Maharashtra Board Class 12 Chemistry Sample Paper Set 3 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 3 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 3 with Solutions

General Instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each.
   Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks
   each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   E.g., (a) ………./(b) …….. /……..(c) …../(d) Only first attempt will be considered for evaluation.
8. Draw well labeled diagrams and write balanced equations wherever necessary.
9. Given data:
   Atomic mass of C = 12, H = 1, O = 16,
   Atomic number (Z): Mn = 25, Fe = 26, Ni = 28, Ar = 18, No = 102.
   R = 8.314 JKT^-1 mol^-1 or 0.083 L bar K^-1 mol^-1
   N_A = 6.022 × 10²³ mol^-1, F = 96500 C

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. The atomicity of sulphur in orthorhombic sulphur is ____.
(A) 8
(B) 6
(C) 4
(D) 2
Answer:
(A) 8

ii. The enthalpy change for the chemical reaction \(\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(l)}\) throw is called enthalpy of ____.
(A) vapourisation
(B) fusion
(C) combustion
(D) sublimation
Answer:
(B) fusion

iii. The species which acts as an conjugate acid as well as base is:
(A) \(\mathrm{HSO}_4^{-}\)
(B) H₂SO₄
(C) NH₄OH
(D) \(\mathrm{CO}_3^{2-}\)
Answer:
(A) HS\(\mathrm{HSO}_4^{-}\)

iv. The CORRECT order of increasing reactivity of C-X bond towards nucleophile in the following compounds is ____.

(A) I < II < IV < III
(B) II < I < III < IV
(C) III < IV < II < I
(D) IV < III < I < II
Answer:
(A) I < II < IV < III

v. In crystal lattice formed by bcc unit cell, the void volume is ____.
(A) 68 %
(B) 74%
(C) 32%
(D) 26 %
Answer:
(C) 32 %

vi. The oxidation number of cobalt in [CoCl₂(en)₂]⁺ is ___.
(A) +1
(B) +3
(C) -1
(D) -3
Answer:
(B) +3

vii. Which of the following information is given by FTIR technique?
(A) Absorption of functional groups
(B) Particle size
(C) Confirmation of formation of nanoparticles
(D) Crystal structure
Answer:
(A) Absorption of functional groups

viii. Open chain formula of glucose does not contain ___.
(A) formyl group
(B) anomeric hydroxyl group
(C) primary hydroxyl group
(D) secondary hydroxyl group
Answer:
(B) anomeric hydroxyl group

ix. In the Wolff-Kishner reduction, alkyl aryl ketones are reduced to alkyl benzenes. During this change, ketones are first converted into ___.
(A) acids
(B) alcohols
(C) hydrazones
(D) alkenes
Answer:
(C) hydrazones

x. Assuming complete dissociation, the value of van’t Hoff factor for Al₂(SO₄)₃ is ___.
(A) 2
(B) 4
(C) 5
(D) 6
Answer:
(C) 5

Question 2.
Answer the following: [8 Marks]

i. What is osmotic pressure?
Answer:
The hydrostatic pressure (on the side of solution) that stops osmosis is called an osmotic pressure of the solution.

ii. Write the relationships between rate constant and half life of first order and zeroth order reactions.
Answer:
a. First order reaction: t_1/2 = \(\frac{0.693}{k}\)
b. Zero order reaction: t_1/2 = \(\frac{\left[A_0\right]}{2 k}\)

iii. What is ferromagnetism?
Answer:
The substances containing large number of unpaired electrons are attracted strongly by magnetic field. Such substances are called as ferromagnetic substances. The property thus exhibited is called ferromagnetism.

iv. Draw the structure of 3-chloro-3-ethylhex-1-ene. ‘
Answer:

v. Write the name of reaction during conversion of phenol to salicylic acid.
Answer:
The name of reaction during conversion of phenol to salicylic acid is Kolbe’s reaction.

vi. Define: Elastomer.
Answer:
Elastomers are elastic polymers having weak van der Waals type of mtermolecular forces which permit them to be stretched.

vii. Write the name and chemical formula of one ore of zinc.
Answer:
Zinc: Name of the ore: Zinc blende
Chemical formula: ZnS.

viii. What is enthalpy of fusion?
Answer:
Enthalpy change that occurs when one mole of a solid is converted into liquid without change in temperature at constant pressure is enthalpy of fusion.

