CBSE Class 12 Chemistry: sample paper 2023-24

1. HCl is a strong electrolyte (completely ionised), so it gives far more ions than weak $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$.

2. Conductivity depends on the number of ions per unit volume, so it falls on dilution. The more concentrated strong acid, 0.1 M HCl, therefore conducts best.

   1 mark

Answer: (B) 0.1 M HCl

1. Chalcone ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}=\mathrm{CHCO}{\mathrm{C}}_6{\mathrm{H}}_5$ is formed by crossed aldol condensation.

2. ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$ (no α-H) reacts with the enolate of ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3$; the aldol then loses water to give chalcone. So A = benzaldehyde and B = acetophenone.

   1 mark

Answer: (C) Benzaldehyde and Acetophenone

1. Fat-soluble vitamins (A, D, E, K) are stored in the liver and adipose tissue. Water-soluble vitamins B and C are excreted in urine, so they are not stored.

   1 mark

Answer: (D) Vitamins A, D, E and K

1. In the iodoform reaction, $\mathrm{RCOC}{\mathrm{H}}_3\to \mathrm{RCOONa}+\mathrm{CH}{\mathrm{I}}_3$, so R = $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)-$.

2. The ketone is $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)\mathrm{COC}{\mathrm{H}}_3$, named 3-methylpent-3-en-2-one.

   1 mark

Answer: (A) 3-Methylpent-3-en-2-one

1. The C–Mg bond is covalent but highly polar, since carbon is more electronegative than Mg. Calling it non-polar is wrong, so statement B is incorrect.

   1 mark

Answer: (B)

1. Zn has no unpaired electrons, so its metallic bonding is weak and its enthalpy of atomisation is lowest: (i)–(r).

2. Mn ($3d^{5}4s^2$) shows +2 to +7, the maximum number of oxidation states: (ii)–(q). Sc³⁺ ($3d^0$) compounds are colourless: (iii)–(p).

   1 mark

Answer: (A) (i)-(r), (ii)-(q), (iii)-(p)

1. Molecularity applies only to elementary steps; it is a whole number, it is theoretical, and termolecular steps are rare and slow.

2. So molecularity has no meaning for a complex reaction as a whole. B is true.

   1 mark

Answer: (B)

1. ${\mathrm{C}}_4{\mathrm{H}}_8\mathrm{N}{\mathrm{H}}_3^{+}{\mathrm{X}}^{-}$ is an ionic ammonium salt, so it dissolves in the polar solvent water.

   1 mark

Answer: (C) water

1. 2-Phenylethanol, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, is a primary alcohol (not benzylic).

2. Primary alcohols give no turbidity with Lucas reagent at room temperature.

   1 mark

Answer: (B) No turbidity

1. For a zero-order reaction, $k=\frac{[R{]}_0-[R]}{t}=\frac{1.5-0.75}{120}=0.00625\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

   1 mark

Answer: (A) 0.00625 mol L⁻¹ s⁻¹

1. The –OCH₃ group releases electrons into the ring by +R effect and strongly activates it, so bromination occurs without a Lewis acid catalyst.

   1 mark

Answer: (A)

1. The y-axis is radius, and it decreases from Sc, levels off around Co/Ni, then rises slightly at Cu and Zn. This is the trend of atomic radii.

   1 mark

Answer: (B) atomic radii

1. A is true: alcohols act as nucleophiles (O–H bond breaks) and as electrophiles (C–O bond breaks, after protonation).

2. As nucleophiles, it is the O–H bond that breaks, not C–O, so R is false.

   1 mark

Answer: (C)

1. KMnO₄ oxidises toluene to benzoic acid, so A is true. Etard's reagent (CrO₂Cl₂) or CrO₃/(CH₃CO)₂O stops the oxidation at benzaldehyde, so R is true.

2. R does not explain why strong oxidants give the acid.

   1 mark

Answer: (B)

1. Both statements are true. Specificity arises from the shape of the active site, not merely from enzymes being biocatalysts, so R is not the correct explanation.

   1 mark

Answer: (B)

1. $E_{\mathrm{C}{\mathrm{u}}^{2+}/Cu}^{\circ }=+0.34$ V is greater than $E_{{\mathrm{H}}^{+}/{\mathrm{H}}_2}^{\circ }=0$ V, so R is true.

