at equilibrium?
Answer:
Increase in the concentration of reactants of a chemical reaction, at equilibrium, favours the forward reaction.

Question 18.
Can catalyst disturb the state of equilibrium?
Answer:
No. When a catalyst is added, the state of equilibrium is not disturbed, but equilibrium is attained quickly. Because a catalyst increases both the rate of forward and backward reaction, to the same extent.

Question 19.
On which factor, the equilibrium constant value changes?
Answer:
Temperature. Increase in temperature favours the endothermic reaction. Hence the equilibrium constant changes.

Question 20.
The equilibrium constants of a reaction at 27°C and at 127°C are 1.6 × 10^-3 and 7.6 × 10^-2 respectively. Is the reaction exothermic or endothermic.
Answer:
K_c = 7.6 × 10^-2, at T27°C
K_c = 1.6 × 10^-3, at 27°C
Thus, the value of equilibrium constant (K_c) is increased with increase in temperature. Hence the reaction is endothermic.

Question 21.
What is the effect of temperature on a system at equilibrium?
Answer:
Increase in temperature of a reaction mixture, at equilibrium, favours the endothermic reaction.

Decrease in temperature of a reaction mixture, at equilibrium, favours the exothermic reaction.

Question 22.
For an exothermic reaction, what happens to the equilibrium constant if temperature is raised?
Answer:
In an exothermic reaction when the temperature is raised, the value of equilibrium constant decreases. .

Question 23.
What kind of equilibrium constant can be calculated from ∆G° value for a reaction involving only gases?
Answer:
We know ∆G = ∆G° + RTlnQ ……….. (1)
At equilibrium, ∆G = 0 and Q = K
∴ from (1), 0 = ∆G° + RT/nK
⇒ ∆G° = -RT/nK

Here, K = Equilibrium constant
Hence, we can calculate the equilibrium constant (K_p), when the reaction is involved only with gases.

Question 24.
What is a Bronsted base? Give one example? [AP 16][TS 19]
Answer:
A chemical substance whose molecules or ions have a tendency to gain proton(s) from a donor, is called a Bronsted base.
Ex: H₂O, NH₃, Cl^–

Question 25.
What is Lewis acid? Give one example? [AP 15,16][TS 18,22]
Answer:
A chemical substance whose molecules or ions, have a tendency to accept an electron pair, to form a coordinate covalent bond with the donor is called Lewis acid.
Ex: BF₃, BCl₃, H⁺

Question 26.
What is meant by ionic product of water? [TS 18][AP 16]
Answer:
The product of the concentrations of H⁺ and OH^– ions, in water, at a given temperature, is called ionic product of water.
Ionic product of water, K_w = [H⁺][OH^–]

Question 27.
What is the value of Kw? What are its units?
Answer:
K_w = 1.0 × 10^-14 mol²/L², at 25°C
Units: mol²/L²

Question 28.
What is the effect of temperature on ionic product of w ater?
Answer:
As the temperature increases, the degree of ionization of water increases. Hence the value of ionic product (K_w) also increases.

Question 29.
H₂O + H₂O ⇌ H₃O⁺ + OH^–
The ionic product of water is 1 × 10^-14 at 25°C and 3 × 10^-14 at 40°C
Is the above process endothermic or exothermic?
Answer:
K_w = 3 × 10^-14 mo²/L², at 40°C
K_w = 1 × 10^-14 mol²/L², at 25°C
Here, the value of K_w is increased with increase of temperature. Hence the given reaction is an endothermic reaction

Question 30.
All Bronsted bases are Lewis bases. Explain.
Answer:
Bronsted base is a proton acceptor.
Lewis base is an electron pair donor.

NH₃ is a Bronsted base it can accept a proton. In accepting a proton, it donates an electron pair.
Thus, a Bronsted base, to accept a proton must donate an electron pair.
So all the Bronsted bases are Lewis bases.

Question 31.
All Lewis acids are not Bronsted acids. Why?
Answer:
Lewis acid is an electron pair acceptor.
Bronsted acid is a Proton donor.
All the Lewis acids may not have hydrogen to donate protons.

Hence all Lewis acids are not Bronsted acids BF₃ is a Lewis acid since it can accept an electron pair from other substances.

But BF₃ can’t donate protons. Hence, It is not a Bronsted acid.

Question 32.
What is degree of ionization?
Answer:
Degree of ionization (α) :
The ratio between the number of molecules ionised and the total number of molecules of an acid or base.

Question 33.
What is the measure of strength of an acid and base?
Answer:
Strength of an acid is measured by the acid dissociation constant (K_a)
Strength of a base is measured by the base . dissociation constant (K_b)
Strength of an acid or base can be measured in terms of p^H

Question 34.
Give two examples of salts whose aqueous solutions are basic.
Answer:
Aqueous solutions of salts of weak acid and strong base, have basic nature.
Ex: Sodium acetate CH₃COONa,
Sodium carbonate Na₂CO₃

Question 35.
Give two examples of salts whose aqueous solutions are acidic.
Answer:
Aqueous solutions of salts of strong acid and weak base, have acidic nature.
Ex: NH₄Cl, CuSO₄, Al₂(SO₄)₃

Question 36.
What equation is used for calculating the pH of an acid buffer?
Answer:
Hendersen’s equation:

Question 37.
Phosphoric acid (H₃PO₄) have three ionization constants Ka₁, Ka₂ and Ka₃. Among these ionization constants which has a lower value? Give reason for it?
Answer:
Among Ka₁, Ka₂, Ka₃ the value of Ka₃ is low.

Reason:
Due to common ion effect. It becomes more difficult to remove a positively charged proton from doubly charged anion HPO^-2₄ when compared to H₂PO^–₄, H₃PO₄

Question 38.
Ice melts slowly at high altitudes. Explain. Why?
Answer:
At higher altitudes, atmospheric pressure is very low. Hence ice melts slowly.

Question 39
Define Basicity of acid and Acidity of base. [AP 18]
Answer:
The number of replaceable H⁺ ions of an acid is known as Basicity of acid.

The number of replaceable OH^– ions of an base is known as Acidity of base.

Short Answer Questions

Question 1.
Write expression for the equilibrium constant K_c for each of the following reactions.

Answer:

Question 2.
Derive the relation between K_P & K_C for the equilibrium reaction: [TS 19,22][AP 15,19]
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
⇒ ∆n = n_P – n_R = 2-(1 + 3) = -2
We know K_p = K_c(RT)^∆n = K_c(RT)^-2
Here K_p < K_c

Question 3.
Define equilibrium constant. Write the equilibrium constant expression for the reaction of H₂(g) + I₂(g) ⇌ 2HI(g) and its reverse reaction. How are the two equilibrium constants related?
Answer:
Equilibrium Constant(K_c):
The ratio of product of the equilibrium concentrations of the products to that of the reactants, with each concentration tenn raised to the power equal to the stoichiometric coefficient of the substance, in the balanced chemical equation, is called Equilibrium Constant.

The given reaction:

∴ Equilibrium constant for a given reaction is the inverse of the equilibrium constant for the reverse reaction.

Question 4.
How does the value of equilibrium constant predict the extent of reaction?
Answer:
Predicting the extent of a reaction:
The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. But equilibrium constant does not give any information about the rate at which the equilibrium is reached.

Values of K_c verses Extent of reactions:
i) If K_c is very high (K_c >10³) then the reaction takes place in the forward direction, as products predominate over reactants. Here the reaction proceeds ‘nearly to completion’.
Ex: H₂(g) + Cl₂(g) ⇌ 2HCl(g)
Here, K_c =4.0 × 10³¹ at 300°C

ii) If K_c is very low (K_c < 10^-3) then the reaction takes place in the backward direction, as reactants predominate over products. Here the reaction proceeds ‘rarely or slow’.
Ex: N₂(g) + O₂(g) ⇌ 2NO(g)
Here, K_c = 4.8 × 10^-31 at 298°C

iii) If K_c is the range of 10^-3 to 10³ then appreciable concentrations of both reactants and products are present in the equilibrium mixture. When K_c = 1 then the reaction proceeds to an extent of 50%
Ex: H₂(g) + I₂(g) ⇌ 2HI(g)
Here, K_c = 57.0 at 700K.

Question 5.
State law of chemical equilibrium? What is K_c for the following equilibrium when the equilibrium concentration of each substance is [SO₂] = 0.60 M, ]O₂] = 0.82 M and [SO₃] =1.90 M?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Answer:
Law of chemical equilibrium :
In a balanced chemical equation, the ratio of the product of molar concentrations of the products raised to the respective stoichiometric coefficients to the product of the concentrations of the reactants raised to their individual stoichiometric coefficients, gives a constant value, at a given temperature. The given equation is

Question 6.
Why sealed soda water bottle on opening shows the evolution of gas with effervescence?
Answer:
The sealed soda water contains CO₂ gas dissolved in water under high pressure. When the bottle is opened the pressure decreases and solubility of CO₂ decreases. So evolution of CO₂ takes place with effervescence.

Question 7.
Explain the significance of
a) a very large value of K,
b) a very small value of K and
c) a value of K of about 1.0
Answer:
a) When K_c is very large, (K_c > 10³), the forward reaction takes place to more extent and backward reaction to less extent. So the net reaction goes to almost completion.

b) If Kc is very small(K_c < 10^-3), the forward reaction takes place to a small extent only. So the net reaction proceeds to a small extent only.

c) If K_c = 1 then the rate constants of both forward backward reactions are equal. This indicates that the reaction proceeds to an extent of 50%.

Question 8.
Why is it useful to compare Q with K? What is the situation when
(a) Q = K (b) Q < K (c) Q > K
Answer:
The values of K and Q are useful in predicting the direction of the reaction.
a) If Q = K, the reaction is at equilibrium and no net reaction occurs.
b) If Q < K, the net reaction takes place in the forward direction.
c) If Q > K, the net reaction takes place in the backward direction.

Question 9.
For the reaction
Cl₂(g) + F₂(g) ⇌ 2ClF(g), K_c = 19.9
What will happen in a mixture originally containing [Cl₂] = 0.4 molL^-1. [F₂] = 0.2 mol L^-1 and [Cl F] = 7.3mol L^-1.
Answer:
Given reaction is

Also we have K_c = 19.9 ⇒ QC > KC
∴ The reaction takes place in the backward direction with reactant side.

Question 10.
Predict which of the following reaction will have appreciable concentration of reactants and products:

Answer:
Among the given equations, the value of K_c is nearer to 1 (in the range of 10^-3 to 10³) in the third equation(c), where K_c = 1.8.

So appreciable concentrations of reactants and products are present in equilibrium mixture of third equation(c).

Question 11.
How to recognise the conditions under which changes in pressure would effect system in equilibrium.
Answer:
Pressure has no effect on reaction involving solids and liquids. It is because the volume of solids and liquids do not be effected by pressure.

Pressure has effect on the equilibrium reactions containing gaseous substances. It depends on the value of (∆n) of the reaction.
(a) ∆n = 0 then pressure has no effect on the equilibrium.
(b) If ∆n ≠ O then
(i) Increase of pressure, shifts the position of equilibrium in the direction of decreasing the number of moles.
Ex: N₂(g) +3H₂(g) ⇌ 2NH₃(g), ∆n < 0
(ii) Decrease of pressure, shifts the position of equilibrium in the direction of increasing the number of moles.
Ex: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ∆n > 0

Question 12.
What property of a reaction can be used to predict the effect of a change in temperature on the magnitude of an equilibrium constant?
Answer:
In general, the temperature dependence of the equilibrium constant depends on the sign of ∆H for a given reaction.

The equilibrium constant for an exothermic reaction decreases as the temperature increases. Here the sign of ∆H is negative.

The equilibrium constant for an endothermic reaction increases as the temperature increases. Here the sign of ∆H is positive. Thus changes in temperature depend upon the sign of ∆H.

Question 13.
Docs the number of moles of reaction products increase, decrease, or remains same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
i) PCl₅(g) ⇌ PCl₃(g) + CI₂(g)
ii) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
Answer:
According to Le Chatelier’s principle, if pressure is decreased on the equilibrium mixture, the equilibrium shifts in the direction so that the decreased pressure is nullified.
i) PCl₅(g) ⇌ PCl₃(g) + CI₂(g)
i) In the above reaction, the number of gaseous molecules of products(2) is more than the number of gaseous molecules of the reactants(l). So the reaction proceeds in the forward direction. So the number of moles of products increases.

ii) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
ii) In the above reaction, gaseous molecule (CO₂) is present in the reactant. When pressure is decreased, the number of CO₂ molecules increases on the reactant side. To increase the number of CO₂ moles, CaCO₃ should decompose. So the number of moles of product decrease.

Question 14.
Which of the following reactions will get affected by increasing the pressure? Also mention whether change will cause the reaction to go into forward or backward direction.
i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
iii) CO₂(g) + C(s) ⇌ 2CO(g)
iv) 4NH₃ (g) + 5O₂ (g) ⇌ 4N0(g) + 6H₂O(g)
Answer:
According to Le Chatelier’s principle, when pressure is increased on the equilibrium mixture, the equilibrium shift in the direction where the number of gaseous molecules is less.
i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
In this reaction, the number of gaseous reactant molecules (1) is less than the total number of gaseous molecules (2) of products. So increase in pressure shifts the equilibrium in the backward direction.

ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
In this reaction, the total number of gaseous reactant molecules (3) is equal to the total number of gaseous product molecules (3). So pressure has no effect on this equilibrium.

iii) CO₂(g) + C(s) ⇌ 2CO(g)
In this reaction, the number of gaseous reactant molecules (1) is less than the total number of gaseous product molecules (2). So increase in pressure, shifts the equilibrium in the backward direction.

iv) 4NH₃ (g) + 5O₂ (g) ⇌ 4N0(g) + 6H₂O(g)
In this reaction, the sum of the number of gaseous reactant molecules (9) is less than the sum of the number of the gaseous product molecules (10). So increase in pressure, shifts the equilibrium in the backward direction.

