Maharashtra Board Class 12 Chemistry 2023 Question Paper with Answer Key And Solutions PDF

In a body-centered cubic (BCC) lattice, the body diagonal equals √3a, where a is
the edge length. The body diagonal equals 4r (two atomic radii from corner to body center).
Thus:
4r = √3a ⇒ r =
√3
4 a.

pH = 11.2, so pOH = 14 − 11.2 = 2.8.
[OH−] = 10−pOH = 10−2.8 = 10−2 × 10−0.8 ≈ 10−2 × 0.1585 = 1.585 × 10−3 mol dm−3.

For a first-order gas phase reaction A → 2B, initial pressure Pi, pressure of A at
time t = PA, total pressure P = PA + PB . Since PB = 2(Pi − PA), total pressure
P = PA + 2Pi − 2PA = 2Pi − PA. Thus, PA = 2Pi − P . Integrated rate law:
k = 2.303
t log [A]0
[A] = 2.303
t log Pi
2Pi − P .

Mn2+ has electron configuration [Ar] 3d5. Number of unpaired electrons n = 5.
Spin-only magnetic moment:
μ = pn(n + 2) = p5(5 + 2) = √35 ≈ 5.916 BM.

Tetraamminedibromoplatinum(IV) indicates Pt4+ with 4 NH3 (ammine) and 2 Br−
(bromo) ligands in the coordination sphere, and bromide counter ions balance the +2 charge
of the complex ion. Thus, [PtBr2(NH3)4]2+ with 2 Br− gives [PtBr2(NH3)4]Br2.

An allylic halide has a halogen atom on a carbon adjacent to a C=C double bond.
Option (D) CH2 CH CH2 X shows the halogen on the CH2 next to the double
bond, making it allylic.

LiAlH4 reduces aldehydes to primary alcohols without affecting the C=C double
bond. The aldehyde CHO becomes CH2OH, resulting in
CH3 CH CH CH2 CH2 OH.

2,3-dimethyl but-2-ene is CH32C CCH32 . Ozonolysis cleaves the C=C bond, each
carbon forming a carbonyl. Both carbons have two methyl groups, yielding two molecules of
acetone (CH32C O).

Natural rubber is a polymer of isoprene (2-methyl-1,3-butadiene), with repeating units forming polyisoprene. Quick Tip: For chemistry MCQs, focus on key concepts like lattice geometry, pH calculations, reaction mechanisms, and molecular structures to select the correct option efficiently.

TEM is commonly used to determine the geometry, size, and shape of
nanoparticles by passing electrons through a thin sample to produce high-resolution images
of its structure.

Acetamide (CH3 C ONH2) is reduced by LiAlH4 in dry ether, converting
the carbonyl group to a methylene group, yielding ethylamine (CH3 CH2 NH2).

Chlorobenzene (C6H5 Cl) reacts with sodium in dry ether via the Wurtz-Fittig
reaction, coupling two phenyl groups to form biphenyl (C6H5 C6H5).

Cryolite is a naturally occurring mineral used in aluminum extraction, with the
chemical formula Na3AlF6 (sodium hexafluoroaluminate).

Cisplatin (cis-[PtCl2(NH3)2]) is a platinum-based coordination complex widely
used as a chemotherapeutic agent for treating various cancers.

For a process to be spontaneous at constant temperature and pressure, the Gibbs
free energy change must be negative: ∆G = ∆H − T ∆S < 0.

Molar conductivity \( \Lambda_m = \frac{\kappa \cdot 1000}{C} \), where \( \kappa = 0.01 \, \Omega^{-1} cm^{-1} \), \( C = 0.5 \, mol dm^{-3} = 0.5 \, mol L^{-1} = 500 \, mol m^{-3} \). Convert units: \[ \Lambda_m = \frac{0.01 \cdot 1000}{0.5} = \frac{10}{0.5} = 20 \, \Omega^{-1} cm^2 mol^{-1}. \] Quick Tip: For short-answer chemistry questions, ensure precise chemical formulas, unit derivations, and awareness of standard techniques or compounds used in specific contexts.

Lanthanides and actinides are f-block elements, but they differ as follows:

1. **Electronic Configuration:**

- Lanthanides: Electrons fill the 4f orbitals ([Xe] 4f\(^{1-14}\) 5d\(^{0-1}\) 6s\(^2\)).

