1. Identify the geometry of each complex

For isomerism in coordination compounds, geometry is crucial.

• $[Pt(NH_3)_2Cl_2]$ : Pt(II) is typically square planar.
• $[Ni(NH_3)_4(H_2O)_2]^{2+}$ : six-coordinate Ni(II), so typically octahedral.
• $[Ni(en)_3]^{2+}$ : three bidentate ligands around Ni(II), hence octahedral.
• $[Ni(NH_3)_2Cl_2]$ : Ni(II) with coordination number $4$; for such a complex, Ni(II) is generally tetrahedral.

2. Check isomerism for each option

A. $[Pt(NH_3)_2Cl_2]$

This is a square planar complex of type $[MA_2B_2]$.

Such complexes show geometrical isomerism:

• cis-$[Pt(NH_3)_2Cl_2]$
• trans-$[Pt(NH_3)_2Cl_2]$

So, A shows isomerism.

B. $[Ni(NH_3)_4(H_2O)_2]^{2+}$

This is octahedral of type $[MA_4B_2]$.

Such complexes show geometrical isomerism:

• cis form
• trans form

So, B shows isomerism.

C. $[Ni(en)_3]^{2+}$

This is octahedral with three bidentate ligands.

Complexes of type $[M(AA)_3]$ show optical isomerism:

• $\Delta$
• $\Lambda$

So, C shows isomerism.

D. $[Ni(NH_3)_2Cl_2]$

Ni(II) with coordination number $4$ is expected to be tetrahedral.

A tetrahedral complex of type $[MA_2B_2]$ does not show geometrical isomerism, because all four positions are equivalent in a tetrahedron. It also does not show optical isomerism here.

So, D is not expected to show isomerism.

3. Final conclusion

The complex which is not expected to show isomerism is:

$\boxed{D\;[Ni(NH_3)_2Cl_2]}$

4. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

So they agree.