Question 15.
How will an increase in pressure affect each of the following equilbria? An increase in temperature?
i) 2NH₃ (g) ⇌ N₂ (g) + 3H₂ (g), ∆H = 92kJ
ii) N₂(g) + O₂(g) ⇌ 2NO(g) ∆H = 181 kJ
iii) 2O₃(g) ⇌ 3O₂(g) ∆H = -285kJ
iv) CaO(S) + CO₂(g) ⇌ CaCO₃(S) ∆H = -176kJ
Answer:
According to Le Chatelier’s principle increase in pressure shifts the equilibrium in the direction where the number of gaseous molecules is less.

Increase in temperature favours the endothermic reaction, while decrease in temperature favours the exothermic reaction.

i) 2NH₃ (g) ⇌ N₂ (g) + 3H₂ (g), ∆H = 92kJ
Here, the number of gaseous reactant molecules (2) is less than the sum of the total number of product molecules (4). So increase in pressure, shifts the equilibrium in the backward direction.
Increase in temperature shifts the equilibrium in the forward direction.

ii) N₂(g) + O₂(g) ⇌ 2NO(g) ∆H = 181 kJ
Here, the sum of the gaseous reactant molecules is equal to the sum of the gaseous product molecules. So pressure has no affect on the equilibrium.
Increase in temperature favours the forward direction.

iii) 2O₃(g) ⇌ 3O₂(g) ∆H = -285kJ
Here, the number of gaseous reactant molecules (2) is less than the number of gaseous product molecules. So increase in pressure shifts the equilibrium in the backward direction.
Increase in temperature shifts the equilibrium in the backward direction.

iv) CaO(S) + CO₂(g) ⇌ CaCO₃(S) ∆H = -176kJ
In this reaction, there is only one gaseous reactant molecule. No gaseous product molecules are present. So increase in the pressure shift the equilibrium in the forward direction.
Increase in temperature shifts the equilibrium in the backward direction.

Question 16.
The dissociation of HI is independent of pressure, while the dissociation of PCl₅ depends upon the pressure applied. Explain.
Answer:
Dissociation of HI:
2HI_(g) ⇌ H_2(g) + I_2(g)
Here n_P = n_R ⇒ ∆n = 0
∴ Pressure has no effect on the above reaction.

Dissociation of PCl₅:
PCl_5(g) ⇌ PCl_3(g) +Cl_2(g)
Here n_P = 2 and n_R = 1 ⇒ n_P ≠ n_R ⇒ ∆n ≠ 0
∴ This reaction is affected by pressure.

So dissociation of PCl₅ depends upon the pressure.

Question 17.
Explain the terms: (i) Electrolyte (ii) Non-electrolyte (iii) Strong & weak electrolytes (iv) ionic equilibrium
Answer:
i) Electrolyte :
The substance which conducts electricity either in aqueous form (or) in fused state is known as Electrolyte. Ex: Aqueous salts, acids and bases.

ii) Non -electrolyte :
The substances which do not conduct electricity in their aqueous solutions are called non-electrolytes.
Ex: Glucose, Urea, Sucrose.

iii) Strong and Weak electrolytes :
The electrolytes which are almost completely ionised, in solutions are called strong electrolytes. The electrolytes which ionise weakly in their solutions are called weak electrolytes.

iv) Ionic Equilibrium :
The equilibrium that exists between neutral molecules and its ions in solution is called ionic equilibrium. Most ionic reactions are acid-base reactions.

Question 18.
Explain the terms:
i) extent of ionization and on what factors it depends (ii) dissociation (iii) ionization
Answer:
i) The extent of ionization is the ratio of number of molecules ionized to the total number of molecules of the substance. It depends on the polarity and strength of the covalent bond.

ii) Dissociation refers to the process of separation of ions in water which are already existing as such, in the solid solute of the given salt.

iii) Ionization refers to the process of splitting a neutral molecule into charged ions, in the solution.

Question 18.
Explain the Arrhenius concept of acids and bases. [TS 22]
Answer:
Arrhenius acid-base theory:

Acid :
A substance which dissociates to give H⁺ or H₃O⁺ ions in aqueous solution is called acid.
Ex: HCl, HBr, HClO₄, HNO₃ etc.

Base :
The substance which dissociates to give OH^– ions in aqueous solution is called a base.
Ex: NaOH, KOH, NH₄OH etc.

Neutralisation :
The process in which the H⁺ ions from acid react with OH^– ions from base to form undissociated water molecule is called neutralisation.
HCl + NaOH → NaCl + H₂O
The basic neutralisation reactions is
H⁺_(aq) + OH^–_(aq) → H₂O_(l)

Limitations:

1. This theory is applicable to only aqueous solutions.
2. It failed to explain acidic nature of CO₂, SO₂, and basic nature of NH₃,CaO etc.

Question 19.
What is a conjugate acid-base pair? Illustrate with examples. [AP 16]
Answer:
An acid-base pair which differ only by proton is called conjugate acid base pair.

A base fonned by the loss of a proton by an acid is called conjugate base of the acid. An acid is formed by the gain of proton, by a base is called conjugate acid of the base. Acid-base pairs such as H₂O & OH^– and NH⁺₄ & NH₃ which are formed by loss or gain of a proton are called conjugate acid-base pair.

In the above given reaction, H₂O & OH^– and NH⁺₄ & NH₃ are conjugate acid-base pairs.

A strong acid would have large tendency to donate proton. Thus conjugate base of a strong acid would be a weak base. Similarly, conjugate base of a weak acid would be a strong base.

Question 20.
Acetic acid is a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in 1M aqueous solution of acetic acid.
Answer:
Aqueous solution of CH₃COOH contains H₂O, CH₃COOH, CH₃COO^–, H₃O⁺, OH^– In this solution ionic species are CH₃COO^–, H₃O⁺, OH^– which are produced from partial ionisation of H₂O & CH₃COOH.

The order of concentration of ionic species and molecular species:
[H₂O]>[CH₃COOH]>[H₃O⁺] = [CH₃COO^–] >[OH^–]

Question 21.
Show by suitable equations that each of the following species can act as a Bronsted acid.
a) H₃O⁺ b) HCl c) NH3 d) HSO^–₄
Answer:
a) H₃O⁺ + OH^– → 2H₂O
Here, H₃O⁺ is donating a proton to OH^– ion.
So it is a Bronsted acid.

b) HCl + H₂O → H₃O⁺ + Cl^–
Here, HCl is donating a proton to H₂O molecule. So it is a Bronsted acid.

c) NH₃ + NH₃ → NH⁺₄+ + NH^–₂
Here, NH₃ is giving H+ion to another NH₃ molecule. So it is a Bronsted acid.

d) HSO^–₄ + H₂O → SO^-2₄2 + H₃O⁺
Here, HSO₄ gives a proton to H₂O, hence it is a Bronsted acid.

Question 22.
Show by suitable equations that each of the following spieces can act as a Bronsted base.
(a) H₂O
(b) OH^–
(c) C₂H₅OH
(d) HPO₄^-2
Answer:
(a) H₂O + H⁺ → H₃O⁺
H₂O is accepting a proton.
SO it is a Bronsted base.

(b) OH^– + H⁺ → H₂O
OH^– ion is accepting H⁺ ion.
So it is a Bronsted base.

(c) C₂H₅OH + H⁺ → C₂H₅O⁺H₂
C₂H₅OH is accepting H⁺ ion.
So it is a Bronsted base.

(d) HPO^-2₄ + H⁺ → H₂PO^–₄
HPO^-2₄ is accepting a proton.
So it is a Bronsted base.

Question 23.
The species : H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base. [AP 18]
Answer:

Question 24.
Write equation that shows H₂PO^–₄ acting both as an acid and as a base.
Answer:
H₂PO^–₄ + H₂O → HPO^2-₄+ H₃O⁺
Here H₂PO^–₄ is giving a proton to H₂O.
So it is an acid.
H₂PO^–₄ + HCl → H₃PO₄ + Cl^–
Here H₂PO^–₄ is taking proton.
So it is a base.

Question 25.
Write the conjugate acid and conjugate base of each of the following: [AP 19,20,22]
a) OH^–
b) H₂O
c) HCO^–₃
d) H₂O₂.
Answer:

Question 26.
Identify and label the Bronsted acid and its conjugate base, the Bronsted base and its conjugate acid in each of the following equations.
a) H₂SO₄ + Cl^– → HCl + HSO₄
b) H₂S + NH^–₂ → HS^– + NH₃
c) CN^– + H₂O → HCN + OH^–
d) O^2- + H₂O → 2OH^–
Answer:

i) H₂SO₄ can donate proton to Cl^–. Hence it is a Bronsted acid and its conjugate base is HSO^–₄
Cl^– can accept a proton from H₂SO₄.
So it is a Bronsted base and its conjugate acid is HCl.

H₂S can donate a proton to NH^–₄
So it is a Bronsted acid and its conjugate base is HS^–.
NH^–₂ can accept a proton from H₂S.
So it is a Bronsted base and its conjugate acid is NH₃

CN^– can accept a proton from H₂O.
So it is a Bronsted base and its conjugate acid is HCN.
H₂O can donate proton from CN^–.
So it is a Bronsted acid and its conjugate base is OH^–.

Base Acid Conjugate conjugate Acid Base
O^2- can accept a proton from H₂O.
So it is a Bronsted base and its conjugate acid is OH^–.
H₂O can donate proton from O^2-.
So it is a Bronsted acid and its conjugate base is OH^–.

Question 27.
Classify the species AlCl₃, NH₃, Mg⁺² and H₂O into Lewis acids and Lewis bases and justify your answer?
Answer:
i) AlCl₃ can accept an electron pair from
NH₃. SO it is a Lewis acid.
AlCl₃ + NH₃ → [H3N → AlCl₃]

ii) NH₃ can donate an electron pair to H⁺. So it is a Lewis base.
NH₃ + H⁺ → [NH₄⁺]

iii) Mg²⁺ can accept an electron pair from H₂O. SO it is a Lewis acid.
Mg²⁺ + 4H₂O → [Mg(H₂O)₄]²⁺

iv) H₂O can donate an electron pair to Cu⁺². So it is a Lewis base.
Cu²⁺ + 4H₂O → [Cu(H₂O)₄]²⁺

Question 28.
What are the strengths of conjugate bases of a strong acid and a weak acid?
Answer:
An acid which gives a proton easily is a strong acid. A conjugate base which cannot hold the proton strongly is a weak base. So the conjugate base of strong acid is weak base.

A strong base is that which can hold the proton strongly. So the conjugate base of a weak acid is a strong base.

Question 29.
What are the strength of conjugate acids of a strong base and weak base?
Answer:
A strong base is that which holds the proton strongly. So its conjugate acid is a weak acid.

A weak base that which cannot hold the proton strongly. So its conjugate is a strong acid.

Question 30.
Define Ionic Product of water. What is the value at room temperature? [APTS 17]
Answer:
The product of the concentrations of H⁺ and OH^– ions, in water, at a given temperature, is called ionic product of water.
Ionic product of water Kw = [H⁺][OH^–]
K_w = 1.0 × 10^-14mol²/lit², at25°C
Units: mol²/lit²

Question 31.
Define pH. pH cannot be calculated directly from the molar concentration of a weak acid or weak base. Why? Derive an equation for the PH of a weak acid.
Answer:
The pH of a solution is defined as the negative value of the logarithm to base 10 of hydrogen ion concentration expressed in moles/litre in a solution.

Mathematically, pH = -log₁₀ [H⁺] = log₁₀\(\frac{1}{[H^+]}\)
pH can not be calculated directly from the molar concentration of weak acid or weak base. Because they do not undergo complete dissociation.

Derivation for the pH of a weak acid:
Let the concentration of a weak monoprotic acid [HA] is ‘c’ moles /litre and the degree of ionisation is ‘α’.

The equilibrium equation:

Question 32.
Write equations to show the step wise ionization of the polyprotic acids H₂SO₄
Answer:
The stepwise ionization of H₂SO₄ :

Question 33.
Explain how acid strength changes among (i) the hydrides of the group elements and (ii) the hydrides in the same row’ of the periodic table.
Answer:
In the hydrides of the elements in a group, as we move down the atomic size increases. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases.

In a period of the Periodic table as we move from left to right, H-A bond polarity increases. As the electronegativity of A increases, the acid strength also increases.

Question 34.
Justify the statement that water behaves like an acid and also like base on the basis of protonic concept.
Answer:
The Bronsted-Lowry reaction:
HCl + H₂O ⇌ H₃O⁺ + Cl^–

When hydrogen chloride is dissolved in water, HCl donates a proton and H₂O accepts a proton. So HCl is an acid and H₂O is a base.

The Bronsted Lowry reaction:
H₂O + NH₃ ⇌ NH₄⁺ + OH^–
When ammonia is dissolved in water, NH₃ accepts a proton and H₂O donates a proton. So NH₃ is a base and H₂O is an acid. So Water behaves like an acid as well a base.

Question 35.
What is common ion effect? liiustrate.
Answer:
Common ion effect :
The solubility of an electrolyte (acid, base, salt) in water decreases by the addition of another electrolyte (acid, base, salt) which has one ion (cation or anion) common with electrolyte. This is known as common ion effect.
Ex: Ionisation of acetic acid in aqueous solution decreases by the addition of sodium acetate.

CH₃COO^– ion acts as common ion and shifts the equilibrium towards left side.

As the concentration of acetate ion increases in aqueous solution, ionisation of acetic acid decreases. This is due to common ion effect (CH₃COO^– ions).