- Actinides: Electrons fill the 5f orbitals ([Rn] 5f\(^{1-14}\) 6d\(^{0-1}\) 7s\(^2\)).

2. **Radioactivity:**

- Lanthanides: Mostly non-radioactive (except promethium).

- Actinides: All are radioactive due to unstable nuclei.

3. **Oxidation States:**

- Lanthanides: Predominantly +3, rarely +2 or +4.

- Actinides: Show variable oxidation states (e.g., +3 to +6 for uranium).

4. **Chemical Reactivity:**

- Lanthanides: Less reactive, similar to group 2 metals.

- Actinides: More reactive, form complexes easily due to higher charge density.



Quick Tip: Focus on orbital filling and radioactivity to distinguish lanthanides from actinides effectively.

Using Raoult’s law for a non-volatile solute: \( P = P_0 x_A \), where \( P_0 = 640 \, mmHg \), \( P = 600 \, mmHg \), \( x_A \) is mole fraction of solvent (benzene).
\[ x_A = \frac{P}{P_0} = \frac{600}{640} = 0.9375. \]
Mole fraction of solute \( x_B = 1 - x_A = 1 - 0.9375 = 0.0625 \).

Answer: Mole fraction of solute = 0.0625.


Quick Tip: Raoult’s law applies to non-volatile solutes; mole fractions sum to 1.

Definition: Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances, promoting sustainability and environmental safety.

Advantages of Nanoparticles and Nanotechnology:

1. **Enhanced Reactivity:** Nanoparticles have a high surface area-to-volume ratio, increasing efficiency in catalysis and drug delivery.

2. **Targeted Applications:** Nanotechnology enables precise targeting (e.g., in medicine for cancer treatment), reducing side effects and improving efficacy. Quick Tip: Green chemistry emphasizes eco-friendly processes; nanoparticles excel due to size-dependent properties.

i. Substitutional Impurity Defect: Occurs when an atom in a crystal lattice is replaced by a different type of atom. For example, in a silicon crystal, a phosphorus atom may substitute a silicon atom, altering electrical properties (used in doping semiconductors).

ii. Interstitial Impurity Defect: Occurs when a foreign atom occupies an interstitial (empty) site in the lattice, not replacing any host atom. For example, carbon atoms in iron form interstitial defects, increasing hardness in steel. Quick Tip: Substitutional defects involve atom replacement, while interstitial defects involve atoms in lattice voids.

i. Chlorobenzene with fuming H\(_2\)SO\(_4\):
\[ \chemfig{C_6H_5-Cl} + \chemfig{H_2SO_4} \xrightarrow{heat} \chemfig{C_6H_5-SO_3H} + \chemfig{HCl} \]
Fuming H\(_2\)SO\(_4\) (containing SO\(_3\)) causes sulfonation, forming benzenesulfonic acid.

ii. Ethyl bromide with silver acetate:
\[ \chemfig{CH_3-CH_2-Br} + \chemfig{CH_3COOAg} \xrightarrow{heat} \chemfig{CH_3-CH_2-OCOCH_3} + \chemfig{AgBr} \]
This is a nucleophilic substitution reaction, forming ethyl acetate and silver bromide precipitate. Quick Tip: Aromatic compounds undergo electrophilic substitution with strong reagents; alkyl halides react with silver salts to form esters.

Definition: An acidic buffer solution is a mixture of a weak acid and its conjugate base (or salt), maintaining a relatively constant pH (below 7) by resisting changes upon addition of small amounts of acid or base. Example: CH\(_3\)COOH and CH\(_3\)COONa.

Relationship for PbI\(_2\): For PbI\(_2 \rightleftharpoons Pb^{2+} + 2I^-\), let solubility = \( S \, mol L^{-1} \). Then, \( [Pb^{2+}] = S \), \( [I^-] = 2S \). Solubility product: \[ K_{sp} = [Pb^{2+}][I^-]^2 = S \cdot (2S)^2 = 4S^3. \]



Quick Tip: Acidic buffers use weak acid-conjugate base pairs; for sparingly soluble salts, express \( K_{sp} \) in terms of solubility S.

