12th Chemistry Question Paper 2024 Maharashtra Board Pdf

Maharashtra State Board Class 12th Chemistry Question Paper 2024 with Solutions Answers Pdf Download.

Class 12 Chemistry Question Paper 2024 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 questions: [10]
(i) The spin only magnetic moment of Cr³⁺ cation is:
(a) 3.742 BM
(b) 3.755 BM
(c) 3.873 BM
(d) 3.633 BM
Answer:
(c) 3.873 BM

Explanation: Electronic configuration of Cr(24)

Number of unpaired electrons = 3
∴ Magnetic moment (μ) =\(\sqrt{n(n+2)}\) BM
Where, n = Number of unpaired electrons
= \(\sqrt{3(3+2)}\)
= \(\sqrt{3 \times 5}\)
= \(\sqrt{15}\)
= 3.873 BM

ii) The linkage present in Lactose is:
(a) α, ß-1,2-glycosidic linkage
(b) α-1,4-glycosidic linkage
(c) ß-1,4-glycosidic linkage
(d) α-1,4-glycosidic linkage
Answer:
(c) ß-1,4-glycosidic linkage

Explanation :
Latcose (C¹²H²²O¹¹) is a disaccharide present in milk. It is formed from two monosaccharide units, namely D-galactose and D-glucose. The glycosidic linkage is formed between C-1 of ß-D-galactose and C-4 of glucose. Therefore the linkage in lactose is called B-1, 4-glycosidic linkage.

iii) The product of the following reaction is

Answer:

Explanation:

(iv) The pH of 0.001M HCl solution is:
(a) 10
(b) 3
(c) 2
(d) 11
Answer:
(b) 3

Explanation:
HCl is strong acid dissociate completely

[H⁺] = 10^-3
pH = – log₁₀ [H⁺]
= – log₁₀ 10^-3
= 3 log₁₀ 10
= 3 × 1 (∵ log10 = 1)
= 3
∴ pH = 3

v) The correct structure of complex having IUPAC name sodium hexanitrocobaltate(III) is :
(a) Na₃ [CO(NO₂)₅]
(b) Na₄[Co(NO₂)₆]
(c) Na₃[Co(NO₂)₆]
(d) Na₄ [CO(NO₂)₅]
Answer:
(c) Na₃[Co(NO₂)₆]

Explanation:
The correct structure of complex having IUPAC name sodium hexanitrocobaltate (iii) is Na3 Na₃[Co(NO₂)₆].

(vi) The number of particles present in Face Centred Cubic Unit Cell is/are:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Explanation:

(vii) The monomer used in preparation of teflon is:
(a) E-caprolactum
(b) vinyl chloride
(c) styrene
(d) tetrafluoroethene
Answer:
(d) tetrafluoroethene

Explanation :
Chemically teflon is polytetrafluoroethylene. The monomer used in preparation of teflon is tetrafluoroethylene, (CF₂ – CF₂) which is a gas at room temperature.

(viii) Among the following vinylic halide is:

Answer:
(b) CH₂=CH-X

Explanation : In Vinyl haldies the carbon that bears the halogen is doubly bonded to another carbon.
Vinylic halid ⇒ halogen atom directly attached to sp2 hybridised carbon
∴ CH₂=CH-X is vinylic halid

(ix) The product of hydrolysis of propyne in the. presence of 1% HgSO₄ and 40% H₂SO₄ is:
(a) methanal
(b) ethanal
(c) propanal
(d) propanone
Answer:
(d) Propanone

(x) If unit of rate constant is mol dm-3 s-1, the order of reaction would be:
(a) zero
(b) 1
(c) 2
(d) 3
Answer:
(a) zero

Units of rate constant of zero order reactions
k = \(\frac{[A]_0-[A]_t}{t}\) = \(\frac{\mathrm{mol} \mathrm{~L}^{-1}}{t}\) = mol dm^-3 t^-1
The units of rate constant of zero order reaction are the same as the rate.
Units of rate constant of zero order reactions
k = \(\frac{[A]_0-[A]_t}{t}\) = \(\frac{\mathrm{mol} \mathrm{~L}^{-1}}{t}\) = mol dm^-3 t^-1
The units of rate constant of zero order reaction are the same as the rate.

Question 2.
Answer the following questions: [8]
(i) Write the name of metal nanoparticle used to remove E. coli bacteria from water.
Answer:
Silver nanoparticles act as highly effective bacterial disinfectant, remove EColi from water.

(ii) Write the name of reduction product formed when ethyl cyanide is treated with sodium and alcohol.
Answer:
Ethyl cyanide on reaction with sodium and ethanol (alcohol) forms propylamine.

