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).
Given below are two statements:
Statement I: A homoleptic octahedral complex, formed using monodentate ligands, will not show stereoisomerism
Statement II: cis- and trans-platin are heteroleptic complexes of Pd.
In the light of the above statements, choose the correct answer from the options given below
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: