CBSE Class 12 Chemistry: 2024 board paper, set 56/1/1

1. Third ionisation removes an electron from the M${}^{2+}$ ion. Mn${}^{2+}$ is $3d^5$, a stable half-filled configuration.

2. Removing an electron from this stable $d^5$ arrangement needs the most energy, so Mn has the highest third ionisation enthalpy.

   1 mark

Answer: Manganese, option (B)

1. Only secondary alcohols are oxidised to ketones. Primary alcohols give aldehydes, and tertiary alcohols resist oxidation.

2. Butan-2-ol, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_3$, is secondary, so CrO${}_3$ oxidises it to butan-2-one.

   1 mark

Answer: Option (D)

1. A nucleotide is made of a sugar, a nitrogenous base and a phosphate group. The third component is therefore phosphoric acid.

   1 mark

Answer: Phosphoric acid, option (B)

1. Association reduces the number of solute particles in solution, so $i<1$.

   1 mark

Answer: Always less than one, option (D)

1. For X + 2Y → P: Rate $=-\frac{d[X]}{dt}=-\frac{1}{2}\frac{d[Y]}{dt}=+\frac{d[P]}{dt}$.

2. Multiplying the Y and P terms by 2 gives $2\frac{d[P]}{dt}=-\frac{d[Y]}{dt}$, which is option (A).

   1 mark

Answer: Option (A)

1. The S${}_N$1 rate depends on how stable the carbocation intermediate is.

2. 3-Bromocyclohexene forms an allylic carbocation, which is stabilised by resonance, so it reacts fastest.

   1 mark

Answer: Option (A)

1. PCl${}_5$ replaces the –OH of the acid by Cl: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{PC}{\mathrm{l}}_5\to \mathrm{C}{\mathrm{H}}_3\mathrm{COCl}+\mathrm{POC}{\mathrm{l}}_3+\mathrm{HCl}$.

   1 mark

Answer: CH${}_3$COCl, option (C)

1. The nucleophile CN${}^{-}$ attacks the carbonyl carbon and HCN adds across the C=O bond. This is nucleophilic addition.

   1 mark

Answer: Nucleophilic addition, option (A)

1. The Arrhenius equation in log form is $\log k=\log A-\frac{E_a}{2.303R}\cdot \frac{1}{T}$.

2. Comparing with $y=mx+c$: slope $=-E_a/2.303R$ and intercept $=\log A$.

   1 mark

Answer: Option (D)

1. R–X + R′ONa → R–O–R′ + NaX is the Williamson synthesis of ethers.

   1 mark

Answer: Williamson synthesis, option (C)

1. Acid dehydration goes through a carbocation, whose stability is 3° > 2° > 1°.

2. So the ease of dehydration is 3° > 2° > 1°, which is the order in option (A).

   1 mark

Answer: Option (A)

1. The C-5 –OH reacts with the C-1 aldehyde group to form a six-membered (pyranose) cyclic hemiacetal.

   1 mark

Answer: Option (B)

1. Any non-volatile solute lowers the freezing point of water, so (A) is true. NaCl does dissociate in water, so (R) is also true.

2. The depression is caused by the presence of solute particles, not by dissociation itself; dissociation only increases its magnitude. So (R) is not the correct explanation of (A).

   1 mark

Answer: Option (B)

1. Lanthanoid contraction gives Zr and Hf almost the same radii, so their chemical properties are nearly identical and they are hard to separate. (R) correctly explains (A).

   1 mark

Answer: Option (A)

1. The –I effect of Cl makes chloroacetic acid the stronger acid, so its pK${}_a$ is lower than that of ethanoic acid. (A) is false and (R) is true.