Section – B

Attempt any EIGHT of the following questions: [16 Marks]

Question 3.
Write reaction showing aldol condensation of cyclohexanone.
Answer:

Question 4.
What are interstitial compounds? Give an example.
Answer:

1. When small atoms like hydrogen, carbon or nitrogen are trapped in the interstitial spaces within the crystal lattice, the compounds formed are called interstitial compounds.
2. Sometimes sulphides and oxides are also trapped in the crystal lattice of transition elements.
3. Steel and cast iron are examples of interstitial compounds of carbon and iron. Due to presence of carbon, the malleability and ductility of iron is reduced while its tenacity increases.

Question 5.
For the reaction,

i. How does reaction rate changes if [OH^–] is decreased by a factor of 5?
ii. What is change in rate if concentrations of both reactants are doubled?
Answer:
For the reaction
CH₃Br(aq) + \(\mathrm{Br}_{(\mathrm{aq})}^{-}\) → CH₃OH(aq) + Br(aq)
Rate = k[CH₃Br][OH^–]
i. If [OH^–] is decreased by a factor of 5, keeping [CH₃Br] constant rate will decrease by a factor of 5.
ii. If concentrations of CH₃Br and OH^– are doubled, rate will increase by a factor of 4.

Question 6.
Calculate the work done and comment on whether work is done on or by the system, for the decomposition of 2 moles of NH₄NO₃ at 100 °C.
NH₄NO_3(s) → N₂O_(g) + 2H₂O(g)
Answer:
Given: Decomposition of 1 mole of NH₄NO₃
Temperature = T = 100 °C = 373 K
To find: Work done and to determine whether work is done on the system or by the system.
Formula: W = -∆n_gRT
Calculation: The given reaction is for 1 mole of NH₄NO₃. For 2 moles of NH₄NO₃, the reaction is given as follows:
2NH₄NO_3(s) → 2N₂O_(g) + 4H₂O_(g)
Now,
∆n_g = (moles of product gases) – (moles of reactant gases)
∆n_g = 6 – 0 = +6 mol (∵ NH₄NO₃ is in solid state)
Hence,
W = -∆n_g RT
= – (+6 mol) × 8.314 J K^-1 mol^-1 × 373 K
= -18606.75 J = -18.61 kJ
Work is done by the system
(since W < 0).
Ans: The work done is -18.61 kj.
The work is done by the system.

Question 7.
Obtain the relationship between ∆H and ∆U for gas phase reactions.
Answer:
i. At constant pressure, ∆H and ∆U are related as
∆H = ∆U + P∆V …….(1)

ii. For reactions involving gases, AV cannot be neglected. Therefore,
∆H = ∆U + P∆V = ∆H + P(V₂ – V₁)
∆H = ∆U + PV₂ – PV₁ …….(2)
where, V₁ is the volume of gas phase reactants and V₂ that of the gaseous products.

iii. We assume reactant and product behave ideally. Applying ideal gas equation, PV = nRT. Suppose that m moles of gaseous reactants produce n2 moles of gaseous products.
Then,
PV₁ = n₁RT and PV₂ = n₂RT …(3)

iv. Substitution of equation (3) into equation (2) yields
∆H = ∆U + n₂RT – n₁RT
= ∆U + (n₂ – n₁) RT
= ∆U + ∆n_g RT …..(4)
where, ∆n_g is difference between the number of moles of products and those of reactants.

