The metal atom present in the complex MABXL (where $\mathrm{A}, \mathrm{B}, \mathrm{X}$ and L are unidentate ligands and M is metal) involves $\mathrm{sp}^{3}$ hybridization. The number of geometrical isomers exhibited by the complex is:

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Coordination Compounds

The metal atom present in the complex MABXL (where

A, B, X

and L are unidentate ligands and M is metal) involves

sp^{3}

hybridization. The number of geometrical isomers exhibited by the complex is:

(A)4
(B)0✓ Correct
(C)2
(D)3

Answer: (B)

Solution

With four different unidentate ligands A, B, X, L on a tetrahedral (

s p^{3}

) metal centre, any arrangement of the four groups at the vertices of a tetrahedron can be rotated into any other arrangement — there is only one distinct spatial structure. Tetrahedral complexes therefore show no geometrical (cis–trans type) isomerism, regardless of how many different ligands are attached, so the number of geometrical isomers is 0.

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option

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