Aqueous Reaction Scheme of Potassium Ferricyanide and Zinc Sulfate

To determine the correctness of each statement, we analyze the reaction sequence step by step:

1. First Reaction: Potassium ferricyanide ($\text{K}_3[\text{Fe}(\text{CN})_6]$) contains iron in the $+3$ oxidation state, which acts as an oxidizing agent in an acidic medium ($\text{dilute } \text{H}_2\text{SO}_4$). When reacted with excess potassium iodide ($\text{KI}$), iodide ions ($\text{I}^-$) are oxidized to iodine ($\text{I}_2$), while ferricyanide is reduced to ferrocyanide ($[\text{Fe}(\text{CN})_6]^{4-}$): $2[\text{Fe}(\text{CN})_6]^{3-} + 2\text{I}^- \longrightarrow 2[\text{Fe}(\text{CN})_6]^{4-} + \text{I}_2$ In the presence of excess $\text{I}^-$, iodine dissolves to form the triiodide ion ($\text{I}_3^-$), which gives the solution its characteristic brownish-yellow color: $\text{I}_2 + \text{I}^- \rightleftharpoons \text{I}_3^-$ Since this reaction involves a change in oxidation states ($\text{Fe}^{3+} \to \text{Fe}^{2+}$ and $2\text{I}^- \to \text{I}_2$), it is a redox reaction.
   Therefore, (A) is true.

2. Precipitation with $\text{ZnSO}_4$: When zinc sulfate ($\text{ZnSO}_4$) is added to the mixture, $\text{Zn}^{2+}$ ions react with the ferrocyanide ions ($[\text{Fe}(\text{CN})_6]^{4-}$) to precipitate zinc ferrocyanide, which is white: $2\text{Zn}^{2+} + [\text{Fe}(\text{CN})_6]^{4-} \longrightarrow \text{Zn}_2[\text{Fe}(\text{CN})_6] \downarrow \quad (\text{or } \text{K}_2\text{Zn}_3[\text{Fe}(\text{CN})_6]_2 \downarrow)$ The precipitate is a salt of ferrocyanide ($[\text{Fe}(\text{CN})_6]^{4-}$), not ferricyanide ($\text{Zn}_3[\text{Fe}(\text{CN})_6]_2$, which contains $[\text{Fe}(\text{CN})_6]^{3-}$).
   Therefore, (B) is false.

3. Filtrate Reaction with Starch: The brownish-yellow filtrate contains the liberated iodine in the form of $\text{I}_3^-$ (which also decolorizes upon the addition of sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$, due to reduction to iodide and formation of tetrathionate): $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \longrightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ Iodine gives an intensely colored blue inclusion complex when added to starch solution.
   Therefore, (C) is true.

4. Solubility of the White Precipitate in $\text{NaOH}$: Zinc is an amphoteric metal. When sodium hydroxide ($\text{NaOH}$) is added to the white precipitate of zinc ferrocyanide, it dissolves to form the soluble sodium tetrahydroxozincate(II) complex along with ferrocyanide ions: $\text{Zn}_2[\text{Fe}(\text{CN})_6]\,(\text{s}) + 8\text{OH}^-\,(\text{aq}) \longrightarrow 2[\text{Zn}(\text{OH})_4]^{2-}\,(\text{aq}) + [\text{Fe}(\text{CN})_6]^{4-}\,(\text{aq})$ Therefore, the precipitate is soluble in $\text{NaOH}$ solution.
   Therefore, (D) is true.

Correct Answer: A, C, D