The orange colour of K _ 2 Cr _ 2 O _ 7 and purple colour of KMnO _...

1. The deep blue colour of the solution ${K}_{2}{Cr}_{2}{O}_{7}$ is because an electron from the Oxygen lone pair is transferred to low lying Chromate ion orbital.

2. Oxidation number of Chromium in ${Cr}_{2}{{O}_{7}}^{2 -} ⇒ 2x + 7 \times \left(- 2\right)x = + 6$. The electronic configuration of Chromium is $1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}4{s}^{1}3{d}^{5}$so it will lose six electrons to form ${Cr}^{+ 6}$.

3. Electronic configuration of ${Cr}^{6 +} = 1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}$since there are no electrons present in the d-orbital, the d-d transition will not occur.

4. The oxygen atom in ${Cr}_{2}{{O}_{7}}^{2 -}$that act as a ligand and due to charge transfer transfers its one electron to chromium and due to which the coloured compound is formed.

5. Similarly, in ${MnO}_{4}^{-}$the oxidation number of Manganese$⇒ x + 4 ( - 2 ) = - 1x = + 7$. The electronic configuration of Manganese is $1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}4{s}^{2}3{d}^{5}$so it will lose seven electrons to form ${Mn}^{+ 7}$

6. Electronic configuration of ${Mn}^{+ 7} = 1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}$, here also no d-orbital is present so no d-d transition will take place.

7. The oxygen atom in ${MnO}_{4}^{-}$ acts as a ligand and due to charge transfer transfers its one electron to manganese due to which a coloured compound is formed.

Therefore, ${K}_{2}{Cr}_{2}{O}_{7}$ and ${KMnO}_{4}$ are coloured compounds due to charge transfer or ligand to metal transfer.