Question 8.
What are isotonic and hypertonic solutions?
Answer:
i. Isotonic solutions: Two or more solutions having the same osmotic pressure are said to be isotonic solutions.
e.g. For example, 0.1 M urea solution and 0.1 M sucrose solution are isotonic because their osmotic pressures are equal. Such solutions have the same molar concentrations but different concentrations in g/L. If these solutions are separated by a semipermeable membrane, there is no flow of solvent in either direction.

ii. Hypertonic solution: If two solutions have unequal osmotic pressures, the more concentrated solution with higher osmotic pressure is said to be hypertonic solution.
e.g. For example, if osmotic pressure of sucrose solution is higher than that of urea solution, the sucrose solution is hypertonic to urea solution.

Question 9.
The normal boiling point of ethyl acetate is 77.06 °C. A solution of 50 g of a nonvolatile solute in 150 g of ethyl acetate boils at 84.27 °C. Evaluate the molar mass of solute if Kb for ethyl acetate is 2.77 °C kg mol^-1.
Answer:
Given: Normal boiling point of ethyl acetate = \(\mathrm{T}_{\mathrm{b}}^0\) = 77.06 °C
Boiling point of solution = T_b = 84.27 °C
Mass of nonvolatile solute = Wz = 50 g
Mass of ethyl acetate = W₁ = 150 g
K_b of ethyl acetate = 2.77 °C kg mol^-1 = 2.77 K kg mol^-1

To find: Molar mass of solute = M₂

Formula: M₂ = \(\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{~W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{~W}_1}\)

Calculation:
For solution of nonvolatile solute and ethyl acetate,
∆T_b = T_b – \(T_b^0\)
= 84.27 °C – 77.06 °C
= 7.21 °C = 7.21 K
K_b = 2.77 °C kg mol^-1 = 2.77 K kg mol^-1
Now, using formula and substituting values,
M₂ = \(\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{~W}_2}{\Delta \mathrm{~T}_{\mathrm{b}} \mathrm{~W}_1}\)
M₂ = \(\frac{1000 \mathrm{~g} \mathrm{~kg}^{-1} \times 2.77 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 50 \mathrm{~g}}{7.21 \mathrm{~K} \times 150 \mathrm{~g}}\) = 128 g mol

Ans: The molar mass of nonvolatile solute is 128 g mol^-1.

Question 10.
Derive an expression for ionic product of water.
Answer:
i. Pure water ionizes to a very small extent. The ionization equilibrium of water is represented as,
H₂O_(l) + H₂O_(l) \(\rightleftharpoons\) \(\mathrm{H}_3 \mathrm{O}_{(a q)}^{+}\) + \(\mathrm{OH}_{(\mathrm{aq})}^{-}\)

ii. The equilibrium constant (K) for the ionization of water is given by
K = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]^2}\) …….. (1)
or K[H₂O]² = [H₃O⁺][OH^–] ……(2)

iii. A majority of H₂O molecules are undissociated, consequently concentration of water [H₂O] can be treated as constant. Then, [H₂O]² = K’.

iv. Substituting this in equation (2), we get,
K × K’ = [H₃O⁺][OH ] ……(3)
K_w = [H₃O⁺][OH^–]
where K_w = KK’ is called ionic product of water.

v. The product of molar concentrations of hydronium (or hydrogen) ions and hydroxyl ions at equilibrium in pure water at the given temperature is called ionic product of water.

Question 11.
Explain the trends in atomic radii of d block elements.
Answer:

1. Atomic radii of the elements of the transition series decrease gradually from left to right along the Periodic Table.
2. As we move across a transition series from left to right, the nuclear charge increases by one unit at a time.
3. The last filled electron enters a penultimate (n-1)d subshell. However, d orbitals in an atom are less penetrating or more diffused and, therefore d electrons offer smaller screening effect.
4. The result is that effective nuclear charge also increases as the atomic number increases along a transition series. Hence, the atomic radii gradually decrease across a transition series from left to right.

