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

<p>To determine the number of geometrical isomers for the given complex $\mathrm{MABXL}$, where $\mathrm{M}$ is a metal and $\mathrm{A}$, $\mathrm{B}$, $\mathrm{X}$, and $\mathrm{L}$ are unidentate ligands, we need to consider the geometry implied by the $\mathrm{sp}^3$ hybridization of the metal. The $\mathrm{sp}^3$ hybridization suggests a tetrahedral geometry for the metal complex.</p> <p>In a tetrahedral complex, geometrical isomerism does not arise. This is because geometrical isomerism (also known as cis-trans isomerism) typically occurs in square planar or octahedral complexes where ligands can be positioned differently around the central metal atom to give distinct isomers. In a tetrahedral complex, every position is equivalent due to the symmetric arrangement of the ligands around the metal center, making it impossible to distinguish between different sides of the molecule as you would with a square planar or octahedral complex.</p> <p>Given that the complex follows tetrahedral geometry because of $\mathrm{sp}^3$ hybridization and that tetrahedral complexes do not exhibit geometrical isomerism, the number of geometrical isomers exhibited by the complex is:</p> <p>Option A: 0</p> <p>This is because in a tetrahedral arrangement of ligands around a metal center, no geometrical isomers can be formed due to the lack of cis or trans arrangements possible within such a symmetric geometry.</p>