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JEE Advanced 2014 · Paper 2, Chemistry question 34

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Paragraph For Question 33 and 34

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below:

SCHEME: M1 with Q (excess) gives a tetrahedral complex and with R (excess) a square planar complex. M2 with Q (excess) or R (excess) gives tetrahedral complexes; M2 with S (stoichiometric amount) gives a white precipitate, which dissolves in excess S.

Reagent S is

Choose one option
Answer and worked solution

Official answer: D

M2 is ZnX2+\ce{Zn^{2+}}, which forms tetrahedral [ZnClX4]X2−\ce{[ZnCl4]^{2-}} and [Zn(CN)X4]X2−\ce{[Zn(CN)4]^{2-}}.

Reagent S. With a little KOH, white Zn(OH)X2\ce{Zn(OH)2} precipitates; excess KOH dissolves it as the zincate:
Zn(OH)X2+2 OHX−→[Zn(OH)X4]X2−.\ce{Zn(OH)2 + 2OH- -> [Zn(OH)4]^{2-}}.
KX4[Fe(CN)X6]\ce{K4[Fe(CN)6]} gives a white precipitate that does not dissolve in excess; NaX2HPOX4\ce{Na2HPO4} and KX2CrOX4\ce{K2CrO4} do not fit either.

(D) KOH

Topics:Cation analysisStability and applications
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