   1 mark

Answer: Option (D)

1. The lone pair on N in aniline is delocalised into the ring by resonance, so it is less available for protonation.

2. Aniline is therefore a weaker base than ammonia. (A) is false and (R) is true.

   1 mark

Answer: Option (D)

1. Electrode reaction: $\mathrm{F}{\mathrm{e}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{Fe}(\mathrm{s})$, so $n=2$. Nernst equation: $E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}=E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}^{\circ }-\frac{0.059}{n}\log \frac{1}{[\mathrm{F}{\mathrm{e}}^{2+}]}$

   ½ mark

2. Substitution: $E=-0.45-\frac{0.059}{2}\log \frac{1}{0.01}=-0.45-0.0295\times \log 100=-0.45-0.0295\times 2$

   1 mark

3. $E=-0.45-0.059=-0.509\text{V}$

   ½ mark

Answer: $E_{\mathrm{F}{\mathrm{e}}^{2+}/Fe}=-0.509$ V

1. Molecularity: the number of reacting species (atoms, ions or molecules) that must collide simultaneously in an elementary reaction to bring about the chemical change.

   1 mark

2. A bimolecular reaction becomes kinetically first order when one of the reactants is present in large excess (e.g. water in the acid hydrolysis of an ester), so its concentration stays practically constant — a pseudo first order reaction.

   1 mark

Answer: Molecularity = number of species colliding simultaneously in an elementary step; a bimolecular reaction is first order when one reactant is in large excess.

1. (a) On prolonged heating with HI, glucose is reduced to n-hexane, showing that its six carbon atoms are in a straight chain: $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{\mathrm{\Delta }}{\overset{\mathrm{HI}}{\to }}\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$

   1 mark

2. (b) Conc. $\mathrm{HN}{\mathrm{O}}_3$ oxidises both the –CHO group and the primary –CH₂OH group to –COOH, giving saccharic acid (glucaric acid): $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{[\mathrm{O}],\mathrm{HN}{\mathrm{O}}_3}{\to }\mathrm{HOOC}-(\mathrm{CHOH}{)}_4-\mathrm{COOH}$

   1 mark

Answer: (a) n-Hexane is formed; (b) Saccharic (glucaric) acid is formed.

1. (i) Electrophilic addition follows Markovnikov's rule; H⁺ adds to the terminal CH₂ giving the more stable benzylic carbocation, so Br attaches to the carbon bearing the ring: ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+\mathrm{HBr}\to {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}(\mathrm{Br})-\mathrm{C}{\mathrm{H}}_3$ (1-bromo-1-phenylethane).

   1 mark

2. (ii) In UV light Br₂ gives free-radical substitution at the allylic position (the ring CH₂ next to the double bond): cyclohexene + Br₂ → 3-bromocyclohexene (six-membered ring with C=C between C1 and C2 and Br on C3) + HBr.

   1 mark

Answer: (i) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{Br})\mathrm{C}{\mathrm{H}}_3$; (ii) 3-bromocyclohexene

1. (i) Grignard reagents are highly reactive and react with even traces of moisture to form alkanes, so the reagent is destroyed: $\mathrm{RMgX}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{RH}+\mathrm{Mg}(\mathrm{OH})\mathrm{X}$. Hence they are prepared under anhydrous conditions.

   1 mark

2. (ii) In aqueous KOH, the OH⁻ ions are highly hydrated and act mainly as nucleophiles, giving substitution (alcohol). In alcoholic KOH, alkoxide ions ($\mathrm{R}{\mathrm{O}}^{-}$) are formed, which are stronger bases than OH⁻; they abstract a β-hydrogen, so dehydrohalogenation (elimination) gives alkenes.

   1 mark

Answer: (i) RMgX reacts with water to give alkane; (ii) alkoxide ion in alcoholic KOH is a strong base causing elimination, while aqueous OH⁻ acts as nucleophile.

1. (a) Clemmensen reduction converts C=O to CH₂: $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Zn}(\mathrm{Hg}),conc.\mathrm{HCl}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$ (butane).

   1 mark

2. (b) Cannizzaro reaction (benzaldehyde has no α-H): $2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}+conc.\mathrm{NaOH}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$ (benzyl alcohol + sodium benzoate).

