Check Statement I

Statement I says: Iron (III) catalyst, acidified $\mathrm{K_2Cr_2O_7}$ and neutral $\mathrm{KMnO_4}$ can independently oxidise $\mathrm{I^-}$ to $\mathrm{I_2}$.

We examine each oxidising agent:

• Acidified $\mathrm{K_2Cr_2O_7}$ is a strong oxidising agent and does oxidise iodide to iodine: $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O}$ So this part is true.

• Neutral $\mathrm{KMnO_4}$ also oxidises iodide to iodine: $\mathrm{2MnO_4^- + 6I^- + 4H_2O \rightarrow 2MnO_2 + 3I_2 + 8OH^-}$ So this part is true.

• Iron(III): $\mathrm{Fe^{3+}}$ can oxidise iodide to iodine because $\mathrm{2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2}$ This is feasible since $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=+0.77\,\text{V}, \qquad E^\circ(\mathrm{I_2/I^-})=+0.54\,\text{V}$ Hence, $E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23\,\text{V} > 0$ Therefore $\mathrm{Fe^{3+}}$ can oxidise $\mathrm{I^-}$ to $\mathrm{I_2}$.

Thus, Statement I is true.

Check Statement II

Statement II says: Manganate ion is paramagnetic in nature and involves $\mathrm{p\pi-d\pi}$ bonding?

But the given statement says it involves $\mathrm{p\pi-p\pi}$ bonding.

For manganate ion, $\mathrm{MnO_4^{2-}}$:

• Oxidation state of Mn is $+6$.
• Mn electronic configuration in ground state: $[Ar]3d^54s^2$
• Therefore $\mathrm{Mn^{6+}}: [Ar]3d^1$

So manganate has one unpaired electron, hence it is paramagnetic.

However, the multiple bonding in oxyanions of transition metals like $\mathrm{MnO_4^{2-}}$ is described as $p\pi-d\pi$ bonding, not $p\pi-p\pi$ bonding.

Therefore, Statement II is false.

Conclusion

• Statement I: True
• Statement II: False

Hence the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

Stored correct answer is B, but our derived answer is C.

So the stored answer appears to be incorrect.