Note: Trends in atomic radii of d block elements is shown in the graph: (Textbook page no. 169)

Question 12.
How will you obtain l-bromo-l-methylcyclohexane from alkene? Write possible structures of alkene and the reaction involved.
Answer:

Question 13.
Explain the effect of dilution of solution on conductivity.
Answer:

1. The electrolytic conductivity is electrical conductance of unit volume (1 cm³) of solution. It depends on the number of current carrying ions present in unit volume of solution
2. On dilution total number of ions increase as a result of increased degree of dissociation.
3. An increase in total number of ions is not in proportion of dilution. Therefore, the number of ions per unit volume of solution decreases.
4. This results in decrease of conductivity with decrease in concentration of solution.

Question 14.
Predict the products for the following reaction.

Answer:
The substrate (A) contains an ioslated C = C and an aldehyde group. H₂/Ni can reduce both these functional groups while LiAlH₄ can reduce only -CHO of the two. Hence,

Section – C

Attempt any EIGHT of the following questions: [24 Marks]

Question 15.
i. Why nobelium is the only actinoid with +2 oxidation state?
ii. The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?
Answer:
i. a. The electronic configuration of nobelium (No) is [Rn] 5f¹⁴6d⁰ 7s².
b. After losing 2 electrons, No forms No²⁺ ion which is stable due to completely filled 5f-orbitals.
c. Such extra stability due to completely filled orbital is not acquired by any other actinoid in their +2 oxidation state.
Hence, nobelium is the only actinoid with +2 oxidation state. [2 Marks]

ii. The electronic configuration of element with atomic number = 90 is [Rn] 5f° 6d² 7s².
There are two unpaired electrons in the 6d orbital of the element. Hence, it is paramagnetic.

Question 16.
How is phenol prepared from the following?
i. Chiorobenzene
ii. Cumene
Answer:
i. Chlorobenzene (Dow process): In this process, chlorobenzene is fused with NaOH at high temperature and pressure (623K and 150 atm) followed by treatment with dilute HCl to obtain phenol.

ii. Cumene: This is the commercial method of preparation of phenol. Cumene
(isopropylbenzene) on air oxidation in presence of Co-naphthenate gives cumene
hydroperoxide, which on decomposition with dilute acid gives phenol with acetone as a valuable by product.

Question 17.
Why is phenol more acidic than ethyl alcohol?
Answer:
Difference in acidic character of phenols and alcohols is due to difference in reactivity of these compounds towards ionization of O-H bond. This can be explained as follows:
i. Ionization of alcohols is represented by the following equilibrium

Electron donating inductive effect (+I effect) of alkyl group destabilizes the alkoxide ion (the conjugate base of alcohol). As a result, alcohol does not ionize much in water, and behaves like neutral compound in aqueous medium.

ii. Ionization of phenol is represented by the following equilibrium

Phenoxide ion, the conjugate base of phenol, is resonance stabilized by delocalization of the negative charge. Therefore, phenol ionizes in aqueous medium to a moderate extent, and thereby shows a weak acidic character.

Note: The reactivity of alcohols and phenols towards ionization of O-H bond in them
is different because aqueous medium phenols show weak acidic character while alcohols are neutral.

Question 18.
How is glucose prepared from the following?
i. Sucrose
ii. Starch
Answer:
i. Preparation of glucose from sucrose:

a. Sucrose is hydrolysed by warming with dilute hydrochloric acid or sulfuric acid for about two hours.
b. This hydrolysis converts sucrose into mixture of glucose and fructose.
c. Glucose is separated from fructose by adding ethanol during cooling.
d. Glucose being almost insoluble in alcohol crystallizes out first. The solution is filtered to obtain crystals of glucose. The reaction can be given as,

ii. Preparation of glucose from starch: Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute sulphuric acid at 393K under 2 to 3 atm pressure.

Question 19.
How is polyacrylonitrile prepared? Mention its uses.
Answer:
i. Preparation: Polyacrylonitrile is prepared by addition polymerization of acrylonitrile by using peroxide initiator.

ii. Uses:
a. Polyacrylonitrile resembles wool and is used as wool substitute.
b. It is used for making orlon or acrilan.

