The wavelength of light absorbed for the following complexes are in the order
(I)$$\left[Co{(NH_{3})}_{6}\right]^{3+}$$
(II)$$\left[Co{(H_{2}O)}_{6}\right]^{3+}$$
(III)$$\left[Co{(CN)}_{6}\right]^{3-}$$
(IV)$$\left[Co(NH_{3})_{5}(H_{2}O)\right]^{3+}$$
(V)$$\left[CoF_{6}\right]^{3-}$$

We need to find the order of wavelength of light absorbed by the given Co(III) complexes.

The wavelength of absorbed light is inversely proportional to crystal field splitting energy ($$\Delta$$): $$\lambda = \frac{hc}{\Delta}$$. A stronger field ligand gives larger $$\Delta$$ and thus shorter (smaller) wavelength of absorbed light.

Order the ligands by field strength (spectrochemical series):

$$F^- < H_2O < NH_3 < CN^-$$

Order the complexes by $$\Delta$$ (crystal field splitting):

(V) $$[CoF_6]^{3-}$$: weakest field ($$F^-$$), smallest $$\Delta$$

(II) $$[Co(H_2O)_6]^{3+}$$: $$H_2O$$

(IV) $$[Co(NH_3)_5(H_2O)]^{3+}$$: mixed, mostly $$NH_3$$ with one $$H_2O$$, slightly less than pure $$NH_3$$

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

(III) $$[Co(CN)_6]^{3-}$$: $$CN^-$$, strongest field, largest $$\Delta$$

Order by wavelength (inversely proportional to $$\Delta$$):

Largest $$\Delta$$ = smallest $$\lambda$$, so:

$$\lambda_{III} < \lambda_{I} < \lambda_{IV} < \lambda_{II} < \lambda_{V}$$

This reads: III < I < IV < II < V

This matches Option C: III < I < IV < II < V.

The correct answer is Option C.