AP Inter 1st Year Chemistry Question Paper May 2019

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AP Inter 1st Year Chemistry Question Paper May 2019

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.
Name the major particulate pollutants present in Tropo-sphere.
Answer:
The major particulate pollutants present in troposphere are dust, mist, fumes, smoke, smog etc.

Question 2.
Why is KO² Paramagnetic?
Answer:
In KO² the O^2– ion have one unpaired Electron so it is para-magnetic.

Question 3.
What is closed system? Give example.
Answer:
Closed system : A system which may exchange energy but nqt matter with surroundings is called a closed system.
Ex : Consider boiling of water in a closed metallic vessel. Here, heat is transferred from the burner (surroundings) to the system. Steam remains inside the vessel. Thus, matter is not exchanged.

Question 4.
Calculate the kinetic energy (in SI units) of 4g of methane at – 73°c.
Answer:
ke’= 3/2 nRT
= 3/2 × 1/4 × 8.314 × 200
R = 623 J

n = 4/16 = 1/4
T = – 73°c = 200°k
R = 8.314 J

Question 5.
What is PAN? What effect is caused by it?
Answer:
Peroxy Acetyl Nitrate is called as PAN.
Peroxy Acetyl Nitrate is (PAN) is a powerful eye irritant.

Question 6.
Describe the important uses of Quicklime.
Answer:
Uses :

1. Quick lime is used in the purification of sugar.
2. Quick lime is used in the manufacture of dyestuffs.
3. Quick lime is used in the manufacture of Na₂CO₃ from NaOH.
4. It is an important material for manufacturing of ce/ment and it is the cheapest form of alkali.

Question 7.
State Le-Chatelier’s principle?
Answer:
Le Chatelier’s Principle – Statement: “If a system at equi-librium is subjected to the change of pressure, temperature (or) concentration, the system is shifted in such a way as to nullify the effect of change”.

Question 8.
Define Normality.
Answer:
Normality : The no.of gram equivalent weights of solute present in one litre of solution.

Question 9.
State Hess law.
Answer:
Hess law : The heat change in a reaction is same whether the reaction takes place in one step (or) in several steps.

Question 10.
Write the I.U.P. A. C names of the following compounds:
a) (CH₃)3C CH₂C(CH₃)₃

Answer:

Question 11.
Define sp² Hybridisation. Explain the structure of Ethylene (C2H4).
Answer:
sp²– hybridisation : Inter mixing of one’s orbital and two ‘p’ orbitals of an atom of identical energy to for three new hybrid or-bitals. This is called sp² hybridisation.
Formation of double bond between the carbon atoms of C²H⁴

1. In ethylene two carbons undergo sp² hybridisation.
2. One of sp² hybrid orbital of carbon overlaps with sp² hybrid orbital of another carbon atom to form C – C sigma bond.
3. The two other sp² hybrid orbitals of each carbon overlap with ‘s’ orbital of hydrogen atoms to form C – H bonds.
4. The unhybridised orbital of one carbon atom overlap side wisely with the similar orbital to form weak n bond.

Question 12.
Write the postulates of Kinetic Molecular Theory of Gases.
Answer:
Assumptions :

1. Gases are composed of minute particles called molecules. All the molecules of a gas are identical.
2. Gaseous molecules are always, at a random movement. The molecules are moving in all possible directions in straight lines with very high velocities. They keep on colliding against each other and against the walls of the vessel at very small intervals of time.
3. The actual volume occupied by the molecules is negligible when compared to the total volume occupied by the gas.
4. There is no appreciable attraction or repulsion between the molecules.
5. There is no loss of kinetic energy when the molecules collide with each other or with the wall of vessel. This is because the molecules are spherical and perfectly elastic in nature.
6. The pressure exerted by the gas is due to the bombardment of the molecules of the gas on the walls of the vessel.
7. The average kinetic energy of the molecules of the gas is directly proportional to the absolute temperature, Average K.E. oc T.
8. The force of gravity has no effect on the speed of gas molecules.

Boyle’s law:
According to kinetic theory of gases, the pressure of a gas is due to collisions of gas molecules on the walls of the vessel. At a particular temperature the molecules make definite number of collisions with the, walls of the vessel; When the volume of the vessel is reduced the molecules have to travel lesser distance only before making collisions on the walls. As a result the number of collisions per unit increases. The pressure then increases, i.e., the pressure increases when the volume is reduced at constant temperature. This explains Boyle’s law.

