JEE Advanced 2024 · Paper 2 · Chemistry Q17

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ChemistryNumerical+3Question 51 of 51

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PARAGRAPH II

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex P is formed. In a strong acidic medium, the equilibrium shifts completely towards P. Addition of zinc chloride to P in a slightly acidic medium results in a sparingly soluble complex Q.

The number of zinc ions present in the molecular formula of Q is _____.

Your answer

Answer and worked solution

Official answer: 2

$\ce{Zn^2+}$ precipitates ferrocyanide in slightly acidic solution as the sparingly soluble white complex potassium zinc ferrocyanide:
$\ce{3Zn^2+ + 2K4[Fe(CN)6] -> K2Zn3[Fe(CN)6]2 v + 6K+}$

Q = $\ce{K2Zn3[Fe(CN)6]2}$ , which contains 3 zinc ions. The key also accepts 2, for $\ce{Zn2[Fe(CN)6]}$ .

Topics:Cation analysis Nomenclature

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JEE Advanced 2024 · Paper 2, Chemistry question 17