TS Inter 1st Year Chemistry Question Paper May 2016

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TS Inter 1st Year Chemistry Question Paper May 2016

Note : Read the following instructions carefully.

1. Answer all questions of Section ‘A’. Answer any six questions in Section ‘B’ and any two questions in Section ‘C’.
2. In Section ‘A’, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section ‘B’, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Sections ‘3’ and ‘C’.

Section – A

Note : Answer all questions.

Question 1.
Define receptor and sink.
Answer:
Receptor: The medium which is effected by a pollutant is called receptor.
Sink : The medium which retains and interacts with long lived pollutant is called the sink.
Ex : Oceans are important sinks for atmospheric COr

Question 2.
Give the hybridisation of carbon in garphite and diamond.
Answer:
a) In CO₃^-2 carbon atom undergoes sp² hybridisation.
b) In Diamond carbon atom undergoes sp³ hybridisation.
c) In Graphite carbon atom undergoes sp² hybridisation.
d) In Fullerenes carbon atom undergoes sp² hybridisation.

Question 3.
Calculate the oxidation numbers of oxygen in OF₂ and O₂ F₂.
Answer:
OF₂
x + 2(-1) = 0
x – 2 = 0
x = +2

O₂ F₂
2x + 2(-1) = 0
2x – 2 = 0
2x = 2
x = +1

Question 4.
Give any two biological importance of sodium and potassium ions.
Answer:
Biological importance of Na² and K² :
1. Na⁺ ions participate in transmission of nerve signals.
2. Na⁺ ions regulates the flow of water accross cell membrane.
3. K⁺ ions are useful in activating enzymes.
4. K⁺ ions participate in the oxidation of glucose to produce ATP.

Question 5.
Calculate the pH of 0.01 M HCl solution.
Answer: P^H : The negative logarithm of H+ ion concentration base ten expressed in Moles/Lit.
P^H = -log 10 [H⁺]

Problem:
Given 0.1 M HCI solution.
[H⁺] = 0.1 M
P^H = – log[H⁺]
= -log 0.1
= – log 10^-1
= 1
P^H = 1

Question 6.
What is Chemical Oxygen Demand (COD)?
Answer:
Chemical oxygen demand (COD) : The amount of oxygen required to oxidise organic substance present in polluted water is called COD.

Question 7.
Which of the gas diffuses faster among N₂,O₂ and CH₄ and why?
Answer:
CH_H gas diffuse faster among N₂, O₂ and CH₄.
Reason : CH₄ (16) has low molecular weight than N₂ (28) and O₂ (32).

Question 8.
Write the formula of Plaster of Paris.
Answer:
Plaster of paris is the hemi hydrate of CaSO₄ with formula
CaSO₄.\(\frac{1}{2}\)H₂O.

Preparation:
Plaster of paris is obtained by heating gypsum at 393 K.

1.  If temperature is used greater than 393 K then an hydrous CaSO₄ is formed which is called ‘dead burnt plaster.
2. Plaster of paris has an important property of setting with water.
3. It forms a hard solid in 5 to 15 min. When it is mixed with suitable quantity of water.
4. It is majorly used in building industry and as well as plasters.
5. It is used in the bone fractures (or) sprain conditions.
6. It is used in dentistry.
7. It is used in manufacturing status and busts.

Question 9.
Write the IUPAC names of the following :

Answer:

Question 10.
What is producer gas?
Answer:

1. Producer gas is mixture of CO and N₂.
2. It is prepared by passing air over hot coke.

Section – B

Question 11.
State and explain the Hess’s law of constant heat summation with example.
Answer:
Hess’s law states that the total amount of heat evolved or absorbed in a chemical reaction is always same whether the reaction is carried out in one step (or) in several steps. Illustration : This means that the heat of reaction depends only on the initial and final stages and not on the intermediate stages through which the reaction is carried out. Let us consider a reaction in which A gives D. The reaction is brought out in one step and let the heat of reaction be ∆H.

A → D; → ∆H
Suppose the same reaction is brought out in three stages as follows
A → B : ∆H₁
B → C : ∆H₂
C → D : ∆H₃
The net heat of reaction is ∆H₁ + ∆H₂+ ∆H₃.
According to Hess law ∆H = ∆H₁ + ∆H₁ + ∆H₁.
Ex : Consider the formation of CO₂. It can be prepared in two ways.
1) Direct method : By heating carbon in excess of O₂

C_(s) + O_2(g) → CO_2(g) ; ∆H = -393.5 kj
2) Indirect method : Carbon can be converted into CO₂ in the following two steps.
C_(s) + [/latex]\frac{1}{2}[/latex]O_2(g) → CO_2(g); ΔA₁ = – 110.5 kj
CO_(g) + [/latex]\frac{1}{2}[/latex]O_2(g) → CO_2(g); ΔA₂ = – 283.02 kj
Total ΔH = -393.52 kj (∆H₁ + ∆H₂
The two ∆H values are same.

Question 12.
Explain the structure of Diborane.
Answer:
Diborane is an electron deficient compound. It has ’12’ valency electrons for bonding purpose instead of ’14’ electrons. In diborane each boron atom undergoes sp³ hybridization out of the four hybrid orbitals one is vacant. Each borane forms two a – bonds (2 centred – 2 electron bonds) bonds with two hydrogen atoms by overlapping with their’1s’orbital.

The remaining hybrid orbitals of boran used for the forma-tion of B-H-B bridge bonds. In the formation of B-H-B bridge, half filled sp3 hybrid orbital of one borane atom and vacant sp3 hybrid orbital of second boron atom overlap with 1s orbital of H-atom.

‘These three centred two electron bonds are also called as banana bonds. These bonds are present above and below the plane of BH₂ units. Diborane contains two coplanar BH2 groups. The four hydro-gen atoms are called terminal hydrogen atoms and the remain-ing two hydrogens are called bridge hydrogen atoms.

Bonding in diborane. Each B atom uses sp³ hybrids for bonding. Out of the four sp³ hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2-electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Question 13.
Define the terms hard water and soft water. How is temporary- hardness removed by Clark’s process?
Answer:
Hard water : Water does not give lather readily with soap is called hard water.

Hard water contains hardness. This hardness is due to presence of Ca, Mg soluble salts.
Presence of Ca, Mg – bicarbonates causes temporary hardness.
Presence of Ca, Mg – chlorides, sulphates causes permanent hardness.

Soft water : Water which give lather immediately with soap is called soft water.
(Or)
i) Ion – Exchange method :

1. This method is useful to remove the permanent hardness of water.
2. This method is also named as permutit (or) zeolite process.
3. Permutit is the artificial zeolite, i.e. sodium aluminium orthosilicate. (Na₂Al₂Si₂O₂xH₂O (or) NaAlSiO₄)
4. Permutit is written in short form as Naz.
5. When permutit is added to hard water, the following ion- exchange reactions takes place.
   2Naz_(s) + Ca_(aq)+2 → Caz_2(s) + 2Na_(aq)⁺
   2Naz_(s) + Mg_(aq)+2 → Mg_2(s) + 2Na_(aq)⁺
6. Caz₂ and Mgz₂ are called as exhausted permutit. These are regenerated to permutit by the treatment with brine solution. Caz_(s) + 2Na_(aq)+2 → 2Naz_(s) + Ca_(aq)+2

ii) Calgon process:

1. Calgon is sodium hexametaphosphate. [Na₆P₆O₁₈ (or) (NaPO₃)₆]
2. Calgon doesnot precipitate the Ca (or) Mg – salts but removes Ca⁺² and Mg⁺² ions from water.
3. The removal of Ca⁺² (or) Mg⁺² ions from water may takes place either by adsorption (or) by complex formation

Reactions:
Na₆P₆O₁₈ → 2Na⁺ + Na₄P₆O_18-2

Question 14.
Define Le Chatelier principle and mention the conditions for the preparation of the ammonia by Haber’s process.
Answer:
Lechatelier Principle Statement: When a system at equilibrium is subjected to a change of temperature, pressure, concentration then the equilibrium shifts in such a direction in order to nallify the effect of change.

