Question

The number of possible optical isomers for the complexes $\left[{\mathrm{MA}}_2{\mathrm{B}}_2\right]\mathrm{with}{\mathrm{sp}}^3\mathrm{or}{\mathrm{dsp}}^2$ hybridized metal atom, respectively, is:

• A. $0\mathrm{and}1$
• B. $0\mathrm{and}0$
• C. $2\mathrm{and}2$
• D. $0\mathrm{and}2$

Answer 1

The correct option is B

$0\mathrm{and}0$

The explanation for the correct option:

B: $0\mathrm{and}0$

Optical Isomers:

• When two isomers exhibit identical molecular weight, chemical, and physical properties, this is known as optical isomerism.
• Their impact on the rotation of polarised light, however, varies.
• When two substances with the same molecular and structural formula cannot be superimposed on one another, optical isomerism mostly arises.

$\left[{\mathrm{MA}}_2{\mathrm{B}}_2\right]$ Complex with hybridization of ${\mathrm{sp}}^3$:

• It will have a tetrahedral shape if $\mathrm{M}\mathrm{is}{\mathrm{sp}}^3$ hybridized.
• Since there will be a plane of symmetry, optical activity is not present.
• 

$\left[{\mathrm{MA}}_2{\mathrm{B}}_2\right]$ Complex with hybridization of ${\mathrm{dsp}}^2$:

• M will take on a square planar geometry if $\mathrm{M}\mathrm{is}{\mathrm{dsp}}^2$ hybridized.
• It doesn't exhibit optical activity since there is a plane of symmetry there.

Final Answer: Therefore, Option B is the correct option.