   1 mark

Answer: (a) Butane; (b) Benzyl alcohol + sodium benzoate (Cannizzaro reaction)

1. Cell constant from KCl: $G^{\ast }=\kappa \times R=1.35\times 10^{-2}\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}\times 100\Omega =1.35\mathrm{c}{\mathrm{m}}^{-1}$

   1 mark

2. Conductivity of AgNO₃: $\kappa =\frac{G^{\ast }}{R}=\frac{1.35}{90}=0.015\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}$

   1 mark

3. Molar conductivity: ${\Lambda }_m=\frac{\kappa \times 1000}{C}=\frac{0.015\times 1000}{0.02}=750\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$

   1 mark

Answer: κ = 0.015 S cm⁻¹; Λm = 750 S cm² mol⁻¹

1. Let rate $=k[\mathrm{NO}{]}^p[\mathrm{B}{\mathrm{r}}_2{]}^q$. Expt 1 ÷ Expt 2: $\frac{1.0\times 10^{-3}}{3.0\times 10^{-3}}={\left(\frac{0.05}{0.15}\right)}^q\Rightarrow \frac{1}{3}={\left(\frac{1}{3}\right)}^q\Rightarrow q=1$. Expt 1 ÷ Expt 3: $\frac{1}{9}={\left(\frac{1}{3}\right)}^p\Rightarrow p=2$.

2. (a) Order with respect to NO = 2; order with respect to Br₂ = 1.

   1 mark

3. (b) From Expt 1: $1.0\times 10^{-3}=k(0.05{)}^2(0.05)$, so $k=\frac{1.0\times 10^{-3}}{1.25\times 10^{-4}}=8{\mathrm{L}}^2\mathrm{mo}{\mathrm{l}}^{-2}{\mathrm{s}}^{-1}$

   1 mark

4. (c) Rate $=8\times (0.4{)}^2\times (0.2)=8\times 0.16\times 0.2=0.256=2.56\times 10^{-1}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$

   1 mark

Answer: (a) 2 w.r.t. NO, 1 w.r.t. Br₂; (b) k = 8 L² mol⁻² s⁻¹; (c) 0.256 mol L⁻¹ s⁻¹

1. (a) ${\mathrm{K}}_2[\mathrm{Zn}(\mathrm{OH}{)}_4]$ — four OH⁻ ligands on Zn²⁺ give a −2 complex ion, balanced by two K⁺.

   1 mark

2. (b) Conductivity depends on the number of ions produced: $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_3\mathrm{C}{\mathrm{l}}_3]$ (0 ions) < $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Cl}]\mathrm{C}{\mathrm{l}}_2$ (3 ions) < $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6]\mathrm{C}{\mathrm{l}}_3$ (4 ions).

   1 mark

3. (c)(i) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{N}{\mathrm{O}}_2{]}^{2+}$: linkage isomerism (ambidentate NO₂⁻ binds through N or O as –ONO). (ii) $[\mathrm{Co}(\mathrm{en}{)}_3]\mathrm{C}{\mathrm{l}}_3$: optical isomerism (non-superimposable mirror images, d and l forms).

   1 mark

Answer: (a) K₂[Zn(OH)₄]; (b) [Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl]Cl₂ < [Cr(NH₃)₆]Cl₃; (c) (i) linkage, (ii) optical

1. (a) Compound (ii) $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{CH}(\mathrm{Br})-\mathrm{C}{\mathrm{H}}_3$ is allylic, as Br is on an sp³ carbon next to the C=C. (Compound (i) is a vinylic halide — Br is on the doubly bonded carbon.)

   1 mark

2. (b) 2,4,6-Trinitrochlorobenzene is more reactive, because the –NO₂ groups at ortho and para positions withdraw electrons and stabilise the carbanion intermediate formed on attack of the nucleophile.

   1 mark

3. (c) $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{Cl}$ (2-chloro-2-methylpropane, tert-butyl chloride) has the lowest boiling point; it is the most branched and nearly spherical, so it has the least surface area and weakest van der Waals forces.

   1 mark

Answer: (a) (ii); (b) 2,4,6-trinitrochlorobenzene (–NO₂ electron withdrawal); (c) tert-butyl chloride

1. (a) Step 1 – protonation: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\ddot{\mathrm{O}}\mathrm{H}+{\mathrm{H}}^{+}\rightleftharpoons \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2$. Step 2 – loss of water (slow): $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$. Step 3 – loss of proton: $\mathrm{H}-\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}^{+}$.

