12th Chemistry Question Paper 2022 Maharashtra Board Pdf

Maharashtra State Board Class 12th Chemistry Question Paper 2022 with Solutions Answers Pdf Download.

Class 12 Chemistry Question Paper 2022 Maharashtra State Board with Solutions

Time: 3 Hrs
Max. Marks: 70

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.
   Only the first attempt will be considered for evaluation.
   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. Given: R = .8.314 J.K^-1mol^-1
   N_A = 6.022 × 10²³
   F = 96500 C

Section-A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: (10)

(i) The co-ordination number of atoms in body centred cubic structure (bcc) is …………..
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(c) 8

ii) In calculating osmotic pressure, the concentration of solute is expressed in ……………
(a) molarity
(b) molality
(c) mole fraction
(d) percentage mass
Answer:
(a) molarity

iii) The enthalpy change for the chemical reaction H₂O(s) → H₂O(l) is called enthalpy of …………..
(a) vapourisation
(b) fusion
(c) combustion
(d) sublimation
Answer:
(b) fusion

iv) Which of the following transition element shows maximum oxidation state?
(a) Sc
(b) Fe
(c) Mn
(d) V
Answer:
(c) Mn

(v) The correct formula for the complex compound, sodium hexacyanoferrate (III) is
(a) Na [Fe(CN)₆]
(b) Na₂[Fe(CN)₆]
(c) Na₃[Fe(CN)₆]
(d) Na₄[Fe(CN)₆]
Answer:
(c) Na₃[Fe(CN)₆]

(vi) Isopropylbenzene on air oxidation followed by decomposition by dilute acid gives
(a) C₆H₅OH
(b) C₆H₅COOCH₃
(c) C₆H₅COOH
(d) C₆H₅CHO
Answer:
(a) C₆H₅OH

(vii) The name of metal nanoparticle which acts as highly effective bacterial disinfectant in water purification process is ……………
(a) carbon black
(b) silver
(c) gold
(d) copper
Answer:
(b) silver

(viii) Acid anhydride on reaction with primary amine gives compound having a functional group
(a) amide
(b) nitrile
(c) secondary amine
(d) imine
Answer:
(a) amide

ix) The standard potential of the cell in the following reaction is ……………

(\(\mathrm{E}_{\text {cell }}^{\circ}\) = – 0.403V, \(\vec{v}\) = 0,334V)
(a) -0.737 V
(b) 0.737 V
(c) – 0.069 V
(d) 0.069 V
Answer:
(b) 0.737 V

\(\mathrm{E}_{\text {cell }}^{\circ}\) = \(\mathrm{E}_{\text {RHE }}^{\circ}\) – \(\mathrm{E}_{\text {LHE }}^{\circ}\)
= 0.334 – (- 0.403 V) = 0.737 V.

(x) The value of [H_n0⁺] in mol lit^-1 of 0.001 M acetic acid solution (Ka = 1.8 × 10^-5) is …………
(a) 1.34 × 10^-1
(b) 1.34 × 10^-2
(c) 1.34 × 10^-3
(d) 1.34 × 1o^-4
Answer:
(d) 1.34 × 1o^-4

Question 2.
Answer the following questions: (8)

(i) Write the product formed when alkyl halide reacts with silver nitrite.
Answer:

(ii) Write the name of product formed, when acetone is treated with 2,4-dinitrophenyl hydrazine.
Answer:

24 – Dinitrophenyl hydrazone is formed when acetone is treated with 2,4 – dinitrophenyl hydrazine.

(iii) Write the name of biodegradable polyamide copolymer.
Answer:
Nylon – 2 – nylon – 6 is the biodegradable polyamide copolymer polyamide.

(iv) Identify the molecularity of following elementary reaction:
Answer:
Molecularity = 2 (Two)
The reaction is bimolecular in nature.

(v) What is the action of selenium on magnesium metal?
Answer:
Mg + Se → MgSe(Maganesium Selenide)
Magnesium selenide is formed when selenium reacts with magnesium metaL

(vi) Write the name of isomerism in the following complexes:
[CU(NH₃)₄] [Ptcl₄] and [Pt(NH₃)₄] [ CuCl₄]
Answer:
Both complexes show coordination isomerism because there is interchange of ligands between cationic and anionic entities of different metal ions present in the complex.

(vii) Write the name of the alloy used in Fischer Tropsch process in the synthesis of gasoline.
Answer:
Co -Th (Cobolt-Thorium) alloy is used in the Fischer Tropsch process in the synthesis of gasoline.

