The number of geometrical isomers found in the metal complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3] and…

🧠 Count Isomers for Each in Order

Given complexes (in order): $[\mathrm{PtCl_2(NH_3)_2}]$, $[\mathrm{Ni(CO)_4}]$, $[\mathrm{Ru(H_2O)_3 Cl_3}]$, $[\mathrm{CoCl_2(NH_3)_4}]^+$.

🗺️ One-by-One

| Complex | Geometry | Type | Geometric isomers | |---|---|---|---| | $[\mathrm{PtCl_2(NH_3)_2}]$ | Square planar (Pt(II) d⁸) | $MA_2 B_2$ | 2 (cis, trans) | | $[\mathrm{Ni(CO)_4}]$ | Tetrahedral (Ni(0) d¹⁰) | $MA_4$ | 0 (no isomers) | | $[\mathrm{Ru(H_2O)_3 Cl_3}]$ | Octahedral $MA_3 B_3$ | $MA_3 B_3$ | 2 (fac, mer) | | $[\mathrm{CoCl_2(NH_3)_4}]^+$ | Octahedral $MA_4 B_2$ | $MA_4 B_2$ | 2 (cis, trans) |

So the count sequence is: 2, 0, 2, 2.

⚡ Geometric Isomer Counts by Type

| Type | Geometry | Count | |---|---|---| | $MA_4$ (Td) | tetrahedral | 0 | | $MA_2 B_2$ (sq pl) | square planar | 2 (cis/trans) | | $MA_4 B_2$ (oct) | octahedral | 2 (cis/trans) | | $MA_3 B_3$ (oct) | octahedral | 2 (fac/mer) | | $MA_5 B$ (oct) | octahedral | 1 | | $MA_6$ (oct) | octahedral | 1 |

⚠️ Tetrahedral $MA_4$ Has Zero Geometric Isomers

In $T_d$, all four positions are equivalent — only one arrangement possible. JEE often pairs Ni(CO)₄ with sq pl Pt complexes to test this distinction.

$\boxed{\text{Answer: (2) 2, 0, 2, 2}}$