The total number of unpaired electrons present in [Co(NH$$_3$$)$$_6$$]Cl$$_2$$ and [Co(NH$$_3$$)$$_6$$]Cl$$_3$$ is ___.

In $$[\text{Co(NH}_3\text{)}_6]\text{Cl}_2$$, cobalt has an oxidation state of +2 (since $$\text{NH}_3$$ is neutral and each Cl is $$-1$$). Co$$^{2+}$$ has the electronic configuration $$[\text{Ar}]\, 3d^7$$. NH$$_3$$ is a strong-field ligand, so in the octahedral crystal field the electrons are arranged as $$t_{2g}^6\, e_g^1$$, giving 1 unpaired electron.

In $$[\text{Co(NH}_3\text{)}_6]\text{Cl}_3$$, cobalt has an oxidation state of +3. Co$$^{3+}$$ has the configuration $$[\text{Ar}]\, 3d^6$$. With NH$$_3$$ as a strong-field ligand, all six d-electrons pair in the $$t_{2g}$$ orbitals: $$t_{2g}^6\, e_g^0$$, giving 0 unpaired electrons.

Therefore the total number of unpaired electrons is $$1 + 0 = 1$$.