Board Question Paper: July 2023
Subject: Chemistry | Max. Marks: 70 | Time: 3 Hrs.
SECTION ‒ A
[10 Marks]
Q.1. Select and write the correct answer for the following multiple choice type of questions:
(i) Anisole on heating with concentrated HI gives _______.
Answer:
(d) Phenol + Iodomethane
Explanation: The bond between oxygen and the methyl group breaks because the phenyl-oxygen bond has partial double bond character due to resonance. Thus, phenol and methyl iodide are formed.
(ii) Which solution shows positive deviation from Raoult’s law?
Answer:
(c) Ethanol and Acetone
Explanation: In pure ethanol, molecules are hydrogen-bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds, increasing vapor pressure (Positive Deviation).
(iii) The coordination number of cobalt in [CoCl2(en)2]+ is
Answer:
(a) 6
Explanation: 'en' (ethylenediamine) is a bidentate ligand (coordinates through 2 sites) and Cl is monodentate. Total coordination = (2 × 2) + (2 × 1) = 6.
(iv)
The name of above reaction is _______. (Reaction shown: Benzene + CO, HCl, Anhydrous AlCl3, High Pressure → Benzaldehyde)
The name of above reaction is _______. (Reaction shown: Benzene + CO, HCl, Anhydrous AlCl3, High Pressure → Benzaldehyde)
Answer:
(d) Gatterman-Koch reaction
(v) Which is an example of thermoplastic polymer?
Answer:
(b) Polystyrene
(vi) Nichrome is an alloy of _______.
Answer:
(c) Ni, Cr
(vii) Identify ‘A’ in the following reaction:
A + 2Na $\xrightarrow{\text{dry ether}}$ Biphenyl + 2NaCl
A + 2Na $\xrightarrow{\text{dry ether}}$ Biphenyl + 2NaCl
Answer:
(d) Chlorobenzene
Explanation: This is the Fittig reaction. The byproduct is 2NaCl, indicating the halogen in reactant 'A' must be Chlorine. Hence, A is Chlorobenzene.
(viii) Which amine does NOT react with Hinsberg reagent?
Answer:
(c) N, N-diethylethanamine
Explanation: Tertiary amines (like N,N-diethylethanamine) do not have replaceable hydrogen atoms on the nitrogen, so they do not react with benzenesulfonyl chloride (Hinsberg reagent).
(ix) The dissociation constant of NH4OH is \(1.8 \times 10^{-5}\). The degree of dissociation in its 0.01 M solution is _______.
Answer:
(a) 0.04242
\(\alpha = \sqrt{\frac{K_b}{C}} = \sqrt{\frac{1.8 \times 10^{-5}}{10^{-2}}} = \sqrt{1.8 \times 10^{-3}} = \sqrt{18 \times 10^{-4}} = 4.242 \times 10^{-2} = 0.04242\)
(x) Half-life of a first order reaction is 30 minutes at 300 K. The value of its rate constant, K is _______.
Answer:
(b) \(0.0231 \text{ min}^{-1}\)
\(K = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \text{ min}^{-1}\)
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Q.2. Answer the following questions: [8 Marks]
(i) Write the name of radioactive element in group 16.
Answer:
Polonium (Po).
(ii) Write the structure of glycine.
Answer:
\( \text{H}_2\text{N} - \text{CH}_2 - \text{COOH} \)
(iii) Write the unit of cell constant.
Answer:
\( \text{m}^{-1} \) or \( \text{cm}^{-1} \).
(iv) Write the number of particles present in FCC per unit cell.
Answer:
4 particles.
(v) Name the \(\gamma\)-isomer of BHC.
Answer:
Lindane (or Gammexane).
(vi) Write the IUPAC name of isobutyraldehyde.
Answer:
2-Methylpropanal.
(vii) Which alloy is used in Fischer Tropsch process in the synthesis of gasoline?
Answer:
Cobalt-Thorium alloy (Co-Th) or Fe-based catalyst.
(viii) Three moles of an ideal gas are expanded isothermally from \(15 \text{ dm}^3\) to \(20 \text{ dm}^3\) at constant external pressure of 1.2 bar. Estimate the amount of work in Joules.
Answer:
\(W = -P_{ext} \Delta V\)
\(P_{ext} = 1.2 \text{ bar}\)
\(\Delta V = V_2 - V_1 = 20 - 15 = 5 \text{ dm}^3\)
\(W = -1.2 \times 5 = -6 \text{ bar dm}^3\)
Convert to Joules: \(1 \text{ bar dm}^3 = 100 \text{ J}\)
\(W = -6 \times 100 = -600 \text{ J}\)
Work done by the system is 600 Joules.
