Idea: An oxidising agent is a species that gets reduced itself.

So, among the given ions, we check which ones can readily go to a lower oxidation state.

Use the stability trend due to inert pair effect for heavier p-block elements:

• For Group 14:
  • Lower oxidation state $+2$ becomes more stable down the group.
  • So for Sn and Pb, the higher oxidation state $+4$ tends to get reduced to $+2$.
• For Group 13:
  • Lower oxidation state $+1$ becomes more stable down the group.
  • So for Tl, the $+3$ state tends to get reduced to $+1$.

Now examine each ion:

(i) $\mathrm{Sn}^{4+}$ $\mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+}$ Since $+2$ is relatively stable, $\mathrm{Sn}^{4+}$ can act as an oxidising agent. ✅

(ii) $\mathrm{Sn}^{2+}$ It is more likely to be oxidised to $\mathrm{Sn}^{4+}$ or act as a reducing agent, not as an oxidising agent. ❌

(iii) $\mathrm{Pb}^{2+}$ For lead, due to strong inert pair effect, $+2$ is more stable than $+4$. So $\mathrm{Pb}^{2+}$ does not tend to get reduced further easily; it is not expected to behave as an oxidising agent in this context. ❌

(iv) $\mathrm{Tl}^{3+}$ $\mathrm{Tl}^{3+} + 2e^- \rightarrow \mathrm{Tl}^{+}$ Since $+1$ is the more stable oxidation state for thallium, $\mathrm{Tl}^{3+}$ acts as an oxidising agent. ✅

(v) $\mathrm{Pb}^{4+}$ $\mathrm{Pb}^{4+} + 2e^- \rightarrow \mathrm{Pb}^{2+}$ Since $+2$ is more stable for lead, $\mathrm{Pb}^{4+}$ is an oxidising agent. ✅

(vi) $\mathrm{Tl}^{+}$ $\mathrm{Tl}^{+}$ is already the more stable oxidation state, so it does not tend to get reduced further; hence not an oxidising agent. ❌

Therefore, oxidising agents are: $\mathrm{Sn}^{4+},\ \mathrm{Tl}^{3+},\ \mathrm{Pb}^{4+}$

Hence, the number of ions behaving as oxidising agents is: $3$

Option matching:

• A: $3$ ✅
• B: $4$ ❌
• C: $2$ ❌
• D: $1$ ❌

Comparison with stored answer: Stored correct answer is C (2), but the correct count is 3.