Significance :

1. Common ion effect is an important phenomenon in qualitative analysis.
2. Common ion effect principle is also used in controlling of H⁺ ion concentration in buffer solution

Question 36.
Define solubility product? Write solubility product expressions for the following:
(i) Ag₂Cr₂O₇
(ii) Zr₃(PO₄)₄.
Answer:
Solubility product(K_sp):
The product of concentrations of the cation and anion of a salt in saturated solution, at constant temperature, is called solubility product.
(i) For Ag₂Cr₂O₇, the solubility product is
K_sp = [Ag⁺]²[Cr₂O^2-₇] – 2Ag⁺ + [Cr₂O^2-₇]

(ii) For Zr₃(PO₄)₄, the solubility product is
K_sp – [Zr⁴⁺]³ [PO^3-₄]⁴ = 3Zr⁴⁺ + 4[PO^3-₄] .

Question 37.
Give the classification of salts. What type of salts undergo hydrolysis?
Answer:
Salts are 4 types.

1. Salt of strong acid and strong base
   Ex: NaCl, K₂SO₄, KNO₃
2. Salt of strong acid and weak base.
   Ex: NH₄Cl, CuSO₄, Mg(NO₃)₂
3. Salt of weak acid and strong base.
   Ex: KCN, Na₂CO₃, CH₃COONa
4. Salt of weak acid and weak base.
   Ex: CaCO₃, CH₃COONH₂, Mg(CN)₂
   Except salts of strong acid & strong base, all the remaining salts undergo salt hydrolysis.

Question 38.
What must be true to the value of ∆G° for a reaction if
a) K > 1 b) K = 1 c) K < 1
Answer:
a) If K > 1 then the value of ∆G° becomes negative. So the reaction is spontaneous or the reaction proceeds in the forward direction in a greater extent.
The products occur predominantly.

b) If K = 1 then the value of ∆G° becomes zero. So the reaction proceed neither in the forward direction nor in the backward direction.

c) If K < 1 then the value of ∆G° becomes positive. This indicates that the reaction is non spontaneous or the reaction proceed in the forward direction to a small extent only. The products occur in small amounts only.

Question 39.
Aqueous solution of NH₄Cl is’acidic. Explain. [TS 16]
Answer:
Ammonium chloride when dissolved in water, reacts with water forming weak base NH₄OH and strong acid HCl.
NH₄Cl + H₂O ⇌ NH₄OH + HCl

Here cationic hydrolysis takes place. HCl is a strong acid and it ionises completely, but NH₄OH being weak base do not ionise completely. So there remains some excess H⁺ ions in the solution. Due to this reason the aqueous solution of ammonium chloride is acidic.

Question 40.
Aqueous solution of CH₃COONa is basic. Explain. [TS 16]
Answer:
Sodium acetate when dissolved in water ionises completely and the ions formed react water forming weak acid (CH₃COOH) and strong base (NaOH).
CH₃COONa + H₂O ⇌ CH₃COOH + NaOH

Here anionic hydrolysis takes place. Acetic acid being weak acid ionise partially but sodium hydroxide is strong electrolyte and ionises completely. So there remains some excess OH^– in the solution. Hence the solution of sodium acetate is basic in nature.

Question 41.
Give reason that acetic acid is less acidic in sodium acetate solution than in sodium chloride solution.
Answer:
The ionic equilibrium reactions of acetic acid, sodium acetate and sodium chloride are as follows.
CH₃COOH ⇌ CH₃COO^– + H⁺
CH₃COONa → CH₃COO^– + Na⁺

By the addition of CH₃COONa to CH₃COOH, the concentration of common ion CH₃COO^– increases. Then according to Lechatelier’s principle, the equilibrium shifts in the backward direction. As a result, the concentration of H⁺ ion decreases.
NaCl → Na⁺ + Cl^–

But in Sodium chloride, there is no common ion. So its equilibrium is not disturbed. Hence acetic acid is less acidic in sodium acetate than in Sodium chloride.

Question 42.
AgCl is less soluble in AgNO₃ solution than in pure water. Explain.
Answer:
Consider the solubility equilibrium of AgCl and AgNO₃.
AgCl ⇌ Ag⁺(aq) + Cl^–(aq)
AgNO₃ ⇌ Ag⁺ + NO^–₃

By the addition of AgNO₃ to AgCl solution the concentration of Ag⁺ increases. Due to common ion effect and Lechatelier’s principle the equilibrium shifts in the backward direction. As a result, the solubility of AgCl is less in AgNO₃ solution.
AgCl ⇌ Ag⁺ (aq) + Cl^– (aq)
H₂O ⇌ H⁺ + OH^–

In pure water, there is no common ion.
Hence AgCl is less soluble in AgNO₃ solution than in pure water.

Question 43.
Predict whether the following reaction will proceed from left to right to any measurable extent:
CH₃COOH(aq) + Cl^– (aq) →
Answer:
CH₃COOH is a weak acid than HCl. This indicates that the conjugate base CH₃COO^– of CH₃COOH is strong and holds the proton strongly. HCl is a strong acid. So its conjugate base Cl^– ion is weak base. So the reaction cannot proceed from left to right.

Question 44.
Aqueous solution of H₂S contains H₂S, HS^–, S^2-, H₃O⁺, OH^– and H₂O in varying concentrations. Which of these species can act only as a base? Which can act only as an acid? Which can act both an acid and as a base?
Answer:
a) S^2- ion can accept proton but cannot donate proton. So, it acts only as a base.

b) H₂S and H₃O⁺ can donate proton but cannot accept proton. So they act only as acids.

c) HS^–, OH^– and H₂O can act as both acids and bases, as they can give H⁺ ions and can accept H⁺ ions.

Long Answer Questions

Question 1.
What are equilibrium processes? Explain equilibrium in Physical and Chemical processes with examples.
Answer:
The state at which the rate of forward reaction is equal to the rate of backward reaction in a reversible reaction is called equilibrium state. This process is called equilibrium process. Equilibrium processes are of two types. They are
1) Physical Equilibrium process :
At equilibrium state if there is a change in the physical state such equilibrium is called physical equilibrium. The most familiar physical equilibria are phase transformation processes.
Ex: Solid ⇌ liquid, liquid ⇌ gas, Solid ⇌ gas
a) Solid ⇌ liquid (Melting or fusion): Ex : Ice and water kept in an insulated vessel at 0°C.
b) liquid ⇌ gas (Vaporisation): Ex: When water is taken in a closed container, an equilibrium exists between water & its vapour
c) Solid ⇌ gas (Sublimation): Ex: When solid iodine is placed in a closed vessel after some time, the vessel gets filled up with violet vapour.

2) Chemical Equilibrium process :
At equilibrium state if it involves a change in the chemical species the equilibrium is called chemical equilibrium. These are of two types.

a) Homogeneous Equilibrium :
If the physical state of reactants and products are same in an equilibrium reaction then the reaction is called homogeneous equilibrium.
Ex: H_2(g) + I_2(g) ⇌ 2HI_(g)

b) Heterogeneous Equilibrium :
If the physical state of reactants and products are different in an equilibrium reaction then the reaction is called heterogeneous equilibrium.
Ex: CO_2(g) + C_(s) ⇌ 2CO_(g)

Question 2.
What is meant by dynamic equilibrium? Explain with suitable examples. [TS 20]
Answer:
Analogous to the physical systems, chemical reactions also attain a state of equilibrium. These reactions can occur both in forward and backward directions. When the rates of the forward and reverse reactions become equal, the concentrations of the reactants and the products remain constant. This is the stage of chemical equilibrium. This equilibrium is dynamic in nature as it consists of a forward reaction in which the reactants give product(s) and reverse reaction in which product(s) gives the original reactants.

The forward and backward reactions of a reversible reaction continuously takes place w ith equal rates simultaneously, at the equilibrium stage also. Hence the equilibrium state is a dynamic equilibrium state. Any chemical equilibrium process is a dynamic equilibrium in nature.
Ex: The dynamic nature of chemical equilibrium is observed in the synthesis of ammonia by Haber’s process.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g) ∆H = -92.0kJ

Question 3.
Give the general characteristics of equilibrium involving physical processes
Answer:
General characteristics of equilibrium involving physical processes:

1. Both forward and backward reactions continue to take place in a reversible reaction.
2. At equilibrium state, the rate of the forward reaction is equal to that of the reverse reaction.
3. All the reactants and all the products of the reaction are present at equilibrium.
4. Properties like concentration, pressure, intensity of colour remain unchanged with time after the equilibrium is established.
5. The same equilibrium can be attained by carrying out the reaction starting with the reactants or starting with the products.
6. The equilibrium is not static. It is dynamic, since forward and backward reactions are not ceased at equilibrium.
7. Change in temperature, pressure or concentration of substances may change the position of equilibrium.
8. Addition of a catalyst to the reaction does not alter the position of equilibrium. It only speeds up the attainment of equilibrium.
9. At equilibrium, change in free energy ∆G = 0.

Question 4.
What arc the important features of equilibrium constant? Discuss any two applications of equilibrium constant.
Answer:

1. The equilibrium constant has a definite unique value for every chemical reaction at a particular temperature.
2. The value of equilibrium constant is independent of initial concentrations of reactants and products.
3. The value of equilibrium constant K changes with change in temperature.
4. Equilibrium constant for a given reaction is the inverse of the equilibrium constant for the reverse reaction.
5. The equilibrium constant is independent of the presence of catalyst.
6. The equilibrium constant is expressed in terms of concentration of reactants and products.

Applications of equilibrium Constant:
1) Predicting of extent of reaction :
The magnitude of equilibrium constant indicates about the extent to which the reactants are converted into the products before the equilibrium is attained.

2) Predicting the direction of the reaction:
From the magnitude of equilibrium constant it is possible to predict the direction in which the net reaction is taking place. For this purpose we have to calculate the reaction quotient Q_c.
If Q_c > K_c then the reaction takes place in the direction of reactants.
If Q_c < K_c then the reaction takes place in the direction of products.
If Q_c = K_c then the reaction is at equilibrium and no net reaction takes place.

Question 5.
What is Le-Chatelier’s principle? Discuss briefly the factors which can influence the equilibrium. [AP 19]
Answer:
Le-Chatelier’s principle :
If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration, the equilibrium position shifts in the direction, in which this change is reduced or nullified.

Factors effecting the equilibrium position :
a) Effect of concentration:

1. Increase in the concentration of reactants in the reaction mixture at equilibrium favours the forward reaction
2. Decrease in the concentration of reactants in the reaction mixture at equilibrium favours the backward reaction
3. Increase in the concentration of products in the reaction mixture at equilibrium favours the backward reaction
4. Decrease in the concentration of products in the reaction mixture at equilibrium favours the forward reaction

b) Effect of pressure :

1. Increase in the pressure of the system at equilibrium, shifts the equilibrium position in that direction in which the volume or the number of moles gets decreased.
2. Decrease in the pressure of the system at equilibrium, shifts the equilibrium position in that direction in which the volume or the number of moles gets increased.

c) Effect of temperature :

1. Increase in the temperature of the system at equilibrium, shifts the equilibrium position in that direction in which temperature decreases or heat is absorbed. Thus endothermic reaction is favoured.
2. Decrease in the temperature of the system at equilibrium, shifts the equilibrium position in that direction in which temperature increases or heat is released. Thus exothermic reaction is favoured.

d) Effect of catalyst :
Catalyst does not change the position of equilibrium, but helps to achieve equilibrium quickly by increasing the rate of both forward and backward reaction to the same extent.

Question 6.
Discuss the application of Le-Chatelier’S principle for the industrial synthesis of Ammonia and Sulphur trioxide. [AP, TS 15,16,18,19,22]
Answer:
Le-Chatelier’s principle :
If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration, the equilibrium position shifts in the direction, in which this change is reduced or nullified.

Synthesis of ammonia – Haber’s process :
Nitrogen and hydrogen combine to form ammonia. This reaction is reversible and exothermic.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g), ∆H = – 92.3 KJ. ”

Effect of temperature :
Formation of ammonia is an exothermic reaction. Hence low tem-perature is favourable for greater yield of NH₃. But at low temperature the reaction between N₂ and H₂ is very slow. Therefore an optimum temperature of 725-775 K is used. To speed up the reaction, finely divided iron is used as catalyst and molybdenum is used as a promoter.

Effect of pressure : N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
no.of moles of reactants = 4, no.of moles of products = 2
The formation of ammonia is accompanied with decrease in volume (or) number of moles. So high pressure is favourable for the higher yield of NH₃. Hence a pressure 200-300 atm is used.

Optimum conditions:
Concentration :
Pure N₂ and H₃ should be mixed in high concentration
Temperature : 725-775 K
Pressure : 200-300 atm
Catalyst: Finely divided iron
Promoter: Mo

Synthesis of SO₃ – Contact process : [TS 18]
Sulphurdioxide and oxygen combine to form SO₃.
2SO_2(g) + O_2(g) ⇌ 2SO_3(g), ∆H = – 189.0KJ

Effect of temperature :
Formation of SO₃ is an exothermic reaction. Low temperature favours for greater yield of SO₃. But at low temperature the reaction between SO₂ and O₂ is very slow. Hence an optimum temperature of 673K is used. To increase the rate of reaction, V₂O₅ (or) platinised asbestos is used as a catalyst.

Effect of pressure: 2SO_2(g) + O_2(g) ⇌ 2SO_3(g)
no.of moles of reactants = 2 + 1 = 3, no. of moles of products = 2
The formation of SO₃ is accompanied with decrease in volume (or) number of moles. Hence high pressure is favourable for the formation of SO₃. But due to the acidic nature of SO₃, the towers used in the synthesis are corroded at high pressure. So an optimum pressure of 1.5 to 1.7 atm is used.

