Indicate the complex/complex ion which did not show any geometrical isomerism :

Solution

Chemistry ma’am · 3 steps

Write each complex as a type like Ma₄b₂. Only Ma₅b has no second place to put the odd ligand.

1. Reduce to a type. All four are octahedral cobalt complexes. Geometrical isomers exist only when the same set of ligands can sit in different relative positions, so first write each complex as a general type.

   Complex	Type	Possible arrangements	Geometrical isomers?
   [CoCl₂(en)₂]	M(AA)₂b₂	two Cl at 90° or at 180°	yes, cis and trans
   [Co(CN)₅(NC)]³⁻	Ma₅b	the single NC can go on any vertex, and every vertex is equivalent	no
   [Co(NH₃)₃(NO₂)₃]	Ma₃b₃	three NO₂ on one face, or along a meridian	yes, fac and mer
   [Co(NH₃)₄Cl₂]⁺	Ma₄b₂	two Cl at 90° or at 180°	yes, cis and trans

2. Why Ma5b fails. In (b) one cyanide binds through N, so it counts as a different ligand b. With only one b, there is nothing for it to be cis or trans to. Every position gives the same complex. This complex can show linkage isomerism, but not geometrical isomerism.
3. So. The only complex with no geometrical isomers is [Co(CN)₅(NC)]³⁻, option (b).

The idea to keep. In an octahedron, Ma₆ and Ma₅b never show geometrical isomerism. Ma₄b₂ and M(AA)₂b₂ give cis/trans, and Ma₃b₃ gives fac/mer.

Answer from the NTA final answer key. Solution by PrepWiser.