(iii) complete the reaction:

Answer:

(iv) Calculate effective atomic number of [Co(NH₃)₆]³⁺ ion.
Answer:
EAN = number of electrons of metal ion + total number of electrons donated by ligands
= atomic number of metal (Z) – number of electrons lost by metal to form the ion (X) + number of electrons donated by ligands (Y).
=Z-X+Y
Consider Co[HN₃]₆\(3 \oplus\)
Oxidation state of Cobalt is +3, six ligands donate 12 electrons.
Z = 27; X = 3; Y = 12
EAN of Co\(3 \oplus\) = 27 – 3 + 12 = 36.

(v) The compounds of Ti⁴⁺ ions are colourless due to …………..
Answer:
The compound Ti⁴⁺ is colourless due because it does not have unpaired electrons in its outer orbital (3d°).

(vi) Write SI unit of molar conductivity.
Answer:
S m² mol^-1 OR ohm^-1 cm² mol^-1

(vii) Write the sign convention of work done during expansion of gas.
Answer:
As a matter of convention, negative work occurs when a system does work on the surroundings. When the gas does work the volume of a gas increase (AV > 0) and the work done is negative.

(viii) Write the condition of reverse osmosis.
Answer:
In the reverse osmosis, the direction of osmosis can be reversed by applying a pressure higher than osmotic pressure is applied on the solution. OR Applied pressure > Osmotic pressure (n).

Section-B
Attempt any Eight of the following questions. [16]

Question 3.
Derive an expression for maximum work obtainable during isothermal reversible expansion of an ideal gas from initial volume (V₁) to final volume (V₂).
Answer:
Total work done for expansion of gas from V₁ to V₂.
W_max = \(-\int_{V_1}^{V_2} P d V\)
W_max = \(-\int_{V_1}^{V_2} n R T \frac{d V}{V}\)
For isothermal expansion of temperature T is constant, hence, we can write
W_max = -nRT ln(\(\frac{V_2}{V_1}\))
W_max = -2.303 nRT log₁₀ (\(\frac{V_2}{V_1}\))
This is a total work done for isothermal reversible expansion of an ideal gas.

Question 4.
What are interhalogen compounds? Write the chemical reaction, when chlorine reacts with dry slaked lime.
Answer:
Interhalogen compound : Due to this difference in electronegativity two or more halogen atoms combine to form species which may be ionic or neutral. The neutral molecules are called Interhalogen compounds.
OR
The halogens react with each other to form Interhalogen compounds.
Reaction of chlorine with dry slaked lime:

Question 5.
What is nano-material? Write the reaction involved in soL-gel process during hydrolysis.
Answer:
Nano-material : The nano-material is a material having structural components with at least one- dimension in the nano-meter scale that is 1-100 nm. Reaction involved in sol-gel process during hydrolysis:

Question 6.
Write classification of proteins with an example.
Answer:
Classification of proteins:
Protein is of two types depending upon their molecular shape.
(a) Globular proteins : Molecules of globular proteins have spherical shape. This shape results from coiling around of the polypeptide chain of protein.
Example : Insulin, egg albumin, serum albumin, legumelin (protein in pulses).

(b) Fibrous proteins : Molecules of fibrous proteins have elongated, rod like shape. This shape is the result of holding the polypeptide chains of protein parallel to each other. Hydrogen bonds and disulfide bonds are responsible for this shape.
Example : keratin (present in hair, nail, wool), myosin (protein of muscles).

Question 7.
Calculate the time required to deposit 2.4 g of Cu, when 2.03 A of current passed through CuSO₄ solution.
(At. mass of Cu = 63.5 g mol^-1)
Answer:

Question 8.
Why amines are basic in nature? Among dimethylamine (pK_b 3.27) and diethylamine (pK_b = 3.0), which one is more basic?
Answer:
Basicity of Amines:
The basic nature of amines is due to presence of a lone pair of electrons on the nitrogen atom. In terms of Lewis theory, amines are bases because they can share a lone pair of electrons on ‘N’ atom with an electron deficient species.
For example: Trimethylamine shares its lone pair of electrons with the electron deficient boron trifluoride.

Basic strength of amines is expressed quantitatively as K_b or pK_b value. In terms of Lowry-Bronsted theory, the basic nature of amines is explained by writing the following equilibrium.

In this equilibrium amine accepts \(H^{\oplus}\), hence an amine is a Lowry-Bronsted base.
Diethylamine (pK_b = 3.0) is more basic than dimethylamine (pKj, = 3.27). The basicity of amine is quantitatively expressed by their pK_b values, with a smaller pK_b value indicating a strong base.
OR
Basic strength: Diethylamine > Dimethylamine.

