Number of Zinc Ions in Molecular Formula of Complex Q

To determine the number of zinc ions present in the molecular formula of complex $\mathbf{Q}$, we analyze the chemical reactions step-by-step:

Step 1: Identification of Complex $\mathbf{P}$

When potassium ferricyanide, $\text{K}_3[\text{Fe(CN)}_6]$, reacts with potassium iodide ($\text{KI}$), a reversible redox reaction occurs: $2\text{K}_3[\text{Fe(CN)}_6] + 2\text{KI} \rightleftharpoons 2\text{K}_4[\text{Fe(CN)}_6] + \text{I}_2$

In this reaction, the ferricyanide ion $[\text{Fe(CN)}_6]^{3-}$ is reduced to the ferrocyanide ion $[\text{Fe(CN)}_6]^{4-}$, and iodide ($\text{I}^-$) is oxidized to iodine ($\text{I}_2$). In a strongly acidic medium, the equilibrium shifts completely forward, forming complex $\mathbf{P}$, which is potassium ferrocyanide, $\text{K}_4[\text{Fe(CN)}_6]$.

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Step 2: Formation and Molecular Formula of Complex $\mathbf{Q}$

When zinc chloride ($\text{ZnCl}_2$) is added to potassium ferrocyanide ($\mathbf{P}$) in a slightly acidic medium, a characteristic sparingly soluble (bluish-white) precipitate of zinc potassium ferrocyanide, $\mathbf{Q}$, is formed according to the balanced equation: $2\text{K}_4[\text{Fe(CN)}_6] + 3\text{ZnCl}_2 \longrightarrow \text{K}_2\text{Zn}_3[\text{Fe(CN)}_6]_2\downarrow + 6\text{KCl}$

Thus, the molecular formula of complex $\mathbf{Q}$ is $\text{K}_2\text{Zn}_3[\text{Fe(CN)}_6]_2$.

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Step 3: Conclusion

From the molecular formula $\text{K}_2\text{Zn}_3[\text{Fe(CN)}_6]_2$, the number of zinc ions ($\text{Zn}^{2+}$) present in one formula unit of complex $\mathbf{Q}$ is 3.