\(P_{ext} = 1.2 \text{ bar}\)
\(\Delta V = V_2 - V_1 = 20 - 15 = 5 \text{ dm}^3\)
\(W = -1.2 \times 5 = -6 \text{ bar dm}^3\)
Convert to Joules: \(1 \text{ bar dm}^3 = 100 \text{ J}\)
\(W = -6 \times 100 = -600 \text{ J}\)
Work done by the system is 600 Joules.
SECTION ‒ B
[16 Marks]
(Attempt any EIGHT of the following questions)
Q.3. Write four postulates of Werner theory of coordination complexes.
Answer:
- Coordination compounds possess two types of valencies: primary (ionizable) and secondary (non-ionizable).
- Primary valency corresponds to the oxidation state and is satisfied by negative ions.
- Secondary valency corresponds to the coordination number and is satisfied by negative ions or neutral molecules.
- Secondary valencies have a fixed spatial arrangement (directionality) around the central metal ion, giving the complex a definite geometry.
Q.4. Why fluorine shows anomalous behaviour?
Answer:
Fluorine shows anomalous behavior due to:
- Extremely small size of the atom and fluoride ion.
- High electronegativity (most electronegative element).
- Absence of d-orbitals in its valence shell, restricting its covalency to 1.
- Low F-F bond dissociation enthalpy.
Q.5. What is the mass of Cu metal produced at the cathode during the passage of 5 ampere current through CuSO4 solution for 6000 seconds. Molar mass of Cu is 63.5 g mol-1.
Answer:
Reaction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
Mole ratio (n) = 2 moles of electrons per mole of Cu.
Using Faraday's Law: \( W = \frac{I \cdot t \cdot M}{n \cdot F} \)
Given: \( I = 5 \text{ A}, t = 6000 \text{ s}, M = 63.5 \text{ g/mol}, F = 96500 \text{ C/mol} \)
\( W = \frac{5 \times 6000 \times 63.5}{2 \times 96500} \)
\( W = \frac{30000 \times 63.5}{193000} \)
\( W = \frac{1905000}{193000} \approx 9.87 \text{ g} \)
Mass of Cu produced = 9.87 g.
Mole ratio (n) = 2 moles of electrons per mole of Cu.
Using Faraday's Law: \( W = \frac{I \cdot t \cdot M}{n \cdot F} \)
Given: \( I = 5 \text{ A}, t = 6000 \text{ s}, M = 63.5 \text{ g/mol}, F = 96500 \text{ C/mol} \)
\( W = \frac{5 \times 6000 \times 63.5}{2 \times 96500} \)
\( W = \frac{30000 \times 63.5}{193000} \)
\( W = \frac{1905000}{193000} \approx 9.87 \text{ g} \)
Mass of Cu produced = 9.87 g.
Q.6. How is glucose prepared from sucrose?
Answer:
Glucose is prepared from sucrose by hydrolysis. Sucrose is boiled with dilute HCl or H2SO4 in alcoholic solution.
\( \underset{\text{Sucrose}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \underset{\text{Glucose}}{\text{C}_6\text{H}_{12}\text{O}_6} + \underset{\text{Fructose}}{\text{C}_6\text{H}_{12}\text{O}_6} \)
Glucose is separated from fructose by crystallization (as glucose is less soluble in alcohol).
Q.7. Derive integrated rate law for zero order reaction.
Answer:
Consider a zero order reaction \( A \rightarrow \text{Products} \).
The rate is given by: \( \text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k \)
Rearranging: \( d[A] = -k dt \)
Integrating both sides within limits \([A]_0\) at \(t=0\) and \([A]_t\) at \(t=t\):
\( \int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \)
\( [A]_t - [A]_0 = -k(t - 0) \)
\( [A]_t = -kt + [A]_0 \) or \( k = \frac{[A]_0 - [A]_t}{t} \)
The rate is given by: \( \text{Rate} = -\frac{d[A]}{dt} = k[A]^0 = k \)
Rearranging: \( d[A] = -k dt \)
Integrating both sides within limits \([A]_0\) at \(t=0\) and \([A]_t\) at \(t=t\):
\( \int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt \)
\( [A]_t - [A]_0 = -k(t - 0) \)
\( [A]_t = -kt + [A]_0 \) or \( k = \frac{[A]_0 - [A]_t}{t} \)
Q.8. The normal boiling point of ethyl acetate is 77.06 °C. A solution of 50 g of a non-volatile 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:
Elevation in boiling point, \( \Delta T_b = T_b - T_b^0 \)
\( \Delta T_b = 84.27 - 77.06 = 7.21^\circ \text{C} \)
Formula: \( M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} \)
Given:
\( W_2 \) (mass of solute) = 50 g
\( W_1 \) (mass of solvent) = 150 g
\( K_b = 2.77 \)
\( \Delta T_b = 7.21 \)
\( M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150} \)
\( M_2 = \frac{138500}{1081.5} \approx 128.06 \text{ g mol}^{-1} \)
\( \Delta T_b = 84.27 - 77.06 = 7.21^\circ \text{C} \)
Formula: \( M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} \)
Given:
\( W_2 \) (mass of solute) = 50 g
\( W_1 \) (mass of solvent) = 150 g
\( K_b = 2.77 \)
\( \Delta T_b = 7.21 \)
\( M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150} \)
\( M_2 = \frac{138500}{1081.5} \approx 128.06 \text{ g mol}^{-1} \)
Q.9. How is phenol prepared from cumene?