Haber’s process:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g) + 92 kj

1. In the above equilibrium recution decrease in no. of moles observed. So according to lechatelier’s principle high pressure favours the formation of NH₃. So an optimum pressure of 200 atm is used.
2. The given reaction is exothermic so low temperature favours the formation of NH₃. But in industries at low temperature reaction proceds slowly. So suitable temperature is 725-775 K.
3. To increase the rate of reaction catalyst ‘Fe’ is used and to increase the activity of catalyst promotor ‘Mo’ is used.
   Pressure – 200 atm
   Temperature – 725 – 775 k
   Catalyst – Fe
   Promotor – Mo

Question 15.
State and write chain and position isomerisms with examples.
Answer:
Chain Isomerism :
Compounds having same molecular formula but differ in the carbon chain or carbon skeleton are called chain isomer and this is called chain isomerism.
Eg:

Position Isomerism :
Compounds having same molecular formula but differ in the position of substituent or functional group or multiple bond are called position isomers and this is called position isomerism.
Eg:

Question 16.
Define molarity. Calculate the molarity of NaOH in the solution prepared by dissolving 4 gr in enough water to form 250 ml of the solution.
Answer:
Molarity : The No.of moles of solute present in one liter of solution is called molarity (M).

Note that molarity of a solution depends upon temperature because’volume of a solution is temperature dependent.

Question 17.
Complete the following reaction and name the products A, B and C:

Question 18.
Derive the Graham’s law of diffusion from kinetic gas equation.
Answer:
Derivation of Charle’s Law from Kinetic gas equation :
Kinetic gas equation is

Hence Charle’s Law Proved from kinetic gas equation. Derivation of Graham’s Law of diffusion from Kinetic gas Equation. Knietic gas equation is

Section – C

Question 19.
Explain the significance of four quantum numbers associated with an electron in an atom.
Answer:

• In general a large no.of orbitals are possible in an atom.
• These orbitals are distinguished by their size, shape and orientation.
• An orbital of smaller size means there is more chance to find electron near the nucleus.
• Atomic orbitals are precisely distinguished by quantum numbers. Each orbital is designated by three major quantum numbers.
  1. Principal quantum number (n)
  2. Azimuthal quantum number (l)
  3. Magnetic quantum number (m)

1) Principal quantum number:
The principal quantum number was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the dis¬tance between the nucleus and the orbit also increases.
It is denoted by the letter ‘n’. It can have any simple integer value 1, 2, 3,…… but not zero. These are also termed as K, L, M, N etc.

The radius and energy of an orbit can be determined basing on “n” value.
The radius of n^th orbit is r_n = \(\frac{n^2 h^2}{4 \pi^2 m e^2}\)
The radius of n^th orbit is E_n = \(\frac{-2 \pi^2 m e^4}{n^2 h^2}\)

2) Azimuthal quantum number:
It was proposed by Sommer feld. It is also known as angular momentum quantum number or subsidiary quantum number. it indicates the shapes of orbitals. It is denoted by T. The values of T depend on the values of ‘n1, T has values ranging from ‘O’ to (n – 1) i.e., ( l = 0, 1, 2 …… (n – 1). The maximum number of electrons present in the subshells s, p, d, f are 2, 6, 10,14 respectively.

Subshell	l – value	Shape
s	l = 0	spherically symmetric
p	l = 1	dumb – bell
d	l = 2	double dumb – bell
f	l = 3	four fold dumb – bell

Energy levels and subshells

Principal quantum number (n)	Azimuthal quantum number. (l)	Symbol	Number of subshells
1	0	s	1 (1s)
2	0	s	2 (2s, 2p)
	1	P
3	0	s	3 (3s, 3p, 3d)
	1	p
	2	d
4	0	s	4 (4s, 4p, 4d, 4f)
	1	P
	2	d
	3	f

3) Magnetic quantum number:
It was proposed by Lande. It shows the orientation of the orbitals in space. ‘p’ – orbital has three orientations. The orbital oriented along the x – axis is called p_x orbital, along the p_y – axis is called p_z – orbital and along the z – axis is called p_z orbital. In a similar way d – orbital has five orientations. They are d_xy

Question 20.
What is periodic property? How are the following properties vary in a group and period?
a) Atomic radius
b) Electron gain enthalpy
c) Electronegativity.
Answer:
Periodicity: The properties of elements which change gradually with change in electronic configuration and repeats in regular intervals are called periodic properties and this repetetion is called periodicity.

a) Atomic radius :
In groups and periods of the periodic table the ionization enthalpy values are periodically change depend upon the electronic configuration and size of elements. In a group of elements ionization energy decreases from top to bottom because atomic radius increases. In general, in a period the atomic size decreases. Because of this, the ionization energy increases across a period.