i. Chloroform and Caustic Potash: Ethyl amine (\( \chemfig{CH_3-CH_2-NH_2} \)) undergoes the carbylamine reaction: \[ \chemfig{CH_3-CH_2-NH_2} + \chemfig{CHCl_3} + 3KOH \xrightarrow{heat} \chemfig{CH_3-CH_2-NC} + 3KCl + 3H_2O. \]
Product: Ethyl isocyanide (foul-smelling).

ii. Nitrous Acid: Ethyl amine reacts with HNO\(_2\) (from NaNO\(_2\) + HCl) to form an alcohol via diazonium ion decomposition: \[ \chemfig{CH_3-CH_2-NH_2} + \chemfig{HNO_2} \to \chemfig{CH_3-CH_2-OH} + N_2 + H_2O. \]
Product: Ethanol. Quick Tip: Primary amines give isocyanides with chloroform/KOH and alcohols with nitrous acid due to diazonium instability.

Cell: \( Cd | Cd^{2+} || Sn^{2+} | Sn \).

Anode: \( Cd \to Cd^{2+} + 2e^- \), cathode: \( Sn^{2+} + 2e^- \to Sn \).
\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.136 - (-0.403) = 0.267 \, V \).

Number of electrons \( n = 2 \).

Gibbs energy: \( \Delta G^\circ = -n F E^\circ_{cell} \), \( F = 96485 \, C mol^{-1} \), T = 25°C = 298 K (not needed for standard conditions).
\[ \Delta G^\circ = -2 \times 96485 \times 0.267 \approx -51529.59 \, J mol^{-1} = -51.53 \, kJ mol^{-1}. \]



Quick Tip: Use \( \Delta G^\circ = -n F E^\circ \) for electrochemical cells; positive \( E^\circ \) indicates spontaneity (\( \Delta G^\circ < 0 \)).

i. Preparation of Glucose from Sucrose: Sucrose (\( C_{12}H_{22}O_{11} \)) is hydrolyzed in the presence of dilute acid (e.g., HCl) or enzyme (invertase) to produce glucose and fructose: \[ \chemfig{C_{12}H_{22}O_{11}} + \chemfig{H_2O} \xrightarrow{H^+ or invertase} \chemfig{C_6H_{12}O_6} (glucose) + \chemfig{C_6H_{12}O_6} (fructose). \]
ii. Structure of D-Ribose: D-Ribose is a pentose sugar with the following open-chain structure: \[ \chemfig{HO-CH_2-CH(OH)-CH(OH)-CH(OH)-CH=O} \]
In its cyclic form (furanose), it is: \[ \chemfig{HO-CH_2-[:30](-[:90]OH)-[:-30](-[:90]OH)-[:-150](-[:90]OH)-[:150]O-[:90]} \] Quick Tip: Sucrose hydrolysis yields equal amounts of glucose and fructose; D-ribose is commonly depicted in its furanose ring form in biochemical contexts.

Definition: An extensive property is a physical property of a system that depends on the amount or extent of the substance, such as mass, volume, or total energy.

Calculation of Work Done: For an ideal gas expanding in a vacuum (free expansion), external pressure \( P_{ext} = 0 \). Work done \( W = -P_{ext} \Delta V \).
\[ \Delta V = 20 \, dm^3 - 10 \, dm^3 = 10 \, dm^3 = 0.01 \, m^3. \] \[ W = -0 \cdot 0.01 = 0 \, J. \]
Answer: Work done = 0 J. Quick Tip: In free expansion (in vacuum), no work is done since \( P_{ext} = 0 \); extensive properties scale with system size.

Nylon 6,6 is a polyamide formed by condensation polymerization of hexamethylenediamine (\( \chemfig{H_2N-(CH_2)_6-NH_2} \)) and adipic acid (\( \chemfig{HOOC-(CH_2)_4-COOH} \)).
\[ n\chemfig{H_2N-(CH_2)_6-NH_2} + n \chemfig{HOOC-(CH_2)_4-COOH} \to \chemfig{[-NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-]_n} + 2n \chemfig{H_2O}. \]
The reaction forms amide bonds, releasing water molecules. Quick Tip: Nylon 6,6 involves condensation of a diamine and a dicarboxylic acid, forming repeating amide linkages.

i. Chloric Acid (\( HClO_3 \)):
\[ \chemfig{Cl(=[:90]O)(-[:30]OH)(-[:270]O)} \]
Chlorine is bonded to one hydroxyl group and two double-bonded oxygens.

ii. Peroxydisulphuric Acid (\( H_2S_2O_8 \)):
\[ \chemfig{HO-S(=[:90]O)(=[:270]O)-O-O-S(=[:90]O)(=[:270]O)-OH} \]
Contains a peroxide (\( -O-O- \)) bond linking two sulfur atoms, each with two double-bonded oxygens and one hydroxyl group. Quick Tip: Use Lewis structures to confirm oxidation states and bonding in oxoacids; peroxydisulphuric acid has a characteristic peroxide linkage.

