Level 5 — Masteryd-Block (Transition Metals) & f-Block

d-Block (Transition Metals) & f-Block

45 marksprintable — key stays hidden on paper

Mastery Examination (Level 5)

Time: 60 minutes Total Marks: 45

Instructions: Answer all questions. Show all reasoning. Use $\mu = \sqrt{n(n+2)}$ BM for spin-only magnetic moments. Atomic numbers where needed are given.

Question 1 — Magnetic moments, oxidation states & a computational model (18 marks)

A researcher tabulates the observed room-temperature magnetic moments (in BM) of aqueous complexes of first-row transition metal ions. She wants to reverse-engineer the number of unpaired electrons and hence the oxidation state.

(a) Derive the spin-only formula $\mu = \sqrt{n(n+2)}$ starting from $\mu_S = g\sqrt{S(S+1)}\,\mu_B$ with $g = 2$ and $S = n/2$ , showing every algebraic step. (3)

(b) For each ion below, give the ground-state $d$ -electron count, the number of unpaired electrons $n$ (assume weak-field / high-spin where a choice exists), and the predicted $\mu$ to 2 d.p.: $\mathrm{Ti^{3+}}$ , $\mathrm{Mn^{2+}}$ , $\mathrm{Fe^{2+}}$ , $\mathrm{Cu^{2+}}$ . (Z: Ti 22, Mn 25, Fe 26, Cu 29.) (6)

(c) A measured moment is $3.87$ BM. Solve $\sqrt{n(n+2)} = 3.87$ for the nearest integer $n$ , and name a first-row ion consistent with this value. (3)

(d) Write a short pseudocode / Python-style function unpaired_from_mu(mu) that returns the integer $n$ minimising $|\sqrt{n(n+2)} - \mu|$ over $n \in \{0,1,2,3,4,5\}$ , and state what unpaired_from_mu(2.83) returns. (3)

(e) Explain in physical terms why $\mathrm{Zn^{2+}}$ salts are diamagnetic and colourless, linking both facts to a single electronic feature. (3)

Question 2 — Oxidising power of $\mathrm{K_2Cr_2O_7}$ / $\mathrm{KMnO_4}$ : stoichiometry & redox proof (15 marks)

(a) Balance the acidified reaction of dichromate with iron(II): $\mathrm{Cr_2O_7^{2-} + Fe^{2+} + H^+ \longrightarrow Cr^{3+} + Fe^{3+} + H_2O}$ Show the half-reactions and prove electron and charge balance. (5)

(b) A $0.250\,\mathrm{g}$ sample of iron ore is dissolved and all iron reduced to $\mathrm{Fe^{2+}}$ . It is titrated against $0.0200\ \mathrm{M}\ \mathrm{K_2Cr_2O_7}$ , requiring $22.4\ \mathrm{mL}$ to reach the end point. Compute the mass percentage of iron in the ore (Fe = 55.85). (5)

(c) Write the balanced equation for the reaction of $\mathrm{MnO_4^-}$ with oxalate ( $\mathrm{C_2O_4^{2-}}$ ) in acid. State how many moles of oxalate are oxidised per mole of permanganate and explain why this titration must be warmed to ~60 °C whereas the dichromate–iron titration need not be — refer to autocatalysis. (5)

Question 3 — Periodicity across f-block: contraction, size argument & actinide comparison (12 marks)

(a) Explain the lanthanide contraction in terms of imperfect shielding by $4f$ electrons. Give two distinct chemical/physical consequences, one of which must involve the anomalous similarity of a 4d and 5d element pair. (5)

(b) The ionic radii ( $\mathrm{Ln^{3+}}$ , in pm) fall roughly linearly: $\mathrm{La^{3+}}=103.2$ , $\mathrm{Lu^{3+}}=86.1$ , across the 15 lanthanides. Estimate the average contraction per unit increase in atomic number and use a linear model to predict $r(\mathrm{Gd^{3+}})$ (Gd is the 8th, i.e. 7 steps from La). (4)

(c) Give one electronic-configuration reason and one reactivity/oxidation-state reason why actinides show more variable oxidation states than lanthanides. (3)

Answer keyMark scheme & solutions

Question 1

(a) [3] Start: $\mu_S = g\sqrt{S(S+1)}\,\mu_B$ . Put $g=2$ : $\mu_S = 2\sqrt{S(S+1)}$ . (1) Each unpaired electron contributes spin $1/2$ , so $S = n/2$ . Substitute: $\mu = 2\sqrt{\tfrac{n}{2}\left(\tfrac{n}{2}+1\right)} = 2\sqrt{\tfrac{n}{2}\cdot\tfrac{n+2}{2}} = 2\cdot\tfrac{1}{2}\sqrt{n(n+2)}.$ (1) $\boxed{\mu = \sqrt{n(n+2)}\ \text{BM}}$ (1)

(b) [6] (½ config + ½ n + ½ μ each ≈)

Ion	$d$ -count	$n$	$\mu=\sqrt{n(n+2)}$
$\mathrm{Ti^{3+}}$	$d^1$	1	$\sqrt3=1.73$
$\mathrm{Mn^{2+}}$	$d^5$	5	$\sqrt{35}=5.92$
$\mathrm{Fe^{2+}}$	$d^6$ (HS)	4	$\sqrt{24}=4.90$
$\mathrm{Cu^{2+}}$	$d^9$	1	$\sqrt3=1.73$

