Question

The correct order for the wavelength of absorption in the visible resion is :

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

Answer 1

The correct option is A $\left[Ni\right(N{O}_2{)}_6{]}^{4-}<\left[Ni\right(N{H}_3{)}_6{]}^{2+}<\left[Ni\right(H_{2}O{)}_6{]}^{2+}$

Option (A) is correct.

The correct order for the wavelength of absorption in the visible region is $\left[Ni\right(N{O}_2{)}_6{]}^{4-}<\left[Ni\right(N{H}_3{)}_6{]}^{2+}<\left[Ni\right(H_{2}O{)}_6{]}^{2+}$.

The central metal ion in all the three complexes is the same.Therefore, absorption in the visible region depends on the ligands.The order in which the $CFSE$ values of the ligands increases in the spectrochemical series is as follows :

$H_{2}O<N{H}_3<N{O}_2$