CBSE Class 12 Chemistry: 2025 board paper, set 56/1/1

1. On association, $n$ solute particles combine into one, so the number of particles falls and every colligative property comes out lower than expected ($i<1$).

2. Molar mass is inversely proportional to the colligative property, so the observed (abnormal) molar mass is greater than the true value. It increases: option (A).

   1 mark

Answer: Increase, option (A)

1. The couple with the lower reduction potential, $\mathrm{C}{\mathrm{r}}^{3+}/Cr$, acts as the anode. $\mathrm{S}{\mathrm{n}}^{4+}/S{\mathrm{n}}^{2+}$ acts as the cathode.

2. $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.15-(-0.74)=+0.89$ V, which is option (B).

   1 mark

Answer: +0.89 V, option (B)

1. The configuration of $\mathrm{C}{\mathrm{r}}^{3+}$ is $[Ar]3d^3$, so it has $n=3$ unpaired electrons.

2. $\mu =\sqrt{n(n+2)}=\sqrt{15}=3.87$ B.M., which is option (B).

   1 mark

Answer: 3.87 B.M., option (B)

1. In acid, $\mathrm{KMn}{\mathrm{O}}_4$ oxidises sulphite to sulphate: $2Mn{O}_4^{-}+5S{O}_3^{2-}+6H^{+}\to 2M{n}^{2+}+5S{O}_4^{2-}+3H_{2}O$. The answer is option (D).

   1 mark

Answer: $\mathrm{S}{\mathrm{O}}_4^{2-}$, option (D)

1. Let the oxidation state of Pt be $x$: $x+2(0)+2(-1)=+2$, so $x=+4$.

2. The complex is a cation, so the metal keeps the name platinum. The name is diamminedichloridoplatinum(IV), option (B).

   1 mark

Answer: Diamminedichloridoplatinum(IV), option (B)

1. For isomers, more branching gives a more compact shape and smaller surface area. This weakens van der Waals forces and lowers the boiling point.

2. tert-Butyl bromide is the most branched and n-butyl bromide is unbranched, so the order is (iii) < (i) < (ii), option (C).

   1 mark

Answer: (iii) < (i) < (ii), option (C)

1. In S${}_N$2 the nucleophile attacks from the side opposite the leaving group in one concerted step. This gives Walden inversion of configuration, option (C).

   1 mark

Answer: Inversion of configuration, option (C)

1. Phenoxide ion is stabilised by resonance, so phenol is the strongest acid of the three.

2. The +I effect of the alkyl group destabilises alkoxide, so an alcohol is weaker than water. The order is $C_6{H}_{5}OH>H_{2}O>ROH$, option (A).

   1 mark

Answer: C₆H₅OH > H₂O > ROH, option (A)

1. $C_3{H}_{7}MgBr+C{O}_2\to C_3{H}_{7}COOMgBr\stackrel{H_3{O}^{+}}{\to }{C}_3{H}_{7}COOH$. The answer is butanoic acid, option (A).

   1 mark

Answer: C₃H₇COOH, option (A)

1. Propanamine must keep all three carbons. $LiAl{H}_4$ reduces $C{H}_{3}C{H}_{2}CON{H}_2$ to $C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$.

2. $B{r}_2$/NaOH (Hofmann degradation) would remove one carbon and give ethanamine. So the best reagent is $LiAl{H}_4$ in ether, option (D).

   1 mark

Answer: LiAlH₄ in ether, option (D)

1. Glucose is an aldohexose, gives n-hexane with HI, and does not give Schiff's test, so statements A, B and D are true.

2. Glucose exists in the six-membered pyranose form, not the furanose form. The statement that is not true is (C).

   1 mark

Answer: Option (C)

1. Water moves from the mango (lower solute concentration) into the concentrated brine by osmosis, so the mango loses water and shrivels. The answer is option (D).

   1 mark

Answer: It loses water due to osmosis, option (D)

1. In both complexes only $H_{2}O$ ligands are bonded to the metal, so both are homoleptic. The assertion is true.

2. Having only one kind of ligand is exactly the definition of a homoleptic complex, so R is true and correctly explains A. The answer is option (A).

