Use the AgNO$_3$ test to find ionizable chloride ions

The complex is of the form: $\mathrm{CoCl_3}\cdot n\mathrm{NH_3}$

On treatment with excess $\mathrm{AgNO_3}$, it gives 2 moles of AgCl.

This means 2 chloride ions are outside the coordination sphere and are present as counter ions, because only ionizable $\mathrm{Cl^-}$ precipitate with $\mathrm{Ag^+}$.

Write the complex accordingly

Since total chloride atoms are 3, and 2 are outside the coordination sphere, 1 chloride must be coordinated to cobalt: $[\mathrm{Co}(\mathrm{NH_3})_n\mathrm{Cl}]\mathrm{Cl}_2$

Use octahedral geometry to find $n$

The complex is octahedral, so coordination number is 6.

Inside the coordination sphere, ligands are:

• $n$ molecules of $\mathrm{NH_3}$
• 1 $\mathrm{Cl^-}$

Therefore, $n+1=6$ $n=5$

Find oxidation state of Co

In $[\mathrm{Co}(\mathrm{NH_3})_5\mathrm{Cl}]\mathrm{Cl}_2$

Let oxidation state of Co be $x$.

• $\mathrm{NH_3}$ is neutral
• coordinated $\mathrm{Cl}$ has charge $-1$
• the complex ion must have charge $+2$ to balance 2 outside $\mathrm{Cl^-}$

So, $x + 5(0) - 1 = +2$ $x-1=2$ $x=3$

Compute $x+n$

$x+n=3+5=8$

Check options

The correct option is: $\boxed{8}$

So, Option D is correct.