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:
- 4
- 0
- 2
- 3
The Correct Option is B
Approach Solution - 1
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).
Approach Solution -2
The complex is MABXL, where A, B, X, and L are unidentate ligands, and the metal (M) shows sp³ hybridization.
Step 2: Identify the geometry for sp³ hybridization.
For sp³ hybridization, the geometry of the complex is tetrahedral.
Step 3: Determine the possibility of geometrical isomerism.
Geometrical isomerism arises when ligands can occupy different spatial positions (for example, cis/trans in square planar or fac/mer in octahedral complexes).
In a tetrahedral complex, all four positions are equivalent. Therefore, interchanging the ligands does not produce a different spatial arrangement.
Step 4: Conclusion.
Since all positions in a tetrahedral geometry are equivalent, the complex does not exhibit geometrical isomerism.
Final Answer: 0
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