Definition: Osmosis is the spontaneous movement of solvent molecules through a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration, tending to equalize concentrations.

Determination of Molar Mass:

The elevation of boiling point (\( \Delta T_b \)) for a non-volatile solute is given by: \[ \Delta T_b = K_b \cdot m, \]
where \( K_b \) is the molal boiling point elevation constant, and \( m \) is molality (\( mol kg^{-1} \)).

Step 1: Measure the boiling point of the pure solvent (\( T_0 \)) and the solution (\( T \)). Calculate \( \Delta T_b = T - T_0 \).

Step 2: Molality \( m = \frac{n_B}{w_A} \), where \( n_B = \frac{w_B}{M_B} \) (moles of solute, \( w_B \) = mass of solute, \( M_B \) = molar mass of solute), and \( w_A \) = mass of solvent in kg.
\[ \Delta T_b = K_b \cdot \frac{w_B}{M_B \cdot w_A}. \] \[ M_B = \frac{K_b \cdot w_B}{\Delta T_b \cdot w_A}. \]
Step 3: Use known \( K_b \), measured \( \Delta T_b \), \( w_B \), and \( w_A \) to calculate \( M_B \). Quick Tip: Use precise temperature measurements and known \( K_b \) values for accurate molar mass determination.

i. Ethyl Alcohol to Ethyl Acetate:
\[ \chemfig{CH_3-CH_2-OH} + \chemfig{CH_3-COOH} \xrightarrow{H_2SO_4, heat} \chemfig{CH_3-C(=O)-O-CH_2-CH_3} + \chemfig{H_2O}. \]
Ethyl alcohol reacts with acetic acid via esterification to form ethyl acetate.

ii. Phenol to Benzene:
\[ \chemfig{C_6H_5-OH} \xrightarrow{Zn dust, heat} \chemfig{C_6H_6} + \chemfig{ZnO}. \]
Phenol is reduced by heating with zinc dust, removing the hydroxyl group to form benzene.

iii. Diethyl Ether to Ethyl Chloride:
\[ \chemfig{CH_3-CH_2-O-CH_2-CH_3} + \chemfig{HCl} \xrightarrow{ZnCl_2, heat} 2 \chemfig{CH_3-CH_2-Cl} + \chemfig{H_2O}. \]
Diethyl ether reacts with concentrated HCl in the presence of ZnCl\(_2\) to form ethyl chloride. Quick Tip: Choose reagents specific to functional group transformations for efficient organic conversions.

For a weak monobasic acid \( HA \rightleftharpoons H^+ + A^- \), the dissociation constant \( K_a = \frac{\alpha^2 C}{1 - \alpha} \), where \( \alpha \) is the degree of dissociation, and \( C \) is concentration.

Step 1: For 0.05 M, \( \alpha = 0.1 \): \[ K_a = \frac{(0.1)^2 \cdot 0.05}{1 - 0.1} = \frac{0.01 \cdot 0.05}{0.9} = \frac{0.0005}{0.9} \approx 5.5556 \times 10^{-4}. \]
Step 2: For 0.15 M, use \( K_a \): \[ 5.5556 \times 10^{-4} = \frac{\alpha^2 \cdot 0.15}{1 - \alpha}. \]
Since \( \alpha \) is small, approximate \( 1 - \alpha \approx 1 \): \[ \alpha^2 \cdot 0.15 \approx 5.5556 \times 10^{-4}, \quad \alpha^2 \approx \frac{5.5556 \times 10^{-4}}{0.15} \approx 3.7037 \times 10^{-3}. \] \[ \alpha \approx \sqrt{3.7037 \times 10^{-3}} \approx 0.06086. \]
Percent dissociation = \( 0.06086 \times 100 \approx 6.09% \). Quick Tip: For weak electrolytes, dilution increases dissociation but percent dissociation decreases with higher concentration due to \( K_a \) constancy.

