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

🧠 Find the One Without Geometric Isomers

Octahedral $MA_5B$ has no cis/trans isomers (all five A positions equivalent).

🗺️ Evaluate Each

| Complex | Type | Geometric isomers? | |---|---|---| | $[\mathrm{Co(NH_3)_4 Cl_2}]^+$ | $MA_4B_2$ | ✓ cis/trans | | $[\mathrm{Co(NH_3)_3(NO_2)_3}]$ | $MA_3B_3$ | ✓ fac/mer | | $[\mathrm{Co(CN)_5(NC)}]^{3-}$ | $MA_5B$ | ✗ none | | $[\mathrm{CoCl_2(en)_2}]$ | $M(AA)_2 B_2$ | ✓ cis/trans |

The CN⁻ vs NC⁻ distinction in $[\mathrm{Co(CN)_5(NC)}]^{3-}$ is linkage (not geometric) — and even then, with 5 CN-bonded and 1 NC-bonded, all 5 CN positions are equivalent: no geometric isomers.

⚡ Linkage ≠ Geometric

Linkage isomerism arises from ambidentate ligands binding through different atoms. Geometric isomerism arises from different spatial arrangements of the same ligand set. The question asks specifically about geometric isomers.

⚠️ Identify the Ligand Type

CN⁻ = cyanido (C-bonded). NC⁻ = isocyanido (N-bonded). Different linkage, but in $MA_5B$ no geometric isomers exist regardless.

$\boxed{\text{Answer: (3) } [\mathrm{Co(CN)_5(NC)}]^{3-}}$