The main consequence of lanthanoid contraction is that:

Zr and Hf have nearly the same radius

KMnO₄ titrations are not done in HCl because:

HCl is oxidised to Cl₂ by KMnO₄

The d- and f-Block Elements

1. Transition Elements

Elements with partly filled d orbitals in the atom or a common ion. General configuration $(n-1)d^{1-10}ns^{1-2}$. Zn, Cd and Hg ($d^{10}$ in atom and ions) are d-block elements but not usually counted as transition elements.

2. Characteristic Properties

Variable oxidation states: both $(n-1)d$ and $ns$ electrons can be used. Mn shows +2 to +7, the widest range in the 3d series.
Coloured ions: d–d transitions absorb visible light. $Sc^{3+}$, $Ti^{4+}$, $Zn^{2+}$ and $Cu^+$ are colourless ($d^0$ or $d^{10}$).
Paramagnetism: unpaired d electrons; $\mu = \sqrt{n(n+2)}$ BM.
Catalytic activity: variable oxidation states and the ability to adsorb reactants. Fe in the Haber process, $V_2O_5$ in the contact process, Ni in hydrogenation.
Interstitial compounds: small atoms (H, C, N) trapped in the metal lattice; hard, high-melting.
Alloys form readily because the atomic radii are similar.
High melting points from strong metallic bonding; Cr, Mo and W are among the highest.

3. Trends in the 3d Series

Atomic radii decrease slightly and then level off, because added d electrons shield poorly.
$E^\circ(M^{2+}/M)$ for Cu is positive (+0.34 V), so Cu does not liberate $H_2$ from dilute acids.
$Mn^{2+}$ ($d^5$) is especially stable, and $Fe^{3+}$ ($d^5$) is more stable than $Fe^{2+}$.

4. Potassium Dichromate and Potassium Permanganate

$K_2Cr_2O_7$ is made from chromite ore ($FeCr_2O_4$). In solution, $Cr_2O_7^{2-}$ (orange) and $CrO_4^{2-}$ (yellow) interconvert with pH: acid favours dichromate, base favours chromate. A strong oxidising agent in acid, used in volumetric analysis.
$KMnO_4$ is made from pyrolusite ($MnO_2$) by fusion with KOH to give green $K_2MnO_4$, followed by oxidation. In acidic medium it oxidises $Fe^{2+}$, oxalate and iodide; it acts as its own indicator (purple to colourless). Titrations use $H_2SO_4$, not HCl, because HCl would be oxidised to $Cl_2$.

5. Lanthanoids (4f)

Common oxidation state +3. Ce shows +4 (noble-gas configuration), Eu and Yb show +2.
Lanthanoid contraction: a steady decrease in atomic and ionic radii from La to Lu, because 4f electrons shield poorly. Consequences: 4d and 5d elements of the same group have almost identical radii (Zr 160 pm, Hf 159 pm), so Zr and Hf are very hard to separate; basicity of $Ln(OH)_3$ decreases from La to Lu.

6. Actinoids (5f)

Show a wider range of oxidation states than lanthanoids, because 5f, 6d and 7s orbitals are close in energy. All are radioactive. The actinoid contraction is greater than the lanthanoid contraction, because 5f electrons shield even more poorly.

Practice questions

Which ion is colourless in aqueous solution?
1. Ti³⁺
2. Cu²⁺
3. Sc³⁺
4. Fe²⁺
Answer
C. Sc³⁺ is 3d⁰, so no d–d transition is possible.
The spin-only magnetic moment of Fe³⁺ (Z = 26) is approximately:
1. 2.83 BM
2. 4.90 BM
3. 5.92 BM
4. 3.87 BM
Answer
C. Fe³⁺ is 3d⁵ with 5 unpaired electrons: √(5 × 7) = √(35) ≈ 5.92 BM.
The main consequence of lanthanoid contraction is that:
1. Zr and Hf have nearly the same radius
2. Lanthanoids are all coloured
3. Actinoids are radioactive
4. La has the largest IE
Answer
A. The poor shielding by 4f electrons cancels the expected increase in size from the 4d to the 5d series.
In acidic medium, the colour change when K₂Cr₂O₇ acts as an oxidant is:
1. Orange to green
2. Yellow to orange
3. Purple to colourless
4. Green to orange
Answer
A. Orange Cr₂O₇²⁻ is reduced to green Cr³⁺.
Which 3d element shows the largest number of oxidation states?
1. Sc
2. Cr
3. Mn
4. Zn
Answer
C. Manganese shows +2 to +7.
KMnO₄ titrations are not done in HCl because:
1. HCl is a weak acid
2. HCl is oxidised to Cl₂ by KMnO₄
3. HCl reduces Fe³⁺
4. HCl precipitates MnO₂
Answer
B. Permanganate would oxidise chloride, consuming extra oxidant and giving wrong results.