2. Because Cu²⁺ is reduced more easily than H⁺, Cu deposits at the cathode and no H₂ is evolved. So A is false.

   1 mark

Answer: (D)

1. (a) Number of half-lives in 2 days = 48 h: for X, $n=\frac{48}{12}=4$, so X left $=1\times {\left(\frac{1}{2}\right)}^4=\frac{1}{16}$ g.

   ½ mark

2. For Y, $n=\frac{48}{16}=3$, so Y left $=1\times {\left(\frac{1}{2}\right)}^3=\frac{1}{8}$ g. Ratio X : Y $=\frac{1}{16}:\frac{1}{8}=1:2$.

   ½ mark

3. (b) Rate $=k[P{]}^{1/2}[Q{]}^0$, so $k=\frac{\text{rate}}{[P{]}^{1/2}}=\frac{{\text{mol L}}^{-1}{\text{s}}^{-1}}{({\text{mol L}}^{-1}{)}^{1/2}}={\text{mol}}^{1/2}{\text{L}}^{-1/2}{\text{s}}^{-1}$.

   1 mark

Answer: (a) X : Y = 1 : 2; (b) mol^{1/2} L^{-1/2} s^{-1}

1. Isotonic solutions have equal osmotic pressure: ${\pi }_1={\pi }_2$, i.e. $i{C}_{1}RT=C_{2}RT$, so $i{C}_1=C_2$.

   ½ mark

2. $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4\cdot 10{\mathrm{H}}_2\mathrm{O}\to 2N{\mathrm{a}}^{+}+\mathrm{S}{\mathrm{O}}_4^{2-}$, so $i=3$; X is a non-electrolyte, $i=1$.

   ½ mark

3. Taking 100 mL of each solution: $3\times \frac{5}{322}=\frac{2}{M}$

   ½ mark

4. $M=\frac{2\times 322}{3\times 5}=42.9{\text{g mol}}^{-1}$

   ½ mark

Answer: Molecular mass of X ≈ 42.9 g mol⁻¹

1. (a) Melting point: m-dichlorobenzene < o-dichlorobenzene < p-dichlorobenzene. The para isomer is symmetrical and packs best in the crystal lattice, so it has the highest melting point.

   ½ mark

2. Boiling point: m-dichlorobenzene < p-dichlorobenzene < o-dichlorobenzene (446 K, 448 K, 453 K). The ortho isomer has the largest dipole moment, so stronger dipole–dipole attractions and the highest boiling point.

   ½ mark

3. (b) The halogen withdraws electrons from the ring by its strong –I effect (stronger than its +R effect), lowering the electron density of the ring. The ring is deactivated compared with benzene, so electrophilic substitution is slower and needs harsher conditions.

   1 mark

Answer: (a) m.p.: meta < ortho < para; b.p.: meta < para < ortho; (b) –I effect of halogen deactivates the ring

1. (a) p-Nitrobenzaldehyde is more reactive. The –NO₂ group withdraws electrons (–I, –R), making the carbonyl carbon more electron-deficient, so nucleophiles attack it more easily. In p-tolualdehyde the –CH₃ group releases electrons (+I, hyperconjugation), which lowers the electrophilicity of the carbonyl carbon.

   1 mark

2. (b) Acetone condenses with 2,4-dinitrophenylhydrazine, losing water, to give acetone 2,4-dinitrophenylhydrazone (orange-red solid): $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{N}-\mathrm{NH}-{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2$, with the two –NO₂ groups at positions 2 and 4 of the benzene ring.

   1 mark

Answer: (a) p-Nitrobenzaldehyde; (b) (CH₃)₂C=N–NH–C₆H₃(NO₂)₂ (acetone 2,4-DNP derivative)

1. (a) Gattermann–Koch reaction: ${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{CO},\mathrm{HCl},anhyd.\mathrm{AlC}{\mathrm{l}}_3/CuCl}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$. Nitration: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3/conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4,273-283\mathrm{K}}{\to }\mathrm{m}\text{-}{\mathrm{O}}_2\mathrm{N}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$. The –CHO group is meta-directing, so m-nitrobenzaldehyde forms.