   1 mark

2. (b)(i) Hydroboration–oxidation gives anti-Markovnikov addition of water: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (butan-1-ol).

   1 mark

3. (b)(ii) Kolbe's reaction: phenol → sodium phenoxide → on CO₂ and acidification gives salicylic acid (2-hydroxybenzoic acid, o-HO–C₆H₄–COOH).

   1 mark

Answer: (a) Protonation → carbocation (loss of H₂O) → loss of H⁺ gives ethene; (b)(i) butan-1-ol; (ii) salicylic acid

1. (a) Peptide linkage is the amide (–CO–NH–) bond formed between the –COOH group of one α-amino acid and the –NH₂ group of another with loss of a water molecule.

   1 mark

2. (b) The two strands of DNA are held together by hydrogen bonds between complementary bases (A=T with two, G≡C with three H-bonds).

   1 mark

3. (c) Starch is the polysaccharide (sucrose is a disaccharide; glucose and fructose are monosaccharides).

   1 mark

4. (d) Water-soluble: vitamin B or C; fat-soluble: vitamin A, D, E or K.

Answer: (a) –CO–NH– bond between amino acids; (b) hydrogen bonds; (c) starch; (d) e.g. vitamin C (water-soluble), vitamin A (fat-soluble)

1. (F) is a monobasic acid of molar mass 60 → $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ (ethanoic acid). So (D), which oxidises to F, is $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$ (ethanal), and (B), which gives D with PCC, is $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (ethanol).

   1½ marks

2. (D) with dil. NaOH undergoes aldol condensation; heating dehydrates the aldol: $2C{\mathrm{H}}_3\mathrm{CHO}\stackrel{dil.\mathrm{NaOH}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}\stackrel{\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHCHO}$, so (E) is but-2-enal (crotonaldehyde).

   ½ mark

3. Catalytic hydrogenation of E: $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHCHO}+2{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, so (C) is butan-1-ol.

   ½ mark

4. (A), C₆H₁₂O₂, gives ethanol and butan-1-ol with LiAlH₄, so it is the ester ethyl butanoate $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5\stackrel{\mathrm{LiAl}{\mathrm{H}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$. (Butyl ethanoate, $\mathrm{C}{\mathrm{H}}_3\mathrm{COOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, also fits the data.)

   ½ mark

Answer: A = ethyl butanoate (or butyl ethanoate); B = ethanol; C = butan-1-ol; D = ethanal; E = but-2-enal; F = ethanoic acid

1. (a) In primary batteries the reaction occurs only once and they cannot be recharged (e.g. dry cell); secondary batteries can be recharged by passing current in the opposite direction and reused (e.g. lead storage battery).

   1 mark

2. (b) The overall reaction $\mathrm{Zn}(\mathrm{Hg})+\mathrm{HgO}(\mathrm{s})\to \mathrm{ZnO}(\mathrm{s})+\mathrm{Hg}(\mathrm{l})$ involves no ion in solution whose concentration changes during use, so the cell potential stays constant at 1.35 V.

   1 mark

3. (c) On recharging, the cell works as an electrolytic cell. At one electrode: $\mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{e}}^{-}\to \mathrm{Pb}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})$; at the other: $\mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})+4{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}$

   1 mark

4. Overall recharging reaction: $2PbS{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Pb}(\mathrm{s})+\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})$

   1 mark

5. (c) OR: Advantages of fuel cells: (i) high efficiency (about 70%) since chemical energy of fuel is directly converted to electricity; (ii) pollution-free — the H₂–O₂ cell produces only water. (They also give continuous supply as long as reactants are fed.)

Answer: (a) Primary – not rechargeable; secondary – rechargeable; (b) no ions in solution whose concentration changes; (c) 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄ / OR high efficiency and pollution-free

1. (a) Crystal field splitting energy (${\Delta }_o$ for octahedral) is the energy difference between the two sets of d-orbitals ($t_{2g}$ and $e_g$) formed when the degenerate d-orbitals split in the presence of ligands arranged in a definite geometry.