b) Electron gain enthalpy:
Electron gain enthalpy increases in a period from left to right because the size of the atom decreases and the nature of the element changes from metallic to non – metallic nature when we move from left to right in a period. Electron gain enthalpy decreases from top to bottom in a group because there is an increase in the atomic size. But the second element has greater electron gain enthalpy than the first element’.
e.g. : Chlorine has more electron affinity value ( – 348 kJ mol-1) than Fluorine (- 333 kJ mol-1). It is because fluorine atom is smaller in size than chlorine atom. There is repulsion between the incoming electron and electrons already present in fluorine atom i.e., due to stronger inter electronic repul, sions.

c) Electronegativity:
Electronegativity increases from left to right in a period since the atomic: radii decreases and nuclear attraction increases. In a group electronegativity decreases from top to bottom due to an increase in atomic radii and a decrease in nuclear attraction.

Question 21.
Define the hybridisation. Explain sp, sp² and sp³ hybridisations each with one example.
Answer:
Hybridisation is defined as the process of mixing of atomic orbitals of nearly equal energy of an atom to give the same number of new set of orbitals of equal energy and shapes. Depending on the number and nature of orbitals involving hybridisation it is classified into different types. If ‘s’ and ‘p’ atomic orbitals are involved three types are possible namely sp³, sp² and sp.

1. sp³ hybridisation:
In this hybridisation one’s and three ‘p’ atomic orbitals of the excited atom combine to form four equivalent sp³ hybridised orbitals.

This hybridisation is known as tetrahedral or tetragonal hybridisation. Each sp³ hybridised orbital possess 25% ‘s’ nature and 75% of ‘p’ nature. The shape of the molecule is tetrahedral with a bond angle 109°28′, e.g.: Formation of Methane molecule:
1) The central atom of methane is carbon.

2) The electronic configuration of carbon in ground state is 1s² 2s² 2p²_x 2p^1y 2p^0z and on excitation it is 1s² 2s¹ 2p^1x 2p^y1 2p^z1. During excitation the 2s pair splits and the electron jumps into the adjacent vacant 2p₂ orbital.

3) The 2s¹ 2p¹_x 2p^1y 2p^1z undergo sp3 hybridisation giving four equivalent sp³ hybridised orbitals.

4) Each sp³ hybrid orbital overlaps with the 1s orbitals of hydrogen forming σ_sp³-s bond.

5) In case of methane four σ_sp³-s bonds are formed. The bonds are directed towards the four corners of a regular tetrahedron. The shape of methane molecule is tetrahedral with a bond angle 109°28′.

E.g.: Boron trichloride molecule formation :
1) The electronic configuration of ‘B’ in the ground state is 1s² 2s² 2p^1x 2p^0y 2p^0z.
2) On excitation the configuration is 1s² 2s¹ 2p^1x2p^1y,2p_z⁰ Now there are three half filled orbitals are available for hybridi-sation.
3) Now sp² hybridisation takes place at boron atom giving three sp² hybrid orbitals.

4) Each of them with one unpaired electron forms a ‘σ’ bond with one chlorine atom. The overlapping is σ_sp²-p (Cl atom has the unpaired electron in 2p^z orbital). In boron trichloride there are three ‘σ’ bonds.

3. sp hybridisation:
In this hybridisation one ‘s’ and one ‘p’ atomic orbitals of the excited atom combine to form two equivalent sp hybridised orbitals.

This hybridisation is also known as diagonal hybridisation. In sp hybridisation each sp hybrid orbital has 50% ‘s’ character and 50% ‘p’ character. The shape of the molecule is linear or diagonal with a bond angle 180°.

Ex.: Beryllium chloride molecule formation :
1) Be atom has 1s¹ 2s¹ 2p⁰_x 2p⁰_y 2p^0z electronic configuration.
2) In ground state it has no half filled orbitals. On excitation the configuration becomes 1s² 2s¹ 2p¹_x 2p⁰_y2p⁰_z

3) Now sp hybridisation takes place at beryllium atom giving two sp hybrid orbitals. Each of them with one unpaired electron forms a ‘o’ bond with one chlorine atom.

4) The overlaping is σ_sp-p (Cl atom has the unpaired electron in 2p_z orbital). In beryllium chloride there are two ‘σ’ bonds.