   1 mark

2. (b) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}\stackrel{\mathrm{Mg},\mathrm{dry}\mathrm{ether}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{MgBr}\stackrel{(\mathrm{i})\mathrm{C}{\mathrm{O}}_2(\mathrm{dry}\mathrm{ice})}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOMgBr}\stackrel{(\mathrm{ii}){\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   1 mark

Answer: (a) Benzene → (CO/HCl, AlCl₃) benzaldehyde → (HNO₃/H₂SO₄) m-nitrobenzaldehyde; (b) Bromobenzene → (Mg, dry ether) PhMgBr → (CO₂, then H₃O⁺) benzoic acid

1. (a) DNA replicates and passes faithfully from parents to offspring, and the sequence of bases in each person's DNA is unique. Matching the child's DNA pattern with that of the parents therefore establishes paternity.

   1 mark

2. (b) A change in pH denatures the protein. Hydrogen bonds are broken, so the globules unfold and the helices uncoil, destroying the secondary and tertiary structures. The primary structure (sequence of amino acids) stays intact, but the protein loses its biological activity.

   1 mark

Answer: (a) Unique base sequence of DNA, passed on by replication from parents; (b) Denaturation – 2° and 3° structures destroyed, 1° structure intact

1. (a) $[\mathrm{Cr}(\mathrm{en}{)}_2(\mathrm{OH}{)}_2]\mathrm{Cl}$, i.e. $[\mathrm{Cr}({\mathrm{H}}_2\mathrm{NC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2{)}_2(\mathrm{OH}{)}_2]\mathrm{Cl}$. Check: Cr is +3, two OH⁻ give –2, so the complex ion is +1 and needs one Cl⁻.

   1 mark

2. (b) No. Ionisation isomers arise when a ligand inside the coordination sphere swaps places with a counter ion outside it. Here the ion outside the sphere is Hg²⁺, a metal cation that cannot act as a ligand, so no such exchange (and no ionisation isomer) is possible.

   1 mark

3. (c) The central metal atom/ion is a Lewis acid. It accepts lone pairs of electrons from the ligands (Lewis bases) into its vacant orbitals, forming coordinate bonds.

   1 mark

Answer: (a) [Cr(en)₂(OH)₂]Cl; (b) No – the counter ion Hg²⁺ cannot exchange with a ligand; (c) Lewis acid (electron-pair acceptor)

1. (a) Yes, provided the ZnSO₄ concentrations in the two half-cells differ (a concentration cell). Then the two electrode potentials are unequal, giving $E_{cell}=\frac{0.059}{2}\log \frac{[Z{n}^{2+}{]}_{cathode}}{[Z{n}^{2+}{]}_{anode}}\ne 0$. With equal concentrations, $E_{cell}=0$.

   1 mark

2. (b) By Kohlrausch's law: ${\Lambda }_m^0(\mathrm{MgC}{\mathrm{l}}_2)={\lambda }_m^0(\mathrm{M}{\mathrm{g}}^{2+})+2{\lambda }_m^0(\mathrm{C}{\mathrm{l}}^{-})$, so $258.6=106+2{\lambda }_m^0(\mathrm{C}{\mathrm{l}}^{-})$ and ${\lambda }_m^0(\mathrm{C}{\mathrm{l}}^{-})=\frac{152.6}{2}=76.3\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$.

   1 mark

3. (c) $\kappa =\frac{G^{\ast }}{R}=\frac{0.146\mathrm{c}{\mathrm{m}}^{-1}}{1000\Omega }=1.46\times 10^{-4}\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}$

   1 mark

Answer: (a) Yes, if ZnSO₄ concentrations differ (concentration cell); (b) 76.3 S cm² mol⁻¹; (c) 1.46 × 10⁻⁴ S cm⁻¹

1. (a) Reimer–Tiemann reaction. Phenol with CHCl₃/NaOH gives a –CHCl₂ intermediate at the ortho position, which is hydrolysed to salicylaldehyde, $\mathrm{o}\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$. IUPAC name: 2-hydroxybenzaldehyde.

   1½ marks

2. (b) Williamson ether synthesis: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{ONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{O}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaBr}$. IUPAC name: 2-ethoxy-3-methylpentane.

   1½ marks

3. (c) Stephen reaction (alternative part, 1.5 marks if chosen): $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CN}\stackrel{\mathrm{SnC}{\mathrm{l}}_2/HCl}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{NH}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$. IUPAC name: propanal.

