Identify the oxidation state of cobalt in each complex

• In $[Co(NH_3)_6]Cl_2$, the complex ion is $[Co(NH_3)_6]^{2+}$ because there are $2$ chloride ions outside the coordination sphere. Since $NH_3$ is a neutral ligand, $\text{Oxidation state of Co} = +2$

• In $[Co(NH_3)_6]Cl_3$, the complex ion is $[Co(NH_3)_6]^{3+}$ because there are $3$ chloride ions outside the coordination sphere. Thus, $\text{Oxidation state of Co} = +3$

Find the electronic configuration of the metal ions

Cobalt has atomic number $27$. $Co: [Ar]3d^74s^2$

• For $Co^{2+}$, remove two electrons from $4s$: $Co^{2+}: [Ar]3d^7$

• For $Co^{3+}$, remove two from $4s$ and one from $3d$: $Co^{3+}: [Ar]3d^6$

Nature of ligand and spin state

$NH_3$ is a strong enough ligand with $Co^{3+}$ to produce a low-spin octahedral complex.

For $Co^{2+}$, $[Co(NH_3)_6]^{2+}$ is generally high-spin octahedral.

Count unpaired electrons

(a) $[Co(NH_3)_6]^{2+}$

This is octahedral $d^7$ high spin: $t_{2g}^5 e_g^2$ Unpaired electrons $= 3$

(b) $[Co(NH_3)_6]^{3+}$

This is octahedral $d^6$ low spin: $t_{2g}^6 e_g^0$ Unpaired electrons $= 0$

Total unpaired electrons

$3 + 0 = 3$

Compare with stored correct answer

Stored correct answer = $3$

This matches the derived answer.