Correct increasing order for the wavelengths of absorption in the visible region for the complexes of $\mathrm{Co}{}^{3+}$ is:

Solution & Explanation

\textbf{Answer:} (A) $[\mathrm{Co}(\mathrm{en}){}_3]{}^{3+}<[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}<[\mathrm{Co}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$ \textbf{Solution:} From the spectrochemical series the field strength of the ligands increases as $\mathrm{H}{}_2\mathrm{O}<\mathrm{NH}{}_3<\text{en}$. A stronger field produces a larger crystal-field splitting ${\Delta }_o$. The energy of the absorbed light is $E=\frac{hc}{\lambda }$, so a larger ${\Delta }_o$ means a larger absorbed energy and therefore a \emph{shorter} wavelength. Thus the wavelength absorbed is inversely related to field strength: field: $[\mathrm{Co}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}<[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}<[\mathrm{Co}(\mathrm{en}){}_3]{}^{3+}$ wavelength: $[\mathrm{Co}(\mathrm{en}){}_3]{}^{3+}<[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}<[\mathrm{Co}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$. This matches option A.