Question

The correct order for the wavelength of absorption in the visible region is

• A. $[Ni(N{O}_2{)}_6{]}^{4-}<[Ni(N{H}_3{)}_6{]}^{2+}<[Ni(H_{2}O{)}_6{]}^{2+}$
• B. $[Ni(N{O}_2{)}_6{]}^{4-}<[Ni(H_{2}O{)}_6{]}^{2+}<[Ni(N{H}_3{)}_6{]}^{2+}$
• C. $[Ni(H_{2}O{)}_6{]}^{2+}<[Ni(N{H}_3{)}_6{]}^{2+}<[Ni(N{O}_2{)}_6{]}^{4-}$
• D. $[Ni(N{H}_3{)}_6{]}^{2+}<[Ni(H_{2}O{)}_6{]}^{2+}<[Ni(N{O}_2{)}_6{]}^{4-}$

Answer 1

Answer: (A)

When central metal ion is same (here $N{i}^{2+}$ ), the absorption of colour depends on the, ligand. According to spectrochemical series, various ligands are as follows
$I^{-}<B{r}^{-}<C{l}^{-}<N{O}_3^{-}<F^{-}<H_{2}O<N{H}_3<N{O}_2^{-}<C{N}^{-}<CO$
Thus, $H_{2}O$ is weakest ligand among these, therefore, the absorbed energy will be lowest in ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$, so it will absorb highest wavelength (red light) $\left(E\propto \frac{1}{\lambda }\right)$
$\therefore$ the order of increasing wavelength is
${\left[Ni{\left(N{O}_2\right)}_6\right]}^{4-}<{\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}<{\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$