Answer: (a) Reimer–Tiemann, 2-hydroxybenzaldehyde; (b) Williamson synthesis, CH₃CH₂CH(CH₃)CH(CH₃)OC₂H₅, 2-ethoxy-3-methylpentane; (c) Stephen reaction, CH₃CH₂CHO, propanal

1. A, B and C give an orange-red precipitate with 2,4-DNP, so they contain a carbonyl group. A and B reduce Tollens' reagent, so they are aldehydes. C does not, so it is a ketone.

   ½ mark

2. B and C give the iodoform test, so both contain a $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group: C is a methyl ketone and B is an aldehyde with a CH₃CO– unit. D gives brisk effervescence (CO₂) with NaHCO₃, so it is a carboxylic acid.

   ½ mark

3. The only aldehyde that gives a positive iodoform test is ethanal, so B $=\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$ (the two-carbon compound). Hence A, C and D have three carbon atoms each.

   ½ mark

4. A = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (propanal): an aldehyde that does not give iodoform. C = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$ (propanone): a 3-C methyl ketone. D = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$ (propanoic acid): a 3-C acid.

   1½ marks

Answer: A = CH₃CH₂CHO (propanal), B = CH₃CHO (ethanal), C = CH₃COCH₃ (propanone), D = CH₃CH₂COOH (propanoic acid)

1. (a) At t = 0 only sucrose is present, which is dextrorotatory (+66.5°). At t = ∞ sucrose is fully hydrolysed to an equimolar mixture of D-(+)-glucose (+52.5°) and D-(–)-fructose (–92.4°). The laevorotation of fructose exceeds the dextrorotation of glucose, so the mixture is laevorotatory (–39.9°).

   1 mark

2. (b) The sign of rotation changes from dextro (+) to laevo (–) on hydrolysis, so sucrose is called invert sugar; this change is called inversion.

   1 mark

3. (c) Glucose reacts with hydroxylamine to form glucose oxime: $\mathrm{CHO}(\mathrm{CHOH}{)}_4\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{N}{\mathrm{H}}_2\mathrm{OH}\to \mathrm{HON}=\mathrm{CH}(\mathrm{CHOH}{)}_4\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+{\mathrm{H}}_2\mathrm{O}$. A = glucose oxime.

   1 mark

Answer: (a) +66.5° is sucrose; –39.9° is the glucose (+52.5°) + fructose (–92.4°) mixture; (b) Invert sugar; (c) A = glucose oxime, HON=CH(CHOH)₄CH₂OH

1. A is optically active, so it must contain a chiral carbon. The only C4H9Br isomer with a chiral carbon is 2-bromobutane: A = CH3CH2CH(Br)CH3, (+)-2-bromobutane.

   1 mark

2. The product is racemic (±)-butan-2-ol, so the reaction is SN1. Step 1 (slow, rate-determining): the C–Br bond ionises to give the planar sp2 secondary carbocation and bromide: CH3CH2CH(Br)CH3 → CH3CH2C⁺HCH3 + Br⁻. Rate = k[RBr].

   1 mark

3. Step 2 (fast): the nucleophile (OH⁻/H2O) attacks the planar carbocation equally from both faces, CH3CH2C⁺HCH3 + OH⁻ → CH3CH2CH(OH)CH3. Attack from the front gives retention and attack from the back gives inversion, so a 50:50 mixture of (+) and (–) butan-2-ol forms, i.e. (±)-butan-2-ol (racemisation).

   1 mark

Answer: A = (+)-2-bromobutane, CH3CH2CH(Br)CH3. The SN1 mechanism has a slow step giving a planar carbocation and a fast step in which OH⁻ attacks from both faces, giving racemic (±)-butan-2-ol.

1. Arrhenius equation: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

   1 mark

2. $\log \frac{0.20}{0.02}=\frac{E_a}{19.15}\left(\frac{500-200}{200\times 500}\right)$, so $1=\frac{E_a}{19.15}\times 3\times 10^{-3}$

   1 mark

3. $E_a=\frac{19.15}{3\times 10^{-3}}=6383.3\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}\approx 6.38\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

   1 mark

Answer: Ea ≈ 6383 J mol⁻¹ (6.38 kJ mol⁻¹)

1. (a) Ligand D causes maximum splitting. $[\mathrm{Cr}{\mathrm{D}}_6{]}^{3-}$ absorbs the highest wavenumber (26280 cm⁻¹), i.e. the highest energy (314 kJ/mol), since $E\propto \bar{\nu }$. So the t₂g → e_g gap (${\Delta }_o$) is largest.