   1 mark

Answer: Option (A)

1. Boiling point rises with the size and mass of the halogen, so RI > RBr > RCl > RF. A is true.

2. R is also true, because the polarity of the C–X bond and the higher molar mass increase intermolecular forces. R does not explain the trend among the halides, so the answer is option (B).

   1 mark

Answer: Option (B)

1. Ethanol has a higher boiling point because of intermolecular hydrogen bonding, so A is true.

2. R wrongly says the hydrogen bonding is intramolecular, so R is false. The answer is option (C).

   1 mark

Answer: Option (C)

1. Every α-amino acid except glycine has a chiral α-carbon, so A is true. Most natural amino acids are L-configured, so R is also true.

2. Optical activity comes from chirality, not from the L-configuration, so R does not explain A. The answer is option (B).

   1 mark

Answer: Option (B)

1. (a) A pressure cooker is sealed, so the pressure above the water becomes higher than atmospheric pressure. Water boils when its vapour pressure equals the external pressure, so its boiling point rises above 100 °C. The food cooks at this higher temperature, which is why cooking is faster.

   1 mark

2. (b) The volume decreases on mixing ($\Delta V_{mix}<0$). This means the X–Y attractions are stronger than the X–X and Y–Y attractions, so the solution shows negative deviation from Raoult's law.

   ½ mark

3. For negative deviation, heat is released on mixing ($\Delta H_{mix}<0$, exothermic). The temperature of the mixture rises.

   ½ mark

Answer: (a) Higher pressure raises the boiling point of water, so food cooks faster; (b) negative deviation; temperature increases

1. Azeotropes are binary liquid mixtures that have the same composition in the liquid phase and in the vapour phase and boil at a constant temperature. They therefore cannot be separated by fractional distillation.

   1 mark

2. A solution showing large negative deviation from Raoult's law forms a maximum boiling azeotrope.

   ½ mark

3. Example: nitric acid and water, about 68% $\mathrm{HN}{\mathrm{O}}_3$ and 32% water by mass, which boils at about 393.5 K.

   ½ mark

Answer: Azeotrope: a constant-boiling mixture with the same composition in liquid and vapour; negative deviation gives a maximum boiling azeotrope, e.g. 68% HNO₃ + 32% H₂O

1. (a) In acidic medium permanganate is reduced to $\mathrm{M}{\mathrm{n}}^{2+}$ and iodide is oxidised to iodine:
   $2\mathrm{Mn}{\mathrm{O}}_4^{-}(\mathrm{aq})+10{\mathrm{I}}^{-}(\mathrm{aq})+16{\mathrm{H}}^{+}(\mathrm{aq})\to 2\mathrm{M}{\mathrm{n}}^{2+}(\mathrm{aq})+5{\mathrm{I}}_2(\mathrm{s})+8{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   1 mark

2. (b) In acidic medium dichromate is reduced to $\mathrm{C}{\mathrm{r}}^{3+}$ and $\mathrm{F}{\mathrm{e}}^{2+}$ is oxidised to $\mathrm{F}{\mathrm{e}}^{3+}$:
   $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}(\mathrm{aq})+6\mathrm{F}{\mathrm{e}}^{2+}(\mathrm{aq})+14{\mathrm{H}}^{+}(\mathrm{aq})\to 2\mathrm{C}{\mathrm{r}}^{3+}(\mathrm{aq})+6\mathrm{F}{\mathrm{e}}^{3+}(\mathrm{aq})+7{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   1 mark

Answer: (a) → 2Mn²⁺ + 5I₂ + 8H₂O; (b) → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

1. Benzaldehyde is less reactive than propanal in nucleophilic addition reactions.

   1 mark

2. In benzaldehyde the C=O group is conjugated with the benzene ring. Resonance releases electrons from the ring towards the carbonyl carbon and reduces the polarity of C=O. Its carbonyl carbon is therefore less electrophilic than that of propanal, so a nucleophile attacks it less easily.

   1 mark

Answer: Less reactive, because resonance with the benzene ring makes its carbonyl carbon less electrophilic

1. (a) Cyanide ion replaces Cl by nucleophilic substitution: A = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CN}$ (propanenitrile).

