JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Coordination Compounds

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:

Reagent S is

(A) $K_{4} [Fe (CN)_{6}]$
(B) $Na_{2} HPO_{4}$
(C) $K_{2} CrO_{4}$
(D)KOH✓ Correct

Answer: (D)

Solution

From the previous question, M2 is

Zn^{2 +}

. When

Zn^{2 +}

reacts with KOH, it forms a white precipitate of

Zn (OH)_{2}

Zn^{2 +} + 2 OH^{-} \to Zn (OH)_{2}

(white precipitate)

In excess KOH, this precipitate dissolves to form the soluble zincate complex:

Zn (OH)_{2} + 2 OH^{-} \to [Zn (OH)_{4}]^{2 -}

(soluble)

This behavior is characteristic of amphoteric hydroxides. Therefore, reagent S is KOH.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Coordination Compounds

Solution

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