Question

(i) Why do Zr and Hf exhibit similar properties ?

(ii) "\( E^\circ \) for \( \mathrm{Mn^{3+}/Mn^{2+}} \) couple is more positive than that for \( \mathrm{Fe^{3+}/Fe^{2+}} \)." Justify the statement.

[Atomic number : Mn = 25, Fe = 26]

Hint:

Think of lanthanoid contraction for Zr and Hf. For Mn and Fe, compare the d-configurations before and after reduction and look for the half-filled d5 set.

Answer 1

Step 1: Understanding the Concept:
Zr (Z = 40) and Hf (Z = 72) are the second and third members of group 4. Their similarity comes from the way atomic size changes down the group. The second part compares the stability of d-configurations after reduction.

Step 2: Part (i) - Why Zr and Hf are similar:
Between La and Hf, fourteen 4f electrons are added. The 4f electrons shield the nuclear charge poorly.
So the effective nuclear charge rises and the atoms shrink. This steady shrinking across the lanthanoids is the lanthanoid contraction.
Because of it, Hf (about 159 pm) has almost the same atomic and ionic radius as Zr (about 160 pm).
Similar size and the same valence electron pattern give them nearly identical chemical properties. They occur together in nature and are very hard to separate.

Step 3: Part (ii) - Write the electronic configurations:
\( \mathrm{Mn^{3+}} \) (Z = 25) is \( 3d^4 \) and \( \mathrm{Mn^{2+}} \) is \( 3d^5 \).
\( \mathrm{Fe^{3+}} \) (Z = 26) is \( 3d^5 \) and \( \mathrm{Fe^{2+}} \) is \( 3d^6 \).

Step 4: Part (ii) - Compare the two reductions:
\[ \mathrm{Mn^{3+}} (3d^4) + e^- \rightarrow \mathrm{Mn^{2+}} (3d^5) \]
Here the product has the extra-stable half-filled \( d^5 \) set. So this reduction is very easy, and \( E^\circ \) is high (+1.57 V).
\[ \mathrm{Fe^{3+}} (3d^5) + e^- \rightarrow \mathrm{Fe^{2+}} (3d^6) \]
Here the reactant already has the stable half-filled \( d^5 \) set, and reduction destroys it. So this reduction is harder, and \( E^\circ \) is lower (+0.77 V).
Hence \( E^\circ \) for \( \mathrm{Mn^{3+}/Mn^{2+}} \) is more positive than for \( \mathrm{Fe^{3+}/Fe^{2+}} \).

Final Answer:
(i) Lanthanoid contraction makes the radii of Zr and Hf almost equal. (ii) \( \mathrm{Mn^{2+}} \) gains the stable half-filled \( 3d^5 \) configuration, while \( \mathrm{Fe^{3+}} \) loses it on reduction.