   ½ mark

2. ${\mathrm{H}}_2/Ni$ reduces $-\mathrm{C}\equiv \mathrm{N}$ to $-\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$: B = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$ (propan-1-amine).

   ½ mark

3. (b) Diazotisation of aniline at 0–5 °C gives A = ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}$ (benzenediazonium chloride).

   ½ mark

4. Coupling with aniline in acidic medium at the para position gives B = ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{N}=\mathrm{N}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{N}{\mathrm{H}}_2$ (p-aminoazobenzene, a yellow dye).

   ½ mark

Answer: (a) A = CH₃CH₂CN, B = CH₃CH₂CH₂NH₂; (b) A = C₆H₅N₂⁺Cl⁻, B = p-aminoazobenzene

1. (a) HCN adds across the aldehyde group of glucose to form glucose cyanohydrin:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{HCN}}{\to }\mathrm{CH}(\mathrm{OH})\mathrm{CN}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$

   1 mark

2. (b) Bromine water, a mild oxidising agent, oxidises the –CHO group to –COOH and gives gluconic acid:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{B}{\mathrm{r}}_2\mathrm{water}}{\to }\mathrm{COOH}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$

   1 mark

Answer: (a) Glucose cyanohydrin; (b) Gluconic acid

1. Elevation in boiling point: $\Delta T_b=T_b-T_b^{\circ }=100.20-100=0.20$ °C, i.e. 0.20 K.

   ½ mark

2. Using $\Delta T_b=K_{b}m$: $m=\frac{\Delta T_b}{K_b}=\frac{0.20}{0.512}$

   ½ mark

3. $m=0.390$ mol kg${}^{-1}$

   ½ mark

4. Depression in freezing point: $\Delta T_f=K_{f}m=1.86\times 0.390=0.726$ K

   1 mark

5. Freezing point of the solution: $T_f=0-0.726=-0.726$ °C (about $-0.73$ °C, or 272.42 K)

   ½ mark

Answer: Freezing point ≈ −0.726 °C (≈ 272.42 K)

1. (a)(i) Kohlrausch's law: the limiting molar conductivity of an electrolyte equals the sum of the individual contributions of its cation and anion, i.e. ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$.

   1 mark

2. (a)(ii) Faraday's first law: the amount of chemical reaction (mass of substance deposited or liberated) at an electrode during electrolysis is directly proportional to the quantity of electricity passed, i.e. $w=ZIt$.

   1 mark

3. (b) X is better for coating iron.

   ½ mark

4. $E_{X^{2+}/X}^{\circ }=-2.36$ V is more negative than $E_{F{e}^{2+}/Fe}^{\circ }=-0.44$ V, so X is oxidised more readily than iron. X corrodes sacrificially and protects iron. Y ($-0.14$ V) is less easily oxidised than Fe and would not give this protection.

   ½ mark

Answer: (a) Statements of Kohlrausch's law and Faraday's first law; (b) X, since it has a more negative E° (higher oxidation potential) than Fe

1. For a first order reaction, $t_{1/2}=\frac{0.693}{k}$, so $k=\frac{0.693}{t_{1/2}}$.

   ½ mark

2. At 300 K: $k_1=\frac{0.693}{20}=3.465\times 10^{-2}$ min${}^{-1}$

   ½ mark

3. At 350 K: $k_2=\frac{0.693}{5}=1.386\times 10^{-1}$ min${}^{-1}$

   ½ mark

4. Arrhenius equation: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
   $\log \frac{0.1386}{0.03465}=\log 4=\frac{E_a}{2.303\times 8.314}\left[\frac{350-300}{350\times 300}\right]$

   ½ mark

5. $0.602=\frac{E_a}{19.15}\times \frac{50}{105000}$
   $E_a=\frac{0.602\times 19.15\times 105000}{50}\approx 24209$ J mol${}^{-1}$ $=24.21$ kJ mol${}^{-1}$

   1 mark

Answer: Eₐ ≈ 24.21 kJ mol⁻¹ (24209 J mol⁻¹)

1. (a) Converting Cu(s) to $\mathrm{C}{\mathrm{u}}^{2+}$(aq) needs a high enthalpy of atomisation plus a high sum of ionisation enthalpies. Its hydration enthalpy is relatively low and does not compensate for this energy, so $E_{C{u}^{2+}/Cu}^{\circ }$ is positive (+0.34 V).