Charles’ law :
According to kinetic theory of gases, the average kinetic energy of the molecules is directly propor-tional to the absolute temperature of the gas.
K.E. ∝ T
but K.E. = \(\frac{1}{2}\) m²

As temperature increases, the velocity of the molecules also increases. As a result the molecules make more number of collisions against the walls of the vessel. This results in an increase of pressure if the volume is kept constant. If the volume is allowed to increase the number of collisions de-crease due to the increased distance between the molecules and the walls of the vessel. The pressure then decreases. In other words, with rise of temperature, the volume should increase in order to keep the pressure constant.
V ∝ T at constant pressure.
This is Charles’ law.

Question 13.
How does Diborane react with the following :
(a) CO
(b) NH₃
Answer:
a) Reaction with carbon monoxide :
Diborane reacts with CO at 100° C and 20 atm. pressure to form borane carbonyl.

b) Reaction with ammonia
Diborane reacts with ammonia at 120° C first forms B₂H₆.2NH₃ (or) [BH₂(2NH₃)] + [BH₂(NH₃)₂]⁺[BH₄]^– and on further heating forms borazole (or) borazine. Which is also called as “In organic ben-zene”. It has iso structural with benzene. Hence it is named as “In-organic Benzene”.

Question 14.
What is meant by Bond order? Calculate the bond orders in the following:
(a) N₂
(b) O₂₊
Answer:
Bond order: The half of the difference between the no.of bond¬ing electrons and anti bonding electrons is known as bond order,
a) N₂: Molecular orbital energy level sequence.

c) O₃⁺: Molecular orbital energy level sequence is

Question 15.
A carbon compound contains 12.8% Carbon, 2.1% Hydro-gen, 85.1% Bromine. The molecular weight of the com-pound is 187.9. Calculate the molecular formula.
Answer:

∴Emperical formula of the compound = C₁ H₂ Br
Molecular formula = n (Emperical formula)
n = \(\frac{\text { Molecular wt }}{\text { Emperical wt }}=\frac{187.9}{94}\)
Given molecular wt = 187.9
∴ Molecular formula = 2 (CH²Br)
Emperical wt = 94 (CH₂Br)
= C₂H₂Br₂

Question 16.
Derive the relation between K_b and K_c for the equilibrium reaction
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Answer:
Given equation (Equilibrium reaction)
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Relation between K_p and K_c is
K_p = K_c(RT)^∆n
∆n = n_p – n_R
= 2 – (1 + 3) = -2
∆n = -2(-ve)
∴ K_p < K_c

Question 17.
Write any two oxidising and two reducing properties of H202 with equations.
Answer:

1. H₂O₂ has the ability to function as an oxidising agent as well as reducing agent in both acid and alkaline solutions.
2. In H₂2_p oxidation state of oxygen is -1. It oxidised to O₂. Here H₂O₂ is reductant.
3. H₂O_p can be reduced to H₂O (or) OH’. Here H₂O₂ is oxidant.

i) Oxidising action in acidic medium :
2Fe_(aq)⁺² + 2H_(aq)⁺² → 2Fe_(aq)⁺³ + 2H₂O_(l)

ii) Reducing action in acidic medium :
2MnO₄^– + 6H⁺ + 5H₂O₂ → 2Mn₊₂ + 8H₂O + 5O₂

iii) Oxidising action in basic medium :
2Fe⁺² + H₂O₂ → 2Fe⁺³ + 2OH^–

iv) Reducing action in basic medium :
2MnO₄^– + 3H₂O₂ → 2MnO₂ + 3O₂ + 2H₂O + 2OH^–

Question 18.
a) SiF₆^-2 is known while SiCl₆^-2 is not. Explain,
a) SiF₆^-2 is known while SiCl₆^-2 is not. Explain,
b) Diamond has high melting point. Why?
Answer:
a) SiF₆^-2 is known while SiCl_6-2 is not because
1. Si⁺⁴ has small size so it cannot be accomodate six large chloride
2. The interaction between lone pairs of Cl_– ion and Si₊₄ is not very strong.

b) 1.  In Diamond each carbon undergoes sp³ hybridisation and it is surrounded by four other carbon atoms with strong a – bonds tetrahedrally.
2. The C – C bond energy in diamond is very high and it has 3 – dimensional structure.
3. Due to these reasons diamond has high melting point.
4. It has melting point 4200 K.