Question 20.
i. Define: Nanotechnology.
ii. Write two disadvantages of nanotechnology.
Answer:
i. Nanotechnology is the design, characterization, production and application of structures, devices and systems by controlling shape and size at nanometer scale.

ii. Disadvantages of nanotechnology:
Despite the possibilities and the advancements that the nanotechnology offers to the world, there also exist certain potential risks involved with it.

i. Nanotechnology has raised the standard of living but at the same time, it has increased the pollution which includes air pollution. The pollution caused by nanotechnology is known as nano pollution. This kind of pollution is very dangerous for living organisms.

ii. Nanoparticles can cause lung damage. Inhaled particulated matter may get deposited throughout the human respiratory tract and then in the lungs.

iii. The characteristics of nanoparticles that are relevant for health effects are size, chemical composition and shape. (Any two disadvantages)

Question 21.
Calculate molar conductivities at zero concentration for CaCl₂ and Na₂SO₄.
Given: Molar ionic conductivities of Ca²⁺, Cl^–, Na⁺ and \(\mathrm{SO}_4^{2-}\) ions are respectively, 104, 76.4, 50.1 and 159.6 Ω^-1 cm² mol^-1.
Answer:
i. Given: \(\lambda_{\mathrm{Ca}^{2-}}^0\) = 104 Ω^-1 cm² mol^-1,
\(\lambda_{\mathrm{Cr}}^0\) = 76.4 Ω^-1 cm² mol^-1,
\(\lambda_{\mathrm{Na}^{+}}^0\) = 50.1 Ω^-1 cm² mol^-1.
\(\lambda_{\mathrm{SO}_4^{2-}}^0\) = 159.6 Ω^-1 cm² mol^-1
To find: Molar conductivity at zero concentration for CaCl₂ and Na₂SO₄‘.

i. ∧0 (CaCl₂) = \(\lambda_{\mathrm{Ca}^{2+}}^0+2 \lambda_{\mathrm{Cl}^{-}}^0\)
= 104Ω^-1 cm² mol^-1 + 2 × 76.4 Ω^-1 cm²mol^-1
= 256.8 Ω^-1 cm²mol^-1

ii. ∧0(Na₂SO₄) = \(2 \lambda_{\mathrm{Na}^{+}}^0+\lambda_{\mathrm{SO}_4^{2-}}^0\)
= 2 × 50.1 Ω^-1 cm² mol^-1 + 159.6 Ω^-1 cm² mol^-1
= 259.8 Ω^-1 cm² mol^-1

Ans:
i. Molar conductivity at zero concentration of CaCl₂ is 256.8 Ω^-1 cm² mol^-1.
ii. Molar conductivity at zero concentration of Na₂SO₄ is 259.8 Ω^-1 cm² mol^-1.

Question 22.
Derive Ostwald’s dilution law for a weak acid.
Answer:
i. Consider an equilibrium of weak acid HA that exists in solution partly as the undissociated species HA and partly H⁺ and A^– ions. Then
HA_(aq) \(\rightleftharpoons\) \(H_{(\mathrm{qq})}^{+}+A_{(q q)}^{-}\)

ii. The acid dissociation constant is given as:
K_a = \(\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}\) ……. (1)

iii. Suppose 1 mol of acid HA is initially present in volume V dm³ of the solution. At equilibrium, the fraction dissociated would be a, where a is degree of dissociation of the acid. The fraction of an acid that remains undissociated would be (1 – a).

iv. Thus, at equilibrium [HA] = \(\frac{1-a}{V}\) mol dm^-3,
[H⁺] = [A^–] = \(\frac{a}{V}\) mol dm^-3

v. Substituting these in equation (1),
K_a = \(\frac{(a / V)(a / V)}{(1-a) / V}\) = \(\frac{a^2}{(1-a) V}\) ……. (2)

vi. If c is the initial concentration of an acid in mol dm^-3 and V is the volume in dm³ mol^-1 then c = 1/V.
Replacing 1/V in equation (2) by c, we get
K_a = \(\frac{a^2 c}{1-a}\) ….. (3)

vii. For the weak acid HA, a is very small, or (1 – a) 1.
With this equation (2) and (3) becomes:
K_a = a³/V and K_a = a²C ………(4)
a = \(\frac{\sqrt{K_a}}{c}\) or a = \(\sqrt{K_a \cdot V}\) …….. (5)

The equation (5) implies that the degree of dissociation of a weak acid is inversely proportional to the square root of its concentration or directly proportional to the square root of volume of the solution containing 1 mol of the weak acid.