   1 mark

2. (b) Ti³⁺ is 3d¹, i.e. $t_{2g}^1{e}_g^0$. The electron absorbs light in the yellow-green region and is promoted from $t_{2g}$ to $e_g$ (d–d transition); the complementary colour, violet, is seen.

   1 mark

3. (c) Cr³⁺ (Z = 24) is 3d³: configuration $t_{2g}^3$ with three unpaired electrons (d²sp³), whatever the ligand strength, so $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$ is paramagnetic.

   1 mark

4. Ni²⁺ (Z = 28) is 3d⁸; CN⁻ is a strong field ligand that pairs up the electrons, leaving one 3d orbital empty for dsp² (square planar) hybridisation; with no unpaired electron, $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ is diamagnetic.

   1 mark

5. (c) OR: Fe³⁺ (Z = 26) is 3d⁵. CN⁻ is a strong field ligand, pairs the electrons into three 3d orbitals leaving two inner (n–1)d orbitals free → d²sp³ hybridisation, so $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$ is an inner orbital complex. H₂O is a weak field ligand, no pairing occurs, so outer 4d orbitals are used → sp³d² hybridisation, making $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$ an outer orbital complex.

Answer: (a) Energy gap between t₂g and e_g sets; (b) d–d transition of the t₂g¹ electron; (c) Cr³⁺ t₂g³ has 3 unpaired e⁻ (paramagnetic), Ni²⁺ with strong-field CN⁻ has all electrons paired, dsp² (diamagnetic) / OR CN⁻ strong → d²sp³ inner; H₂O weak → sp³d² outer

1. (i) ${\mathrm{O}}_2$ has the higher $K_H$. By Henry's law $p=K_{H}x$, so at a given pressure, the higher the $K_H$, the lower the solubility; O₂ is less soluble than CO₂.

   1 mark

2. (ii) The solution (> 0.9% NaCl) is hypertonic, so water moves out of the blood cells by osmosis and the cells shrink.

   1 mark

3. (iii) ${\mathrm{A}}_2{\mathrm{B}}_3\to 2{\mathrm{A}}^{3+}+3{\mathrm{B}}^{2-}$, so n = 5 ions. $\alpha =\frac{i-1}{n-1}\Rightarrow 0.6=\frac{i-1}{4}$

   ½ mark

4. $i=1+2.4=3.4$

   1 mark

5. $\Delta T_b=i{K}_{b}m=3.4\times 0.52\mathrm{K}\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}\times 1\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1}=1.768\text{K}$

   1 mark

6. $T_b=373.15+1.768=374.918$ K (or 374.768 K taking the boiling point of water as 373 K).

   ½ mark

Answer: (i) O₂ (higher K_H → lower solubility); (ii) cells shrink; (iii) T_b = 374.918 K (374.768 K if T_b° = 373 K)

1. (i) $x_A=0.4,x_B=0.6$. By Raoult's law: $p_A=p_A^{\circ }{x}_A=75\times 0.4=30$ mm Hg; $p_B=p_B^{\circ }{x}_B=25\times 0.6=15$ mm Hg; $P_{total}=30+15=45$ mm Hg

   1 mark

2. Mole fraction of B in vapour (Dalton's law): $y_B=\frac{p_B}{P_{total}}=\frac{15}{45}=0.33$

   1 mark

3. (ii) A colligative property is one that depends only on the number of solute particles in a given amount of solution, not on their nature.

   1 mark

4. Osmotic pressure is preferred for macromolecules (measured at room temperature, large value even for dilute solutions).

   1 mark

5. (iii) NaCl dissociates in water into Na⁺ and Cl⁻ (i ≈ 2), whereas glucose does not dissociate (i = 1). Since $\pi =iCRT$, the NaCl solution has nearly double the osmotic pressure, so equimolar solutions are not isotonic.

   1 mark

Answer: (i) y_B = 0.33; (ii) property depending on number of solute particles; osmotic pressure; (iii) NaCl dissociates (i = 2), glucose does not (i = 1)

1. (a) In N,N-diethylbenzenesulphonamide, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{S}{\mathrm{O}}_2\mathrm{N}({\mathrm{C}}_2{\mathrm{H}}_5{)}_2$, there is no hydrogen on the nitrogen atom, so it is not acidic and does not form a salt with alkali; hence it is insoluble.