19. What are Quantum Numbers? Explain the significance of Quantum Numbers.
Answer:

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

1) Principal quantum number:
The principal quantum num-ber was introduced by Neils Bohr. It reveals the size of the atom (main energy levels). With increase in the value of ‘n’ the distance 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 r_n = \(\frac{n^2 h^2}{4 \pi^2 m^2}\)
The radius of n^th orbit is E_n = \(\frac{-2 \pi^2 \cdot m e^4}{n^2 h^2}\)

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 ‘n’, 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 respec-tively

Subshell	l – value	Shape
s	l = 0	spherical 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 (0)	Symbol	Number of subshells
1	0	s	1 (1s)
2.	1	s	2 (2s,2p)
	0	p
3.	0	s	3 (3s,3p,sd)
	1	p
	2	d
4.	0	s	4 (4s, 4p, 4d, 4f)
	1	p
	2	d
	3	f

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 y-axis is called p_y -orbited and along the z-axis is called p_z orbital. In a similar way d – orbital has five orientations. They are d_xy, d_yz, d_zx, d_x²–y² and d_z². It is denoted by’m’. Its values depends on azimuthal quantum number, ‘m’ can have all the integral values from -1 to +1 including zero. The total number of’m’ values are (2l + 1).

Question 20.
Define IE₁, and IE₂. Why IE₂ > IE₁ for a given atom? Discuss the factors that affect IE of an element.
Answer:
1) Ionization energy is the amount of energy required to re¬move the most loosely held electron from isolated a neutral gaseous atom to convert it into gaseous ion. It is also known as first ionization energy because it is the energy required to remove the first electron from the atom.
It is denoted as I₁ and is expressed in electron volts per atom, kilo calories (of) kilo joules per mole.
M(g) + l₁ → M_(g) + e^– I₁ is first ionization potential.

2) The energy required to remove another electron from the unipositive ion is called the second ionization energy. It is denoted as I₂.
M_(g)⁺ + I₂ → M_(g)²⁺ + e^–

3) The second ionization potential is greater than the first ionization potential. On removing an electron from an atom, the unipositive ion formed will have more effective nuclear charge than the number of electrons. As a result the effective nuclear charge increases over the outermost electrons. Hence more energy is required to remove the second electron. This shows that the second ionization potential is greater than the first ionization potential. For sodium, I₁ is 5.1 eV and I₂ is 47.3 eV I₁ < I₂ , I₃ ……. I_n Factors affecting ionization potential:

1. Atomic radius :
As the size of the atom increases the distance between the nucleus and the outermost electrons increases. So the effective nuclear charge on the outermost electrons decreases. In such a case the energy required to remove the electrons also decreases. This shows that with an increase in atomic radius the ionization energy decreases.

2. Nuclear charge :
As the positive charge of the nucleus increases its attraction increases over the electrons. So it becomes more difficult to remove the electrons. This shows that the ionization energy increases as the nuclear charge increases.

3. Screening effect or shielding effect:
In multielectron atoms, valence electrons are attracted by the nucleus as well as repelled by electrons of inner shells. The electrons present in the inner shells screen the electrons present in the outermost orbit from the nucleus. As the number of electrons in the inner orbits increases, the screening effect increases. This reduces the effective nuclear charge over the outermost electrons. It is called screening or shielding effect. With the increase of screening effect the ionization potential decreases. Screening efficiency of the orbitals falls off in the order s > p > d > f.
(Magnitude of screening effect) ∝ \(\frac{1}{\text { (Ionization enthalpy) }}\)

TREND IN A GROUP:
The ionisation potential decreases in a group, gradually from top to bottom as the size of the elements increases down a group.

TREND IN A PERIOD:
In a period from left to right I.P. value increases as the size of the elements decreases along the period.

Question 21.
a) Describe any two methods of preparation of Benzene with corresponding equations.
b) How benzene reacts with the following :
(i) CH₃Cl/Anhy. AlCl₃
(ii) H₂/Ni
Answer: Preparations of Benzene :
From Acetvle Acetyle undergo polymerisation in red hot metal tubes to form benzene

From Decarboxylation : Benzene is prepared by the decarboxylation of sodium benzoate

Reaction with halogen
Benzene reacts with halogens in presence of law is acid to for halobenzene.
C₆H₆ + X₂ C₂H₅ X + HX
Reaction with Acetyl Chloride : Benzene reacts with acetyl chloride inpresence of A1C1₃ to for Acetophenone.
i) Benzene reacts with CH₂C1 and A1CI₃ to for methyl benzene (Toluene)

ii) Benzene reacts with hydrogen in presence of Ni to for Cyclohexane.