Optimum conditions :
Concentration : High concentration of SO₂ and O₂.
Temperature : 673 K
Pressure : 1.5 – 1.7 atm
Catalyst: V₂O₅ (or) platinised asbestos.

Question 7.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction.
CH_4(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
a) Write an expression for K_p for the above reaction.
b) How will the values of K_p and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst?
Answer:

b) i) If the pressure is increased, according to Lechatelier’s principle, the equilibrium favours the backward direction. As a result, more of reactants (CH₄, H₂O) will be formed.
Hence the value of K_p decreases.

ii) If temperature is increased, according to Lecbatelier’s principle, the forward reaction will be favoured, as it is endothermic. Therefore, the equilibrium favours the forward direction As a result, more products (CO, H₂) will be formed. Hence the value of K_p increases.

iii) Catalyst does not influence the value of K_p in the equilibrium mixture.

Question 8.
Describe the effect of:
a) addition of H₂
b) addition of CH₃OH
c) removal of CO
d) removal of CH₃OH on the equilibrium of the reaction 2H₂(g) + CO(g) ⇌ CH₃OH(g)
Answer:
a) Addition of H₂:
According to Le Chatelier’s principle, increase in the concentration of reactants shifts the equilibrium in the forward direction, which decreases the concentration of H₂.

b) Addition of CH₃OH:
According to Le Chatelier’s principle, increase in the concentration of the products shifts the equilibrium in the backward direction. This decreases the concentration of CH₃OH.

c) Removal of CO:
According to Le Chatelier’s principle, decrease in the concentration of reactants shifts the equilibrium in the backward direction. This increases the concentration of CO.

d) Removal of CH₃OH:
According to Le Chatelier’s principle, decrease in the concentration of products shifts the equilibrium in the forward direction. This increases the concentration of CH₃OH.

Question 9.
At 473K equilibrium constant K_c for the decomposition of phosphorous pentachloride, PCl₅ is 8.3 × 10^-3 if the decomposition is depicted as:
PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) ∆H = 124.0 kJmol^-1
a) Write an expression of K_c for the reaction
b) What is the value of K_c for the reverse reaction at the same temperature?
c) What would be effect on K_c if
(i) more PCl₅ is added (ii) pressure is increased (iii) the temperature is increased.
Answer:

c) i) With increase in concentration of PCl₅ , according to Lechatelier’s principle forward reaction is favoured. So K_c remains constant.
ii) With increase in pressure according to Lechatelier’s principle backward reaction takes place i.e., decrease in number of moles. So K_c remains constant.
iii) By increasing the temperature, the forward reaction will be favoured. Since it is endothermic in nature, the value of K_c increases.

Question 10.
Explain the concept of Bronsted acids and Bronsted bases. Illustrate the answer with suitable examples. [TS 17,20]
Answer:
Bronsted-Lowry theory :
Acidic or basic nature of substances is explained interms of acceptance and donation of proton(s) of the molecules.

Bronsted acid :
Any substance that can loose a proton or protons is an acid.
Ex: HCl, H₂SO₄, H₃PO₄, CH₃COOH.

Bronsted base :
Any substance that can gain a proton or protons is a base.
Ex: NH₃, H₂O, OH^– etc.

When an acid react with base neutralization takes place. As per Bronsted Lowry theory proton transfer from acid to base is neutralization.
Ex: HCl + H₂O ⇌ H₃O⁺ + Cl^–
The above reaction is reversible. In this reaction,
i) HCl donates proton to H₂O. So HCl is a Bronsted acid.
ii) H₂O gains the proton. So it is a Bronsted base.
iii) H₃O⁺ donates a proton to Cl^–. So H₃O^– is a Bronsted acid.
iv) Cl^– gains the proton. So it is a Bronsted base.

The acid base pair which differ by a single proton is said to be conjugate acid base pair.

In the above reaction HCl & Cl^– and H₃O⁺ and H₂O are conjugate acid-base pairs. Every Bronsted acid has its conjugate base and every Bronsted base has a conjugate protonic acid.

According to Bronsted-Lowry theory, greater the tendency to donate proton, stronger is the acid. Higher the tendency of the base to accept protons, stronger is the base. In the above reaction, HCl gives proton easily than H₃O⁺. So HCl is stronger acid than H₃O⁺. H₂O accepts the proton more easily than Cl^–. So H₂O is a stronger base than Cl^–.

A stronger acid has a weaker conjugate base and a stronger base has weaker conjugate acid.

Advantages of Bronsted-Lowry theory:

1. This theory explains the behaviour of acids and bases in both aqueous and non-aqueous solvents.
2. It explains the behaviour of NH₃, CaO etc. as bases and CO₂, SO₂ etc., as acids.

Drawbacks of Bronsted-Lowry theory:

1. Proton donation or acceptance happens only in the presence of other substances.
2. It could not explain the acidic nature of electron deficient molecules like AlCl₃, BCl₃.

Question 11.
Explain Lewis acid-base theory with suitable example. Classify the following species into Lewis acids and Lewis bases and show these act as Lewis acid/base. [TS 18]
a) OH^–
b) F^–
c) H⁺
d) BCl₃
Answer:
Lewis acid-base theory :
Acidic or basic nature of substances is explained in-terms of acceptance and donation of electron pair of the molecules.

Lewis acid :
The substance which can accept an electron pair to form a coordinate covalent bond is called Lewis acid.

Types of Lewis acids:
1) All transition metal cations and some cations of other metals.
Ex: Ag⁺, CO⁺³, Cu⁺², Fe⁺³.etc.,

2) Proton and protonated species.
Ex: H⁺, PH₄⁺etc.,

3) All electron deficient molecules.
Ex: BF₃, BCl₃, AlCl₃, FeCl₃.

4) Compounds in which central atom has vacant ‘d’ orbitals and which can expand its octet.
Ex: SiF₄, SnCl₄, SF₄, TeF₄, FeCl₃.

5) Non metallic oxides. Molecules having multiple bonds between atoms of different electronegativities.
Ex: CO₂, SO₂, SO₃, NO₂, Cl₂O₇, P₄O₁₀.

6) Elements with an electron sextet. Ex: S, O.

Lewis base :
The substance which donates electron pair to form a coordinate covalent bond is called Lewis base.

Types of Lewis bases:

1. All anions act as Lewis bases.
   Ex: Cl^–, OH^–, CN^–, NH^–₂, F^–, SCN^–.
2. Molecules with one or two lone pairs on the central atom.
   EX: H₂O, NH₃, R – OH, R – NH₂, R – O – R, C₅H₅N.
3. Molecules with multiple bonds.
   Ex: CO, NO, HC ≡ CH, H₂C = CH₂
4. All metallic oxides. Ex: Na₂O, K₂O MgO, CaO etc.
   a) OH^– is Lewis base, because it can donate a electron pair.
   b) F^– is Lewis base, because it can donate any one of its four electron lone pairs.
   c) H⁺ is Lewis acid, because it can accept a lone pair of electrons.
   d) BCl₃ is Lewis acid, because it can accept a lone pair of electrons.

Question 12.
What is degree of ionisation in respect of weak acids and weak base? Derive the relation between the degree of ionisation (α) and the ionisation constant (Ka) for the weak acid HX.
Answer:
The ratio of number of moles of weak acid or weak base ionised, to the total number of moles, is called its degree of ionisation (α).

Ionisation of a weak acid :
Let the concentration of a weak monoprotic acid [HX] is ‘c’ moles /litre and the degree of ionisation is ‘α’.

The equilibrium equation:

Question 13.
Define pH. What is buffer solution? Derive Henderson-Hasselbalch equation for calculating the pH of an acid buffer solution.
Answer:
The pH of a solution is defined as, the negative value of the logarithm to base 10 of hydrogen ion concentration, expressed in moles/litre in a solution.
Mathematically, pH = -log₁₀[H⁺] = log₁₀ \(\frac{1}{[H^+]}\)

Buffer solution :
A solution which can resist the change in pH, on dilution or on addition of small amount of acid or base is called Buffer solution.

Acid Buffer solution :
An acid buffer consists of a weak acid (HA) and its salt with a strong base (A^–).

Derivation of Henderson-Hasselbalch equation:
The equilibrium equation of acid buffer solution:

This is called Henderson’s equation.
Since most of the conjugate base [A^–] comes from the ionisation of salt of the acid,
we have pH = pK_a + log\(\frac{[Salt]}{[Acid]}\)

Question 14.
Explain the term “Hydrolysis of salts” with examples. Discuss the pH of the following types of salt solutions. [TS 16]
i) Saits of weak acid and strong base
ii)Salts of strong acid and weak base.
Answer:
Salt Hydrolysis :
The reaction of a salt with water to form an acid and a base is called Salt hydrolysis. Hydrolysis of salts is the reverse process of neutralisation.
Salt + Water ⇌ Acid + Base

The phenomenon in which the anion or cation or both of a salt react with water producing excess of OH^– ions or H⁺ ions or both in an aqueous solution is called salt hydrolysis. Salt hydrolysis may be cationic hydrolysis or anionic hydrolysis.

Cationic hydrolysis :
The cation in a salt reacts with water to give excess H⁺ ions. As a result, the aqueous solution of such salt is acidic in nature. That is why, it is called cationic hydrolysis.

Anionic hydrolysis :
The anion in a salt reacts with water to give excess OH^– ions. As a result, the aqueous solution of such salt is basic in nature. That is why, it is called anionic hydrolysis.

Types of Salt and their hydrolysis:
i) Salts of weak acid and strong base: CH₃COONa, Na₂CO₃, NaCN etc.
The anion of these salts undergo hydrolysis.
Ex: CH₃COONa
Sodium acetate is a salt of weak acid(CH₃COOH) and strong base(NaOH).
In aqueous solution, the acetate ions undergo hydrolysis resulting basic solution.
CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–
Here, [OH^–] >[H⁺] ⇒ [OH^–] > 1.0 × 10^-7 moles/lit ⇒ pH > 7. So the solution is basic.

ii) Salts of strong acid and weak base: NH₄Cl, CuSO₄, FeSO₄, FeCl₃ etc.
The cations of these, salts undergo hydrolysis.
Ex: NH₄Cl NH4CI ⇌ NH₄⁺ + CI^–
H₂O ⇌ H⁺ + OH^–
NH₄Cl is a salt of strong acid(HCl) and weak base(NH₄OH).

In aqueous solution, the ammonium ions undergo hydrolysis resulting an acidic solution.
NH₄⁺+ H₂O ⇌ NH₄OH + H⁺
Here, [H⁺] > [OH^–] ⇒ [H⁺] > 1.0 × 10^-7 moles/lit ⇒ pH < 7. So the solution is acidic.

Question 15.
What is solubility product? Explain the common ion effect on solubility of ionic salts.
Answer:
Solubility product(K_sp):
The product of concentration of cation and anion of a salt present in a saturated aqueous solution is called Solubility product.
Units: (Mol/L)ⁿ
Where n = Total number of ions that can be given by salt per molecule on ionization.

Factors effecting K_sp:
a) nature of substance b) temperature

Significance :
Solubility product is very important in preferential precipitation process of quantitative analysis.

The solubility’ of an electrolyte in water decreases by the addition of another electrolyte which has one ion common with the electrolyte.

For example, the solubility of NaCl decreases on the addition of HCl to the solution of NaCl in water. This is known as salting out of water.

Common ion effect :
The solubility of an electrolyte (acid, base, salt) in water decreases by the addition of another electrolyte (acid, base, salt) which has one ion (cation or anion) common with electrolyte. This is known as common ion effect.
Ex: Acetic acid is weak acid and it undergoes partial ionization as follows.
CH₃COOH ⇌ CH₃COO^– + H⁺

In case of such electrolytes there exists an equilibrium in between ions and molecular form. In such a condition, the addition of electrolyte with common ions shifts the equilibrium in backward direction which causes the decrease in solubility.

Significance:

1. Common ion effect is an important phenomenon in quantitative analysis.
2. The common ion effect principle is also used in controlling the concentration of H⁺ions in buffer solution.

Question 16.
Write notes on
i) Common ion effect.
ii) The relation between K_sp and solubility (S) of a sparingly soluble salt BaSO₄.
Answer:
Common ion effect :
The solubility of an electrolyte (acid, base, salt) in water decreases by the addition of another electrolyte (acid, base, salt) which has one ion (cation or anion) common with electrolyte. This is known as common ion effect.
Ex: Acetic acid is weak acid and it undergoes partial ionization as follows.
CH₃COOH ⇌ CH₃COO^– + H⁺

In case of such electrolytes there exists an equilibrium in between ions and molecular lorm. In such a condition, the addition of electrolyte with common ions shifts the equilibrium in backward direction which causes the decrease in solubility.

Significance:

1. Common ion effect is an important phenomenon in quantitative analysis.
2. The common ion effect principle is also used in controlling the concentration of H+ions in buffer solution.

ii) The relation between and solubility (S) Of a sparingly soluble salt BaSO₄:
The sparing soluble salt like BaSO₄, when added to water form saturated solution There exists an equilibrium between undissolved solid and the ions in saturated solution.

