Among (a)-(d), the complexes that can show geometrical isomerism are: (a) [Pt(NH3)3Cl]+ (b) [Pt(NH3)Cl5]- (c)…

🧠 Geometrical Isomerism Needs at Least 2 Distinguishable Positions

For a Pt(IV) octahedral or Pt(II) square-planar complex to show cis/trans (geometrical) isomerism, you need at least two different ligand types arranged in cis vs trans positions.

🗺️ Test Each

| Complex | Layout | Geometrical isomerism? | |---|---|---| | (a) $[\mathrm{Pt(NH_3)_3Cl}]^+$ | $\mathrm{[MA_3B]}$ octahedral — 3 NH₃ + 1 Cl + need 2 more sites! Wait — only 4 ligands listed (CN=4). Square-planar $\mathrm{[MA_3B]}$ → only 1 arrangement | ✗ | | (b) $[\mathrm{Pt(NH_3)Cl_5}]^-$ | octahedral $\mathrm{[MAB_5]}$ — 1 distinct + 5 identical → only 1 arrangement | ✗ | | (c) $[\mathrm{Pt(NH_3)_2Cl(NO_2)}]$ | square-planar $\mathrm{[MA_2BC]}$ — NH₃ pair can be cis or trans | ✓ | | (d) $[\mathrm{Pt(NH_3)_4ClBr}]^{2+}$ | octahedral $\mathrm{[MA_4BC]}$ — Cl/Br can be cis or trans | ✓ |

So (c) and (d) show geometrical isomerism → option (3).

⚡ The "Single Identical Ligand" Death Knell

If five (or more) ligands are identical and only one is different ($\mathrm{[MAB_5]}$), there's only one possible arrangement — no geometrical isomerism. Same for $\mathrm{[MA_3B]}$ in any geometry.

⚠️ Always Identify the Geometry First

Pt(II) (4-coordinate) is square-planar; Pt(IV) (6-coordinate) is octahedral. The isomer count rules are very different. (b) is octahedral, (c) is square-planar — always check geometry before counting.

$\boxed{\text{Answer: (3) c and d}}$