1. Identify the geometry and ligand type in each complex

We check whether each complex can show geometrical isomerism.

Geometrical isomerism in coordination compounds depends on the arrangement of ligands in space, typically in octahedral or square planar complexes.

2. Option A: $[CoCl_2(en)_2]$

• Here, $en$ = ethylenediamine, a bidentate ligand.
• Coordination number of Co = $2$ (from two Cl) $+ 2+2$ (from two $en$ ligands) $= 6$.
• So the complex is octahedral.

Type: $[M(AA)_2B_2]$, where $AA$ is bidentate and $B$ is monodentate.

Such octahedral complexes show cis-trans geometrical isomerism:

• cis: both Cl adjacent
• trans: both Cl opposite

So, A shows geometrical isomerism.

3. Option B: $[Co(CN)_5(NC)]^{3-}$

• This is an octahedral complex with six coordination sites occupied by:
  • five cyanide ligands written as $CN$
  • one ligand written as $NC$
• Cyanide is an ambidentate ligand:
  • $CN$ means C-bonded cyano
  • $NC$ means N-bonded isocyano

So this complex exhibits linkage isomerism, not geometrical isomerism.

Also, since there is only one distinct ligand position for the single $NC$ ligand in an octahedral $[MA_5B]$ type complex, no cis/trans or fac/mer arrangement is possible.

So, B does not show geometrical isomerism.

4. Option C: $[Co(NH_3)_3(NO_2)_3]$

• Coordination number = $6$, so octahedral.
• Type: $[MA_3B_3]$.

Such complexes show fac-mer geometrical isomerism:

• fac: three identical ligands on one face
• mer: three identical ligands arranged meridionally

Hence, C shows geometrical isomerism.

5. Option D: $[Co(NH_3)_4Cl_2]^+$

• Coordination number = $6$, so octahedral.
• Type: $[MA_4B_2]$.

Such complexes show cis-trans geometrical isomerism:

• cis: two Cl adjacent
• trans: two Cl opposite

Hence, D shows geometrical isomerism.

6. Final conclusion

Among the given complexes, the only one that does not show geometrical isomerism is:

$[Co(CN)_5(NC)]^{3-}$

So the correct option is B.