Multiple Choice Questions

Question 1.
1 mole of PCl₅ is heated in a closed Vessel of 1 litre capacity. At equilibrium 0.4 moles of chlorine is found. Calculate the equilibrium constant.
1) 0.257
2) 0.276
3) 0.286
4) 0.267
Answer:
4) 0.267
Solution:

Question 2.
Nitrogen dioxide forms dinitrogen tetroxide according to the equation. 2NO₂(g) ⇌ N₂O₄(g) when 0.1 mole of NO₂ is added to a 1 litre flask at 25°C, the concentration changes so that at equilibrium [NO₂] = 0.016M and [N₂O₄] = 0.042M.
a. What is the value of the reaction quotient before any reaction occurs.
b. What is the value of the equilibrium constant for the reaction.
1) 0,1.61 × 10²
2) 0,1.64 × 10²
3) 0,1.62 × 10²
4) 0,1.63 × 10²
Answer:
2) 0,1.64 × 10²
Solution :
(a) The value of reaction quotient before any reaction occur is zero.
(b) Given reaction: 2NO_2(g) ⇌ N₂O_4(g)

Question 3.
The equilibrium constant for the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 725K is 6.0 × 10^-2. At equilibrium, [H₂] = 0.25 molL^-1 and [NH₃] = 0.06molL^-1. Calculate the equilibrium concentration of N₂.
1) 3.84 mole L^-1
2) 3.74 mole L^-1
3) 3.64 mole L^-1
4) 3.54 mole L^-1
Answer:
1) 3.84 mole L^-1
Solution :

Question 4.
At certain temperature, K_c for the reaction SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g) is 16. If initially one mole each of all the four gases are taken in one vessel. What are the equilibrium concentrations of NO and NO₂.
1) 15 mol/L, 0.3 mol/L
2) 1.6 mol/L, 0.4 mol/L
3) 1.4 mol/L, 0.2 mol/L
4) 1.3 mol/L, 0.1 mol/L
Answer:
2) 1.6 mol/L, 0.4 mol/L
Solution :

Question 5.
Under certain conditions, the equilibrium, constant for the decomposition of PCl₅(g) into PCl₃(g) and Cl₂(g) is 0.0211 mol L^-1. What are the equilibrium concentrations of PCl₅, PCl₃ and Cl₂, if the initial concentration of PCl₅ was LOOM?
1) 0.685, 0.153, 0.135
2) 0.885, 0.135, 0.135
3) 0.865, 0.135, 0.135
4) 0.865, 0.125, 0.133
Answer:
3) 0.865, 0.135, 0.135
Solution :

Question 6.
For the reaction A + B ⇌ 3C at 25°C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C respectively predict the direction of reaction if
a. K_c for the reaction is 10
b. K_c for the reaction is 15
c. K_c for the reaction is 10.66
1) forward, backward, equilibrium
2) backward, forward, equilibrium
3) backward, forward, backward
4) backward, forward, forward
Answer:
2) backward, forward, equilibrium
Solution :
a) For the reaction K_c = 10. ∴ Q > K
Hence the reaction will be in the backward direction

b) For the reaction K_c = 15 ∴ Q < K
Hence the reaction will be in the forward direction

c) For the reaction K_c = 10.66 ∴ Q = K
Hence the reaction is at equilibrium.

Question 7.
A mixture of H₂, N₂ and NH₃ with molar concentrations 5.0 × 10^-3 mol L^-1, 4. 0 × 10^-3 mol L^-1 and 2.0 × 10^-3 mol L^-1 respectively was prepared and heated to 500K. The value of K_c for the reaction: 3H_2(g) + N_2(g) ⇌ 2 NH_3(g) at this temperature is 60. Predict whether ammonia tends to form or decompose at this stage of concentration.
1) decompose
2) form
3) neutral
4) None of these
Answer:
1) decompose
Solution :
[NH₃] = 2.0 × 10^-3mol L^-1
[N₂] = 4.0 × 10^-3mol L^-1
[H₂] = 5.0 × 10^-3mol L^-1
Given reaction: N_2(g) + 3H_2(g) ⇌ 2NH₃(g)

The given value of K_c = 60
∴ Q_c > K_c. Hence the the reaction will be in backward direction and hence ammonia tends to decompose.

Question 8.
At 5000K, K_p value for the reaction
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is 2.5 × 10¹⁰.
Find the value of K_p for each of following reactions at the same temperature.
a. SO₂(g) + 1/2O₂ (g) ⇌ SO₃(g)
b. SO₃(g) ⇌ SO₂(g) + 1/2O₂(g)
c. 3SO₂(g) + 3/2O₂(g) ⇌ 3SO₃(g)
1) 1.48 × 10⁵, 6.3 × 10⁶, 3.95 × 10¹⁵
2) 1.48 × 10⁵, 6.2 × 10⁶, 3.59 × 10¹⁵
3) 1.58 × 10⁵, 6.3 × 10⁶, 3.95 × 10¹⁵
4) 1.58 × 10⁵, 6.2 × 10⁶, 3.59 × 10¹⁵
Answer:
3) 1.58 × 10⁵, 6.3 × 10⁶, 3.95 × 10¹⁵
Solution :

Question 9.
K_c for the reaction N₂O₄(g) ⇌ 2NO₂(g) is 4.63 × 10^-3 at 25°C.
a) What is the value of K_p at this temperature.
b) At 25°C, if the partial pressure of N₂O₄(g) at equilibrium is 0.2atm. Calculate equilibrium pressure of NO₂(g)
1) 0.1132 atm, 0.15 atm
2) 0.1122 atm, 0.15 atm
3) 0.1132 atm, 0.11 atm
4) 0.1123 atm, 0.15 atm
Answer:
1) 0.1132 atm, 0.15 atm
Solution :

Question 10.
At 27°C, K_p value for the reversible reaction PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) Calculate K_c.
1) 0.0367 mol/L
2) 0.0463 mol/L
3) 0.0264 mol/L
4)0.0162 mol/L
Answer:
3) 0.0264 mol/L
Solution :
K_p = 0.65; R = 0.0821 lit. atm. moL^-1. K^-1;
T = 27 + 273 = 300K
PCl_5(g) ⇌ PCl_3(g) +Cl_2(g)
⇒ ∆n – n_P – n_R = 2 – 1 = 1
K_p = K_c(RT)^∆n
⇒ 0.65 = K_c (0.0821 × 300)¹
⇒ K_c = \(\frac{0.65}{0.0821\times300}\) = 0.0264 mol/L

Question 11.
K_c for the reaction N_2(g) + 3H_2(g) ⇌ 2NH_3(g) is 0.5 at 400K, find K_p.
1) 4.63 × 10^-4
2) 5.64 × 10^-4
3) 6.65 × 10^-4
4) 7.66 × 10^-4
Answer:
1) 4.63 × 10^-4
Solution :
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Given k_c = 0.5, T = 400K, ∆n = 2 – 4 = -2
K_p = k_c(RT)^∆n
= 0.5 × (0.0821 × 400)^-2;
= 0.5 × (8.21 × 4)^-2 = 0.5 × (32.84)^-2
= \(\frac{0.5}{1078.46}\) = 4.63 × 10^-4

Question 12.
1 mole of A and 1 mole of B are taken in a 5 litre flask, 0.5 mole of C is formed in the equilibrium of. A + B ⇌ C + D
What is molar concentration of each species if the reaction is carried with 2 mole of A 1 mole of B in a 5 litre flask at the same temperature.
1) 0.266 mol/ L, 0.066 mol/L, 0.134 mol/L 0.134 mol/L
2) 0.265 mol/ L, 0.066 mol/L, 0.134 mol/L 0.124 mol/L
3) 0.266 mol/ L, 0.006 mol/L, 0.134 mol/L 0.134 mol/L
4) 0.266 mol/L, 0.066 mol/L, 0.124 mol/L 0.134 mol/L
Answer:
1) 0.266 mol/ L, 0.066 mol/L, 0.134 mol/L 0.134 mol/L
Solution :

Question 13.
For the following reaction
PCl_5(g) ⇌ PCl_3(g) + Cl_2(g) 0.4 mol of PCl₅, 0.2 mole of PCl₃ and 0.6 mole of Cl₂ are taken in a 1 litre flask, if K_c = 0.2. Predict the direction in which reaction proceeds.
1) forward
2) backward
3) neutral
4) None
Answer:
2) backward
Solution :

Question 14.
In an equilibrium A + B ⇌ C + D ; A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the attainment of equilibrium concentration of C was thrice concentration of B. Calculate K_c.
1) 1.6
2) 1.7
3) 1.8
4) 1.9
Answer:
3) 1.8
Solution :

Question 15.
A mixture of SO₂, SO₃, and O_c, gases are maintained at equilibrium in 10 litre flask at a temperature at which K_c for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is 100. At equilibrium.
a) If no. of moles of SO₃ and SO₂ in flask are same, how many moles of O₂, are present.
b) If no. of moles of SO₃, in flask is twice the no.of moles of SO₂, how many moles of O₂, are present.
1) 0.1 mole, 0.4 mole
2) 0.2 mole, 0.4 mole
3) 0.3 mole, 0.6 mole
4) 0.1 mole, 0.5 mole
Answer:
1) 0.1 mole, 0.4 mole
Solution :

Question 16.
For A + B ⇌ C, the. equilibrium concentrations of A and B at a temperature are 15 mol L^-1. When volume is doubled the reaction has equilibrium concentration of A as 10 molL^-1. Calculate a) K_c b) Concentration of C in original equilibrium.
1) 0.1
2) 0.2
3) 0.3
4) 0.4
Answer:
2) 0.2
Solution :

Question 17.
A vessel at 100 K contains CO₂ with a pressure of 0.5 atm. Some of the CO₂ is converted into CO on addition of graphite. Calculate the value of K, if total pressure of equilibrium is 0.8 atm.
1) 1.6 atm
2) 1.7 atm
3) 1.8 atm
4) 1.9 atm
Answer:
3) 1.8 atm
Solution :

Question 18.
The K_p value for the rieaytion H₂(g) + I₂(g) ⇌ 2HI(g) at 460°C is 49. If the initial pressure of H₂ and I₂ are 0.5atm respectively, determine the partial pressure of each gases at equilibrium.
1) 0.112 atm, 0.111 atm, 0.788 atm
2) 0.110 atm, 0.110 atm, 0.778 atm
3) 0.111 atm, 0.111 atm, 0.778 atm
4) 0.112 atm, 0.112 atm, 0.778 atm
Answer:
4) 0.112 atm, 0.112 atm, 0.778 atm
Solution :

Question 19.
0.5 mol of H₂ and 0.5 mole of I₂ react in 10 litre flask at 448°C. The equilibrium constant K_c is 50 for H₂(g) + I₂(g) ⇌ 2HI(g)
a) What is the value of K_p.
b) Calculate the mole of I₂ at equilibrium.
1) 50,0.001
2) 50,0.111
3) 50,0.101
4) 50,0.119
Answer:
2) 50,0.111
Solution :

Question 20.
How much PCl₅ must be added to a one little vessel at 250°C in order to obtain a concentration of 0.1 mole of Cl₂ at equilibrium. K_c for PCl₅(g) ⇌ PCl₃(g) + Cl₂(R) is 0.0414M.
1) 0.3215 moles
2) 0.3425 moles
3) 0.3446 moles
4) 0.3415 moles
Answer:
4) 0.3415 moles
Solution :

Question 21.
K_p for the reaction N_2(g) + 3H_2(g) ⇌ 2NH_3(g) at 400°C is 1.64 × 10^-4. a) Calculate the K_c b) Calculate ∆G° value using K_c value.
1) 0.4006, 3872.25 J
2) 0.5006, 3872.25 J
3) 0.5006, 3822.25 J
4) 0.5506, 3872.25 J
Answer:
2) 0.5006, 3872.25 J
Solution :
a) N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
∆n = n_P – n_R = 2 – 4 = -2
R = 0.0821 lit. atm. mol^-1 .K^-1
T = 400+ 273 = 673 K .
K_p = 1.64 v 10^-4
Now K_p = K_c (RT)^∆n
⇒ 1.64 × 10^-4 = K_c (0.0821 × 673)^-2
⇒ K_c = 1.64 × 10^-4 (0.0821 × 673)² = 0.5006

b) ∆G° = – 2.303 RT log K_c
= – 2.303 × 8.314 × 673 log (0.5006)
= 3872.251

Question 22.
Calculate the pH of [TS 15]
(a) 10^-3 M HCl
(b) 10^-3 M H₂SO₄
(c) 10^-6 M HNOa
(d) 0.02 M H₂SO₄
1) 3, 2.699, 6, 1.3979
2) 3, 2.999, 6, 1.3679
3) 3, 2.669, 6, 1.3679
4) 3, 2.589, 6, 1.3979
Answer:
1) 3, 2.699, 6, 1.3979
Solution :

a) HCl is a strong and monobasic acid.
∴ [H⁺] = Molarity × Basicity
= 10^-3 × 1 = 10^-3
∴ pH = -log₁₀[H⁺]
= log₁₀10^-3 = – (-3) log₁₀10 = 3

b) H₂SO₄ is a strong and dibasic acid.
N = m × n = 10^-3 × 2
∴ pH = -log[H⁺] = -log(2 × 10^-3)
= 3-log2 = 3 – 0.3010 = 2.699

c) HNO3 is a strong and monobasic acid.
∴ [H⁺] = Molarity x Basicity
= 10^-6 × 1 =10^-6
∴ pH =-log10[H⁺]
= -log₁₀10^-6 = -(-6) log₁₀10 = 6

b) H₂SO₄ is a strong and dibasic acid.
∴ [H⁺] = Molarity × Basicity
= 0.02 × 2 = 4 × 10^-2
∴ pH = -log[H⁺] = -log₁₀(4 × 10^-2)
= -log 10⁴-log 10^-2
= -0.6021 – (-2)log₁₀10
= 2 – 0.6021 = 1.3979

