d-Block (Transition Metals) & f-Block

Q1 [8]

Cr (Z=24): naïve = $[\text{Ar}]3d^4 4s^2$; actual = $[\text{Ar}]3d^5 4s^1$. [1] Cu (Z=29): naïve = $[\text{Ar}]3d^9 4s^2$; actual = $[\text{Ar}]3d^{10} 4s^1$. [1]

Why (exchange energy + symmetry): The 3d and 4s orbitals are very close in energy. A half-filled ($d^5$) or completely-filled ($d^{10}$) set has:

• Symmetrical charge distribution (extra stability), and
• Maximum exchange energy — the number of exchange pairs of parallel-spin electrons is maximised. Exchange energy is a stabilising (negative) contribution that scales with the number of same-spin electron pairs $\binom{k}{2}$. Promoting one 4s electron to give $d^5$ (5 parallel) or $d^{10}$ costs little (4s↔3d near-degenerate) but gains substantial exchange stabilisation. [3 for the exchange + symmetry reasoning]

General formula: $(n-1)d^{1-10}\,ns^{0-2}$. [1] Permitted because the (n−1)d subshell (ℓ=2, 5 orbitals holding up to 10 e⁻) lies energetically near the ns; incomplete d-subshells define the block. [1]

Q2 [10]

(a) In transition metals the (n−1)d and ns electrons have comparable energies, so both sets can participate in bonding; successive ionisation enthalpies increase only gradually. Hence a range of oxidation states differing by 1 occurs. In s-block, only ns electrons are available and their removal to reach noble-gas core is heavily favoured → single (or two) fixed states. [3]

(b) Mn: $[\text{Ar}]3d^5 4s^2$. Maximum OS = (5 d) + (2 s) = +7, using all 3d + 4s electrons; configuration fully utilised is $3d^5 4s^2 →$ loses 7 e⁻. [3]

(c) In acidic solution the +2 state of Mn is most stable (half-filled $3d^5$ configuration, extra stability); +7 (MnO₄⁻) is a strong oxidant → readily reduced. [2]

• Mn²⁺ ($d^5$) is unusually stable → Mn³⁺ ($d^4$) is a strong oxidant (unstable). [1]
• Fe³⁺ ($d^5$, half-filled) is more stable than Fe²⁺ ($d^6$); hence Fe²⁺ is oxidisable. Anomaly explained by the special stability of the half-filled $d^5$ set at Mn²⁺ and Fe³⁺. [1]

Q3 [9]

(a) For $n$ unpaired electrons, total spin $S = n/2$. Ignoring orbital contribution, $\mu = g\sqrt{S(S+1)}$ BM with $g=2$, i.e. $\mu = 2\sqrt{S(S+1)} = \sqrt{4S(S+1)}$. Substituting $S=n/2$: $\mu = \sqrt{4\cdot\tfrac{n}{2}\left(\tfrac{n}{2}+1\right)} = \sqrt{n(n+2)}\ \text{BM}.$ [3]

(b)

• Ti³⁺: $d^1$, n=1 → $\sqrt{3}=1.73$ BM. [1]
• Fe³⁺: $d^5$, n=5 → $\sqrt{35}=5.92$ BM. [1.5]
• Ni²⁺: $d^8$, n=2 → $\sqrt{8}=2.83$ BM. [1.5]

(c) Fe³⁺ gives 5.92 BM. First-row ions: orbital angular momentum is largely quenched by the ligand field / crystal field, so spin-only ≈ experimental. Lanthanides: the 4f orbitals are buried/shielded and orbital angular momentum is not quenched; $L$ contributes, so one must use $\mu = g_J\sqrt{J(J+1)}$ (total $J$), and spin-only fails. [2]

Q4 [8]

(a) Cause: across the 4f series, added electrons enter the inner 4f orbitals which shield the nucleus poorly (diffuse, poor shielding); effective nuclear charge felt by outer electrons rises steadily → steady decrease in size of atoms/ions (lanthanide contraction). [2] Consequences: [1.5 each]

1. The 5d elements (2nd & 3rd transition series) end up with nearly identical atomic/ionic radii to the 4d ones (e.g. Zr≈Hf, Nb≈Ta) → very similar chemistry, hard to separate.
2. Basicity of Ln(OH)₃ decreases Ce→Lu: as ionic radius falls, charge density/covalent character rises, so La(OH)₃ is most basic, Lu(OH)₃ least basic (most acidic).

(b) Total contraction = $103 - 86 = 17$ pm over Ce→Lu (13 steps, 14 elements). [1] Average $= 17/13 \approx 1.31$ pm per step (≈1.2 pm if divided by 14). [1] The actual per-step contraction is not uniform — larger for early members and smaller later. [1]

Q5 [9]

(a) $\text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O$. n = 5 electrons → n-factor 5; equivalent weight $=M/5 = 158/5 = 31.6\ \text{g/eq}$. [3]

(b) $2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16H^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8H_2O$. [3] (balance: 2×5=10 e⁻ = 5×2)

(c) $\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14H^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7H_2O$. [2] n-factor of dichromate = 6. [1]

Q6 [6]

Why greater variability: In early actinides the 5f, 6d, 7s energies are close and 5f electrons less tightly bound / less shielded than 4f, so more f electrons can participate in bonding → higher/variable oxidation states (e.g. U⁶⁺). [3]

[
 {"claim":"Fe3+ (d5) spin-only = sqrt(35) ≈ 5.92","code":"mu=sqrt(5*(5+2)); result = abs(float(mu)-5.916)<0.01"},
 {"claim":"Ni2+ (d8, n=2) spin-only = sqrt(8) ≈ 2.83","code":"mu=sqrt(2*(2+2)); result = abs(float(mu)-2.828)<0.01"},
 {"claim":"Ti3+ (d1) spin-only = sqrt(3) ≈ 1.73","code":"mu=sqrt(1*(1+2)); result = abs(float(mu)-1.732)<0.01"},
 {"claim":"MnO4- to Mn2+ transfers 5 electrons; eq wt = 158/5 = 31.6","code":"result = Rational(158,5)==Rational(316,10)"},
 {"claim":"Avg lanthanide contraction per step = (103-86)/13 ≈ 1.31 pm","code":"result = abs(float((103-86)/13)-1.3077)<0.01"}
]