Consider the following metal complexes:
$$[Co(NH_3)]^{3+}$$
$$[CoCl(NH_3)_5]^{2+}$$
$$[Co(CN)_6]^{3-}$$
$$[Co(NH_3)_5(H_2O)]^{3+}$$
The spin-only magnetic moment value of the complex that absorbs light with shortest wavelength is ______ B.M. (Nearest integer)

We need to find the spin-only magnetic moment of the complex that absorbs light with the shortest wavelength among the given cobalt complexes.

Energy of absorbed light: $$ E = \frac{hc}{\lambda} $$. Shorter wavelength means higher energy, which corresponds to a larger crystal field splitting energy ($$ \Delta_0 $$).

The spectrochemical series for the ligands present is:

$$Cl^- < H_2O < NH_3 < CN^-$$

Ranking the complexes by their overall crystal field strength:

$$[CoCl(NH_3)_5]^{2+} < [Co(NH_3)_5(H_2O)]^{3+} < [Co(NH_3)_6]^{3+} < [Co(CN)_6]^{3-}$$

The complex $$ [Co(CN)_6]^{3-} $$ has the strongest field ligand (CN⁻), so it has the largest $$ \Delta_0 $$ and absorbs light with the shortest wavelength.

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

CN⁻ is a strong field ligand, so $$ [Co(CN)_6]^{3-} $$ is a low-spin octahedral complex. The $$ 3d^6 $$ electrons fill as:

$$t_{2g}^6 \, e_g^0$$

All 6 electrons are paired in the $$ t_{2g} $$ orbitals, giving 0 unpaired electrons.

$$\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \text{ B.M.}$$

The spin-only magnetic moment is 0 B.M.