(viii) Henry’s law constant for CH₃ Br_(g)) is 0.159 mol dm^-3 bar_-1 at 25°C. What is solubility of CH₃ Br_(g) in water at same temperature and partial pressure of 0.164 bar?
Answer:
S= K_HP
= 0.159 × 0.164
= 0.0261
The Solubility of CH₃Br_(g) in water at same temperature and partial pressure of 0.164 bar will be 0. 0261.

Section-B
Attempt any EIGHT of the following questions: (16)

Question 3.
Explain pseudo-first order reaction with suitable example.
Answer:
The reactions which have higher order and true rate law but are found to behave as first order kinetics are called pseudofirst order reactions.
For Example: Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH₃COOCH₃ (aq) + H₂O(l) → CH₃COOH (aq) + CH₃OH (aq)
rate = K'[CH₃COOCH₃] [H₂O]

Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be,
Rate = k'[CH₃COOCH₃] × [H₂O] …..(i)
Hence, the reaction is expected to follow second order kinetics. However, experimentally, it is found that the reaction follows first order kinetics.
This is because solvent water being in a large excess, its concentration remains constant
Thus, [H₂O] = constant = k” …(ii)
On substitution of equation (ii) in (i)
Rate = k[CH₃COOCH₃] × [H₂O]
= k[CH₃COOCH₃] × k”
Then, Rate = k’ × k” × [CH₃COOCH₃]
If k’ × k”= k,
then Rate = k[CH₃COOCH₃],
This indicates that second-order true rate law is forced into first order rate law. Therefore, this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question 4.
Write the consequences of Schottky defect with reasons.
Answer:
Consequences of Schottky defect are:
1. As the number of ions decreases, mass decreases. However, Volume remains unchanged. Hence, the density of a substance decreases.
2. The number of missing cations and anions is equal, the electrical neutrality of the compound is preserved.
3. This defect arises in ionic crystals like NaCl, AgBr, KCl etc.

Question 5.
What is the action of following on ethyl bromide:
(i) Na in dry ether
(ii) Mg in dry ether
Answer:

Question 6.
Explain formation of peptide linkage in protein with an example.
Answer:
Proteins are polymer of a-Amino acid and large number of a-amino acid connects to each other with peptide bond (amide bond).

A peptide bond is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amino group of the other molecule, releasing a molecule of water (H₂O). This is a dehydration synthesis reaction (also known as a condensation reaction), and usually occurs between amino acids.
For Example

Question 7.
Derive an expression to calculate molar mass of non-volatile solute by osmotic pressure measurement.
Answer:
For every dilute solution
where π = \(\frac{n_2 R T}{V}\)
π = Osmotic pressure
n₂ = Number of moles of solute
R = Gas constant
T = Absolute temperature
V = Volume of solution
n₂ = \(\frac{w_2 \text { (Mass of solute) }}{M_2 \text { (Molar mass of solution) }}\)
M₂ = \(=\frac{w_2 R T}{\pi V}\)

Question 8.
Explain monodentate and ambidentate ligands with example.
Answer:
Monodentate ligand: A monodentate ligand is the one where a single donor atom shares an electron pair to form a coordinate bond with central metal ion.
For Example: \(Cl^{\ominus}\), \(\mathrm{OH}^{\ominus}\) etc.
Ambidentate ligands: The ligands which have two donor atoms and use the electron pair of either donor atom to form a coordinate bond with the metal ion. For Example:

Question 9.
Explain the trends in the following atomic properties of group 16 elements:
(i) Atomic radii
(ii) Ionisation enthalpy
(iii) Electronegativity
(iv) Electron gain enthalpy
Answer:
(i) Atomic radii: Top to bottom in a group number of shells increases, distance from nucleus to outermost shell increases therefore atomic radii also increases.
(ii) Ionisation enthalpy: From top to bottom in group 16 atomic radii increases, therefore, nuclear attraction force decreases to outermost shell hence, ionisation enthalpy decreases.
(iii) Electronegativity: When We move from top to bottom in a group, then nuclear attraction force decreases and atomic radii increases, therefore electronegativity decreases.
(iv) Electron gain enthalpy: When we move from top to bottom in a group, the atomic size increases and less amount of energy is released due to addition of extra electron, therefore, electron gain enthalpy decreases.

