Question

The $ {\left[ {{\text{Pt}}\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)\left( {{\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}} \right)\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)\left( {{\text{py}}} \right)} \right]^{\text{ + }}} $ will form how many geometrical isomers?
(A) 2
(B) 3
(C) 4
(D) 5

Answer 1

Hint: To answer this question, you must recall the concept of isomerism in coordination compounds. Isomerism shown by coordinate compounds are broadly of two types, namely, structural isomerism and stereoisomerism. Geometrical isomerism falls under the category of stereoisomerism.

Complete step by step solution
First we must understand the concept of geometrical isomerism. Geometrical isomerism is said to be exhibited when molecules have the same empirical formula but the way in which the atoms are arranged is different. They generally differ in the atom to atom bonds. It arises in coordination compounds when they are heteroleptic, i.e. they have more than one type of ligands.
The given compound is attached to four ligands and assumes square planar geometry. All the ligands are different and such a compound shows geometrical isomerism. Assuming no structural isomerism, the number of geometrical isomers formed by the given compound are three.
The correct answer is (B).

Note
Structural isomers are those isomers which have similar molecular formula but different structural formula. The arrangement of the atoms differs in structural isomers having no kind of reference with the spatial arrangement of the molecule. On the other hand, stereoisomers are those isomers which have the same molecular formula, similar structural formula but different spatial arrangement of the bonded atoms.
In the given compound, we have an ambidentate ligand, i.e. the nitrogen dioxide molecule. It can form the bond from either the nitrogen atom or the oxygen atom. Considering structural isomers as well, we would have gotten the number of isomers as 6.