The complex cation which has two isomers is:

We need to identify the complex cation that has two isomers.

Option A: $$[\text{Co}(\text{H}_2\text{O})_6]^{3+}$$

This complex has only one type of ligand (H$$_2$$O). It has the form [MA$$_6$$], which has only one possible structure. No isomers.

Option B: $$[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+}$$

This is of the type [MA$$_5$$B]. In an octahedral complex, there is only one way to arrange 5 identical ligands and 1 different ligand. No geometric isomers. Also no linkage isomers since Cl coordinates only through Cl.

Option C: $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$

This is of the type [MA$$_5$$B]. While there are no geometric isomers, the NO$$_2^-$$ ligand is an ambidentate ligand — it can coordinate through nitrogen (nitro, -NO$$_2$$) or through oxygen (nitrito, -ONO). This gives two linkage isomers:

1. $$[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$$ (nitro — N-bonded)

2. $$[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$$ (nitrito — O-bonded)

Option D: $$[\text{Co}(\text{NH}_3)_5\text{Cl}]^{+}$$

Same type as Option B. Only one geometric arrangement and Cl is not ambidentate. No isomers.

Only $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$ has exactly two isomers (linkage isomers).

The correct answer is Option C: $$[\text{Co}(\text{NH}_3)_5\text{NO}_2]^{2+}$$.