The number of geometric isomers that can exist for square planar $\left[\mathrm{Pt}(\mathrm{Cl})(\mathrm{py})(\mathrm{NH}_{3})(\mathrm{NH}_{2} \mathrm{OH})\right]^{+}$ is (py = pyridine):

JEE Main · 2015 · 04 Apr 2015 · Offline · Coordination Compounds

The number of geometric isomers that can exist for square planar

[Pt (Cl) (py) (NH_{3}) (NH_{2} OH)]^{+}

is (py = pyridine):

(A)3✓ Correct
(B)4
(C)6
(D)2

Answer: (A)

Solution

A square-planar complex

[Mabcd]

carrying four different ligands has three geometric isomers, one for each choice of which ligand lies trans to a chosen reference. Taking

Cl

as that reference, either

py

N H_{3}

N H_{2} OH

can sit opposite it, and each choice fixes a distinct arrangement. That gives three geometric isomers.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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