The complex that can show optical activity is:

First, recall the basic rule for optical activity (chirality) in coordination compounds. A complex will be optically active if it does not possess any internal element of symmetry - that is, it must lack a plane of symmetry, a centre of symmetry and an improper rotation axis. In octahedral complexes this usually happens when:

$$\text{(i)}$$ at least two bidentate (chelating) ligands are present, because the left-handed ($$\Lambda$$) and right-handed ($$\Delta$$) chelate orientations then become non-superimposable mirror images, and

$$\text{(ii)}$$ the ligands are arranged in the cis fashion; a trans arrangement normally introduces a mirror plane or a centre of symmetry, destroying chirality.

With this in mind we examine each option.

Option A: trans - $$[Cr(Cl_2)(ox)_2]^{3-}$$

The complex is octahedral with two bidentate $$ox^{2-}$$ ligands and two monodentate $$Cl^-$$ ligands placed opposite each other (trans). Because the two $$Cl^-$$ ions are exactly opposite, we can draw a mirror plane passing through both $$Cl^-$$ ions and bisecting the complex. Hence a plane of symmetry exists, so

$$\Rightarrow \text{optically inactive.}$$

Option B: trans - $$[Fe(NH_3)_2(CN)_4]^-$$

All six ligands here are monodentate. With four $$CN^-$$ and two $$NH_3$$ ligands in a trans arrangement, the complex certainly has a centre of symmetry lying at the metal ion because each ligand is matched by an identical ligand directly opposite to it:

$$\Rightarrow \text{optically inactive.}$$

Option C: cis - $$[Fe(NH_3)_2(CN)_4]^-$$

Although this is a cis isomer, all ligands are still monodentate. We can place a vertical mirror plane that passes through the two $$NH_3$$ ligands and bisects two opposite $$CN^-$$ ligands, so the complex possesses a plane of symmetry:

$$\Rightarrow \text{optically inactive.}$$

Option D: cis - $$[CrCl_2(ox)_2]^{3-}$$

Here we have two bidentate $$ox^{2-}$$ ligands and two monodentate $$Cl^-$$ ligands arranged cis to each other. Because the oxalate ligands wrap around the metal centre in two possible screw-like fashions, we obtain a pair of mirror images - the $$\Lambda$$ (left-handed) and $$\Delta$$ (right-handed) forms - which cannot be superimposed on one another. The cis disposition removes any centre or plane of symmetry, so

$$\Rightarrow \text{the complex is optically active.}$$

Comparing all four cases, only Option D satisfies the criteria for optical activity.

Hence, the correct answer is Option 4.