Which of the following cannot give iodometric titrations

The correct option is Pb²⁺.

To understand why ${\mathrm{Pb}}^{2+}$ cannot undergo iodometric titration, let us analyze the principles behind iodometric titrations and the chemical behavior of each ion provided in the options.

What is Iodometric Titration?
Iodometric titration is an indirect titration method that involves a redox reaction. In this process, an oxidizing agent is reacted with an excess of iodide ions ($I^{-}$) to liberate free iodine ($I_2$). The liberated iodine is then titrated against a standard reducing agent, typically sodium thiosulfate (${\mathrm{Na}}_2{S}_2{O}_3$). For a metal ion to participate in an iodometric titration, it must be a strong enough oxidizing agent to oxidize iodide ions to free iodine.

Analyzing the Given Ions:
1. Fe³⁺: The ferric ion (${\mathrm{Fe}}^{3+}$) is a good oxidizing agent. It readily oxidizes iodide ions to iodine while being reduced to ferrous ions (${\mathrm{Fe}}^{2+}$) according to the reaction:
$2{\mathrm{Fe}}^{3+}+2I^{-}\to 2{\mathrm{Fe}}^{2+}+I_2$
Thus, ${\mathrm{Fe}}^{3+}$ can be estimated using iodometric titration.

2. Cu²⁺: The cupric ion (${\mathrm{Cu}}^{2+}$) reacts with iodide ions to form a precipitate of cuprous iodide (${\mathrm{Cu}}_2{I}_2$) and liberate iodine:
$2{\mathrm{Cu}}^{2+}+4I^{-}\to {\mathrm{Cu}}_2{I}_2\downarrow +I_2$
Because iodine is quantitatively liberated, ${\mathrm{Cu}}^{2+}$ is commonly determined using iodometric titration.

3. Ag⁺: The silver ion (${\mathrm{Ag}}^{+}$) is capable of oxidizing iodide ions to iodine under suitable conditions, allowing it to undergo iodometric determination.

4. Pb²⁺: The lead(II) ion (${\mathrm{Pb}}^{2+}$) does not possess sufficient oxidizing power to oxidize iodide ions to free iodine. Instead of a redox reaction, when ${\mathrm{Pb}}^{2+}$ is mixed with iodide ions ($I^{-}$), a simple precipitation reaction occurs, yielding yellow lead iodide precipitate without liberating any iodine:
${\mathrm{Pb}}^{2+}+2I^{-}\to {\mathrm{PbI}}_2\downarrow$
Since no iodine is liberated during this reaction, ${\mathrm{Pb}}^{2+}$ cannot be determined via iodometric titration.