   1 mark

2. (b) Chromium: $\mathrm{C}{\mathrm{r}}^{2+}$ is a strong reducing agent.

   ½ mark

3. $\mathrm{C}{\mathrm{r}}^{2+}$ ($d^4$) readily loses an electron to become $\mathrm{C}{\mathrm{r}}^{3+}$ ($d^3$). $d^3$ is a stable configuration with a half-filled $t_{2g}$ level.

   ½ mark

4. (c) $\mathrm{Z}{\mathrm{n}}^{2+}$ has the configuration $3d^{10}$, with completely filled d-orbitals and no unpaired electrons. No d–d transition is possible, so no visible light is absorbed and its salts are colourless.

   1 mark

Answer: (a) High ΔaH and ionisation enthalpy not balanced by hydration enthalpy; (b) Cr, since Cr²⁺ (d⁴) → Cr³⁺ (d³, stable half-filled t₂g); (c) 3d¹⁰, so no d–d transition

1. (a) $\mathrm{B}{\mathrm{r}}_2$ with heat brings about free-radical substitution at the benzylic carbon, which forms the most stable radical. Product: $\mathrm{Cl}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{CH}(\mathrm{Br})-\mathrm{C}{\mathrm{H}}_3$ [1-bromo-1-(4-chlorophenyl)ethane].

   1 mark

2. (b) HBr adds according to Markovnikov's rule: H goes to the =CH– carbon and Br to the methyl-bearing carbon, via a tertiary carbocation. Product: 1-bromo-1-methylcyclohexane, a cyclohexane ring with Br and $\mathrm{C}{\mathrm{H}}_3$ on the same carbon.

   1 mark

3. (c) Only the benzylic –OH is replaced by Cl. The phenolic –OH has partial C=O double bond character because of resonance and is not substituted. Product: $\mathrm{HO}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$ [4-(chloromethyl)phenol].

   1 mark

Answer: (a) Cl–C₆H₄–CHBr–CH₃; (b) 1-bromo-1-methylcyclohexane; (c) HO–C₆H₄–CH₂Cl

1. (a) Fittig reaction, heating chlorobenzene with sodium in dry ether:
   $2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Cl}+2\mathrm{Na}\stackrel{\text{dry ether}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{C}}_6{\mathrm{H}}_5+2\mathrm{NaCl}$

   1 mark

2. (b) Anti-Markovnikov addition of HBr in the presence of peroxide, then the Finkelstein reaction:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\stackrel{HBr/peroxide}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}\stackrel{\mathrm{NaI},\mathrm{dry}\mathrm{acetone}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{I}+\mathrm{NaBr}$

   1 mark

3. (c) Dehydrohalogenation with alcoholic KOH on heating; by Saytzeff's rule the more substituted alkene is the major product:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{Br})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3\stackrel{alc.\mathrm{KOH},\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHC}{\mathrm{H}}_3+\mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (a) Fittig reaction with Na/dry ether; (b) HBr/peroxide, then NaI/acetone; (c) alcoholic KOH, heat

1. (a) $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}<\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2<\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. A 2° amine has only one N–H and forms weaker hydrogen bonds than a 1° amine. O–H forms stronger hydrogen bonds than N–H because O is more electronegative.

   1 mark

2. (b)(i) Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, because the C–X bond has partial double bond character. So aromatic primary amines cannot be made by this route.

   1 mark

3. (b)(ii) In amides the lone pair on N is delocalised onto the carbonyl oxygen by resonance ($\mathrm{R}-\mathrm{C}(=\mathrm{O})-\mathrm{N}{\mathrm{H}}_2\leftrightarrow \mathrm{R}-\mathrm{C}({\mathrm{O}}^{-})=\mathrm{N}{\mathrm{H}}_2^{+}$). The lone pair is therefore less available for donation to a proton, so amides are less basic than amines.