Dehydrohalogenation of 2-Chlorobutane: Dehydrohalogenation involves the removal of HX from an alkyl halide using a strong base (e.g., alcoholic KOH) to form an alkene. For 2-chlorobutane (\( \chemfig{CH_3-CH(Cl)-CH_2-CH_3} \)): \[ \chemfig{CH_3-CH(Cl)-CH_2-CH_3} \xrightarrow{KOH (alc.), heat} \chemfig{CH_3-CH=CH-CH_3} + \chemfig{HCl}. \]
Major product (Saytzeff rule): 2-butene (more substituted alkene). Minor product: 1-butene (\( \chemfig{CH_2=CH-CH_2-CH_3} \)).

Use of CFC: Chlorofluorocarbons (CFCs) are used as refrigerants, propellants in aerosols, and solvents due to their stability and low toxicity.

Environmental Effect of CFC: CFCs deplete the ozone layer by releasing chlorine radicals upon UV exposure, catalyzing ozone (O\(_3\)) breakdown, leading to increased UV radiation reaching Earth. Quick Tip: Saytzeff rule favors the more substituted alkene; CFCs are phased out due to ozone depletion.

For isothermal reversible expansion, work done \( W = -n R T \ln \left( \frac{V_f}{V_i} \right) \).
\( n = 2000 \, mmol = 2 \, mol \), \( T = 300 \, K \), \( V_i = 20 \, L \), \( V_f = 30 \, L \), \( R = 8.314 \, J K^{-1} mol^{-1} \).
\[ W = -2 \cdot 8.314 \cdot 300 \cdot \ln \left( \frac{30}{20} \right) = -2 \cdot 8.314 \cdot 300 \cdot \ln (1.5). \] \[ \ln (1.5) \approx 0.405465, \quad W \approx -2 \cdot 8.314 \cdot 300 \cdot 0.405465 \approx -2022.77 \, J. \]
Answer: Work done \( \approx -2022.77 \, J \). Quick Tip: For isothermal reversible expansion, work depends on the volume ratio; negative sign indicates work done by the system.

Interstitial Compounds: These are compounds formed when small atoms (e.g., H, C, N) occupy interstitial sites in the lattice of transition metals (e.g., Fe, Ti). Example: TiC (titanium carbide), where carbon atoms fit in the metal lattice, enhancing hardness.

Classification of Alloys:

1. **Substitutional Alloys:** Atoms of one metal replace atoms of another in the lattice. Example: Brass (Cu and Zn).

2. **Interstitial Alloys:** Small atoms occupy interstitial sites in a metal lattice. Example: Steel (Fe with C).

3. **Intermetallic Alloys:** Specific stoichiometric compounds of metals. Example: Ni\(_3\)Al (used in jet engines). Quick Tip: Interstitial compounds enhance material properties; alloys are classified based on atomic arrangement.

Diagram:



Reactions: Anode: \( 2H_2 \to 4H^+ + 4e^- \), Cathode: \( O_2 + 4H^+ + 4e^- \to 2H_2O \).

Applications:

1. Power generation in spacecraft (e.g., Apollo missions).

2. Electric vehicle power sources for clean energy. Quick Tip: Fuel cells convert chemical energy directly to electrical energy; H\(_2\)-O\(_2\) cells produce water as a byproduct.

For [CoF\(_6\)]\(^{3-}\), Co is in +3 oxidation state (\( Co^{3+} \), d\(^6\)).

i. Hybridisation: F\(^-\) is a weak-field ligand, so high-spin complex. Co\(^{3+}\) uses 3d, 4s, and 4p orbitals for d\(^2\)sp\(^3\) hybridisation (outer orbital complex).

ii. Magnetic Properties: d\(^6\) high-spin: 4 unpaired electrons (\( t_{2g}^4 e_g^2 \)). Magnetic moment: \[ \mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 \, BM. \]
iii. Inner/Outer Complex: Uses 4s and 4p orbitals, so it’s an outer orbital complex (d\(^2\)sp\(^3\)).

iv. Geometry: Octahedral, as six F\(^-\) ligands coordinate around Co\(^{3+}\). Quick Tip: Weak-field ligands like F\(^-\) cause high-spin complexes; check ligand field strength for hybridisation.

