1. We need to identify which complex cation has two isomers.

2. Check each option for the possibility of isomerism.

Option A: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\mathrm{Cl}\right]^+$

Let oxidation state of Co be $x$.

$x + 5(0) + (-1) = +1$ $x-1=+1 \Rightarrow x=+2$

So this is $\left[\mathrm{Co}^{2+}(\mathrm{NH}_3)_5\mathrm{Cl}\right]^+$.

This is a pentaamminechloro complex of type $[MA_5B]$. Such a complex does not show geometrical isomerism.

Also, $\mathrm{Cl}^-$ is not an ambidentate ligand, so no linkage isomerism.

Hence, only one form exists.

Option B: $\left[\mathrm{Co}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{3+}$

This is of type $[MA_6]$.

All six ligands are identical, so neither geometrical nor linkage isomerism is possible.

Hence, only one form exists.

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

Let oxidation state of Co be $x$.

$x + 5(0) + (-1) = +2$ $x-1=+2 \Rightarrow x=+3$

This is $\left[\mathrm{Co}^{3+}(\mathrm{NH}_3)_5\mathrm{NO}_2\right]^{2+}$.

Here $\mathrm{NO}_2^-$ is an ambidentate ligand. It can coordinate through:

• N atom: nitro form
• O atom: nitrito form

So this complex shows linkage isomerism, giving two isomers:

$\left[\mathrm{Co}(\mathrm{NH}_3)_5(\mathrm{NO}_2)\right]^{2+}$ (N-bonded, nitro)

and

$\left[\mathrm{Co}(\mathrm{NH}_3)_5(\mathrm{ONO})\right]^{2+}$ (O-bonded, nitrito)

Hence, this complex has two isomers.

Option D: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\mathrm{Cl}\right]^{2+}$

Let oxidation state of Co be $x$.

$x + 5(0) + (-1) = +2$ $x-1=+2 \Rightarrow x=+3$

So this is $\left[\mathrm{Co}^{3+}(\mathrm{NH}_3)_5\mathrm{Cl}\right]^{2+}$.

Again, this is of type $[MA_5B]$, which does not show geometrical isomerism.

Also, $\mathrm{Cl}^-$ is not ambidentate, so no linkage isomerism.

Hence, only one form exists.

3. Therefore, the only complex cation having two isomers is:

$\boxed{\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\mathrm{NO}_2\right]^{2+}}$

So, Option C is correct.