1. Identify coordination number and geometry for each complex

To have a trans-isomer, the complex must have a geometry where two identical ligands can be placed opposite to each other.

• In square planar complexes: cis/trans is possible.
• In octahedral complexes: cis/trans is possible in suitable cases.
• In tetrahedral complexes: trans-isomerism is not possible.
• Also, chelating ligands may force adjacent positions, affecting possibility of trans arrangement.

2. Check each option

Option A: $[Cr(en)_2(ox)]^+$

• $en$ is bidentate, so $2(en)$ occupies $4$ coordination sites.
• $ox$ is also bidentate, occupying $2$ coordination sites.
• Total coordination number $= 6$, so geometry is octahedral.

Now all three ligands are bidentate. Such a complex is of the type $[M(AA)_2(BB)]$ in octahedral geometry.

A bidentate ligand must occupy adjacent (cis) positions. Hence the two donor atoms of $ox$ must be adjacent, and similarly for each $en$.

In this arrangement, a usual cis/trans description does not arise. Instead, such complexes generally show optical isomerism ($\Delta/\Lambda$), not trans-isomerism.

So, Option A cannot have a trans-isomer.

Option B: $[Pt(en)Cl_2]$

• Pt in such complexes is typically square planar with coordination number $4$.
• $en$ is bidentate and occupies two adjacent positions in a square plane.
• The remaining two positions are occupied by the two $Cl^-$ ligands.

Since $en$ must bind adjacent positions, the two chlorides are forced to occupy the remaining two adjacent positions as well.

Thus only the cis arrangement is possible; trans is not possible.

So, Option B cannot have a trans-isomer.

Option C: $[Pt(en)_2Cl_2]^{2+}$

• Here there are two $en$ ligands and two $Cl^-$ ligands.
• Total coordination number $= 2\times 2 + 2 = 6$, so geometry is octahedral.

This is of the form $[M(AA)_2a_2]$ in octahedral geometry.

For such complexes, the two identical monodentate ligands ($Cl^-$) can be:

• adjacent $\Rightarrow$ cis
• opposite $\Rightarrow$ trans

Therefore, trans-isomerism is possible.

So, Option C can have a trans-isomer.

Option D: $[Zn(en)Cl_2]$

• Zn(II) with coordination number $4$ commonly forms tetrahedral complexes.
• In tetrahedral geometry, cis/trans isomerism does not exist.

So, Option D cannot have a trans-isomer.

3. Conclusion

Only Option C can show trans-isomerism.

$\boxed{[Pt(en)_2Cl_2]^{2+}}$