Question 9.
Explain buffer action of sodium acetate-acetic acid buffer.
Answer:
Buffer action : Let us consider sodium acetate-acetic acid buffer. Here sodium acetate is a strong electrolyte which dissociates completely in water producing large concentration of \(\mathrm{CH}_3 \mathrm{COO}^{\ominus}\) as follows:

On the other hand since the acetic acid is a weak acid, the concentration of undissociated CH₃COOH molecules is usually high. If a strong acid is added to this solution the added \(H^{\oplus}\) ions will be consumed by the conjugate base \(\mathrm{CH}_3 \mathrm{COO}^{\ominus}\) present in large concentration. Similarly, if small amount of base is added, the added \(OH^{\ominus}\) ions will be neutralized by the large concentration of acetic acid as shown in the following reactions:

The acid or base added thus can not change the \(H^{\oplus}\) or \(OH^{\ominus}\) concentrations and, pH of the buffer remains unchanged. Dilution does not have any effect on pH of buffer.

Question 10.
Write preparation of (a) diethyl ether (b) ethyl cyanide from ethyl bromide.
Answer:

Question 11.
Henry’s constant for CH₃Br_(g) is 0.159 mol dm^-3 bar^-1 at 25°C. Calculate its solubility in water at 25°C, if its partial pressure is 0.164 bar.
Answer:
According to Herny’s law
S = K_HP
given, K_H = 0.159
P = 0.164
S = 0.159 × 0.164
S = 0.026076 mol L^-1

Question 12.
Write the structure and name of monomer of
(a) Nylon-6
(b) Natural rubber
Answer:

Question 13.
Define Lanthanide contraction. Write the balanced chemical equations when acidified K₂Cr₂O₇ reacts with H₂S.
Answer:
Lanthanoid contraction : The atomic size or the ionic radii of tri positive lanthanide ions decrease steadily from La to Lu due to increasing nuclear charge and electrons entering inner (n – 2)f orbital. This gradual decrease in the size with an increasing atomic number is called lanthanide contraction.
Reaction of K₂Cr₂O₇ react with H₂S:

K₂Cr₂O₇ + 4H₂SO₄ + H₂S → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3S

Question 14.
Derive the relationship between molar mass, density of the substance and unit cell edge length.
Answer:
Mass of unit cell = m × n
m = mass of one particle
n = particles per unit cell Mass
Density = \(\frac{\text { Mass }}{\text { Volume }}\) = \(\frac{m×n}{a^3}\) ………..(1)
M = mass of one particle x number of particles per mole
Molar mass (M) = m × N_A
Combining equations (1) and (2), gives
Mass M × n
Density or ρ or d = \(\frac{\text { Mass }}{\text { Volume }}\)
= \(\frac{m×n}{a^3×N_A}\)

Section-C
Attempt any Eight of the following Questions [24]

Question 15.
What is osmotic pressure? How will you determine molar mass of solute from osmotic pressure?
Answer:
Osmotic pressure : The minimum pressure required to prevent the inward flow of a solution’s pure solvent through a semipermeable membrane is known as the osmotic pressure.

Question 16.
Write chemical reactions involved in:
(a) Rosenmund reduction.
(b) Gattermann-Koch formylation.
(c) Cannizaro reaction of methanal
Answer:

Question 17.
Calculate the standard enthalpy of combustion of methane, if the standard enthalpy of formation of methane, carbon dioxide and water are -74.8, -393.5 and -285.8 kJ mol^-1 respectively.
Answer:
The equation for combustion of CH₄
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

∆_cH° = ?

∆_cH° = [∆_fH° + 2∆_fH°(OH)] – [∆_fH°(CH₄) + 2∆_fH°(O₂)]

∆_cH° = [1×(-393.5)+2×(-285.8)] – [1×(-74.8)+2×0]

∆_cH° = -889.7 KJ

Question 18.
What is the action of following on ethyl bromide ?
(a) silver nitrite
(b) Mg in dry ether
(c) alcoholic sodium hydroxide
Answer:

Question 19.
For the reaction A + B → P.
If [B] is doubled at constant [A], the rate of reaction doubled. If [A] is triple and [B] is doubled, the rate of reaction increases by a factor of 6. Calculate the rate law equation.
Answer:
Rate = K[A]^x[B]^y ……(1)
Conditions 1; Rate × 2 = K[A]^x[2B]^y …….(2)
Conditions 2; Rate × 6 = K[3A]^x[2B]^y ……..(3)

