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:

The complex is MABXL where A, B, X, and L are unidentate ligands and M is the metal atom with $$sp^3$$ hybridization.

Key concept: $$sp^3$$ hybridization gives a tetrahedral geometry.

Tetrahedral complexes do not exhibit geometrical (cis-trans) isomerism. This is because in a tetrahedron, all positions are equivalent with respect to each other — every pair of ligands is adjacent, and there are no "opposite" positions as in square planar or octahedral geometries.

Therefore, the complex MABXL with tetrahedral geometry has 0 geometrical isomers.

(Note: Tetrahedral complexes can show optical isomerism when they have 4 different ligands, but the question specifically asks about geometrical isomers.)

The correct answer is Option (2): 0.