Pseudo First Order Reaction: A reaction that is second order (or higher) but appears first order due to one reactant in large excess, keeping its concentration nearly constant. Example: Hydrolysis of ester in water (\( H_2O \) in excess).

Zero Order Integrated Rate Law: For a zero-order reaction \( A \to products \), rate = \( k \).
\[ -\frac{d[A]}{dt} = k. \]
Integrate: \( \int_{[A]_0}^{[A]} d[A] = -k \int_0^t dt \).
\[ [A] - [A]_0 = -kt \quad \Rightarrow \quad [A] = [A]_0 - kt. \] Quick Tip: Pseudo first order simplifies kinetics; zero-order reactions have constant rates, common in catalyzed reactions.

Aldol condensation involves the reaction of two aldehyde molecules (with \( \alpha \)-H) in the presence of a base to form a \( \beta \)-hydroxy aldehyde, which may dehydrate to an unsaturated aldehyde. For ethanal (\( \chemfig{CH_3-CHO} \)): \[ 2 \chemfig{CH_3-CHO} \xrightarrow{dil. NaOH} \chemfig{CH_3-CH(OH)-CH_2-CHO} \xrightarrow{-H_2O} \chemfig{CH_3-CH=CH-CHO}. \]
Step 1: Enolate ion forms from one ethanal, attacking the carbonyl of another, forming 3-hydroxybutanal (aldol).

Step 2: Dehydration yields crotonaldehyde (\( \chemfig{CH_3-CH=CH-CHO} \)). Quick Tip: Aldol condensation requires \( \alpha \)-hydrogen; dehydration yields conjugated products.

i. Atomicity: Oxygen exists as a diatomic gas (\( O_2 \)), whereas other group 16 elements (S, Se, Te) form polyatomic structures (e.g., S\(_8\)) due to stronger tendency for catenation.

ii. Magnetic Property: O\(_2\) is paramagnetic due to two unpaired electrons in its molecular orbitals (\( \pi^*_{2p} \)), while others like S\(_8\) are diamagnetic.

iii. Oxidation State: Oxygen commonly shows -2 (e.g., H\(_2\)O), rarely +2 (e.g., OF\(_2\)) due to high electronegativity. Others (S, Se, Te) show +4, +6 due to d-orbital availability. Quick Tip: Oxygen’s anomalies stem from its small size, high electronegativity, and lack of d-orbitals.

i. Acetic Acid to Acetic Anhydride:
\[ 2 \chemfig{CH_3-COOH} \xrightarrow{P_2O_5, heat} \chemfig{CH_3-C(=O)-O-C(=O)-CH_3} + \chemfig{H_2O}. \]
P\(_2\)O\(_5\) dehydrates two acetic acid molecules.

ii. Acetic Acid to Ethyl Alcohol:
\[ \chemfig{CH_3-COOH} \xrightarrow{i) LiAlH_4, ii) H_3O^+} \chemfig{CH_3-CH_2-OH}. \]
LiAlH\(_4\) reduces the carboxyl group to a primary alcohol.

Methylphenylamine:

IUPAC Name: N-Methylaniline.

Structure: \( \chemfig{C_6H_5-NH-CH_3} \). Quick Tip: Use strong reducing agents like LiAlH\(_4\) for carboxylic acid reductions; IUPAC names prioritize functional group hierarchy.

i. Time for First Order Reaction: For a first-order reaction, the rate law is \( \ln \frac{[A]_0}{[A]} = kt \).

For 90 percent completion, 10% of reactant remains: \( \frac{[A]}{[A]_0} = 0.1 \).
\[ \ln \frac{1}{0.1} = kt_{90%} \quad \Rightarrow \quad kt_{90%} = \ln 10 \approx 2.3026. \]
For 99.9% completion, 0.1% remains: \( \frac{[A]}{[A]_0} = 0.001 \).
\[ \ln \frac{1}{0.001} = kt_{99.9%} \quad \Rightarrow \quad kt_{99.9%} = \ln 1000 \approx 6.9078. \]
Ratio: \[ \frac{t_{99.9%}}{t_{90%}} = \frac{\ln 1000}{\ln 10} = \frac{3 \ln 10}{\ln 10} = 3. \]
Thus, \( t_{99.9%} = 3 t_{90%} \).

ii. Electronic Configuration of Gd (Z = 64): Gadolinium: [Xe] 4f\(^7\) 5d\(^1\) 6s\(^2\).

iii. Nanostructured Material in Car Tyres: Carbon black nanoparticles.

