JEE Main 2024 — Chemistry Inorganic Chemistry

The deep blue colour of the solution ${K}_{2}{\mathrm{Cr}}_{2}{O}_{7}$K2​Cr2​O7​ is because an electron from the Oxygen lone pair is transferred to low lying Chromate ion orbital.

Oxidation number of Chromium in ${\mathrm{Cr}}_{2}{{O}_{7}}^{2-}\Rightarrow 2x+7\times (-2) x=+6$Cr2​O7​2−⇒2x+7×(−2)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}$1s22s22p63s23p64s13d5so it will lose six electrons to form ${\mathrm{Cr}}^{+6}$Cr+6.

Electronic configuration of ${\mathrm{Cr}}^{6+}=1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}$Cr6+=1s22s22p63s23p6since there are no electrons present in the d-orbital, the d-d transition will not occur.

The oxygen atom in ${\mathrm{Cr}}_{2}{{O}_{7}}^{2-}$Cr2​O7​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.

Similarly, in ${\mathrm{MnO}}_{4}^{-}$MnO4−​the oxidation number of Manganese$\Rightarrow x+4(-2)=-1 x=+7$⇒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}$1s22s22p63s23p64s23d5so it will lose seven electrons to form ${\mathrm{Mn}}^{+7}$Mn+7

Electronic configuration of ${\mathrm{Mn}}^{+7}=1{s}^{2}2{s}^{2}2{p}^{6}3{s}^{2}3{p}^{6}$Mn+7=1s22s22p63s23p6, here also no d-orbital is present so no d-d transition will take place.

The oxygen atom in ${\mathrm{MnO}}_{4}^{-}$MnO4−​ acts as a ligand and due to charge transfer transfers its one electron to manganese due to which a coloured compound is formed.

Therefore, ${K}_{2}{\mathrm{Cr}}_{2}{O}_{7}$K2​Cr2​O7​ and ${\mathrm{KMnO}}_{4}$KMnO4​ are coloured compounds due to charge transfer or ligand to metal transfer.