Question 23.
Calculate of pH for
(a) 0.001 M NaOH
(b) 0.01 M Ca(GH)₂
(c) 0.0008 M Ba(OH)₂
(d) 0.004M NaOH
1) 11, 12.3010, 11.1041, 11.5021
2) 11, 12.3000, 11.2041, 11.6021
3) 11, 12.3010, 11.2041, 11.6021
4) 11, 12.3000, 11.2241, 11.0021
Answer:
3) 11, 12.3010, 11.2041, 11.6021
Solution :

a) NaOH is a strong and monoacidic base.
∴ [OH^–] = Molarity × Acidity
= 0.001 × 1 = 10^-3
∴ pOH = -log₁₀[OH^–]
= – log ₁₀10^-3 = -(-3) log₁₀10 = 3

But, pH + pOH = 14
∴ pH = 14 – pOH = 14 – 3 = 11

b) Ca(OH)₂ is a strong and diacidic base.
∴ [OH^–] = Molarity × Acidity
= 0.01 × 2 = 2 × 10^-2
∴ pOH = -log₁₀[OH^–]
= -log₁₀(2 × 10^-2)
= -log₁₀2 – log₁₀10^-2
= -0.3010 – (-2) log₁₀10
= 2 – 0.3010 = 1.6990
∴ pH = 14 – pOH = 14 – 1.6990 = 12.3010

c) Ba(OH)₂ is a strong and diacidic base.
∴ [OH^–] = Molarity × Acidity
= 0.0008 × 2 = 16 × 10^-4
∴ pOH = -log₁₀[OH^–]
= -log₁₀(16 × 10^-4)
= -log₁₀16- log₁₀10^-4
= -1.2041 – (-4) log 10
= 4 – 1.2041 = 2.7959
pH = 14 – pOH = 14 – 2.7959 = 11.2041

d) NaOH is a strong and monoacidic base.
∴ [OH^–]= Molarity × Acidity
= 0.004 × 1 = 4 × 10^-3
pOH = -log₁₀[OH^–]
= -log₁₀(4 × 10^-3)
= – log₁₀4 – log₁₀10^-3
= -0.6021 – (-3)log₁₀10
= 3-0.6021 = 2.3979
∴ pH = 14 – pOH =14- 2.3979 = 11.6021

Question 24.
The pH of a solution is 3.6. Calculate H₃O⁺ ion concentration.
1) 2.532 × 10^-4
2) 2.512 × 10^-4
3) 2.522 × 10^-4
4) 2.555 × 10^-4
Answer:
2) 2.512 × 10^-4
Solution :
Given pH = 3.6
⇒ pH= -log[H⁺]
⇒ log[H⁺]= -pH = 3.6 = -3 – 0.6 + 1 – 1 = 4.4
∴ [H₃O⁺] = [H⁺] = Anti log of [4.4]
= Anti log of 0.4 × 10^-4 = 2.512 × 10^-4

Question 25.
The pH of a solution is 8.6. Calculate OH^– ion concentration. .
1) 3.961 × 10^-6 moles/L
2) 3.971 × 10^-6 moles/L
3) 3.991 × 10^-6 moles/L
4) 3.981 × 10^-6 moles/L
Answer:
4) 3.981 × 10^-6 moles/L
Solution :

Question 26.
What is [H⁺] for a solution in which
a) pH = 3 b) pH = 4.75 c) pH = 4.4
1) 10^-5M, 1.778 × 10^-5M, 3.981 × 10^-5M
2) 10^-5M, 1.777 × 10^-5M, 3.981 × 10^-5M
3) 10^-5M, 1.778 × 10^-5M, 3.991 × 10^-5M
4) 10^-5M, 1.778 × 10^-5M, 3.881 × 10^-5M
Answer:
2) 10^-5M, 1.777 × 10^-5M, 3.981 × 10^-5M
Solution :

Question 27.
A solution of 0.005M H2S04 is diluted 100 times. Calculate the pH of diluted solution?
1) 2
2) 16
3) 4
4) 8
Answer:
3) 4
Solution :
Given that [H⁺] = 0.005 × 2 = 0.01
After 100 times dilution
[H⁺] = 10^-4
[H⁺] = \(\frac{0.01}{100}\) = 0.0001
pH = -log[H⁺] = -log 0.0001 = -log10^-4 = 4

Question 28.
A solution of HCl has a pH = 3, If one ml of it is diluted to 1 litre, what will be the pH of the resulting solution?
1) 1
2) 3
3) 5
4) 6
Answer:
4) 6
Solution :
Given pH = 3 for HCl solution
∴ [H⁺] = 10^-3 M
1 ml is diluted to 1 litre
∴ [H⁺] = \(\frac{10^{-3}}{10^3}\) = 10^-6
∴ pH = -log[H⁺] = -log[10^-6] = 6

Question 29.
Find the pH the 10^-8 M HCl solution,
1) 6.9486
2) 6.9586
3) 6.9686
4) 6.9786
Answer:
2) 6.9586
Solution :

Formula : pH = -log[H⁺] [TS 15]
The given acid solution is very dilute. Hence H⁺ ion obtained from the acid and water must be taken into consideration.

H⁺ ion concentration of acid is 10^-8M
H⁺ ion concentration of water is 10^-7M
Total [H⁺] = [H⁺] from acid + [H⁺] from water
= 10^-8 + 10^-7
= 0.1 × 10^-7 M+ 1 × 10^-7M
= 1.1 × 10^-7 M
∴ pH = -log[1.1 × 10^-7]
= -(0.0414-7) = 6.9586.

Question 30.
Calculate the pH of the following basic solutions. [AP 18]
(a) [OH^–] = 0.05M
(b) [OH^–] = 2 × 10^-4M
1) 12.999, 10.3010
2) 12.699, 11.3010
3) 12.699, 10.2110
4) 12.699, 10.3010
Answer:
4) 12.699, 10.3010
Solution :
(a) [OH^–] = 0.05M
pOH = -log(0.05)= -log 5 × 10^-2
= -log5 + 2log10
= 2 – log 5 = 1.3010
pOH = 1.3010
pH = 14 -pOH = 14 -1.3010= 12.699

(b) [OH^–] = 2 × 10^-4M
pOH = -log 2 × 10^-4
= -log2 + log10
= 4 – log 2 = 4 – 0.3010 = 3.699
pOH = 14 – 3.699 = 10.3010.

Question 31.
2g of NaOH is dissolved in water to give 1 litre solution. What is the pH of the solution?
1) 12.6990
2) 12.9990
3) 12.6660
4) 12.8880
Answer:
1) 12.6990
Solution :

Question 32.
Calculate the pH of the following solutions.
a) 0.37g of Ca(OH)₂ dissolved in water to give 500ml solution.
b) 0.3g of NaOH dissolved in water to give 200ml solution.
c) 0.1825% HCl aqueous solution.
d) 1ml of 13.61VI HCl is diluted with water to give 1 litre solution.
1) 12.3000, 12.564, 1.308, 1.87
2) 12.3010, 12.574, 1.308, 1.95
3) 12.3010, 12.574, 1.308, 1.87
4) 12.3310, 12.564, 1.208, 1.87
Answer:
3) 12.3010, 12.574, 1.308, 1.87
Solution :

[H⁺] = 0.05N
∴ pH =-log[H⁺] = -log[0.05]
= -log[ 5 × 10^-2]
= 2 – log 5 = 2 – 0.692 =1.308
∴ pH = 1.308

d) [H⁺] = 13.6
1 ml is diluted with water in to 1 litre solution
∴ [H⁺] = \(\frac{13.6}{1000}\) = 0.0136
pH= -log[H⁺] = -log[0.0136] = 1.87

Question 33.
How many grams of NaOH are present in 100ml solution if pH of the solution is 10?
1) 4 × 10^-4
2) 5 × 10^-4
3) 4 × 10^-2
4) 5 × 10^-2
Answer:
1) 4 × 10^-4
Solution :
Given pH = 10
pOH= 14 – 10 = 4
N = 10^-4

Question 34.
The value of K_w is 9.55 × 10^-14 at certain temperature. Calculate the pH of water at this temperature.
1) 6.56
2) 6.51
3) 6.55
4) 6.54
Answer:
2) 6.51
Solution :

pH = -log(3.09 × 10^-7)
= -[log3.09 + log 10^-7] = -[0.49 – 7]
= 7 – 0.49 = 6.51

Question 35.
Calculate the pH of 10^-8M NaOH.
1) 7.0314
2) 7.0419
3) 7.0444
4) 7.0414
Answer:
4) 7.0414
Solution :
Base is very dilute. So OH^– concentration in water must be taken into consideration
∴ Total [OH^–]
= [OH^–] from base + [OH^–] from water
= 10^-8 + 10^-7 = 1.1 × 10^-7
pOH = log₁₀[OH^–] = log10 1.1 × 10^-7
= 7 – 0.0414 = 6.9586
∴ pH = 14 – pOH= 14 – 6.9586 = 7.0414

Question 36.
150 ml of 0.5 M HCI and 100 ml of 0.2M HCI are mixed. Find the pH of the resulting solution.
1) 0.4202
2) 0.4222
3) 0.4002
4) 0.2202
Answer:
1) 0.4202
Solution :

Question 37.
Calculate the pH of solution obtained by mixing 10ml of 0.1M HCl and 40ml of 0.21M H₂SO₄.
1) 0.4666
2) 0.4886
3) 0.4686
4) 0.4888
Answer:
3) 0.4686
Solution :

Question 38.
100ml of pH = 4 solution is mixed with 100ml of pH = 6 solution. What is the pH of resulting solution.
1) 4.2999
2) 4.2967
3) 4.2667
4) 4.6467
Answer:
2) 4.2967
Solution :
V₁ = Volume of first solution = 100ml
N₁ = Proton concn.of 1^st solution = 10-pH = 10^-4M
V₂ = Volume of second solution = 100ml
N₂ = Proton concn.of 2^nd solution = 10-pH = 10^-6M

Proton concentration of mixed solution is

pH = -log₁₀[H⁺] = -log₁₀(5.05 × 10^-5)
= 5 – log5.05 = 5 – 0.7033 = 4.2967

Tip : If equal volumes of two same solutions whose pH values are less than 7, then Final
pH = lower pH + 0.3 = 4 + 0.3 = 4.3

Question 39.
Equal volumes of 0.5 M NaOH and 0.3M KOH are mixed in an experiment. Find the pOH and pH of the resulting solution.
1) 0.3999, 13.5021
2) 0.3979, 13.6021
3) 0.3969, 13.6121
4) 0.3979, 13.5021
Answer:
2) 0.3979, 13.6021
Solution :

∴ [OH^–] = N = 4 × 10^-1
pOH = -log₁₀[OH^–]= -log10 (4 × 10^-1)
= 1 – log₁₀ 4 – log₁₀ 10^-1
= 1 – log.4 = 1 – 0.6021 = 0.3979
pH = 14 – pOH = 14 – 0.3979 = 13.6021

Question 40.
60ml of 1M HCl is mixed with 40ml of 1M NaOH. What is the pH of resultant solutions?
1) 0.6660
2) 0.5890
3) 0.6990
4) 0.5770
Answer:
3) 0.6990
Solution :

Now [H⁺] = N = 2 × 10^-1
∴ pH = -log[H⁺] = -log(2 × 10^-1)
= 1 – log₁₀2 = 1 – 0.3010 = 0.6990

Question 41.
Calculation the pH of a solution which contains 100ml of 0.1 M HCI and 9.9ml of 1.0M NaOH.
1) 3.0016
2) 3.0416
3) 3.0146
4) 3.0616
Answer:
2) 3.0416
Solution :

Question 42.
What will be the resultant pH when 200ml of an aqueous solution of HCl having PH = 2 is mixed with 300ml of an aqueous solution of NaOH having pH = 12?
1) 11.3666
2) 11.3118
3) 11.3777
4) 11.3010
Answer:
4) 11.3010
Solution :

[OH⁺] = 2 × 10^-3
∴ pOH = -log(2 × 10^-3)
= 3 – log2 = 3 – 0.3010 = 2.6990
∴ pH = 14-2.6990 = 11.3010

Question 43.
43. 50 ml of 0.2 M HCI is added to 30 ml of 0.1 M KOH solution. Find the pH of the solution.
1)1.055
2) 1.085
3) 1.058
4) 1.008
Answer:
3) 1.058
Solution :
Milliequivalents of acid (HCl)= N_aV_a(ml)
= M_a × Basicityx V_a (ml)
= 0.2 × 1 × 50 = 10

Milliequivalents of base (KOH)= N_bV_b (ml)
= M_b × Acidity x V_b(ml)
= 0.1 × 1 × 30 = 3

Here, N_aV_a > N_bV_b .So the mixture is acidic.
∴ N = [H⁺] =

Question 44.
40ml of 0.2M HNO₃ when reacted with 60ml of 0.3M NaOH, gave a mixed solution. What is the pH of the resulting solution.
1) 13
2) 14
3) 15
4) 16
Answer:
3) 15
Solution :

Question 45.
50ml of 0.1M H₂SO₄ were added to 100ml of 0.2M HNO₃. Then the solution is diluted to 300ml. What is the pH of the solution.
1) 2
2) 1
3) 4
4) 3
Answer:
2) 1
Solution :

Question 46.
What is the K_w value in an a aqueous solution of pK_w = 13.725
1) 1.854 × 10^-4
2) 1.864 × 10^-4
3) 1.874 × 10^-4
4) 1.884 × 10^-4
Answer:
4) 1.884 × 10^-4
Solution :
Given that pK_w = 13.725
pK_w = -log K_w
⇒ logK_w = -pK_w = -13.725
⇒ K_w = anti log (14.275) = 1.884 × 10^-4

Question 47.
The ionic product of water at 80°C is 2.44 × 10^-13. What are the concentrations of hydronium ion and the hydroxide in pure water at 80°C?
1) 4.74 × 10^-7mol/L, 4.74 × 10^-7mol/L
2) 4.64 × 10^-7mol/L, 4.64 × 10^-7mol/L
3) 4.94 × 10^-7mol/L, 4.94 × 10^-7mol/L
4) 4.84 × 10^-7mol/L, 4.84 × 10^-7mol/L
Answer:
3) 4.94 × 10^-7mol/L, 4.94 × 10^-7mol/L
Solution :