Question 10.
Write preparation of phenol from aniline.
Answer:
Aniline is diazotized by treatment with nitrous acid (NaNO₂ and HCl) under ice-cold conditions to form benzene diazonium chloride which on hydrolysis forms phenoL

Question 11.
Write chemical reactions to prepare ethanamine from:
(i) acetonitrile (ii) nitroethane
Answer:

Question 12.
Identify A and B from the following reaction:

Answer:

Question 13.
One mole of an ideal gas is expanded isothermally and reversibly from 10 Lto 15 L at 300 K. Calculate the work done in the process.
Answer:
W_max = -2.303 × nRT \(\frac{1}{2}\)
W_max = -2.303 × 1 × 8.314 × 300 × \(\frac{1}{2}\)
= -2.303 × 1 × 8.314 × 300 × \(\frac{1}{2}\)
= -2.303 × 1 × 8.314 × 300 × 0.176
= -1010.96 J
W_max = -1.011 KJ

Question 14.
How many moles of electrons are required for reduction of 2 moles of Zn²⁺ to Zn? How many Faradays of electricity will be required?
Answer:
The balanced equation for the reduction of Zn²⁺ to Zn is:
Zn²⁺ + (aq) + 2e^– → Zn(s)
1 mole of Zn²⁺ is reduced to Zn by 2 mole of electron. Hence, for the reduction of 2 moles of Zn²⁺, 4 moles of electron will be required.
As 1 mole of electron carries’ 1 Faraday of electricity. Hence, 4 Faradays of electricity will be required to form 2 mole of Zn from Zn²⁺.

Section-C
Attempt any EIGHT of the following questions: (24)

Question 15.
Write chemical composition of haematite. Write the names and electronic configurations of first two elements of group 17.
Answer:
Haematite = Fe₂O₃
First two elements of group 17
1. Fluorine (F) (9) = 1s² 2s² 2p⁵
2. Chlorine (Cl) (17) = 1s² 2s² 2p⁶ 3s² 3p⁵

Question 16.
Write classification of polymers on the basis of structure.
Answer:
On the basis of structure polymers are classified in three types:
1. Linear or straight chain polymers: When the monomer molecules are joined together in a linear arrangement, the resulting polymer is straight chain polymer.
For Example: High density polymer (Polythene), PVC (Polyvinylchloride).

2. Branched chain polymers: Monomer having 3 functional groups or already having side chains give rise to branched chain polymers.
For Example: Low density polythene.

3. Cross linked polymers: This type of polymers are produced due to cross links of linear chains to form network like structure.
For Example: Bakelite, melamine etc.

Question 17.
Define green chemistry. Write two disadvantages of nanotechnology.
Answer:
Green Chemistry: Green chemistry is an approach to chemistry that aims to maximise efficient use of renewable resources and minimise hazardous effect on human health and environment.
Disadvantages of Nanotechnology:
1. It has increased the pollution which includes air pollution (nano pollution).
2. Nanoparticles can cause lung damage. Inhaled particulate matter can be deposited throughout the human respiratory tract and then deposits in the lungs.

Question 18.
Write commercial method for preparation of glucose. Write structure of adipic acid.
Answer:
Commercially glucose is obtained by hydrolysis of starch by boiling it with dilute sulphuric acid at 393 K and 2 to 3 atm. pressure.

Question 19.
Write chemical reactions of following reagents on methoxyethane:
(i) hot HI
(ii) PCl₅
(iii) dilute H₂SO₄
Answer:

Question 20.
Explain cationic, anionic and neutralsphere complexes with example.
Answer:
1. Cationic Complexes: A coordination compound having a positively charged coordination sphere is called cationic sphere complex.
For Example:

2. Anionic Sphere Complexes: A Coordination compound having negatively charged coordination sphere is called anionic complex.
For Example:

3. Neutral Sphere Complexes: A neutral coordination complex does not possess cationic or anionic sphere.
For Example: [FT (NH₃)₂ Cl₂], [Ni (Co)₄]

Question 21.
Calculate spin only magnetic moment of divalent cation of transition metal with atomic number 25.
Salts of Ti₄₊ are colourless. Give reason.
Answer:
Electronic configuration of divalent metal ion.
M (25) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁵
M2 + = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
n = number of unpaired electrons
i.e., μ = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5 + 2)}\)
= 5.92 BM.
Ti⁴⁺ is colourless
Ti(22) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Ti⁴⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s⁰
No unpaired electron is present in Ti⁴⁺, no d – d Transition takes place.
Thus Ti⁴⁺ salts are colourless

Question 22.
What is lanthanoid contraction?
Write preparation of acetic acid from:
(i) dry ice
(ii) acetyl chloride
Answer:
Lanthanoid contraction: In Lanthanoid series from ” left to right atomic number increases gradually but atomic and ionic radii decreases. This gradual decrease in the atomic and ionic size of lanthanoids with an increase in atomic number is called lanthanoid contraction.
Causes of Lanthanide contraction are:
1. With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell.
2. The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f sub-shell is lesser than in d subshell.
3. With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus.
Preparation of Acetic acid:
i) From dry ice:

ii) From Acetyl chloride:

Question 23.
Write the classification of aliphatic ketones with example. What is the action of sodium hypoiodite on acetone?
Answer:
Classification of aliphatic ketones:
1. Simple or symmetrical ketones: The ketones in which both the alkyl groups attached to carbonyl carbon are identical
For Example:

Mixed or unsymmetrical Ketones: The Ketones in which two alkyl groups attached to carbonyl carbon are different are called mixed ketones.
For Example:

Action of sodium hypoiodite on acetone

Question 24.
Define half life of first order reaction. Obtain the expression for half life and rate constant of the first order reaction.
Answer:
Half life of first order reaction: Time in which concentration of reactant become half of its initial concentration is called half life. It is denoted by t_1/2.
Expression:

Question 25.
Calculate the standard enthalpy of formation of CH₃ OH(l) from the following data:

Answer:
Required equation

The enthalpy of formation of CH₃OH(l) is – 239 KJ

Question 26.
Calculate the pH of buffer solution composed of 0.01 M weak base BOH and 0.02 M of its salt BA.
[K_b= 1.8 × 10^-5 for weak base]
Answer:
pOH = pK_b + log₁₀ \(\frac{\text { [Salt] }}{[\text { base] }}\)
Now,
pK_b = -log₁₀ K_b
= -log₁₀ (1.8 × 10^-5)
= 5 – log₁₀ 1.8
= 5 – 0.2553
pK_b = 4.7447
Then, pOH = 4.7447 + log₁₀ \(\frac{0.02}{0.01}\)
= 4.7447 + log₁₀ 2
= 4.7447 + 0.3010
pOH = 5.0457
∵ pH + pOH = 14
∴ pH= 14 – pOH
= 14 – 5.0457
= 8.9543 or 8.9
pH =8.95
The pH of the buffer solution will be 8.95.

Section-D
Attempt any THREE of the following questions: (12)

Question 27.
Define the following terms:
(i) Isotonic solution
(ii) Osmosis
Gold crystallises into face-centred cubic cells. The edge length of unit cell is 4.08 × 10^-8 cm. Calculate the density of gold.
[Molar mass of gold = 197g mol^-5]
Answer:
(i) Isotonic Solution: Two or more than two solution having same osmotic pressure are called isotonic solution.
(ii) Osmosis: Process of flow of solvent from lower concentration to higher concentration through semipermeable membrane is called osmosis.
Given, The edge length (a) of the unit cell = 4.08 x 10^-8 cm
It crystallises in Face – centred cubic cells.
Density (ρ) = \(\frac{n M}{a^3 N_A}\)
where n = number of particals
For FCC, n = 4
M = Molar mass
a = edge length
N_A = 6.022 × 10²³
ρ = \(\frac{197 \times 4}{\left(4.08 \times 10^{-8}\right)^3 \times 6.022 \times 10^{23}}\)
ρ = 19.27 g cm^-3

Question 28.
Write the mathematical equation for the first law of thermodynamics for:
(i) isothermal process
(ii) adiabatic process
Derive the relationship between pH and pOH.
Answer:
First law of thermodynamic
∆U = Q + W
For isothermal process
∆T = 0
∴ ∆U= 0
∴ Q = -W
For Adiabatic process
Q = 0
∆U = W
⇒ Relationship between pH and pOH
we know that the ionic product of water is given as:
K_W = [H₃O⁺] [OH^–]
Now, K_W = [H₃O⁺] [OH^–] = 1 × 10^-14
Taking log on both sides, we write
\(\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) + \(\log _{10}\left[\mathrm{OH}^{-}\right]\) = \(\log _{10}\left(1 \times 10^{-14}\right)\) = -14
Multiply by (- 1), we get
–\(\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) – \(\log _{10}\left[\mathrm{OH}^{-}\right]\) = 14

Now pH = \(\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\) and pOH = \(\log _{10}\left[\mathrm{OH}^{-}\right]\)
∴ pH + pOH = 14.

Question 29.
Define reference electrode. Write functions of salt bridge.
Draw neat, labelled diagram of standard hydrogen electrode (SHE).
Answer:
Reference electrode: An electrode whose potential is arbitrarily taken as zero or is exactly known is called Reference electrode.
For example: Standard Hydroge Electrode (SHE), Calomel electrode etc.
Functions of Salt Bridge:
1. It provides an electrical contact between two solutions and thereby completes the electrical circuit.
2. It prevents mixing of two solutions.
3. It maintains electrical neutrality in both the solutions by transfer of ions.
Diagram of Standard Hydrogen Electrode (SHE):

Maharashtra Board Class 12 Chemistry Previous Year Question Papers