   1 mark

2. (a) OR: A is the weak field ligand. $[\mathrm{Cr}{\mathrm{A}}_6{]}^{3-}$ absorbs the lowest energy (13640 cm⁻¹, 163 kJ/mol), so the splitting it causes is least. (Alternative part, 1 mark)

3. (b) $\lambda =1/\bar{\nu }$. For $[\mathrm{Cr}{\mathrm{B}}_6{]}^{3+}$: $\lambda =\frac{1}{17830}$ cm $=5.61\times 10^{-5}$ cm ≈ 560 nm, which lies in 560–570 nm, so it appears violet. For $[\mathrm{Cr}{\mathrm{A}}_6{]}^{3-}$: $\lambda =\frac{1}{13640}$ cm ≈ 733 nm, so it is not violet.

   1 mark

4. (c) Spectrochemical order: Cl⁻ < H₂O < NH₃ < CN⁻. A (weakest, and anionic since the complex is 3–) = Cl⁻; B = H₂O; C = NH₃; D = CN⁻. (i) $[\mathrm{CrC}{\mathrm{l}}_6{]}^{3-}$: hexachloridochromate(III) ion.

   1 mark

5. (ii) $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$: hexaamminechromium(III) ion.

   1 mark

Answer: (a) D (highest energy absorbed); OR A is weak field; (b) [CrB₆]³⁺ (absorbs ≈560 nm); (c) (i) [CrCl₆]³⁻, hexachloridochromate(III) ion; (ii) [Cr(NH₃)₆]³⁺, hexaamminechromium(III) ion

1. (a) Overall reaction: $\mathrm{Pb}+\mathrm{Pb}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to 2PbS{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$. 2 mol e⁻ (2 F) are transferred while 2 mol H₂SO₄ are consumed, so 1 mol H₂SO₄ needs 1 F = 96500 C.

   1 mark

2. (b) $W_{max}=-\Delta G^{\circ }=nFE=2\times 96500\times 2.0=386000\mathrm{J}=386\mathrm{kJ}$ (per mole of cell reaction).

   1 mark

3. (c) For: Hydrogen burns (or reacts in a fuel cell) to give only water, with no CO₂ or other pollutants, and it is abundant and renewable. Against: today it is obtained by electrolysis using electricity from fossil fuels or by separation from fossil fuels (via ammonia/methanol). These steps use more energy than the hydrogen gives and add carbon emissions. View: hydrogen is green at the point of use, but not yet green overall. It becomes truly green only when extracted by clean methods such as solar-powered electrolysis, so research in this area is needed.

   2 marks

4. (c) OR (either choice with justification earns credit): (i) Safe recycling of lead batteries — they are the most widely used rechargeable batteries (UPS, vehicles, grid storage), but lead and H₂SO₄ dumped in landfills pollute soil, water and air. Research into safe recycling would cut these hazards and recover lead. OR (ii) Extraction of hydrogen — a fuel cell is a clean energy source giving only water. Its main barrier is clean, low-energy extraction of hydrogen, and research here would give a sustainable zero-carbon fuel. (Alternative part: 1 mark importance + 1 mark need for research)

Answer: (a) 96500 C (1 F); (b) 386000 J (386 kJ); (c) Hydrogen is clean in use, but its present extraction is not green — justified view

1. (a) Ti³⁺ (3d¹) and Cu²⁺ (3d⁹) each have one unpaired electron, so $\mu =\sqrt{n(n+2)}=\sqrt{3}=1.73$ BM. Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) have none, and Ti²⁺ (3d²) gives 2.83 BM.

   1 mark

2. (b) Zn. Its E° (–0.76 V) is more negative than that of Fe (–0.44 V), so Zn is oxidised in preference and protects iron. Ni (–0.25 V) is less negative than Fe, so iron would corrode instead.

   1 mark

3. (c) Cr⁺ is 3d⁵, a stable half-filled configuration, so removing a second electron from it is difficult. Mn⁺ is 3d⁵4s¹, and losing the 4s electron leaves a stable 3d⁵ ion. Hence the second ionisation enthalpy of Mn is lower.