Question 23.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
i. Hess’s law of constant heat summation: Hess’s law of constant heat summation States that, “Overall the enthalpy change for a reaction is equal to sum of enthalpy changes of individual steps in the reaction”.

ii. Illustration:

a. The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy is state function. The enthalpy change of a reaction depends only on the initial and final states and not on the path by which the reaction occurs.

b. To determine the overall equation of reaction, reactants and products in the individual steps are added or subtracted like algebraic entities.

c. Consider the synthesis of NH₃,

The sum of the enthalpy changes for steps (1) and (2) is equal to enthalpy change for the overall reaction.

iii. Application of Hess’s law: The Hess’s law has been useful to calculate the enthalpy changes for the reactions with their enthalpies being not known experimentally.

Question 24.
Write Arrhenius equation. Derive an expression for temperature variations.
Answer:
i. Arhennius equation is:
k = \(A e^{-E_0 / R T}\)
OR log<sub10 K = log₁₀ A = \(\frac{E_a}{2.303 R T}\)

ii. For two different temperatures T₁ and T₂

Question 25.
Draw the structures of
i. ICI
ii. CIF₃
iii. XeO₃
Answer:
i. The structure of ICI:

ii. The structure of ClF₃:

iii. The structure of XeO₃:

Question 26.
A compound forms hcp structure. What is the number of
(i) octahedral voids
(ii) tetrahedral voids
(iii) total voids formed in 0.4 mol of it?
Answer:
given: Compound forms hcp structure.
Amount = 0.4 mol

To find:
i. Number of octahedral voids
ii. Number of tetrahedral voids
iii. Total number of voids

Calculation:
Number of atoms in 0.4 mol = 0.4 × N_A
= 0.4 × 6.022 × 10²³ = 2.4098 × 10²³

i. Number of octahedral voids = Number of atoms = 2.4098 × 10²³
ii. Number of tetrahedral voids = 2 × Number of atoms
= 2 × 2.4098 × 10² = 4.818 × 10²³
iii. Total number of voids = Number of octahedral voids + Number of tetrahedral voids
= 2.409 × 10²³ + 4.818 × 10²³ = 7.227 × 10²³

Ans:
i. Number of octahedral voids is 2.4098 × 10²³.
ii. Number of tetrahedral voids is 4.818 × 10²³.
iii. The total number of voids is 7.227 × 10²³

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
Based on V.B.T., explain the structure of square planar complex: [Ni(CN)₄]^2-.
Answer:
i. Oxidation state of nickel is +2

ii. Valence shell electronic configuration of Ni²⁺

iii. Number of CN^– ligands is 4. so, the number of vacant metal ion orbitals required for bonding with Iigarids would be four.

iv. Complex is square planar, so Ni²⁺ ion uses dsp² hybrid orbitals.

v. 3d electrons are paired prior to the hybridisation and electronic configuration of Ni²⁺ becomes:

vi. Orbitals available for hybridsation are one 3d, one 4s and two 4p which give dsp² hybridization.

vii. Four vacant dsp² hybrid orbitals of Ni²⁺ overlap with four orbitals of CN^– ions to form Ni-CN coordinate bonds.

Configuration after the complex formation becomes.

viii. The complex has no unpaired electrons and hence, diamagnetic. The structure of the complex is

Question 28.
i. Explain Gatterman-Koch formylation of arene.
ii. Explain: Etard reaction
Answer:
i. Gatter man-Koch formylation of arene:
Treatment of benzene or substituted benzene under high pressure with carbon monoxide and hydrogen chloride gives benzaldehyde or substituted benzaldehyde.