Answer:
This is the commercial method. It involves two steps:
- Oxidation: Cumene (isopropylbenzene) is oxidized by air in the presence of Co-naphthenate catalyst to form Cumene hydroperoxide.
- Decomposition: Cumene hydroperoxide is treated with dilute sulfuric acid to decompose into Phenol and Acetone.
\( \text{Cumene} + \text{O}_2 \rightarrow \text{Cumene hydroperoxide} \xrightarrow{\text{dil H}_2\text{SO}_4} \text{Phenol} + \text{Acetone} \)
Q.10. Why do d-block elements form coloured compounds?
Answer:
d-block elements form colored compounds due to:
- Presence of unpaired d-electrons: Ions with partly filled d-orbitals (d1 to d9 configuration).
- d-d transition: When ligands approach the metal ion, the d-orbitals split into two sets of different energies (crystal field splitting). Electrons absorb energy from the visible region to jump from lower energy d-orbitals to higher energy d-orbitals. The complementary color of the absorbed light is observed.
Q.11. Write a note on: Wolf-Kishner reduction reaction.
Answer:
Wolf-Kishner reduction is used to convert the carbonyl group (\(>\text{C=O}\)) of aldehydes and ketones into a methylene group (\(>\text{CH}_2\)).
The aldehyde or ketone is heated with hydrazine (\(\text{NH}_2\text{NH}_2\)) and potassium hydroxide (KOH) in a high boiling solvent like ethylene glycol.
The aldehyde or ketone is heated with hydrazine (\(\text{NH}_2\text{NH}_2\)) and potassium hydroxide (KOH) in a high boiling solvent like ethylene glycol.
\( >\text{C=O} + \text{NH}_2\text{NH}_2 \xrightarrow{-\text{H}_2\text{O}} >\text{C=N-NH}_2 \text{ (Hydrazone)} \xrightarrow{\text{KOH, glycol, heat}} >\text{CH}_2 + \text{N}_2 \)
Q.12. How is Nylon 6, 6 prepared?
Answer:
Nylon 6, 6 is prepared by the condensation polymerization of Hexamethylenediamine and Adipic acid.
Equimolar amounts of both monomers are mixed to form 'Nylon salt', which upon heating under high pressure and temperature loses water molecules to form Nylon 6, 6.
Equimolar amounts of both monomers are mixed to form 'Nylon salt', which upon heating under high pressure and temperature loses water molecules to form Nylon 6, 6.
\( n(\text{HOOC}-(\text{CH}_2)_4-\text{COOH}) + n(\text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2) \rightarrow [-\text{CO}-(\text{CH}_2)_4-\text{CONH}-(\text{CH}_2)_6-\text{NH}-]_n + 2n\text{H}_2\text{O} \)
Q.13. Derive Ostwald’s dilution law equation for weak acid.
Answer:
Consider a weak acid HA. Let C be the total concentration and \(\alpha\) be the degree of dissociation.
Equilibrium: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \)
Initial conc: \( C \quad 0 \quad 0 \)
Equilibrium conc: \( C(1-\alpha) \quad C\alpha \quad C\alpha \)
Applying Law of Mass Action: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)
\( K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \)
For weak acids, \(\alpha\) is very small (\(\alpha \ll 1\)), so \(1-\alpha \approx 1\).
\( K_a = C\alpha^2 \) or \( \alpha = \sqrt{\frac{K_a}{C}} \)
Equilibrium: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \)
Initial conc: \( C \quad 0 \quad 0 \)
Equilibrium conc: \( C(1-\alpha) \quad C\alpha \quad C\alpha \)
Applying Law of Mass Action: \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \)
\( K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \)
For weak acids, \(\alpha\) is very small (\(\alpha \ll 1\)), so \(1-\alpha \approx 1\).