   1 mark

4. (d) Both show only one oxidation state (Sc +3, Zn +2). Their ions (Sc³⁺ 3d⁰, Zn²⁺ 3d¹⁰) have no unpaired electrons, so their compounds are colourless and diamagnetic.

   1 mark

5. (e) Actinoid contraction is the steady decrease in atomic and ionic (M³⁺) radii from Th to Lr as atomic number increases. It is caused by the poor shielding of the nuclear charge by 5f electrons, so the increasing nuclear charge pulls the electrons closer.

   1 mark

6. (f) Cr is in the +6 oxidation state in both chromate ($\mathrm{Cr}{\mathrm{O}}_4^{2-}$) and dichromate ($\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$). (Alternative part, 1 mark if chosen)

7. (g) $2Mn{\mathrm{O}}_4^{-}+10{\mathrm{I}}^{-}+16{\mathrm{H}}^{+}\to 2M{\mathrm{n}}^{2+}+5{\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$ (Alternative part, 1 mark if chosen)

Answer: (a) Ti³⁺ and Cu²⁺; (b) Zn; (c) Cr⁺ is stable 3d⁵, Mn⁺ loses its 4s¹ electron easily; (d) single oxidation state; colourless, diamagnetic compounds; (e) poor 5f shielding; (f) +6 in both; (g) 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

1. (a) Dissolving glucose in water is endothermic. By Le Chatelier's principle, raising the temperature favours the endothermic dissolution, so the solubility of glucose increases.

   1 mark

2. (b) Q (freezing at –1.3 °C) is ocean water. Its dissolved salts depress the freezing point below 0 °C ($\Delta T_f=i{K}_{f}m$), while fresh water P freezes at 0 °C.

   1 mark

3. (c) ${\mathrm{K}}_3[\mathrm{Fe}(\mathrm{CN}{)}_6]\to 3{\mathrm{K}}^{+}+[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$, so $n=4$ ions.

   ½ mark

4. $i=1+(n-1)\alpha =1+(4-1)(0.852)=3.556$

   1 mark

5. $\Delta T_b=i{K}_{b}m=3.556\times 0.52\times 1=1.85\mathrm{K}$

   1 mark

6. $T_b=100+1.85=101.85{}^{\circ }\mathrm{C}$ (≈ 375.0 K)

   ½ mark

Answer: (a) Solubility increases with temperature; (b) Q is ocean water; (c) i = 3.556, boiling point = 101.85 °C

1. (a) Phenol and aniline show negative deviation from Raoult's law. Strong hydrogen bonding between phenol's –OH and aniline's –NH₂ makes the A–B attractions stronger than the A–A and B–B attractions, so escaping tendency decreases.

   1 mark

2. The net volume decreases ($\Delta V_{mix}<0$, and also $\Delta H_{mix}<0$). Graph: plot vapour pressure against mole fraction (x₁ = 1 to x₂ = 1). The total vapour pressure curve lies below the straight dashed line of Raoult's law and passes through a minimum. The partial pressure curves of each component also lie below their ideal lines.

   1 mark

3. (b) Relative lowering of vapour pressure: $\frac{p^{\circ }-p}{p^{\circ }}=x_2=\frac{n_2}{n_1+n_2}$

   ½ mark

4. $n_1=\frac{100}{18}=5.556$ mol, $n_2=1$ mol, so $x_2=\frac{1}{5.556+1}=0.1525$

   ½ mark

5. $\frac{23.80-p}{23.80}=0.1525$

   ½ mark

6. $23.80-p=3.63$

   ½ mark

7. $p=23.80-3.63=20.17$ mm Hg

   1 mark

Answer: (a) Negative deviation; volume decreases (ΔV < 0); total vapour pressure curve below the Raoult line with a minimum; (b) p ≈ 20.17 mm Hg

1. Identify A. ${\mathrm{C}}_7{\mathrm{H}}_7\mathrm{N}{\mathrm{O}}_2$ is $\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_2$ (nitrotoluene), which has o-, m- and p-isomers. The para isomer is the most symmetrical and packs best in the crystal lattice, so it has the highest melting point.

   A = p-nitrotoluene (4-methylnitrobenzene): a benzene ring with $\mathrm{C}{\mathrm{H}}_3$ at C-1 and $\mathrm{N}{\mathrm{O}}_2$ at C-4.

