1. Identify the geometries from hybridization

For a 4-coordinate complex of type $MA_2B_2$:

• $sp^3$ hybridization $\Rightarrow$ tetrahedral geometry
• $dsp^2$ hybridization $\Rightarrow$ square planar geometry

We must find the number of optical isomers in each case.

2. Case 1: $sp^3$ hybridized metal atom ($MA_2B_2$ tetrahedral)

A tetrahedral complex with two identical ligands $A$ and two identical ligands $B$ is of the type $M(A)_2(B)_2$.

For optical isomerism, the molecule must be chiral, i.e. it should not possess a plane of symmetry or center of symmetry.

In tetrahedral $MA_2B_2$:

• the arrangement has symmetry,
• it is superimposable on its mirror image,
• therefore it is achiral.

Hence, number of optical isomers for tetrahedral $MA_2B_2$ is:

$0$

3. Case 2: $dsp^2$ hybridized metal atom ($MA_2B_2$ square planar)

Square planar complexes of type $MA_2B_2$ can show geometrical isomerism:

• cis-$MA_2B_2$
• trans-$MA_2B_2$

Now check optical activity:

(i) cis form

The cis square planar complex has the molecular plane itself as a plane of symmetry. So it is achiral.

(ii) trans form

The trans form also has symmetry elements (plane/center of symmetry), so it is also achiral.

Thus, square planar $MA_2B_2$ shows no optical isomerism.

Hence, number of optical isomers for square planar $MA_2B_2$ is:

$0$

4. Final derived result

So, for $sp^3$ and $dsp^2$ hybridized metal atoms respectively, the number of possible optical isomers is:

$0 \text{ and } 0$

This corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.