The number of isomers possible for $\left[\mathrm{Pt}(\mathrm{en})\left(\mathrm{NO}_{2}\right)_{2}\right]$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Coordination Compounds

The number of isomers possible for

[Pt (en) (NO_{2})_{2}]

is:

(A)4
(B)3✓ Correct
(C)1
(D)2

Answer: (B)

Solution

[Pt (en) (NO X_{2}) X_{2}]

the platinum(II) centre is square planar and ethylenediamine is a bidentate ligand that locks onto two adjacent sites, so no geometrical isomerism is possible. The variety comes instead from the ambidentate

NO X_{2}

groups, which can bind through nitrogen (nitro) or through oxygen (nitrito): both as nitro, both as nitrito, or one of each. That gives three linkage isomers in all.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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