   1 mark

Answer: (a) (CH₃)₂NH < CH₃CH₂NH₂ < CH₃CH₂OH; (b)(i) aryl halides do not undergo SN with phthalimide anion; (ii) resonance with C=O reduces N lone pair availability

1. (a) A native protein has its natural three-dimensional (secondary and tertiary) structure intact and is biologically active. In a denatured protein this structure is destroyed (only the primary structure remains), and it is biologically inactive.

   1 mark

2. (b) Lactose is a disaccharide (β-D-galactose + β-D-glucose). Glucose and fructose are monosaccharides, and amylose is a polysaccharide.

   1 mark

3. (c) Vitamin K is responsible for the coagulation of blood.

   1 mark

Answer: (a) Native: 3D structure intact, biologically active; denatured: 3D structure destroyed, inactive; (b) Lactose; (c) Vitamin K

1. (a)(i) The rate determining step is the slowest step in the mechanism of a reaction. It controls the overall rate.

   1 mark

2. (a)(ii) A complex reaction takes place through a series of two or more elementary steps rather than in a single step.

   1 mark

3. (b) The rate constant increases with increase in temperature, roughly doubling for every 10 K rise ($k=A{e}^{-E_a/RT}$).

   1 mark

4. (b) OR: Molecularity is the number of species colliding in a single elementary step, so it has meaning only for an elementary reaction. In a complex reaction each step may have a different molecularity. Order is found experimentally from the rate law, which is decided by the slowest step, so it applies to both elementary and complex reactions. (Alternative to (b) above; carries the same 1 mark.)

5. (c) For second order, Rate $=k[X{]}^2$. If $[X]$ becomes $3[X]$, Rate${}^{\prime }=k(3[X]{)}^2=9k[X{]}^2$. The rate of formation of Y becomes 9 times.

   1 mark

Answer: (a)(i) Slowest step; (ii) reaction proceeding in two or more elementary steps; (b) k increases with temperature / molecularity defined only for elementary steps, order is experimental; (c) 9 times

1. (a)(i) With $\mathrm{B}{\mathrm{r}}_2$ in $\mathrm{C}{\mathrm{S}}_2$ (low polarity solvent, low temperature), monobromination occurs and gives a mixture of 2-bromophenol and 4-bromophenol (para is the major product).

   1 mark

2. (a)(ii) With concentrated $\mathrm{HN}{\mathrm{O}}_3$, phenol is nitrated at all ortho and para positions to give 2,4,6-trinitrophenol (picric acid).

   1 mark

3. (b) The lone pair of electrons on the oxygen of phenol is delocalised into the benzene ring by resonance. It is therefore not easily available for accepting a proton, so phenol is not readily protonated.

   1 mark

4. (c) Phenol is the stronger acid. The $\mathrm{C}{\mathrm{H}}_3$ group in cresol is electron releasing (+I effect). It intensifies the negative charge on the phenoxide ion and destabilises it, so cresol is less acidic.

   1 mark

5. (c) OR: The Reimer–Tiemann reaction of phenol ($\mathrm{CHC}{\mathrm{l}}_3$/aq. NaOH, then ${\mathrm{H}}^{+}$) gives salicylaldehyde, IUPAC name 2-hydroxybenzaldehyde. (Alternative to (c) above; carries the same 1 mark.)

Answer: (a)(i) 2-bromophenol + 4-bromophenol; (ii) 2,4,6-trinitrophenol; (b) O lone pair delocalised into the ring; (c) phenol, since CH₃ (+I) destabilises cresoxide / 2-hydroxybenzaldehyde

1. (a) Sn is the anode and the hydrogen electrode is the cathode. Cell reaction (n = 2):
   $\mathrm{Sn}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to \mathrm{S}{\mathrm{n}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

2. Nernst equation:
   $E_{cell}=(E_c^{\circ }-E_a^{\circ })-\frac{0.059}{2}\log \frac{[\mathrm{S}{\mathrm{n}}^{2+}]}{[{\mathrm{H}}^{+}{]}^2}$
   $=[0-(-0.14)]-\frac{0.059}{2}\log \frac{0.004}{(0.02{)}^2}$

   1 mark

3. $=0.14-0.0295\log 10=0.14-0.0295$
   $E_{cell}=0.1105$ V

   1 mark

4. (b)(i) Oxygen evolution at the anode has a high overpotential, so in practice $\mathrm{C}{\mathrm{l}}^{-}$ is oxidised in preference to water and $\mathrm{C}{\mathrm{l}}_2$ is liberated.

