Question

The correct increasing order of wavelengths of absorption in the visible region for the complexes of $C{o}^{3+}$ is:

• A. $\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$
• B. $\left[Co\right(H_{2}O{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• C. $\left[Co\right(H_{2}O{)}_6{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+}$
• D. $\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$

Answer 1

The correct option is A $\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$

The order of the ligand in the spectrochemical series is $H_{2}O<N{H}_3<en$

Hence, the wavelength of the light observed will be in the order: $\left[Co\right(H_{2}O{)}_6{]}^{3+}$ < $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ < $\left[Co\right(en{)}_3{]}^{3+}$

Thus, the wavelength absorbed will be in the opposite order.
i.e., $\left[Co\right(en{)}_3{]}^{3+},\left[Co\right(N{H}_3{)}_6{]}^{3+},\left[Co\right(H_{2}O{)}_6{]}^{3+}$