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:

M1, Q and R, respectively are

(A) $Zn^{2 +}, KCN$ and HCl
(B) $Ni^{2 +}, HCl$ and KCN✓ Correct
(C) $Cd^{2 +}, KCN$ and HCl
(D) $Co^{2 +}, HCl$ , and KCN

Answer: (B)

Solution

M1 forms both tetrahedral (with Q) and square planar (with R) complexes.

Ni^{2 +}

is known to form tetrahedral complexes with

Cl^{-}

(

[NiCl_{4}]^{2 -}

) and square planar complexes with

CN^{-}

(

[Ni (CN)_{4}]^{2 -}

).

M2 always forms tetrahedral complexes and gives a white precipitate with S that dissolves in excess. This is characteristic of

Zn^{2 +}

, which forms tetrahedral complexes with both

Cl^{-}

and

CN^{-}

.

Therefore, M1 =

Ni^{2 +}

, Q = HCl (source of

Cl^{-}

), and R = KCN (source of

CN^{-}

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

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