   1 mark

5. (b)(ii) Conductivity is the conductance of unit volume of solution. On dilution the number of ions carrying current per unit volume decreases, even though ionisation of $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ increases. So conductivity decreases.

   1 mark

Answer: Sn + 2H⁺ → Sn²⁺ + H₂; E_cell = 0.1105 V; (b)(i) overpotential of O₂; (ii) fewer ions per unit volume on dilution

1. (a) Anode: $\mathrm{Pb}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2e^{-}$

   ½ mark

2. Cathode: $\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})+4{\mathrm{H}}^{+}(\mathrm{aq})+2e^{-}\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   ½ mark

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

   1 mark

4. (b) Nernst equation for the half cell (n = 6):
   $E=E^{\circ }-\frac{0.059}{6}\log \frac{[\mathrm{C}{\mathrm{r}}^{3+}{]}^2}{[\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}][{\mathrm{H}}^{+}{]}^{14}}$

   1 mark

5. $E=1.33-\frac{0.059}{6}\log \frac{(10^{-2}{)}^2}{(10^{-2})(10^{-4}{)}^{14}}=1.33-\frac{0.059}{6}\log 10^{54}$

   1 mark

6. $E=1.33-\frac{0.059\times 54}{6}=1.33-0.531=0.799$ V

   1 mark

Answer: Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; E = 0.799 V

1. (a) In tetrahedral complexes the crystal field splitting is small (${\Delta }_t\approx \frac{4}{9}{\Delta }_o$). It is never large enough to exceed the pairing energy and force electrons to pair, so low spin tetrahedral complexes are not formed.

   1 mark

2. (b) $\mathrm{C}{\mathrm{o}}^{2+}$ is $3d^7$. With a strong field ligand, ${\Delta }_o$ is large and one electron goes into a high-energy $e_g$ orbital, which it easily loses. This gives $\mathrm{C}{\mathrm{o}}^{3+}$ ($3d^6$, $t_{2g}^6{e}_g^0$), which has a very high crystal field stabilisation energy and is more stable. So $\mathrm{C}{\mathrm{o}}^{2+}$ is readily oxidised.

   1 mark

3. (c) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6][\mathrm{Cr}(\mathrm{CN}{)}_6]$ shows coordination isomerism, because the ligands can be interchanged between the cationic and anionic complex entities, e.g. $[\mathrm{Cr}(\mathrm{N}{\mathrm{H}}_3{)}_6][\mathrm{Co}(\mathrm{CN}{)}_6]$.

   1 mark

4. (d) Ni²⁺ is $3d^8$. ${\mathrm{H}}_2\mathrm{O}$ is a weak field ligand, so $[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ has 2 unpaired electrons, and d–d transitions absorb red light, so the solution looks green. $\mathrm{C}{\mathrm{N}}^{-}$ is a strong field ligand that pairs all the electrons (dsp², square planar). $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ has no unpaired electrons and no visible d–d absorption, so it is colourless.

   1 mark

5. (e) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{C}{\mathrm{O}}_3)]\mathrm{Cl}$: pentaamminecarbonatocobalt(III) chloride

   1 mark

Answer: (a) Δt < pairing energy; (b) strong field gives stable d⁶ Co³⁺; (c) coordination isomerism; (d) unpaired electrons in [Ni(H₂O)₆]²⁺, none in [Ni(CN)₄]²⁻; (e) Pentaamminecarbonatocobalt(III) chloride

1. (a) Chelate effect: A complex formed by a chelating (polydentate) ligand, which binds the metal through two or more donor atoms to form a ring, is more stable than a similar complex formed by comparable monodentate (non-chelating) ligands. This extra stability is called the chelate effect.

   1 mark

2. Example: $[\mathrm{Co}(\mathrm{en}{)}_3{]}^{3+}$ is more stable than $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$, because each ethane-1,2-diamine (en) forms a five-membered chelate ring with Co. (Another example: $[\mathrm{Ni}(\mathrm{en}{)}_3{]}^{2+}$ compared with $[\mathrm{Ni}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{2+}$.)