Question 48.
The ionization constant for water is 2.9 × 10^-14 at 40°C. Calculate [H₃O⁺], [OH], pH and pOH for pure water at 40°C.
1) 6.7689
2) 6.7779
3) 6.7669
4) 6.9999
Answer:
1) 6.7689
Solution :
Given that Ka = 2.9 × 10^-14
[H⁺] = \(\sqrt{\mathrm{k}_{\mathrm{a}}}=\sqrt{2.9 \times 10^{-14}}\) = 1.703 × 10^-7
For pure water, [H⁺] = [OH^–]
∴ [H⁺] = 1.703 × 10^-7
[OH^–] = 1.703 × 10^-7
pH = -log[H⁺] = -log[1.703 × 10^-7]
= 7-log 1.7= 6.7689
Hence, pOH = 6.7689

Question 49.
Calculate the pH of
a) 0.002 M aceticacid having 2.3% dissociation.
b) 0.002M NH₄OH having 2.3%; dissociation.
1) 4.3999, 9.998
2) 4.3372, 9.6628
3) 4.8765, 9.2345
4) 4.0001, 9.0123
Answer:
2) 4.3372, 9.6628
Solution :
a) c = 0.002 M = 2 × 10^-3, α = 2.3%
[H⁺] = cα = 2 × 10^-3 × \(\frac{2.3}{100}\) = 4.6 × 10^-5
pH = -log[H⁺] = -log[4.6xlO-5]
= 5 – log 4.6= 5-0.6628 = 4.3372

b) c = 0.002M = 2 × 10^-3, α = 2.3%
[OH^–] = Cα
= 2 × 10^-3 × \(\frac{2.3}{100}\) = 4.6 × 10^-5
pOH = -log[OH^–] = -log[4.6 × 10^-5]
= 5-log 4.6 = 5-0.6628 = 4.3372
pOH = 4.3372
∴ pH= 14 – pOH = 14 – 4.3372 = 9.6628

Question 50.
Calculate K_a of acetic acid from equilibrium concentration given below. [H₃O⁺]=[CH₃COO^–] = 1.34 × 10^-3M, [CH₃COOH] = 9.866 × 10^-2M
1) 1.02 × 10^-5
2) 1.92 × 10^-5
3) 1.22 × 10^-5
4) 1.82 × 10^-5
Answer:
4) 1.82 × 10^-5
Solution :

Question 51.
Calculate pH of 0.1M acetic acid haying K_a = 1.8 × 10^-5M
1) 2.8712
2) 2.1111
3) 2.9912
4) 2.7712
Answer:
1) 2.8712
Solution :

Question 52.
The pH of 0.1M Solution of weak mono protic acid is 4.0. Calculate its [H⁺] and K_a.
1) 10^-7
2) 10^-9
3) 10^-5
4) 10^-8
Answer:
1) 10^-7
Solution :
Given c = 0.1 M; pH = 4
a) [H⁺] = 10^-pH = 10^-4

Question 53.
K_a of 0.02M CH₃COOH is 1.8 × 10^-5. Calculate
a) [H₃O⁺]
b) % ionisation
c) pH
1) 5 × 10^-4, 3%, 3.2218
2) 6 × 10^-4, 2%, 2.2218
3) 6 × 10^-4, 3%, 3.2218
4) 5 × 10^-4, 2%, 2.2218
Answer:
3) 6 × 10^-4, 3%, 3.2218
Solution :

Question 54.
Calculate the pH of 0.01M solution of CH₃COOH. K_a for CH₃COOH at 298K is 1.8 × 10^-5.
1) 3.7724
2) 3.3724
3) 3.7324
4) 3.3334
Answer:
2) 3.3724
Solution :

Question 55.
The pH of 0.1M solution of an organic acid is 4.0 calculate the dissociation constant of the acid.
1) 10^-8
2) 10^-9
3) 10^-5
4) 10^-7
Answer:
4) 10^-7
Solution :

Question 56.
Three ionization constants of HF, HCOOH and HCN at 298K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
1) 1.49 × 10^-11, 1.59 × 10^-11, 2.09 × 10^-6
2) 1.47 × 10^-11, 1.66 × 10^-11, 2.09 × 10^-6
3) 1.47 × 10^-11, 1.56 × 10^-11, 2.08 × 10^-6
4) 1.57 × 10^-11, 1.66 × 10^-11, 2.08 × 10^-6
Answer:
3) 1.47 × 10^-11, 1.56 × 10^-11, 2.08 × 10^-6
Solution :

Question 57.
Find the concentration of hydroxide ion in a 0.25M solution of trimethylamine, a weak base.(eH₃)₃N + H₂O ⇌ (CH₃)₃N⁺H+OH ; K_b = 7.4 × 10^-5
1) 4.3 × 10^-3M
2) 4.6 × 10^-6M
3) 4.5 × 10^-4M
4) 4.8 × 10^-3M
Answer:
1) 4.3 × 10^-3M
Solution :

Question 58.
The 0.005 M monobasic acid has a pH of 5. What is the extent of ionization?
1) 0.1%
2) 0.2%
3) 0.3%
4) 0.4%
Answer:
2) 0.2%
Solution :

Question 59.
50 ml of 0.1M NH₄OH, 25ml of 2M NH₄Cl are used to make buffer. What is the pH of the solution? p^K_b of NH₄OH is 4.8.
1) 8.2
2) 8.3
3) 8.4
4) 8.5
Answer:
1) 8.2
Solution :

Question 60.
The pH of a buffer prepared by mixing 50ml of 0.2M CH₃COOH and 25ml of CH₃COONa is 4.8. If the pK_a is 4.8. What is the strength of CH₃COONa.
1) 0.6 M
2) 0.5 M
3) 0.4 M
4) 0.7 M
Answer:
3) 0.4 M
Solution :

Question 61.
50ml of 0.1 sodium acetate, 25ml of 0.2m acetic acid are added together to or buffer solution. pK_a of acetic acid is 4.8. Find the pH of the solution.
1) 4.6
2) 4.8
3) 4.9
4) 4.2
Answer:
2) 4.8
Solution :

Question 62.
When 20ml of 0.1M NH₄OH are added to 20ml of 1M NH₃Cl solution, the pH of the buffer formed is 8.2. What is the
pK_b of NH₄OH?
1) 4.6
2) 4.2
3) 4.9
4) 4.8
Answer:
4) 4.8
Solution :

Question 63.
One litre of buffer solution contains 0.1 mole of aceticacid add 1 mole of sodium acetate. Find its pH if pK_a of CH₃COOH is 4.8.
1) 5.2
2) 5.4
3) 5.6
4) 5.8
Answer:
4) 5.8
Solution :

Question 64.
50ml of 1M CH₃COOH solution, when added to 50ml of 0.5M NaOH gives a solution with a pH value ‘X’. Find the value of ‘X’. pK_a of acetic acid is 4.8.
1) 4.6
2) 4.7
3) 4.8
4) 4.2
Answer:
3) 4.8
Solution :

Question 65.
The solubility product of AgCl is 1.6 × 10¹⁰ moles²/lit². What is its solubility?
1) 1.26 × 10^-5
2) 1.36 × 10^-5
3) 1.46 × 10^-5
4) 1.56 × 10^-5
Answer:
1) 1.26 × 10^-5
Solution :

Question 66.
The solubility product of Zr(OH)₂ is 4.5 × 10^-17 mol³L³. What is solubility?
1) 2.4 × 10^-6
2) 2.7 × 10^-6
3) 2.6 × 10^-6
4) 2.9 × 10^-6
Answer:
3) 2.6 × 10^-6
Solution :

Question 67.
The solubility of Ag₂CrO₄ is 1.3 × 10^-4 molL^-1 what is the solubility product?
1) 9.52 × 10^-12
2) 9.62 × 10^-12
3) 9.72 × 10^-12
4) 9.82 × 10^-12
Answer:
2) 9.62 × 10^-12
Solution :
Ag₂ CrO₄ ⇌ 2Ag⁺ + Ag₂CrO₄^2-
K_sp = 4S² × S = 4S³
∴ K_sp = 4 × S³ = 4 × (1.3 × 10^-4)³ = 9.62 × 10^-12

Question 68.
The solubility of A₂B is 2 × 10^-3 moles/ lit. What is its solubility product ?
1) 3.2 × 10^-8
2) 3.3 × 10^-8
3) 3.4 × 10^-8
4) 3.5 × 10^-8
Answer:
1) 3.2 × 10^-8
Solution :
A₂B ⇌ 2A⁺ + B^-2
K_sp = [A⁺]²[B^-2] = [2S]²[S] = 4S² × S = 4S³
= 4(2 × 10^-3) = 4 × 8 × 10^-9 = 32 × 10^-9
= 3.2 × 10^-8

Question 69.
The solubility product of a salt AB = 10^-10,0mol²L^-2. What is solubility?
1) 10^-4
2) 10^-5
3) 10^-6
4) 10^-7
Answer:
2) 10^-5
Solution :
For AB, k_sp = 10^-10 mol² L^-2
k_sp = [A] [B] = [S]²
⇒ [S] = \(\sqrt{\mathrm{k}_{\mathrm{sp}}}=\sqrt{10^{-10}}\) = 10^-5 mol.L^-2

Question 70.
PQ and RS₂ are two sparingly soluble salts. Their solubility products are equal and each equal to 4.0 × 10^-18 Which salt is more soluble?
1) RS₂
2) PQ
3) PQ & RS₂
4) None of these
Answer:
1) RS₂
Solution :
For PQ salt, K_sp = [P⁺][Q^–]
If x is the solubility then 4 × 10^-18 = x²
⇒ x = 2.0 × 10^-9
For RS₂ salt, K_sp = [R⁺²][S^-1]²
If y is the solubility then 4 × 10^-18 = 4y³
⇒ y = 1 × 10^-6
Since y is more than x, solubility of RS₂ salt is more.

Question 71.
In a 0.1 M Solution, acetic acid is 1.34% ionized. Calculate [H⁺][CH₃COO^–] and [CH₃COOH] in the solution and calculate K_a of acetic acid.
1) 1.39 × 10^-3, 1.34 × 10^-3, 0.099, 1.7 × 10^-5
2) 1.30 × 10^-3, 1.34 × 10^-3, 0.079, 1.8 × 10^-5
3) 1.34 × 10^-3, 1.34 × 10^-3, 0.099, 1.8 × 10^-5
4) 1.34 × 10^-3, 1.30 × 10^-3, 0.089, 1.8 × 10^-5
Answer:
3) 1.34 × 10^-3, 1.34 × 10^-3, 0.099, 1.8 × 10^-5
Solution :

Question 72.
The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500K.
[N₂] = 1.5 × 10^-2M. [H₂] = 3.0 × 10^-2M and [NH₃] = 1.2 × 10^-2M. Calculate equilibrium constant.
1) 3.54 × 10²
2) 3.55 × 10²
3) 3.56 × 10²
4) 3.57 × 10²
Answer:
2) 3.55 × 10²
Solution :

Question 73.
At equilibrium, the concentrations of N₂ = 3.0 × 10^-3M, O₂ = 4.2 × 10^-3M and NO = 2.8 × 10^-3M, in a sealed vessel at 800K. What will be K_c for the reaction N₂(g) + O₂(g) ⇌ 2NO(g)
1) 0.422
2) 0.522
3) 0.622
4) 0.722
Answer:
3) 0.622
Solution :

Question 74.
PCl₅, PCl₃ and Cl₂ are at equilibrium at 500 K and having concentration 1.59M PCl₃, 1.59M Cl₂ and 1.41 M PCl₅. Calculate Kt. for the reaction
PCl₅ ⇌ PCl₃ + Cl₂
1) 1.69
2) 1.79
3) 1.89
4) 1.99
Answer:
2) 1.79
Solution :

Question 75.
The value of K_C = 4.24 at 800K for the reaction,
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) Calculate equilibrium concentrations of CO₂, H₂, CO and H₂O at 800K, if only CO and HO are present initially at concentrations of 0.10 M each.
1) 0.067 M, 0.033 M
2) 0.057 M, 0.023 M
3) 0.069 M, 0.034 M
4) 0.062 M, 0.021 M
Answer:
1) 0.067 M, 0.033 M
Solution :
Let x mole per litre of the product to be formed

⇒ x² =4.24(0.01 + x² – 0.2x)
= 0.0424 + 4.24x² – 0.848x
= 3.24x² – 0.848x+0.0424
By solving the above quadratic equation we get x = 0.067
∴ [CO₂] = [H₂] = x = 0.067M
[CO] = [H₂O] = 0.1 – 0.067 = 0.033M

Question 76.
For the equilibrium
2NOCl(g) ⇌ 2NO(g) + Cl₂(g) the value of the equilibrium constant, K_c is 3.75 × 10^-6 at 1069 K. Calculate the K_p for the reaction at this temperature.
1) 3.33 × 10^-4
2) 3.36 × 10^-4
3) 3.36 × 10^-9
4) 3.33 × 10^-8
Answer:
1) 3.33 × 10^-4
Solution :
The given reaction:
2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
⇒ ∆n = (2 + 1) – 2 = 1
Now , K_p = K_c (RT)^∆n
= 3.75 × 10^-6 (0.0821 × 1069)
= 333.12 × 10^-6 = 3.33 × 10^-4

Question 77.
The value of K_p for-the reaction. CO₂(g) + C(s) ⇌ 2C0(g) is 3.0 at 1000 K. If initially P_CO2 = 0-48 bar and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.
1) 0.99, 0.14
2) 0.44, 0.12
3) 0.66, 0.15
4)0.56, 0.15
Answer:
3) 0.66, 0.15
Solution :
Let ‘x’ be the decrease in pressure of CO₂

⇒ 4x² = 3(0.48 – x) ⇒ 4x² = 1.44 – x
⇒ 4x² + 3x – 1.44 = 0
Solving this quadratic equation,
we get x = 0.33
The equilibrium partial pressures are,
p_CO = 2x = 2 × 0.33 = 0.66 bar
p_CO2 = 0.48 – x = 0.48 – 0.33 = 0.15 bar

Question 78.
The value of K_c for the reaction 2A ⇌ B + C is 2 × 10^-3. At a given time the composition of reaction mixture is [A] = [B] = [C] = 3 × 10^-4 M, In which direction the reaction will proceed?
1) forward
2) reverse
3) neutral
4) None of these
Answer:
2) reverse
Solution :

It is given that, K_c = 2 × 10^-3
∴ Q_c > K_c.
Hence, the reaction will proceed in the reverse direction.

