The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and M is metal) involves sp³ hybridization. The number of geometrical isomers exhibited by the complex is :

Solution

Chemistry ma’am · 4 steps

sp3 means tetrahedral, and in a tetrahedron every corner touches every other, so cis and trans cannot exist.

1. Read the hybridisation. sp³ hybridisation with four unidentate ligands gives a tetrahedral complex, with all bond angles about 109.5°.
2. Why no geometrical isomers. Geometrical (cis–trans) isomers need some positions to be adjacent and others to be opposite. In a tetrahedron all four corners are equally adjacent to each other, so any two ligands always sit at the same angle. Every arrangement of A, B, X and L has the same spatial relationships, so no geometrical isomer is possible.
3. What it does show. Because all four ligands are different, the metal is a chiral centre. So MABXL shows optical isomerism (a pair of enantiomers), but that is a different kind of isomerism.
4. So. The number of geometrical isomers is 0, option (a).

The idea to keep. Tetrahedral complexes never show geometrical isomerism, whatever the ligands. If all four are different, look for optical isomers instead.

Answer from the NTA final answer key. Solution by PrepWiser.