The reaction is carried out in presence of anhydrous aluminium chloride or cuprous chloride.

ii. Etard reaction:
Methyl group in toluene (i.e., methyl benzene/methyl arene) is oxidized by oxidizing agent chromyl chloride (CrO₂Cl₂) in carbon disulfide as solvent, to form a chromium complex, from which the corresponding benzaldehyde is obtained on acid hydrolysis. This reaction is known as Etard reaction.

Question 29.
i. Explain ammonolysis of alkyl halides.
ii. Write resonance structures of aniline.
Answer:
i. a. When alkyl halide is heated with alcoholic solution of excess ammonia, it undergoes nucleophilic substitution reaction in which halogen atom is replaced by an amino (-NH₂) group to form primary amine.
b. This process of breaking of C – X bond by ammonia is known as ammonolysis.
c. The reaction is also known as alkylation of ammonia. The reaction is carried out in a sealed tube at 373 K.
d. The primary amine obtained in the 1^st step is stronger nucleophile than ammonia. Hence, it further reacts with alkyl halide to form secondary and tertiary amines and finally quaternary ammonium salt if NH₃ is not used in large excess.

e. The order of reactivity of alkyl halides with ammonia is R-I > R-Br > R-Cl.

ii. Aniline is resonance stabilized by the following five resonance structures.

Question 30.
i. What is the action of chlorine (Cl₂) on the following:
(a) Cold and dilute caustic soda
(b) Hot and concentrated caustic soda
Answer:
i. (a) Action of Cl₂ on cold and dil. NaOH:
When chlorine is passed through cold and dilute NaOH solution, a mixture of sodium hypochlorite and sodium chloride is formed.

(b) Action of Cl₂ on hot and conc. NaOH: Chlorine reacts with hot and conc.
NaOH to produce sodium chlorate and sodium chloride.

ii. Write any two’ uses of ‘chlorine’.
Answer:
Uses of chlorine:

1. For purification (sterilizing) of drinking water.
2. For bleaching wood pulp required for manufacture of paper and rayon, bleaching cotton and textiles.
3. For extraction of metals like gold and platinum.
4. In the manufacture of dyes, drugs and organic compounds such as CCl₄, CHCl₃, refrigerants, etc.
5. In the preparation of poisonous gases such as phosgene (COCl₂), tear gas (CCl₃NO₂), mustard gas (ClCH₂CH₂SCH₂CH₂Cl).

Question 31.
i. What current strength in amperes will be required to produce 2.4 g of Cu from CuSO₄ solution in 1 hour? Molar mass of Cu = 63.5 g mol^-1.
ii. Write any two applications of electrochemical series.
Answer:
i. Given: Mass of Cu = 2.4 g, Molar mass of Cu = 63.5 g mol^-1
1 hours = 1 × 60 × 60 s = 3600 s
To find: Current strength (in amperes)

Formulae.
i.

ii. W \(\frac{I(A) \times+(s)}{96500\left(C / \mathrm{mole}^{-}\right)}\) × mole ratio × molar mass

Calculation:

Ans: Current strength in amperes required to produce 2.4 g of Cu from CuSO₄ is 2.03 A. [2 Marks]

ii. Applications of electrochemical series:

i. Relative strength of oxidising agents:

a. The species on the left side of half reactions in electrochemical series are oxidizing agents.
b. E° value measures the strength of the substances as oxidising agents. Larger the E° value greater is the strength of oxidising agent.
c. The species in the top left side of half reactions are strong oxidising agents. As we move down the electrochemical series, E° value and strength of oxidisingagents decreases.

ii. Relative strength of reducing agents:

a. The species on the right side of half reactions are reducing agents.
b. The species appearing at the bottom right side of half reactions associated with large negative E° values are the effective electron donors. They serve as strong reducing agents.
c. The strength of reducing agents Increases from top to bottom as E° values decrease.

Maharashtra Board Class 12 Chemistry Previous Year Question Papers