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Step-by-Step Explanation
Step 1: Formation of the Group-IV Sulfide Precipitate
When H2S is passed through a solution containing certain metal cations under neutral or slightly acidic conditions, group-IV cations (such as Zn2+, Mn2+, Co2+, and Ni2+) form sulfide precipitates. In this question, the unknown cation gives a precipitate with H2S, placing it in group IV of the classical qualitative analysis scheme.
Step 2: Dissolving the Sulfide in Dilute HCl
The sulfide precipitate is then dissolved in dilute HCl, producing a clear solution of the cation in acidic medium.
Step 3: Reaction with NaOH
When this acidic solution is treated with NaOH solution, it yields a white precipitate. For Zn2+, the relevant reaction is:
$ \text{Zn}^{2+} + 2 \text{OH}^{-} \rightarrow \text{Zn(OH)}_{2} \downarrow $
Zinc hydroxide is white, confirming the presence of Zn2+.
Step 4: Reaction with Basic Potassium Ferrocyanide
The solution of the cation forms a bluish-white precipitate when treated with potassium ferrocyanide (K4[Fe(CN)6]) under basic conditions. The compound formed is zinc ferrocyanide, which is characteristically bluish-white:
$ \text{Zn}^{2+} + \text{K}_{4}[\text{Fe(CN)}_{6}] \rightarrow \text{Zn}_{2}[\text{Fe(CN)}_{6}] \downarrow $
Step 5: Conclusion
Both the white precipitate with NaOH and the bluish-white precipitate with potassium ferrocyanide are hallmark tests for Zn2+. Therefore, the cation in the given question is Zn2+.
Answer: Zn2+
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