\( K_a = C\alpha^2 \) or \( \alpha = \sqrt{\frac{K_a}{C}} \)
Q.14. What is Grignard reagent? How it is prepared?
Answer:
Definition: Grignard reagent is an organometallic compound with the general formula \( \text{R-Mg-X} \) (Alkyl magnesium halide).
Preparation: It is prepared by the reaction of an alkyl halide (RX) with pure magnesium metal (Mg) in the presence of dry ether.
Preparation: It is prepared by the reaction of an alkyl halide (RX) with pure magnesium metal (Mg) in the presence of dry ether.
\( \text{R-X} + \text{Mg} \xrightarrow{\text{dry ether}} \text{R-Mg-X} \)
SECTION ‒ C
[24 Marks]
(Attempt any EIGHT of the following questions)
Q.15. Calculate the standard enthalpy of the reaction:
\(\text{SiO}_{2(s)} + 3\text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)} + 2\text{CO}_{(g)}\)
From the following reactions:
(i) \(\text{Si}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SiO}_{2(s)}, \Delta_r H^\circ = -911 \text{ kJ}\)
(ii) \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}, \Delta_r H^\circ = -221 \text{ kJ}\)
(iii) \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}, \Delta_r H^\circ = -65.3 \text{ kJ}\)
From the following reactions:
(i) \(\text{Si}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SiO}_{2(s)}, \Delta_r H^\circ = -911 \text{ kJ}\)
(ii) \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}, \Delta_r H^\circ = -221 \text{ kJ}\)
(iii) \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}, \Delta_r H^\circ = -65.3 \text{ kJ}\)
Answer:
To get the target equation:
1. Reverse eq (i): \(\text{SiO}_{2(s)} \rightarrow \text{Si}_{(s)} + \text{O}_{2(g)}\) ; \(\Delta H = +911 \text{ kJ}\)
2. Keep eq (ii) as is: \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}\) ; \(\Delta H = -221 \text{ kJ}\)
3. Keep eq (iii) as is: \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}\) ; \(\Delta H = -65.3 \text{ kJ}\)
Add these three modified equations:
\((\text{SiO}_2) + (2\text{C} + \text{O}_2) + (\text{Si} + \text{C}) \rightarrow (\text{Si} + \text{O}_2) + (2\text{CO}) + (\text{SiC})\)
Cancelling common terms (\(\text{Si}, \text{O}_2\)):
\(\text{SiO}_{2(s)} + 3\text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)} + 2\text{CO}_{(g)}\)
\(\Delta_r H^\circ = (+911) + (-221) + (-65.3)\)
\(\Delta_r H^\circ = 911 - 286.3 = +624.7 \text{ kJ}\)
1. Reverse eq (i): \(\text{SiO}_{2(s)} \rightarrow \text{Si}_{(s)} + \text{O}_{2(g)}\) ; \(\Delta H = +911 \text{ kJ}\)
2. Keep eq (ii) as is: \(2\text{C}_{(graphite)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{(g)}\) ; \(\Delta H = -221 \text{ kJ}\)
3. Keep eq (iii) as is: \(\text{Si}_{(s)} + \text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)}\) ; \(\Delta H = -65.3 \text{ kJ}\)
Add these three modified equations:
\((\text{SiO}_2) + (2\text{C} + \text{O}_2) + (\text{Si} + \text{C}) \rightarrow (\text{Si} + \text{O}_2) + (2\text{CO}) + (\text{SiC})\)
Cancelling common terms (\(\text{Si}, \text{O}_2\)):
\(\text{SiO}_{2(s)} + 3\text{C}_{(graphite)} \rightarrow \text{SiC}_{(s)} + 2\text{CO}_{(g)}\)
\(\Delta_r H^\circ = (+911) + (-221) + (-65.3)\)
\(\Delta_r H^\circ = 911 - 286.3 = +624.7 \text{ kJ}\)
Q.16. Write a note on Hofmann bromamide degradation. Convert benzene diazonium chloride into benzene.
Answer:
Hofmann Bromamide Degradation:
This reaction is used to prepare primary amines from amides. The amide is treated with bromine and aqueous or ethanolic solution of NaOH or KOH. The amine formed has one carbon atom less than the original amide.
Benzene diazonium chloride is reduced to benzene by treating it with hypophosphorous acid (H3PO2) or ethanol.
This reaction is used to prepare primary amines from amides. The amide is treated with bromine and aqueous or ethanolic solution of NaOH or KOH. The amine formed has one carbon atom less than the original amide.
