The order of energy absorbed which is responsible for the colour of complexes: (A) $[\mathrm{Ni}(\mathrm{H}{}_2\mathrm{O}){}_2(\mathrm{en}){}_2]{}^{2+}$ (B) $[\mathrm{Ni}(\mathrm{H}{}_2\mathrm{O}){}_4(\mathrm{en})]{}^{2+}$ (C) $[\mathrm{Ni}(\mathrm{en}){}_3]{}^{2+}$

Solution & Explanation

\textbf{Answer:} (C) C > A > B \textbf{Solution:} The colour of a complex arises from a $d$-$d$ transition; the energy absorbed equals the crystal-field splitting ${\Delta }_o$. Since en is a stronger-field ligand than $\mathrm{H}{}_2\mathrm{O}$, more en ligands give a larger ${\Delta }_o$ and hence a larger energy of absorption. Counting en ligands: (C) has 3 en, (A) has 2 en, (B) has 1 en. Therefore ${\Delta }_o$ and the absorbed energy follow: $[\mathrm{Ni}(\mathrm{en}){}_3]{}^{2+}>[\mathrm{Ni}(\mathrm{H}{}_2\mathrm{O}){}_2(\mathrm{en}){}_2]{}^{2+}>[\mathrm{Ni}(\mathrm{H}{}_2\mathrm{O}){}_4(\mathrm{en})]{}^{2+}$, i.e. C > A > B (option C).