Solution: Carbon black nanoparticles are used in car tyres to enhance durability, improve traction, and increase resistance to wear. Quick Tip: For first-order reactions, use \( \ln \frac{[A]_0}{[A]} = kt \); nanomaterials like carbon black improve mechanical properties in applications.

i. Relationship between \( \Delta H \) and \( \Delta U \): Enthalpy \( H = U + PV \). For a gaseous reaction at constant T, \( \Delta H = \Delta U + \Delta (PV) \). Using ideal gas law, \( PV = nRT \), so: \[ \Delta H = \Delta U + \Delta (n_g RT) = \Delta U + \Delta n_g RT, \]
where \( \Delta n_g \) is the change in moles of gas, \( R = 8.314 \, J K^{-1} mol^{-1} \), T is temperature in K.

ii. Vulcanization: The process of heating natural rubber with sulfur to form cross-links between polymer chains, improving elasticity, strength, and durability.

iii. Peptide Bond: A covalent bond (\( -CO-NH- \)) formed between the carboxyl group of one amino acid and the amino group of another, with loss of water, linking amino acids in proteins. Quick Tip: Use \( \Delta n_g \) to relate \( \Delta H \) and \( \Delta U \); vulcanization enhances rubber properties; peptide bonds are amide linkages in proteins.

i. Density of Silver: For FCC, number of atoms per unit cell \( Z = 4 \). Edge length \( a = 400 \, pm = 4 \times 10^{-8} \, cm \), atomic mass \( M = 108 \, g mol^{-1} \), Avogadro’s number \( N_A = 6.022 \times 10^{23} \, mol^{-1} \).

Density \( \rho = \frac{Z \cdot M}{N_A \cdot a^3} \).
\[ a^3 = (4 \times 10^{-8})^3 = 6.4 \times 10^{-23} \, cm^3. \] \[ \rho = \frac{4 \cdot 108}{6.022 \times 10^{23} \cdot 6.4 \times 10^{-23}} \approx \frac{432}{3.85408 \times 10^1} \approx 11.21 \, g cm^{-3}. \]
ii. Haloform Reaction: Compounds with a methyl ketone group (\( \chemfig{CH_3-C(=O)-} \)) or acetaldehyde, when treated with halogens (e.g., I\(_2\)) and a base (e.g., NaOH), form haloforms (e.g., \( \chemfig{CHI_3} \)) and a carboxylate. Example: \[ \chemfig{CH_3-CO-CH_3} + 3I_2 + 4NaOH \to \chemfig{CHI_3} + \chemfig{CH_3COONa} + 3NaI + 3H_2O. \]
Used in iodoform test to detect methyl ketones or acetaldehyde. Quick Tip: FCC density uses \( Z = 4 \); haloform reaction is diagnostic for compounds with \( CH_3CO- \).

i. Dow Process for Phenol:
\[ \chemfig{C_6H_5-Cl} \xrightarrow{NaOH, 350°C, high pressure} \chemfig{C_6H_5-ONa} \xrightarrow{HCl} \chemfig{C_6H_5-OH} + \chemfig{NaCl}. \]
Chlorobenzene reacts with aqueous NaOH to form sodium phenoxide, which is acidified to yield phenol.

ii. Action of Bromine Water on Phenol: Phenol reacts with Br\(_2\) in water to form 2,4,6-tribromophenol (white precipitate): \[ \chemfig{C_6H_5-OH} + 3\chemfig{Br_2} \to \chemfig{C_6H_2(Br)_3-OH} + 3\chemfig{HBr}. \]
iii. Reason for Lower Ionisation Enthalpy in Group 16: Group 16 elements (e.g., O, S) have \( np^4 \) configuration, with paired electrons in one p-orbital, causing electron-electron repulsion, making it easier to remove an electron compared to group 15’s stable half-filled \( np^3 \) configuration.

iv. Uses of Dioxygen:

1. Respiration in living organisms.

2. Combustion and industrial processes (e.g., steel production). Quick Tip: Dow process uses high temperature/pressure; phenol’s high reactivity with Br\(_2\) is due to ring activation; group 16’s paired electrons lower ionisation energy.