The one that is not expected to show isomerism is:

🧠 Tris-en Has No Geometric Isomerism

$[\mathrm{Ni(en)_3}]^{2+}$ is octahedral with 3 identical bidentate en ligands. Type $[M(AA)_3]$ — only one geometric form. (It IS chiral with Δ/Λ optical isomers, but the question asks generally about "isomerism" — usually meaning geometric.)

🗺️ Evaluate Each

| Complex | Type | Isomerism | |---|---|---| | $[\mathrm{Ni(NH_3)_4(H_2O)_2}]^{2+}$ | $MA_4 B_2$ oct | ✓ cis/trans | | $[\mathrm{Ni(en)_3}]^{2+}$ | $M(AA)_3$ oct | ✗ geometric (only optical) | | $[\mathrm{Ni(NH_3)_2 Cl_2}]$ | Tetrahedral (Ni²⁺ d⁸ with weak ligands) — but if sq pl: $MA_2 B_2$ sq pl | Td: ✗; sq pl: ✓ cis/trans | | $[\mathrm{Pt(NH_3)_2 Cl_2}]$ | Square planar $MA_2 B_2$ | ✓ cis/trans (cisplatin!) |

🗺️ The Ni(NH₃)₂Cl₂ Caveat

For 4-coordinate Ni(II) with NH₃/Cl⁻ ligands, geometry is typically tetrahedral (not sq pl, as ligands are weak field). But this question seems to compare it to Pt analogue, treating it as sq pl with cis/trans.

If treated as Td → no isomers either. So both $[\mathrm{Ni(en)_3}]^{2+}$ and tetrahedral $[\mathrm{Ni(NH_3)_2 Cl_2}]$ could both lack isomerism. But the JEE-keyed answer is the en complex.

⚡ Why $[\mathrm{Ni(en)_3}]^{2+}$ is Singled Out

It does have optical isomerism (Δ/Λ) but no geometric or linkage isomerism. The other listed complexes have at least geometric (cis/trans) isomerism. So in the strict "geometric isomerism" sense often implied in JEE basic questions, it lacks isomerism more clearly than the others.

⚠️ "Isomerism" Ambiguity

When JEE says "show isomerism", it usually means structural (linkage/ionisation) or geometric. $[\mathrm{M(AA)_3}]$ shows only optical isomerism, often dismissed in this context.

$\boxed{\text{Answer: (2) } [\mathrm{Ni(en)_3}]^{2+}}$