Question

The wavelength of absorbed electromagnetic radiation for CoBr₄^2- is $\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{m}$. Will the complex ion CoBr₄^2-absorb electromagnetic radiation having a wavelength longer or shorter than$\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{m}$? Explain.

Short answer

CoBr₄^2-will absorb a wavelength shorter than$3.4\times {10}^{-6}\mathrm{m}$

Step-by-step solution

The reason behind the absorption of shorter wavelength by CoBr42-

CoBr₄^2- being an octahedral ligand, has a higher CFSE value, thus, electron requires higher energy to get excited. From the equation $\mathrm{E}=\mathrm{hc}/\mathrm{\lambda }$, we know that energy is inversely proportional to wavelength. Thus, CoBr₄^2-being an octahedral will require absorption of a shorter wavelength (i.e., higher energy)

Why octahedral splitting has a higher CFSE value than tetrahedral splitting?

In tetrahedral complexes, the ligands come in direct contact with ${\mathrm{d}}_{\mathrm{xy}},{\mathrm{d}}_{\mathrm{yz}},{\mathrm{d}}_{\mathrm{zx}}$ligands, which cause them to split to a higher level. These orbitals are located in between theaxes, which makes them repel the ligands with lesser energy compared to that octahedral orbitals. In Octahedral complexes, the ligands come in direct contact with ${\mathrm{d}}_{{\mathrm{z}}^2}\mathrm{and}{\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$orbitals which are located along the axes and cause them to repel the ligands with higher energy. Thus, octahedral CFSE is always larger than tetrahedral CFSE.