d-Block (Transition Metals) & f-Block

Q1 (10)

(a) Colourless octahedral complex requires no possible d–d transition, which happens when the d-subshell is either empty ($d^0$) or completely full ($d^{10}$). Diamagnetic requires zero unpaired electrons. Both conditions are simultaneously satisfied only by $d^0$ and $d^{10}$. (2 marks conditions; 2 marks the two configs)

(b) $Sc^{3+}$ is $d^0$, $Zn^{2+}$ is $d^{10}$ — both colourless & diamagnetic. The clues "coloured metal vapour" and "coloured lower oxidation states" point to a metal with several accessible d-populated oxidation states. Sc has essentially only +3 (no coloured lower states in solution chemistry), whereas... actually the decisive clue is "coloured lower oxidation states": Zn has no lower coloured states either, BUT the metal capable of coloured lower states with $d^{10}$ as the referenced ion is copper: $Cu^{2+}(d^9)$ is blue. The consistent identity is $Cu^{+}$ ($d^{10}$, colourless) whose lower... — accept: the ion is $d^{10}$ = Cu⁺, since $Cu^{2+}$ (higher, $d^9$) is coloured, and Cu shows variable coloured states; Sc⁺/Sc²⁺ do not exist stably → Sc problematic. (3: identify $d^{10}$/Cu⁺ reasoning 2, Sc problem 1)

(c) $d^5$, weak field octahedral → high spin, $n=5$ unpaired. $\mu = \sqrt{n(n+2)} = \sqrt{5\times7} = \sqrt{35} = 5.92\ \text{BM}$ Coloured: yes — partially filled d-orbitals permit d–d transitions. (calc 2, colour 1)

Q2 (12)

(a) $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$ (3: Cr balance, Fe stoich 6, charge/H₂O)

(b) Moles $Cr_2O_7^{2-} = 0.0184 \times 0.0200 = 3.68\times10^{-4}$ mol. Ratio $Fe^{2+}:Cr_2O_7^{2-} = 6:1$ → moles $Fe^{2+} = 6\times3.68\times10^{-4} = 2.208\times10^{-3}$ mol. $M = \frac{2.208\times10^{-3}}{0.0250} = 0.0883\ \text{M}$ (setup 2, answer 2)

(c) $MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$ Moles $MnO_4^-$ needed $= \dfrac{2.208\times10^{-3}}{5} = 4.416\times10^{-4}$ mol. $V = \frac{4.416\times10^{-4}}{0.0200} = 0.02208\ \text{L} = 22.08\ \text{mL}$ (eqn 1.5, volume 1.5)

(d) $KMnO_4$ is its own indicator: excess purple $MnO_4^-$ gives a persistent pink endpoint; dichromate's orange→green colour change is not sharp, so a redox indicator (e.g. diphenylamine) is required. (2)

Q3 (10)

(a) From 4d to 5d, the intervening lanthanide contraction (filling of 4f orbitals between La and Hf) causes a steady size decrease that offsets the normal size increase expected on going down a group. Poor shielding by 4f electrons increases effective nuclear charge, pulling the outer electrons in, so Hf ≈ Zr and Au ≈ Ag in size. (effect named 2, mechanism/f-shielding 2)

(b) $Zr^{4+}$ and $Hf^{4+}$ are nearly identical in size (both from contraction), so their salts have almost identical solubilities and crystallise together → hard to separate (both equally). Consequence: they form near-identical coordination compounds (same geometry, similar stability constants), always occurring together in ores. (3)

(c) Ce can access the +4 state by losing its single 4f electron to attain the stable $4f^0$ (empty) configuration; the partial adoption of $Ce^{4+}$ character / greater nuclear pull on remaining electrons contracts its radius relative to the smooth trivalent trend. (3)

Q4 (10)

(a) Step 1: $SO_2 + V_2O_5 \rightarrow SO_3 + V_2O_4$ (V: +5 → +4) Step 2: $2V_2O_4 + O_2 \rightarrow 2V_2O_5$ (V: +4 → +5, regenerated) Variable oxidation state is essential because the catalyst must be reduced then re-oxidised, cycling between V(V) and V(IV) to transfer oxygen; a fixed-oxidation-state metal cannot complete the cycle. (steps + OS 2, reasoning 2)

(b) $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$ Initially slow (kinetic barrier at electrode of like-charged $MnO_4^-$/$C_2O_4^{2-}$); once $Mn^{2+}$ forms it catalyses further reaction (Mn²⁺ mediates electron transfer via intermediate Mn(III)) → autocatalysis, catalyst = $Mn^{2+}$. (balance 2, autocatalysis + species 2)

(c) Mn: $[Ar]3d^5 4s^2$, 7 valence electrons → highest OS +7, occurring in $KMnO_4$ (or $Mn_2O_7$). (2)

Q5 (8)

(a) $Eu$: $[Xe]4f^7 6s^2$ — half-filled $4f^7$ is extra stable. $Gd$: $[Xe]4f^7 5d^1 6s^2$ — the added electron enters 5d rather than pairing in 4f, preserving the stable half-filled $4f^7$ core. (each config 1, stability reasoning 2)

(b) Reasons: (i) In actinides the 5f, 6d, 7s orbitals lie close in energy so more electrons are available for bonding; (ii) 5f electrons are less shielded / more spatially extended than 4f, so they participate in bonding and give a wider range of oxidation states. Uranium shows +6 (e.g. $UO_2^{2+}$/$UF_6$), not accessible to lanthanide analogues (which are mostly +3). (2 reasons 2, U state 2)

[
  {"claim":"Q1c spin-only moment for d5 high spin = sqrt(35) ~ 5.92 BM","code":"n=5; mu=sqrt(n*(n+2)); result = abs(float(mu)-5.916) < 0.01"},
  {"claim":"Q2b Fe2+ molarity = 0.0883 M","code":"nCr=0.0184*0.0200; nFe=6*nCr; M=nFe/0.0250; result = abs(M-0.08832) < 0.001"},
  {"claim":"Q2c KMnO4 volume = 22.08 mL","code":"nFe=6*0.0184*0.0200; nMn=nFe/5; V=nMn/0.0200; result = abs(V*1000-22.08) < 0.05"},
  {"claim":"Q4b oxalate redox electron balance: 2*5 e = 5*2 e","code":"result = (2*5)==(5*2)"}
]