\( \text{R-CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{R-NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O} \)
Conversion of Benzene Diazonium Chloride to Benzene:
Benzene diazonium chloride is reduced to benzene by treating it with hypophosphorous acid (H3PO2) or ethanol.
\( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{H}_3\text{PO}_3 + \text{HCl} \)
Q.17. Write any three advantages and disadvantages of nanoparticles and nanotechnology.
Answer:
Advantages:
- Revolutionized electronics (faster, smaller devices).
- Advances in medicine (targeted drug delivery, cancer treatment).
- Development of self-cleaning surfaces and stronger, lighter materials.
- Potential health hazards (lung damage similar to asbestos) if inhaled.
- Environmental pollution (nanopollutants are hard to remove).
- High cost of production and lack of knowledge regarding long-term effects.
Q.18. Write molecular formula and structure of: (i) Sulphuric acid (ii) Peroxy monosulphuric acid (iii) Thiosulphuric acid
Answer:
(i) Sulphuric Acid: \( \text{H}_2\text{SO}_4 \)
Structure: S is central atom bonded to two -OH groups and double bonded to two Oxygen atoms. (Tetrahedral geometry).
(ii) Peroxy monosulphuric acid (Caro's Acid): \( \text{H}_2\text{SO}_5 \)
Structure: Contains a peroxy linkage (-O-O-). S=O (two), S-OH (one), S-O-OH (one).
(iii) Thiosulphuric acid: \( \text{H}_2\text{S}_2\text{O}_3 \)
Structure: Similar to sulphuric acid but one double bonded Oxygen is replaced by Sulphur (S=S).
Structure: S is central atom bonded to two -OH groups and double bonded to two Oxygen atoms. (Tetrahedral geometry).
(ii) Peroxy monosulphuric acid (Caro's Acid): \( \text{H}_2\text{SO}_5 \)
Structure: Contains a peroxy linkage (-O-O-). S=O (two), S-OH (one), S-O-OH (one).
(iii) Thiosulphuric acid: \( \text{H}_2\text{S}_2\text{O}_3 \)
Structure: Similar to sulphuric acid but one double bonded Oxygen is replaced by Sulphur (S=S).
Q.19. Explain optical activity of 2-chlorobutane.
Answer:
2-chlorobutane (\(\text{CH}_3-\text{CH(Cl)}-\text{CH}_2\text{CH}_3\)) contains an asymmetric (chiral) carbon atom, which is attached to four different groups: -H, -Cl, -\(\text{CH}_3\), and -\(\text{C}_2\text{H}_5\).
Due to the presence of the chiral center and the lack of a plane of symmetry, the molecule is non-superimposable on its mirror image.
Therefore, it exists in two enantiomeric forms (dextro and levo) which rotate plane-polarized light in opposite directions, making it optically active.
Due to the presence of the chiral center and the lack of a plane of symmetry, the molecule is non-superimposable on its mirror image.
Therefore, it exists in two enantiomeric forms (dextro and levo) which rotate plane-polarized light in opposite directions, making it optically active.
Q.20. Write different oxidation states of manganese. Why +2 oxidation state of manganese is more stable?
Answer:
Oxidation states: Manganese (Mn, Z=25) exhibits oxidation states from +2 to +7 (+2, +3, +4, +5, +6, +7).
Stability of +2 state: The electronic configuration of Mn is \([\text{Ar}] 3d^5 4s^2\). By losing 2 electrons from the 4s orbital, it forms \( \text{Mn}^{2+} \) which has the configuration \([\text{Ar}] 3d^5\). The 3d orbital is exactly half-filled, which provides extra stability due to symmetry and high exchange energy. Hence, the +2 state is very stable.
Stability of +2 state: The electronic configuration of Mn is \([\text{Ar}] 3d^5 4s^2\). By losing 2 electrons from the 4s orbital, it forms \( \text{Mn}^{2+} \) which has the configuration \([\text{Ar}] 3d^5\). The 3d orbital is exactly half-filled, which provides extra stability due to symmetry and high exchange energy. Hence, the +2 state is very stable.
Q.21. Prepare the following by using methyl magnesium iodide: (i) Ethanol (ii) Propan-2-ol (iii) 2-methylpropan-2-ol
Answer:
Methyl magnesium iodide is \( \text{CH}_3\text{MgI} \).
(i) Ethanol: React with Methanal (Formaldehyde) followed by hydrolysis.
(i) Ethanol: React with Methanal (Formaldehyde) followed by hydrolysis.
\(\text{HCHO} + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{OH}\)
(ii) Propan-2-ol: React with Ethanal (Acetaldehyde) followed by hydrolysis.