Question 79.
13.8g of N₂O₄ was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium N₂O₄g) ⇌ 2NO₂(g). The total pressure at equilibrium was found to be 9,15 bar. Calculate K_c, K_p and parital pressure at equilibrium.
1) 2.9, 88.87
2) 2.7, 86.87
3) 2.6, 85.87
4) 2.3, 89.87
Answer:
3) 2.6, 85.87
Solution :
We know pV = nRT
Total volume (V) = 1L
Molecular mass of NO = 92g
Number of moles (n) = \(\frac{13.8}{92}\) = 0.15
Gas constant(R) = 0.083 bar L mol^-1 L^-1
Temperature (T) = 400K
Now pV = nRT
⇒ p × 1L = 0.15 × 0.083 × 400 = 4.98
The given reaction:
N₂O₄(g) ⇌ 2NO₂(g)

Question 80.
3.00 mol of PCl₅ kept in 1 L closed reaction vessel was allowed to attain equilibrium at 380 K. Calculate composition of the mixture at equilibrium K_c = 1.80.
1) 1.41, 1.59
2) 1.42, 1.57
3) 1.43, 1.58
4) 1.44, 1.56
Answer:
1) 1.41, 1.59
Solution :

Solving this quadratic equation, we get x = 1.59
∴ [PCl₅] = 3.0 – x = 3 – 1.59 = 1.41 M
[PCl₃] = [Cl₂] = x = 1.59 M

Question 81.
The value of ∆G° for the phosphorylation of glucose in glycolysis is 13.8 KJ/mol. Find the value of K_c at 298K.
1) 3.51 × 10^-3
2) 3.81 × 10^-3
3) 3.61 × 10^-3
4) 3.91 × 10^-3
Answer:
2) 3.81 × 10^-3
Solution :
∆G°= 13.8 KJ/mol = 13.8 × 10^-3 J/mol
Also ∆G° = – RT In K_c
∴ In K_c = – 13.8 × 10³ / (8.314 × 298 K) = -5.569
⇒ K_c = e^-5.569 = 3.81 × 10^-3

Question 82.
Hydrolysis of sucrose gives,
Sucrose + H₂O ⇌ Glucose + Fructose
Equilibrium constant K_c for the reaction is 2 × 10¹³ at 300 K. Calculate ∆G° at 300K.
1) -7.64 × 10⁴
2) -7.44 × 10⁴
3) -7.54 × 10⁴
4) -7.67 × 10⁴
Answer:
1) -7.64 × 10⁴
Solution :
Formula: ∆G° = – RT In K_c
∴ ∆G°= -8.314 × 300 xln(2 × 10¹³)
= -7.64 × 10⁴ J/mol^-1

Question 83.
What will be the conjugate bases for the following Bronsted acids:
HF, H₂SO₄, HCO^–₃
1) F^–, HSO^–₄, CO^4-₂
2) F^–, HSO^–₂, CO^2-₃
3) F, HSO^–₂, CO^4-₃
4) F^–, HSO^–₄, CO^2-₃
Answer:
4) F^–, HSO^–₄, CO^2-₃
Solution :
The conjugate bases should have ohe proton less in each and therefore the corresponding conjugate bases are :
F^–, HSO^–₄ and CO^2-₃ respectively.

Question 84.
Write the conjugate acids for the following Bronsted bases:
NH^–₂, NH₃ and HCOO^–
1) NH₃, NH⁺₄, HCOH
2) NH₃, NH⁺₄, HCOOH
3) NH₃, NH⁺₃, HCOOH
4) NH₃, NH⁺₃, HCOH
Answer:
2) NH₃, NH⁺₄, HCOOH
Solution :
The conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are:
NH3, NH⁺₄ and HCOOH respectively.

Question 85.
The species: H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. Among the following which is incorrect.

Answer:
4
Solution :

Question 86.
Classify the following species into Lewis acids and Lewis bases and show how these act as such: (a) HO^– (b) F^– (c) H⁺ (d) BCl₃ Among the following which is incorrect.
1) Hydroxyl ion is a Lewis base, as it can donate an electron lone pair (:OH^–)
2) Fluoride ion acts as a Lewis base, as it can donate any one of its 4 electron pairs.
3) A proton is a Lewis acid, as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
4) BCl₃ acts as a Lewis base, as it can accept a lone pair of electrons from species like ammonia or amine molecules.
Answer:
4) BCl₃ acts as a Lewis base, as it can accept a lone pair of electrons from species like ammonia or amine molecules.
Solution :
(a) Hydroxyl ion is a Lewis base, as it can donate an electron lone pair (:OH^–)
(b) Fluoride ton acts as a Lewis base, as it can donate any one of its 4 electron pairs.
(c) A proton is a Lewis acid, as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl₃ acts as a Lewis acid, as it can accept a lone pair of electrons from species like ammonia or amine molecules.

Question 87.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH?
1) 2.42
2) 3.42
3) 4.42
4) 5.42
Answer:
1) 2.42
Solution :
pH = -log[3.8 × 10^-3]
= – {log [3.8] +log [10^-3]}
= – {(0.58) + (-3.0) } = -{-2.42} = 2.42
∴ pH of the given soft drink is 2.42.

Question 88.
Calculate pH ofa 1.0 × 10^-8 M solution of HCl.
1) 5.98
2) 7.98
3) 6.98
4) 8.98
Answer:
3) 6.98
Solution :
2H₂O(l) ⇌ H₃O⁺(aq) + OH^–(aq)
K_w = [OH^–][H₃O⁺] = 10^-14
Let x = [OH^–] = [H₃O⁺] from H₂O.
The H₃O⁺ concentration is generated
(i) from the ionization of HCl dissolved.
HCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + Cl^–(aq),
(ii) from ionization of H₂O
In these very dilute solutions, both sources of H30+ must be considered.
[H₃O⁺] = 10^-8+ x
K_w = (10^-8 + x)(x) = 10^-14
⇒ x² + 10^-8 x – 10^-14 = 0
∴ [OH^–] = x = 9.5 × 10^-8
⇒ pOH = 7.02 and pH = 6.98

Question 89.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.
1) 7.005
2) 8.005
3) 4.005
4) 9.005
Answer:
1) 7.005
Solution :

Question 90.
Which of the following is not a general characteristic of equilibria involving physical processes?
1) Equilibrium is possible only in a closed system at a given temperature.
2) All measurable properties of the system remain constant.
3) All the physical processes stop at equilibrium.
4) The opposing processes occur at the same rate and there is dynamic but stable condition.
Answer:
3) All the physical processes stop at equilibrium.

Question 91.
We know that’the relationship between K_c and K_p is K_p = K_c (RT)^∆n
What would be the value of n for the reaction NH₄Cl(s) ⇌ NH₃(g) + HCl(g)
1) 1
2) 0.5
3) 1.5
4) 2
Answer:
4) 2

Question 92.
The equilibrium constants of the following are
The equilibrium constant (K) of the reaction:
2NH3 +|o2 ^=±2N0 + 3H20 will be
1) K₂K³₃ / K₁
2) K₂K₃ / K₁
3) K³₂K₃ / K₁
4) K₁K³₃ / K₂
Answer:
1) K₂K³₃ / K₁

Question 93.
For the reaction H₂(g) + I₂(g) ⇌ 2HI(g) the standard free energy is ∆G° > 0.The equilibrium constant (K) would be
1) K = 0
2) K > 1
3) K = 1
4) K < 1
Answer:
4) K < 1

Question 94.
Hydrolysis of sucrose is given by the following reaction:
Sucrose + H₂O ⇌ Glucose + Fructose
If the equilibrium constant (K_C) is 2 × 10¹³ at 300K, the value of ∆rG° at the same temperature will be
1) -8.314 Jmol^-1 K^-1 × 300K × ln(2 × 10¹³)
2) 8.314Jmol^-1K^-1 × 300K × ln(2 × 10¹³)
3) 8.314 Jmol^-1K^-1 × 300K × ln(3 × 10¹³)
4) -8 314 Jmol^-1 K^-1 × 300K × ln(4 × 10¹³)
Answer:
1) -8.314 Jmol^-1 K^-1 × 300K × ln(2 × 10¹³)

Question 95.
For the reversible reaction,
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat
The equilibrium shifts in forward direction
1) by increasing the concentration of NH₃(g)
2) by decreasing pressure
3) by decreasing the concentrations of N₂(g) and H₂(g)
4) by increasing pressure and decreasing temperature
Answer:
4) by increasing pressure and decreasing temperature

Question 96.
Which one of the following conditions will favour maximum formation of the product in the reaction
A₂ (g) + B₂ (g) ⇌ X₂(g), ∆_rH = -XkJ?
1) Low temperature and high pressure
2) Low temperature and low pressure
3) High temperature and high pressure
4) High temperature and low pressure
Answer:
1) Low temperature and high pressure

Question 97.
Acidity of BF₃ can be explained on the basis of which of the following concepts?
1) Arrhenius concept
2) Bronsted Lowry concept
3) Lewis concept
4) Bronsted Low ry as well as Lewis concept
Answer:
3) Lewis concept

Question 98.
Conjugate base for Bronsted acids H₂O and HF are
1) H₃O⁺ and H₂F⁺, respectively
2) OH^– and H₂F⁺, respectively
3) H₃O⁺ and F^–, respectively
4) OH^– and F⁺, respectively
Answer:
4) OH^– and F⁺, respectively

Question 99.
Which of the following cannot act both as Bronsted acid and as Bronsted base?
1) HSO^–₃
2) NH₃
3) HCl
4) HSO^–₄
Answer:
3) HCl

Question 100.
The pK_b of dimethylamine and pK_a of acetic acid are 3.27 and 4.77 respectively at T(K) .The correct option for the pH of dimcthylammonium acetate solution is
1) 6.25
2) 8.50
3) 5.50
4) 7.75
Answer:
4) 7.75

Question 101.
Find out the solubility of Ni(OH)₂ in 0.1M NaOH. Given that the ionic product of Ni(OH)₂ is 2 × 10^-15
1) 2 × 10^-13M
2) 2 × 10^-8M
3) 1 × 10^-13M
4) 1 × 10^-8M
Answer:
1) 2 × 10^-13M

Question 102.
Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations:
pH of which one of them will be equal to 1?
1) B
2) A
3) D
4) C
Answer:
4) C

Question 103.
pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (K_sp) of Ca(OH)₂ is
1) 0.5 × 10^-10
2) 0.5 × 10^-15
3) 0.25 × 10^-10
4) 0.125 × 10^-15
Answer:
2) 0.5 × 10^-15

Question 104.
K_a for CH₃COOH is 1.8 × 10^-5 and K_b for NH₄OH is 1.8 × 10^-5. The pH of ammonium acetate will be
1) 7.005
2) 4.75
3) 7.0
4) Between 6 and 7
Answer:
3) 7.0

Question 105.
What will be the value of pH of 0.01 mol dm^-3 CH₃COOH(K_a = 1.74 × 10^-5)?
1) 3.4
2) 3.6
3) 3.9
4) 3.0
Answer:
1) 3.4

Question 106.
Which of the following will produce a buffer solution when mixed in equal volumes?
1) 0.1 mol dm^-3 NH₄OH and 0.1 mol dm^-3 HCl
2) 0.05 mol dm^-3 NH₄OH and 0.1 mol dm^-3 HCl
3) 0.1 mol dm^-3 NH₄OH and 0.05 mol dm^-3 HCl
4) 0.1 mol dm^-3 CH₄COONa & 0.1 mol dm^-3 NaOH
Answer:
3) 0.1 mol dm^-3 NH₄OH and 0.05 mol dm^-3 HCl

Question 107.
In which of the following solvents is silver chloride most soluble?
1) 0.1 mol dm^-3 AgNO₃ solution
2) 0.1 mol dm^-3 HCl solution
3) H₂O
4) Aqueous ammonia
Answer:
4) Aqueous ammonia

Question 108.
The solubility of BaSO₄ in water is 2.42 × 10^-3gL^-1 at 298K. The value of its solubility product (K_sp) will be
(Given molar mass of BaSO₄ = 233 gmoH)
1) 1.08 × 10^-10mol²L^-2
2) 1.08 × 10^-12mol²L^-2
3) 1.08 × 10^-14mol²L^-2
4) 1.08 × 10^-8 mol²L^-2
Answer:
1) 1.08 × 10^-10mol²L^-2

Question 109.
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10^-4molL^-1. Solubility product of ₂C₂O₄ is
1) 2.66 × 10^-12
2) 4.5 × 10^-11
3) 5.3 × 10^-12
4) 2.42 × 10^-8
Answer:
3) 5.3 × 10^-12