The correct order of the wavelength of light absorbed by the following complexes is: A. $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}$ B. $[\mathrm{Co}(\mathrm{CN}){}_6]{}^{3-}$ C. $[\mathrm{Cu}(\mathrm{H}{}_2\mathrm{O}){}_4]{}^{2+}$ D. $[\mathrm{Ti}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$

Solution & Explanation

\textbf{Answer:} (D) B < A < D < C \textbf{Solution:} The wavelength of light absorbed is inversely related to the crystal-field splitting ${\Delta }_o$, since $\lambda =\frac{hc}{{\Delta }_o}$ (larger ${\Delta }_o\Rightarrow$ shorter $\lambda$). Comparing the field splittings: B $[\mathrm{Co}(\mathrm{CN}){}_6]{}^{3-}$ — strongest field ($\mathrm{CN}{}^{-}$, $\mathrm{Co}{}^{3+}$) $\Rightarrow$ largest ${\Delta }_o$. A $[\mathrm{Co}(\mathrm{NH}{}_3){}_6]{}^{3+}$ — strong field $\mathrm{NH}{}_3$, next largest. D $[\mathrm{Ti}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$ — $\mathrm{H}{}_2\mathrm{O}$ with $\mathrm{Ti}{}^{3+}$. C $[\mathrm{Cu}(\mathrm{H}{}_2\mathrm{O}){}_4]{}^{2+}$ — weakest field $\Rightarrow$ smallest ${\Delta }_o$. So ${\Delta }_o$: B > A > D > C, and therefore the wavelength absorbed increases in the reverse order: B < A < D < C (option D).