\(\text{CH}_3\text{CHO} + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{CH(OH)}-\text{CH}_3\)
(iii) 2-methylpropan-2-ol: React with Propanone (Acetone) followed by hydrolysis.
\(\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgI} \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_3\text{C-OH}\)
Q.22. Define: Ebullioscopic constant. Derive the relation between freezing point depression and molar mass of solute.
Answer:
Definition: Ebullioscopic constant (Kb) is defined as the elevation in boiling point produced when 1 mole of a non-volatile solute is dissolved in 1 kg (1000 g) of solvent.
Derivation (Freezing Point Depression):
The depression in freezing point (\(\Delta T_f\)) is directly proportional to the molality (m) of the solution.
\( \Delta T_f \propto m \)
\( \Delta T_f = K_f \cdot m \) ...(1) (where \(K_f\) is Cryoscopic constant)
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W_2/M_2}{W_1/1000} = \frac{W_2 \times 1000}{M_2 \times W_1} \)
Substituting m in eq (1):
\( \Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1} \)
Rearranging for molar mass \( M_2 \):
\( M_2 = \frac{K_f \times W_2 \times 1000}{\Delta T_f \times W_1} \)
Derivation (Freezing Point Depression):
The depression in freezing point (\(\Delta T_f\)) is directly proportional to the molality (m) of the solution.
\( \Delta T_f \propto m \)
\( \Delta T_f = K_f \cdot m \) ...(1) (where \(K_f\) is Cryoscopic constant)
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W_2/M_2}{W_1/1000} = \frac{W_2 \times 1000}{M_2 \times W_1} \)
Substituting m in eq (1):
\( \Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1} \)
Rearranging for molar mass \( M_2 \):
\( M_2 = \frac{K_f \times W_2 \times 1000}{\Delta T_f \times W_1} \)
Q.23. Define: Buffer solution. Write any four applications of buffer solution.
Answer:
Definition: A buffer solution is a solution which resists drastic change in its pH when a small amount of strong acid or strong base is added to it or upon dilution.
Applications:
Applications:
- Biological Systems: Blood pH is maintained at ~7.36-7.42 by bicarbonate buffer system.
- Agriculture: Soil pH is maintained for specific crop growth using buffers (carbonates, phosphates).
- Industry: Used in paper, dye, ink, and paint industries to maintain specific pH during manufacturing.
- Medicine: Used in the preparation of injections (e.g., Penicillin preparations are stabilized by buffers).
Q.24. An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell?
Answer:
Formula: \( \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \)
Given:
\( M = 27 \text{ g/mol} \)
\( a = 405 \text{ pm} = 4.05 \times 10^{-8} \text{ cm} \)
\( \rho = 2.7 \text{ g/cm}^3 \)
\( N_A = 6.022 \times 10^{23} \)
Step 1: Calculate \( a^3 \)
\( a^3 = (4.05 \times 10^{-8})^3 \approx 66.4 \times 10^{-24} \text{ cm}^3 \)
Step 2: Solve for Z
\( Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \)
\( Z = \frac{2.7 \times (66.4 \times 10^{-24}) \times (6.022 \times 10^{23})}{27} \)
\( Z = \frac{2.7 \times 66.4 \times 6.022 \times 0.1}{27} \)
\( Z = \frac{107.96}{27} \approx 3.99 \approx 4 \)
Since Z = 4, the cubic unit cell is Face Centered Cubic (FCC) or ccp.
Given:
\( M = 27 \text{ g/mol} \)
\( a = 405 \text{ pm} = 4.05 \times 10^{-8} \text{ cm} \)
\( \rho = 2.7 \text{ g/cm}^3 \)
\( N_A = 6.022 \times 10^{23} \)
Step 1: Calculate \( a^3 \)
\( a^3 = (4.05 \times 10^{-8})^3 \approx 66.4 \times 10^{-24} \text{ cm}^3 \)
Step 2: Solve for Z
\( Z = \frac{\rho \cdot a^3 \cdot N_A}{M} \)
\( Z = \frac{2.7 \times (66.4 \times 10^{-24}) \times (6.022 \times 10^{23})}{27} \)
\( Z = \frac{2.7 \times 66.4 \times 6.022 \times 0.1}{27} \)
\( Z = \frac{107.96}{27} \approx 3.99 \approx 4 \)
Since Z = 4, the cubic unit cell is Face Centered Cubic (FCC) or ccp.
Q.25. On the basis of valence bond theory explain the nature of bonding in [Ni(Cl)4]2- complex ion.
