Question

In which of the following compound(s), the colour is due to charge transfer spectra?

• A. $KMn{O}_4$
• B. $Cr{O}_3$
• C. $C{u}_{2}O$
• D. All of the above

Answer 1

The correct option is D All of the above

In $KMn{O}_4$, Mn has a oxidation state of $+7$ therefore doesn't have any electron in d orbital which implies there is no d-d transition in $KMn{O}_4$ and colour will appear due to ligand to metal charge transfer spectra.
Similarly, in $Cr{O}_3,Cr$ has a oxidation state of $+6$ therefore doesn't have any electron in d orbital which implies there is no d-d transition in $Cr{O}_3$ and colour will appear due to ligand to metal charge transfer spectra.

In last option $C{u}_{2}O$ has oxidation state of $+1$ which recommend $d^{10}$ electronic configuration for $C{u}_{2}O$ and therefore there is no d-d transition in $C{u}_{2}O$.
Hence colour is arised in A, B, C due to charge transfer spectra