Question

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:

• A. 4
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (B)

0

Answer 2

Step 1: Understanding $\text{sp}^3$ hybridization
In a complex with $\text{sp}^3$ hybridization, the geometry is tetrahedral. Tetrahedral complexes do not exhibit geometrical isomerism because all positions around the central atom are equivalent in three-dimensional space.
Step 2: Analysis of the given complex
The complex $MABXL$ has four unidentate ligands arranged tetrahedrally around the metal center. Since the tetrahedral geometry does not allow for distinct arrangements of ligands that result in different spatial configurations, geometrical isomerism is not possible.
Conclusion:
The number of geometrical isomers for the complex $MABXL$ is:
$0.$
Final Answer: (2).