Answer:
1. Oxidation state: Ni is in +2 oxidation state. Configuration: \([\text{Ar}] 3d^8 4s^0\).
2. Ligand nature: Cl- is a weak field ligand, so it does not cause pairing of electrons in 3d orbitals.
3. Hybridization: To accommodate 4 ligands, Ni2+ uses one 4s and three 4p orbitals to undergo sp3 hybridization.
4. Geometry: sp3 hybridization corresponds to Tetrahedral geometry.
5. Magnetic property: There are 2 unpaired electrons in 3d orbitals, so the complex is Paramagnetic.
2. Ligand nature: Cl- is a weak field ligand, so it does not cause pairing of electrons in 3d orbitals.
3. Hybridization: To accommodate 4 ligands, Ni2+ uses one 4s and three 4p orbitals to undergo sp3 hybridization.
4. Geometry: sp3 hybridization corresponds to Tetrahedral geometry.
5. Magnetic property: There are 2 unpaired electrons in 3d orbitals, so the complex is Paramagnetic.
Q.26. Convert: (i) Acetic acid to acetamide (ii) Acetyl chloride to acetic anhydride (iii) Sodium acetate to methane
Answer:
(i) Acetic acid to Acetamide:
Reaction with ammonia followed by heating.
\( \text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{COONH}_4 \xrightarrow{\Delta, -\text{H}_2\text{O}} \text{CH}_3\text{CONH}_2 \)
(ii) Acetyl chloride to Acetic anhydride:
Reaction with sodium acetate.
\( \text{CH}_3\text{COCl} + \text{CH}_3\text{COONa} \rightarrow (\text{CH}_3\text{CO})_2\text{O} + \text{NaCl} \)
(iii) Sodium acetate to Methane:
Decarboxylation using soda lime (NaOH + CaO) and heat.
\( \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO, }\Delta} \text{CH}_4 + \text{Na}_2\text{CO}_3 \)
SECTION ‒ D
[12 Marks]
(Attempt any THREE of the following questions)
Q.27. Define isomorphism. Write Arrhenius equation. Derive an expression to determine activation energy for two different temperatures T1 and T2.
Answer:
Isomorphism: The phenomenon where two or more different substances exist in the same crystalline structure is called isomorphism (e.g., NaF and MgO).
Arrhenius Equation: \( k = A \cdot e^{-E_a/RT} \)
(Where k = rate constant, A = frequency factor, Ea = activation energy, R = gas constant, T = temperature).
Derivation: Taking natural log on both sides: \( \ln k = \ln A - \frac{E_a}{RT} \)
At temperature T1: \( \ln k_1 = \ln A - \frac{E_a}{RT_1} \) ...(1)
At temperature T2: \( \ln k_2 = \ln A - \frac{E_a}{RT_2} \) ...(2)
Subtract eq (1) from eq (2): \( \ln k_2 - \ln k_1 = (-\frac{E_a}{RT_2}) - (-\frac{E_a}{RT_1}) \)
\( \ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2}) \)
Or in log base 10: \( \log_{10}(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} [\frac{T_2 - T_1}{T_1 T_2}] \)
Arrhenius Equation: \( k = A \cdot e^{-E_a/RT} \)
(Where k = rate constant, A = frequency factor, Ea = activation energy, R = gas constant, T = temperature).
Derivation: Taking natural log on both sides: \( \ln k = \ln A - \frac{E_a}{RT} \)
At temperature T1: \( \ln k_1 = \ln A - \frac{E_a}{RT_1} \) ...(1)
At temperature T2: \( \ln k_2 = \ln A - \frac{E_a}{RT_2} \) ...(2)
Subtract eq (1) from eq (2): \( \ln k_2 - \ln k_1 = (-\frac{E_a}{RT_2}) - (-\frac{E_a}{RT_1}) \)
\( \ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{1}{T_1} - \frac{1}{T_2}) \)
Or in log base 10: \( \log_{10}(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} [\frac{T_2 - T_1}{T_1 T_2}] \)
Q.28. What are interhalogen compounds? Write any two general characteristics of interhalogen compounds. Draw the Fischer projection formula for \(\alpha-D-(+)\) glucose. Write reaction involved in the formation of Teflon.
Answer:
Interhalogen Compounds: Compounds formed by the combination of two or more different halogen atoms (e.g., ICl, BrF3).
Characteristics: 1. They are essentially covalent compounds.
2. They are generally more reactive than pure halogens (except Fluorine) because the X-X' bond is weaker than the X-X bond.
Teflon Formation: Polymerization of Tetrafluoroethylene.
(Structure description: C1 has OH on right, C2 OH on right, C3 OH on left, C4 OH on right, C5 OH on right, C6 is CH2OH).
