JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

The given complex is K [ M(NCS)(NO _2 )(gly) ]. The complex anion is [ M(NCS)(NO _2 )(gly) ]^ - . The ligands present in the complex are: 1. gly ^- (glycinate): An unsymmetrical bidentate ligand coordinating through nitrogen (N) and oxygen (O). Let this be represented as (AB). 2. NCS ^-: An ambidentate monodentate ligand that can coordinate through nitrogen ( -NCS ) or sulfur ( -SCN ). 3. NO _2^-: An ambidentate monodentate ligand that can coordinate through nitrogen ( -NO _2) or oxygen ( -ONO ). First, we find the number of linkage isomers due to the ambidentate ligands: NCS ^- has 2 linkage modes (N or S). NO _2^- has 2 linkage modes (N or O). Number of linkage combinations = 2 2 = 4. For each linkage combination, the complex is of the type [ M (a)(b)(AB)], where a and b are monodentate ligands and AB is an unsymmetrical bidentate ligand. In a square planar geometry, the bidentate ligand (AB) must occupy adjacent (cis) positions. The monodentate ligands a and b will occupy the remaining two positions. This gives rise to 2 geometrical isomers for each linkage combination: 1. Ligand a is trans to A (and b is trans to B). 2. Ligand a is trans to B (and b is trans to A). Since square planar complexes of this type possess a molecular plane of symmetry, they do not exhibit optical isomerism. Total number of isomers = (Number of linkage combinations) (Number of geometrical isomers per combination) Total isomers = 4 2 = 8. Answer: 8