(c) [3] $n(n+2)=3.87^2=14.98\approx15 \Rightarrow n=3$ (since $3\cdot5=15$ ). (2) Ion: $\mathrm{Cr^{3+}}$ ( $d^3$ ) or $\mathrm{V^{2+}}$ ( $d^3$ ) or $\mathrm{Ni^{2+}}$ ( $d^8$ , $n=2$ ? no — must have $n=3$ ). Correct: $\mathrm{Cr^{3+}}$ / $\mathrm{V^{2+}}$ . (1)

(d) [3]

import math
def unpaired_from_mu(mu):
    return min(range(6), key=lambda n: abs(math.sqrt(n*(n+2)) - mu))

(2) unpaired_from_mu(2.83) → 2 (since $\sqrt{8}=2.83$ ). (1)

(e) [3] $\mathrm{Zn^{2+}}$ is $d^{10}$ — the $d$ subshell is completely filled. (1) All electrons paired ⇒ no net spin ⇒ diamagnetic. (1) With no empty/partially-filled $d$ orbital of the correct energy, no $d$ – $d$ electronic transition is possible, so no visible light is absorbed ⇒ colourless. Single feature: full $d^{10}$ . (1)

Question 2

(a) [5] Reduction: $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O}$ (2) Oxidation: $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$ (×6) (1) Add: $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \to 2Cr^{3+} + 6Fe^{3+} + 7H_2O}$ (1) Charge check LHS $=-2+14+12=+24$ ; RHS $=+6+18=+24$ ✓; electrons 6=6 ✓. (1)

(b) [5] Moles dichromate $=0.0200\times0.0224 = 4.48\times10^{-4}$ mol. (1) Ratio Fe:Cr₂O₇ = 6:1 ⇒ mol Fe $=6\times4.48\times10^{-4}=2.688\times10^{-3}$ . (1) Mass Fe $=2.688\times10^{-3}\times55.85 = 0.1501\,\mathrm g$ . (1) % Fe $=0.1501/0.250\times100 = 60.05\%$ . (2)

(c) [5] $\mathrm{2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O}$ (2) Ratio: 5 mol oxalate per 2 mol $\mathrm{MnO_4^-}$ ⇒ 2.5 mol oxalate per mole permanganate. (1) Warming needed: the permanganate–oxalate reaction is intrinsically slow at room T; it is autocatalysed by the $\mathrm{Mn^{2+}}$ produced, so heating to ~60 °C starts the reaction fast enough before enough catalyst forms. (1) The dichromate–Fe²⁺ reaction is fast at room T and not autocatalytic, so no heating is required. (1)

Question 3

(a) [5] As atomic number rises across the 4f series, each added $4f$ electron shields the increasing nuclear charge poorly (diffuse, penetrating shape), so effective nuclear charge felt by outer electrons rises steadily and the radius contracts — the lanthanide contraction. (2) Consequences (2 needed): (i) Zr and Hf (and Nb/Ta, Mo/W) have almost identical radii and very similar chemistry, making separation difficult — this is the required 4d/5d pair similarity. (2) (ii) Steady decrease in basicity of $\mathrm{Ln(OH)_3}$ / difficulty of separating adjacent lanthanides. (1)

(b) [4] Total contraction $=103.2-86.1=17.1$ pm over 14 steps (La→Lu). (1) Per-step $=17.1/14 = 1.221$ pm. (1) $r(\mathrm{Gd^{3+}}) = 103.2 - 7\times1.221 = 103.2 - 8.55 = 94.65\,\mathrm{pm}$ . (2) (≈94.6 pm; literature ≈93.8 pm — good agreement.)

(c) [3] Electronic: in actinides the $5f$ , $6d$ and $7s$ levels are close in energy so more electrons are available for bonding ⇒ many oxidation states (e.g. U: +3,+4,+5,+6). (1.5) Reactivity: $5f$ electrons are less well shielded / more "available" (larger, more diffuse) than the deeply buried $4f$ , so actinides readily lose variable numbers of electrons, unlike lanthanides that are near-uniformly +3. (1.5)

[
  {"claim":"mu for n=5 (Mn2+) is sqrt35 ~5.92","code":"result = round(float(sqrt(35)),2)==5.92"},
  {"claim":"mu for n=4 (Fe2+ HS) is sqrt24 ~4.90","code":"result = round(float(sqrt(24)),2)==4.90"},
  {"claim":"n=3 gives mu 3.87","code":"result = round(float(sqrt(3*(3+2))),2)==3.87"},
  {"claim":"iron percentage in ore is 60.05","code":"mol_cr=0.0200*0.0224; mol_fe=6*mol_cr; mass=mol_fe*55.85; pct=mass/0.250*100; result = abs(pct-60.05)<0.1"},
  {"claim":"Gd3+ radius linear estimate ~94.65 pm","code":"r=103.2-7*(103.2-86.1)/14; result = abs(r-94.65)<0.05"}
]