1. Identify the coordination entities and their geometries

We check each complex for the possibility of geometrical isomerism.

Geometrical isomerism commonly occurs in:

• square planar complexes of type $[MA_2B_2]$, $[MABCD]$
• octahedral complexes of type $[MA_4B_2]$, $[MA_3B_3]$, $[M(AA)_2ab]$, etc.

It does not occur when all positions are equivalent in a way that prevents distinct cis/trans or fac/mer arrangements.

2. Option A: aquachlorobis(ethylenediamine)cobalt(II) chloride

The coordination complex is: $[\text{Co}(en)_2(H_2O)Cl]Cl_2$ where $en$ = ethylenediamine, a bidentate ligand.

• Coordination number of Co = $2\times 2 + 1 + 1 = 6$
• So the complex is octahedral.
• Type of complex: $[M(AA)_2ab]$

For octahedral complexes of type $[M(AA)_2ab]$, geometrical isomerism is possible depending on whether the two monodentate ligands $a$ and $b$ are arranged differently relative to the chelate rings.

Hence, A shows geometrical isomerism.

3. Option B: pentaaquachlorochromium(III) chloride

The coordination complex is: $[\text{Cr}(H_2O)_5Cl]Cl_2$

• Coordination number = $6$
• Octahedral complex of type $[MA_5B]$

Complexes of type $[MA_5B]$ do not show geometrical isomerism because all five positions occupied by $A$ are equivalent; only one arrangement exists.

Hence, B does not show geometrical isomerism.

4. Option C: potassium amminetrichloroplatinate(II)

The coordination complex is: $K[\text{Pt}(NH_3)Cl_3]$ so the anion is $[\text{Pt}(NH_3)Cl_3]^-$

• Pt(II) generally forms square planar complexes.
• Type: $[MA B_3]$

Square planar geometrical isomerism requires patterns like $[MA_2B_2]$ or $[MABCD]$. For $[MAB_3]$, only one arrangement is possible.

Hence, C does not show geometrical isomerism.

5. Option D: potassium tris(oxalato)chromate(III)

The complex is: $K_3[\text{Cr}(C_2O_4)_3]$

• Oxalate is a bidentate ligand.
• Type: $[M(AA)_3]$
• Octahedral complex

Such complexes do not show geometrical isomerism, though they can show optical isomerism ($\Delta$ and $\Lambda$ forms).

Hence, D does not show geometrical isomerism.

6. Final conclusion

Only Option A shows geometrical isomerism.

Therefore, the correct answer is: $\boxed{A}$

7. Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

They match.