   1 mark

2. A → B (reduction of –NO₂ to –NH₂).

   $\mathrm{p}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_2\stackrel{Sn/HCl(\mathrm{or}Fe/HCl)}{\to }\mathrm{p}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{H}}_2$

   B = p-toluidine (4-methylaniline, p-methylbenzenamine), ${\mathrm{C}}_7{\mathrm{H}}_9\mathrm{N}$.

   1 mark

3. B → C (diazotisation at 273–278 K).

   $\mathrm{p}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{H}}_2+\mathrm{NaN}{\mathrm{O}}_2+2HCl\stackrel{0-5{}^{\circ }\mathrm{C}}{\to }\mathrm{p}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}+\mathrm{NaCl}+2{\mathrm{H}}_2\mathrm{O}$

   C = p-methylbenzenediazonium chloride ($\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{N}}^{+}\equiv \mathrm{N}\mathrm{C}{\mathrm{l}}^{-}$).

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4. C → D (the –N₂⁺ group is replaced by H).

   $\mathrm{p}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_3+{\mathrm{N}}_2+{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3+\mathrm{HCl}$

   D = toluene (methylbenzene), ${\mathrm{C}}_7{\mathrm{H}}_8$.

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5. D → E (Etard reaction). Chromyl chloride oxidises the $-\mathrm{C}{\mathrm{H}}_3$ group to a chromium complex, which on hydrolysis gives the aldehyde.

   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_3+\mathrm{Cr}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{C}{\mathrm{S}}_2}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{OCrOHC}{\mathrm{l}}_2{)}_2\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$

   E = benzaldehyde, ${\mathrm{C}}_7{\mathrm{H}}_6\mathrm{O}$.

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Answer: A = p-nitrotoluene; B = p-toluidine (4-methylaniline); C = p-methylbenzenediazonium chloride; D = toluene; E = benzaldehyde

1. (a)(i) In N-ethylbenzenesulphonamide, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{S}{\mathrm{O}}_2\mathrm{NH}{\mathrm{C}}_2{\mathrm{H}}_5$, the H on nitrogen is acidic. This is because the strongly electron-withdrawing sulphonyl group ($-\mathrm{S}{\mathrm{O}}_2-$) pulls electron density away from N.

   It therefore reacts with alkali to form a water-soluble salt, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{S}{\mathrm{O}}_2\mathrm{NH}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{S}{\mathrm{O}}_2\stackrel{-}{\mathrm{N}}{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{H}}_2\mathrm{O}$, so it dissolves.

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2. (a)(ii) During Fe/HCl reduction, the $\mathrm{FeC}{\mathrm{l}}_2$ formed is hydrolysed and releases HCl again. So only a small amount of HCl is needed to start the reaction, which makes the process cheaper. Sn/HCl needs a large amount of acid.

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3. **(b)(i) Decreasing $p{K}_b$ (increasing basic strength):**

   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2>{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NHC}{\mathrm{H}}_3>\mathrm{N}{\mathrm{H}}_3>{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2>\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$

   Reason: in the aryl amines, the lone pair on N is delocalised into the ring, so they are the weakest bases. In benzylamine, N is not attached to the ring, and alkyl groups (+I effect) raise basicity above that of $\mathrm{N}{\mathrm{H}}_3$.

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4. (b)(ii) Increasing solubility in water:

   ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}<{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2<{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

   Reason: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ cannot form H-bonds with water. O is more electronegative than N, so the alcohol forms stronger H-bonds with water than the amine does.

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5. (b)(iii) Decreasing boiling point:

   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}>{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}>\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2>\mathrm{C}{\mathrm{H}}_3\mathrm{OC}{\mathrm{H}}_3$

   Reason: carboxylic acids form dimers through strong H-bonding. O–H···O bonds are stronger than N–H···N bonds. The ether has no H-bonding at all, so it boils lowest.

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Answer: (a)(i) The N–H is acidic owing to the –SO₂– group, so the compound forms a soluble salt with alkali; (ii) FeCl₂ hydrolyses to regenerate HCl, so little acid is needed. (b)(i) C₆H₅NH₂ > C₆H₅NHCH₃ > NH₃ > C₆H₅CH₂NH₂ > CH₃NH₂; (ii) C₂H₅Cl < C₂H₅NH₂ < C₂H₅OH; (iii) CH₃COOH > C₂H₅OH > CH₃NH₂ > CH₃OCH₃