   1 mark

2. (b) Aniline, a Lewis base, reacts with the catalyst AlCl₃ (Lewis acid) to form a salt; the nitrogen acquires a positive charge, which strongly deactivates the ring, so the Friedel–Crafts reaction does not occur.

   1 mark

3. (c) Azo dye test: treat each with NaNO₂ + dil. HCl at 273–278 K and then with alkaline β-naphthol (or phenol). Aniline forms benzenediazonium chloride which couples to give an orange/red dye; methylamine gives methanol with evolution of N₂ and no dye.

   1 mark

4. (d) Gabriel phthalimide synthesis: phthalimide $\stackrel{\text{KOH (ethanolic)}}{\to }$ potassium phthalimide $\stackrel{R-X}{\to }$ N-alkylphthalimide $\stackrel{\text{NaOH(aq)}}{\to }$ $\mathrm{R}-\mathrm{N}{\mathrm{H}}_2$ + sodium phthalate. i.e. ${\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{NK}+\mathrm{RX}\to {\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{NR}+\mathrm{KX}$; ${\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{NR}+2NaOH\to {\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{COONa}{)}_2+\mathrm{RN}{\mathrm{H}}_2$

   1 mark

5. (e) Protect –NH₂ by acetylation, brominate, then hydrolyse: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2\stackrel{(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O},\mathrm{pyridine}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NHCOC}{\mathrm{H}}_3\stackrel{\mathrm{B}{\mathrm{r}}_2/C{\mathrm{H}}_3\mathrm{COOH}}{\to }\mathrm{p}\text{-}\mathrm{Br}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{NHCOC}{\mathrm{H}}_3\stackrel{{\mathrm{H}}^{+}\mathrm{or}\mathrm{O}{\mathrm{H}}^{-},{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{p}\text{-}\mathrm{Br}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{H}}_2$

   1 mark

6. (f) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}\stackrel{\mathrm{HB}{\mathrm{F}}_4}{\to }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{B}{\mathrm{F}}_4^{-}\stackrel{\mathrm{NaN}{\mathrm{O}}_2/Cu,\mathrm{\Delta }}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_2+{\mathrm{N}}_2+\mathrm{NaB}{\mathrm{F}}_4$ (nitrobenzene).

   1 mark

7. (g) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\stackrel{\mathrm{N}{\mathrm{H}}_3,\mathrm{\Delta }}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CON}{\mathrm{H}}_2$ (A, benzamide) $\stackrel{\mathrm{B}{\mathrm{r}}_2+\mathrm{NaOH}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$ (B, aniline, by Hoffmann bromamide degradation).

Answer: (a) No H on N, not acidic; (b) forms salt with AlCl₃, ring deactivated; (c) azo dye test – aniline gives orange dye; (d) phthalimide → K-phthalimide → N-alkylphthalimide → RNH₂; (e) acetylation, bromination, hydrolysis; (f) nitrobenzene; (g) A = benzamide, B = aniline

1. (i)(1) Zn, Cd and Hg have a completely filled $(n-1)d^{10}$ configuration. Their d-electrons are not available for metallic bonding, so the metallic bonding between their atoms is weak. This gives them low melting and boiling points.

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2. (i)(2) $\mathrm{C}{\mathrm{r}}^{2+}$ ($3d^4$) readily loses an electron to become $\mathrm{C}{\mathrm{r}}^{3+}$ ($3d^3$). That ion has the stable half-filled $t_{2g}^3$ configuration, so $\mathrm{C}{\mathrm{r}}^{2+}$ is a strong reducing agent.

   $\mathrm{M}{\mathrm{n}}^{3+}$ ($3d^4$) readily gains an electron to become $\mathrm{M}{\mathrm{n}}^{2+}$ ($3d^5$). That ion has the extra-stable half-filled $d^5$ configuration, so $\mathrm{M}{\mathrm{n}}^{3+}$ is a strong oxidising agent.