   1 mark

3. (b) In $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$: $x+6(-1)=-4$, so $x=+2$. Fe ($Z=26$) is $[\mathrm{Ar}]3d^{6}4s^2$, so $\mathrm{F}{\mathrm{e}}^{2+}$ is $[\mathrm{Ar}]3d^6$. $\mathrm{C}{\mathrm{N}}^{-}$ is a strong field ligand, so the six 3d electrons pair up in three 3d orbitals, leaving two 3d orbitals empty. Two 3d, one 4s and three 4p orbitals mix to accept six electron pairs from $\mathrm{C}{\mathrm{N}}^{-}$. Hybridisation: $d^{2}s{p}^3$ (inner orbital, octahedral).

   1 mark

4. All electrons are paired ($t_{2g}^6{e}_g^0$), with no unpaired electron, so the complex is diamagnetic.

   1 mark

5. (c) No precipitate of AgCl forms with $\mathrm{AgN}{\mathrm{O}}_3$. So no $\mathrm{C}{\mathrm{l}}^{-}$ is present outside the coordination sphere, and both Cl atoms are bonded to Pt along with the two $\mathrm{N}{\mathrm{H}}_3$. Formula: $[\mathrm{Pt}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{C}{\mathrm{l}}_2]$.

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Answer: (a) Extra stability of complexes with chelating ligands over those with non-chelating ligands, e.g. [Co(en)3]3+; (b) d2sp3, diamagnetic; (c) [Pt(NH3)2Cl2]

1. (a)(i) Ethanal → But-2-enal (aldol condensation): $2\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}\stackrel{\text{dil. NaOH}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (3-hydroxybutanal). On heating, this loses water: $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}\stackrel{\Delta ,-{\mathrm{H}}_2\mathrm{O}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHCHO}$ (but-2-enal).

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2. (a)(ii) Propanoic acid → Ethane (decarboxylation): $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$. Then $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COONa}\stackrel{\text{NaOH + CaO},\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_3+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$.

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3. (b) B gives a positive Fehling test, so it is an aldehyde. It also gives the iodoform test, so it contains a $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$ group. Hence B is ethanal, $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$.

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4. C does not give the Fehling test but forms iodoform, so it is a methyl ketone. The carbons left from ${\mathrm{C}}_5$ after removing ethanal's two carbons give C as propanone, $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$.

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5. Joining the two carbonyl carbons by a C=C gives A as 2-methylbut-2-ene, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3$ (${\mathrm{C}}_5{\mathrm{H}}_{10}$). Ozonolysis: $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{CHC}{\mathrm{H}}_3\stackrel{\text{(i) }{O}_3,{\text{(ii) Zn/H}}_2\text{O}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$.

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Answer: (a)(i) aldol condensation then dehydration; (ii) sodium salt + soda lime, heat. (b) A = 2-methylbut-2-ene (CH3)2C=CHCH3; B = ethanal CH3CHO; C = propanone CH3COCH3

1. C dehydrates to but-1-ene, so C is a straight-chain four-carbon primary alcohol. Oxidation of C gives the acid B, so B has the same four-carbon chain. Hence C = butan-1-ol, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$.

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2. B = butanoic acid, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$.

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3. A (${\mathrm{C}}_8{\mathrm{H}}_{16}{\mathrm{O}}_2$) is the ester made from B (4 C) and C (4 C). So A = butyl butanoate, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$.

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4. Hydrolysis: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COO}{\mathrm{C}}_4{\mathrm{H}}_9+{\mathrm{H}}_2\mathrm{O}\stackrel{\text{dil. }{H}_{2}S{O}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$

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5. Oxidation: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{chromic acid }(\mathrm{Cr}{\mathrm{O}}_3/{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4)}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}$

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6. Dehydration: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{conc. }{H}_{2}S{O}_4,\Delta }{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

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Answer: A = butyl butanoate (C3H7COOC4H9); B = butanoic acid (C3H7COOH); C = butan-1-ol (C4H9OH)