Total number of unpaired electrons in the complex ions $$[Co(NH_3)_6]^{3+}$$ and $$[NiCl_4]^{2-}$$ is _____

We need to find the total number of unpaired electrons in $$[Co(NH_3)_6]^{3+}$$ and $$[NiCl_4]^{2-}$$.

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

Cobalt is in +3 oxidation state: $$Co^{3+}$$ has configuration $$[Ar]3d^6$$.

$$NH_3$$ is a strong field ligand, so this is a low-spin octahedral complex.

In a strong octahedral field, the 6 d-electrons fill the $$t_{2g}$$ orbitals completely: $$t_{2g}^6 e_g^0$$.

Number of unpaired electrons = 0.

For $$[NiCl_4]^{2-}$$:

Nickel is in +2 oxidation state: $$Ni^{2+}$$ has configuration $$[Ar]3d^8$$.

$$Cl^-$$ is a weak field ligand, so this is a tetrahedral complex (common for $$Ni^{2+}$$ with weak field ligands).

In a tetrahedral field with $$d^8$$ configuration, the splitting is: $$e^4 t_2^4$$.

Number of unpaired electrons = 2.

Total unpaired electrons = $$0 + 2 = 2$$.

The answer is $$\boxed{2}$$.