Key idea: The wavelength of light absorbed depends on the crystal field splitting energy $\Delta_o$.

$E = h\nu = \frac{hc}{\lambda}$

Hence,

$\lambda \propto \frac{1}{\Delta_o}$

So, larger $\Delta_o$ means smaller absorbed wavelength, and smaller $\Delta_o$ means larger absorbed wavelength.

Compare ligands using spectrochemical series:

$\mathrm{CN^-} > \mathrm{NH_3} > \mathrm{H_2O} > \mathrm{Cl^-}$

Stronger ligand $\Rightarrow$ larger $\Delta_o$.

Now compare the given complexes:

• $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ (C): all ligands are $\mathrm{CN^-}$, the strongest ligand here. So this has the largest $\Delta_o$, hence smallest $\lambda$ absorbed.

• $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$ (B): all ligands are $\mathrm{NH_3}$, stronger than $\mathrm{H_2O}$ and $\mathrm{Cl^-}$.

• $\left[\mathrm{Co}(\mathrm{NH}_3)_5(\mathrm{H_2O})\right]^{3+}$ (A): one $\mathrm{NH_3}$ in B is replaced by weaker ligand $\mathrm{H_2O}$. Therefore, $\Delta_o(B) > \Delta_o(A)$ so $\lambda_A > \lambda_B$

• $\left[\mathrm{CoCl}(\mathrm{NH}_3)_5\right]^{2+}$ (D): one $\mathrm{NH_3}$ is replaced by even weaker ligand $\mathrm{Cl^-}$. Therefore this has the smallest $\Delta_o$ among A, B, D, so it absorbs the largest wavelength.

**Overall order of $\Delta_o$: **

$C > B > A > D$

**Therefore, order of absorbed wavelength $\lambda$: **

$D > A > B > C$

Match with options: This corresponds to Option C.

Comparison with stored answer: Stored correct answer = C. Our derived answer = C. So they agree.