Question 20.
Arrange the following in the increasing order of the property mentioned:
(i) HOCI, HCIO₂, HCIOJ, HCIO₄ (acidic strength)
(ii) MF, MCI, MBr, Ml (ionic character)
(iii) HF, HCI HBr, HI (thermal stability)
Answer:

Question 21.
Explain Wolf-Kishner reduction reaction. Write preparation of propanone by using ethanoyl chloride and dimethyl cadmium.
Answer:

Question 22.
Write postulates of Werner theory of coordination complexes. Write the name of a hexadentate ligand.
Answer:
Werner theory of coordination complexes:
(i) Unlike metal salts, the metal in a complex possesses two types of valencies primary (ionisable) valency and secondary (nonionisable) valency.

(ii) The ionisable sphere consists of entities which satisfy the primary valency of the metaL Primary valencies are generally satisfied by anions.

(iii) The secondary coordination sphere consists of entities which satisfy the secondary valencies and are non-ionisable. The secondary valencies for a metal ion are fixed and satisfied by either anions or neutral ligands. Number of secondary valencies is equal to the coordination number.

(iv) The secondary valencies have a fixed spatial arrangement around the metal ion.
Hexadentate ligand : Ethylenediaminetetraacetate ion (EDTA)⁴⁺.

Question 23.
Define electrochemical series and write its two applications.
Answer:
EMF series is defined as the arrangement of electrode with the electrode half reaction in order of decreasing standard potentials.
Applications of electrochemical series:
(1) Relative strength of oxidizing agents : The species on the left side of half reactions are oxidizing agents. E° value is a measure of the tendency of the species to accept electrons and get reduced In other words, E° value measures the strength of the substances as oxidising agents. Larger the E° value greater is the oxidising strength. The species in the top left side of half 25. reactions are strong oxidising agents. As we move down the table, E° value and strength of oxidising agents decreases from top to bottom.

(2) Relative strength of reducing agents : The species of right side of half reactions are reducing agents.
The half reactions at the bottom of the table with large negative E° values have a little or no tendency to’occur in the forward direction as written. They tend to favour the reverse direction.
It follows, that 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. The strength of reducing agents increases from top to bottom as E° values decrease.

(3) Spontaneous of redox reactions : A redox reaction in galvanic cell is spontaneous only if the species with higher E° value is reduced (accepts electrons) and that with lower E° value is oxidised (donates electrons).
The standard cell potential must be positive for a cell reaction to be spontaneous under the standard conditions. Noteworthy applications of electromotive series is predicting spontaneity of redox reactions from the knowledge of standard potentials.
Suppose, we ask a question : At standard conditions would \(A g^{\oplus}\) ions oxidise metallic magnesium ? To answer this question, first we write oxidation of Mg by \(A g^{\oplus}\).

Question 24.
Identify ‘A’, ‘B’ and ‘C’ in following chain reaction and rewrite the chemical reactions:

Answer:

Question 25.
Define acids and bases according to Bronsted-Lowry theory. Derive relationship between pH and pOH.
Answer:
Bronsted – Lowry theory:
Acid : Acid is a substance that donates a proton (Hv) to another substance.
Base: Base is a substance that accepts a proton (Hv) from another substance.
Relationship between pH and pOH:
The ionic product of water is
K_W = \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\)\(\left[\mathrm{OH}^{\ominus}\right]\)
Now, K_W = 1 × 10^-14 at 298 K
Thus [H₃O⁺] [OHe] = 1.0 × 10^-14
Taking logarithm of both the sides, we write
\(\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]-\log _{10}\left[\mathrm{OH}^{\ominus}\right]\) = -14
\(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]+\left\{-\log _{10}\left[\mathrm{OH}^{\ominus}\right]\right\}\) = 14
pH + pOH = 14

Question 26.
Write the preparation of potassium dichromate from chromite ore.
Answer:
Preparation of potassium dichromate using chromite are:
In the industrial production, finely powdered chromite are (FeOCr₂O₃) is heated with anhydrous sodium carbonate (Na₂O₃) and a flux of lime in air in a reverbatory furnace.

Sodium chromate (Na₂CrO₄) formed in this reaction is then extracted with water and treated with concentrated sulphuric acid to get sodium dichromate and hydrate sodium sulphate:

Addition of potassium chloride to concentrated solution of sodium dichromate precipitates less soluble orange-red coloured potassium dichromate, K₂Cr₂O₇.

Maharashtra Board Class 12 Chemistry Previous Year Question Papers