Characteristics: 1. They are essentially covalent compounds.
2. They are generally more reactive than pure halogens (except Fluorine) because the X-X' bond is weaker than the X-X bond.
Teflon Formation: Polymerization of Tetrafluoroethylene.
\( n(\text{CF}_2=\text{CF}_2) \xrightarrow{\text{Catalyst, High P}} -(\text{CF}_2-\text{CF}_2)_n- \)
Fischer Projection of \(\alpha-D-(+)\) Glucose:(Structure description: C1 has OH on right, C2 OH on right, C3 OH on left, C4 OH on right, C5 OH on right, C6 is CH2OH).
Q.29. Describe the construction and working of Standard Hydrogen Electrode. Write any two difficulties in setting SHE.
Answer:
Construction:
It consists of a platinum plate coated with platinum black (to increase surface area). This plate is suspended in a solution of 1 M H+ ions (1 M HCl). Pure dry hydrogen gas is bubbled through the solution at 1 bar pressure at 298 K. The glass tube has a side arm for H2 inlet and exit.
Working: It acts as a reference electrode with potential defined as 0.0 V.
Working: It acts as a reference electrode with potential defined as 0.0 V.
- If it acts as Anode (Oxidation): \( \text{H}_{2(g)} \rightarrow 2\text{H}^+_{(aq)} + 2e^- \)
- If it acts as Cathode (Reduction): \( 2\text{H}^+_{(aq)} + 2e^- \rightarrow \text{H}_{2(g)} \)
Q.30. Write any two statements of first law of thermodynamics. For a certain reaction \(\Delta H^\circ\) is -224 kJ and \(\Delta S^\circ\) is -153 Jk-1. At what temperature the change over from spontaneous to nonspontaneous will occur?
Answer:
Statements of First Law:
1. Energy can neither be created nor destroyed, but can be converted from one form to another.
2. The total energy of the universe remains constant.
Calculation: Given: \( \Delta H = -224 \text{ kJ} = -224000 \text{ J} \), \( \Delta S = -153 \text{ J K}^{-1} \). The changeover occurs at equilibrium where \( \Delta G = 0 \). \( \Delta G = \Delta H - T\Delta S \)
\( 0 = \Delta H - T\Delta S \)
\( T = \frac{\Delta H}{\Delta S} \)
\( T = \frac{-224000}{-153} \)
\( T \approx 1464 \text{ K} \)
Calculation: Given: \( \Delta H = -224 \text{ kJ} = -224000 \text{ J} \), \( \Delta S = -153 \text{ J K}^{-1} \). The changeover occurs at equilibrium where \( \Delta G = 0 \). \( \Delta G = \Delta H - T\Delta S \)
\( 0 = \Delta H - T\Delta S \)
\( T = \frac{\Delta H}{\Delta S} \)
\( T = \frac{-224000}{-153} \)
\( T \approx 1464 \text{ K} \)
Q.31. Define: (i) Gangue (ii) Ionization isomer (iii) Aromatic ketones. Write the use and environmental effect of methylene chloride.
Answer:
Definitions:
(i) Gangue: The earthy impurities (sand, clay, rock, silica, etc.) associated with the mineral/ore in the earth's crust.
(ii) Ionization Isomer: Isomers that have the same molecular formula but give different ions in solution (exchange of ligands between coordination sphere and ionization sphere).
(iii) Aromatic Ketones: Ketones in which the carbonyl carbon is attached to at least one aryl (aromatic) group (e.g., Acetophenone).
Methylene Chloride (Dichloromethane, CH2Cl2):
Use: Commonly used as a solvent in paint removers, propellant in aerosols, and process solvent in drug manufacturing.
Environmental Effect: It harms the human central nervous system. Direct contact causes skin burns. It is a potential occupational carcinogen. High levels in air cause dizziness and nausea.
(i) Gangue: The earthy impurities (sand, clay, rock, silica, etc.) associated with the mineral/ore in the earth's crust.
(ii) Ionization Isomer: Isomers that have the same molecular formula but give different ions in solution (exchange of ligands between coordination sphere and ionization sphere).
(iii) Aromatic Ketones: Ketones in which the carbonyl carbon is attached to at least one aryl (aromatic) group (e.g., Acetophenone).
Methylene Chloride (Dichloromethane, CH2Cl2):
Use: Commonly used as a solvent in paint removers, propellant in aerosols, and process solvent in drug manufacturing.
Environmental Effect: It harms the human central nervous system. Direct contact causes skin burns. It is a potential occupational carcinogen. High levels in air cause dizziness and nausea.
End of Question Paper Solution
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