Question

What is meant by lanthanoid contraction? (ii) Why do transition metals form coloured compounds? (iii) Why are $E^\circ_{M^{2+}/M}$ values for Mn and Zn more negative than expected? (iv) Which is the most stable oxidation state of Cu and why? (v) Why is Ce$^{4+}$ in aqueous solution a good oxidising agent?

Hint:

Important facts: \[ \boxed{4f\text{ electrons show poor shielding effect}} \] causing: \[ \boxed{\text{Lanthanoid contraction}} \] Colour in transition compounds: \[ \boxed{d-d\text{ transition}} \] Stable configurations: \[ \boxed{d^5=\text{half-filled, }d^{10}=\text{fully filled}} \] Both provide extra stability.

Answer 1

Step 1: Lanthanoid contraction and coloured compounds.
As atomic number increases across lanthanoids, $4f$ electrons are added but shield poorly. Rising nuclear charge pulls outer electrons inward, causing a steady decrease in ionic radius from $La^{3+}$ to $Lu^{3+}$ called lanthanoid contraction. Transition metal compounds are coloured because ligand fields split $d$ orbitals; electrons absorb specific visible-light wavelengths during $d$-$d$ transitions and the complementary colour is observed.
Step 2: Negative $E^\circ$ of Mn and Zn.
$Mn^{2+}$ retains a stable half-filled $3d^5$ configuration and $Zn^{2+}$ has a stable full-filled $3d^{10}$ configuration. These extra stabilities increase ionisation energy, making the metals harder to oxidise and resulting in more negative $E^\circ_{M^{2+}/M}$ values than the general trend predicts.
Step 3: Stable oxidation state of Cu.
$Cu^+$ ($3d^{10}$) disproportionates in water: $2Cu^+ \rightarrow Cu^{2+} + Cu$. $Cu^{2+}$ is stabilised in aqueous solution by its high hydration enthalpy, which more than compensates the second ionisation energy. Hence $Cu^{2+}$ is the most stable oxidation state.
Step 4: Oxidising nature of $Ce^{4+}$.
$Ce^{4+}$ reduces readily to $Ce^{3+}$ with a high positive reduction potential, meaning it accepts electrons easily from other species. \[ \boxed{Ce^{4+} + e^- \rightarrow Ce^{3+}} \]