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3. (i)(3) Converting $\mathrm{Cu}(\mathrm{s})$ to $\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})$ requires a high enthalpy of atomisation (${\Delta }_a{H}^{\circ }$) and a large sum of first and second ionisation enthalpies.

   The hydration enthalpy (${\Delta }_{hyd}{H}^{\circ }$) of $\mathrm{C}{\mathrm{u}}^{2+}$ is not large enough to compensate for this. Hence $E^{\circ }(\mathrm{C}{\mathrm{u}}^{2+}/Cu)$ is positive (+0.34 V), and Cu does not liberate ${\mathrm{H}}_2$ from acids.

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4. (ii)(1) On heating, potassium permanganate decomposes (at about 513 K):

   $2\mathrm{KMn}{\mathrm{O}}_4\stackrel{\Delta }{\to }{\mathrm{K}}_2\mathrm{Mn}{\mathrm{O}}_4+\mathrm{Mn}{\mathrm{O}}_2+{\mathrm{O}}_2$

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5. (ii)(2) In acid solution, dichromate oxidises iodide to iodine and is itself reduced to $\mathrm{C}{\mathrm{r}}^{3+}$:

   $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+6{\mathrm{I}}^{-}+14{\mathrm{H}}^{+}⟶2\mathrm{C}{\mathrm{r}}^{3+}+3{\mathrm{I}}_2+7{\mathrm{H}}_2\mathrm{O}$

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Answer: (i)(1) Fully filled d¹⁰, so weak metallic bonding; (2) Cr²⁺→Cr³⁺ (t2g³ stable), Mn³⁺→Mn²⁺ (d⁵ stable); (3) high ΔaH° + IE not compensated by ΔhydH°. (ii)(1) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂; (2) Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O

1. (i) $\mathrm{CuC}{\mathrm{l}}_2$ is more stable in aqueous solution.

   The hydration enthalpy of $\mathrm{C}{\mathrm{u}}^{2+}$ is much more negative than that of $\mathrm{C}{\mathrm{u}}^{+}$, and this more than compensates for the second ionisation enthalpy. Therefore $\mathrm{C}{\mathrm{u}}^{+}$ disproportionates in water:

   $2\mathrm{C}{\mathrm{u}}^{+}(\mathrm{aq})\to \mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

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2. (ii) The general electronic configuration of f-block elements is:

   $(n-2)f^{1-14}(n-1)d^{0-1}n{s}^2$

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3. (iii) $\mathrm{F}{\mathrm{e}}^{3+}$ is coloured in aqueous solution.

   • $\mathrm{F}{\mathrm{e}}^{3+}$ is $3d^5$. It has unpaired electrons, so d–d transitions occur by absorbing visible light.
   • $\mathrm{S}{\mathrm{c}}^{3+}$ ($3d^0$) and $\mathrm{Z}{\mathrm{n}}^{2+}$ ($3d^{10}$) have no unpaired d-electrons, so no d–d transition is possible and they are colourless.
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4. (iv) Step 1: acidify sodium chromate with sulphuric acid to convert it to sodium dichromate:

   $2\mathrm{N}{\mathrm{a}}_2\mathrm{Cr}{\mathrm{O}}_4+2{\mathrm{H}}^{+}\to \mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{H}}_2\mathrm{O}$

   Step 2: treat the sodium dichromate solution with KCl. The less soluble potassium dichromate crystallises out:

   $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{KCl}\to {\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7+2\mathrm{NaCl}$

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5. (v) Transition metals and their compounds act as catalysts for these reasons:

   • They can show variable oxidation states.
   • They can form complexes or unstable intermediates with the reactants.
   • They provide a large surface area for the reactants to be adsorbed on.
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Answer: (i) CuCl₂ (higher hydration enthalpy of Cu²⁺; Cu⁺ disproportionates); (ii) (n−2)f¹⁻¹⁴(n−1)d⁰⁻¹ns²; (iii) Fe³⁺ (3d⁵, unpaired electrons, d–d transition); (iv) Acidify Na₂CrO₄ to Na₂Cr₂O₇, then add KCl to get K₂Cr₂O₇; (v) Variable oxidation states, complex formation, large surface area