1. Recall the condition for optical isomerism in coordination compounds

   Optical isomerism arises when a complex is chiral, i.e. it has no plane of symmetry, no center of symmetry, and no improper axis.

   In coordination chemistry, for octahedral complexes, optical isomerism is commonly shown by:

   • $[M(AA)_3]$
   • cis-$[M(AA)_2a_2]$ where $AA$ is a bidentate ligand such as en (ethylenediamine).

2. Examine each option

Option A: $[Co(en)(NH_3)_2]^{2+}$

• Here, only one bidentate ligand $en$ is present.
• The coordination number is also unclear as written, but such a species does not generate the usual chiral octahedral arrangement.
• With just one $en$ and identical monodentate ligands, the complex retains symmetry.

So, A is not optically active.

Option B: $[Co(H_2O)_4(en)]^{3+}$

• This is an octahedral complex of type $[M(AA)a_4]$.
• One bidentate ligand $en$ occupies two adjacent positions, and four identical $H_2O$ ligands occupy the remaining four positions.
• Such a complex has a plane of symmetry and is therefore achiral.

So, B does not show optical isomerism.

Option C: $[Co(en)_2(NH_3)_2]^{3+}$

• This is of type $[M(AA)_2a_2]$.
• Such complexes can exist as cis and trans forms.
• The cis form is chiral and shows optical isomerism.
• The trans form is achiral.

Since the formula corresponds to a complex that can have an optically active cis form, this option has optical isomer.

So, C shows optical isomerism.

Option D: $[Co(NH_3)_3Cl]^+$

• As written, this does not correspond to a coordination environment capable of optical isomerism.
• It lacks the kind of bidentate ligand arrangement needed for chirality here.

So, D does not show optical isomerism.

3. Final conclusion

The complex that has an optical isomer is:

$\boxed{[Co(en)_2(NH_3)_2]^{3+}}$

So the correct option is C.