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

K M n O_{4}

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C r O_{3}

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C u C l_{2}

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C u_{2} O

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Solution

The correct option is D $C u_{2} O$
In $K M n O_{4}$ , 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 $K M n O_{4}$ and colour will appear due to ligand to metal charge transfer spectra.
Similarly, in $C r O_{3}, C r$ has a oxidation state of $+ 6$ therefore doesn't have any electron in d orbital which implies there is no d-d transition in $C r O_{3}$ and colour will appear due to ligand to metal charge transfer spectra.

In case of $C u C l_{2}, C u$ has a oxidation state of $+ 2$ and $C u$ has $d^{9}$ electronic configuration and therefore d-d transition can take place.

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, D due to charge transfer spectra.

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