Question

${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$are colored compounds. Justify your statement.

Answer 1

Charge Transfer:

1. The deep blue color of the solution ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is because an electron from the Oxygen lone pair is transferred to low lying Chromate ion orbital.
2. Oxidation number of Chromium in${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2}$$\Rightarrow 2\mathrm{x}+7\times -2=-2,\mathrm{x}=+6$. The electronic configuration of Chromium is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{1}3{\mathrm{d}}^5$ so it will lose six electrons to form ${\mathrm{Cr}}^{+6}$
3. Electronic configuration of ${\mathrm{Cr}}^{+6}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^6$since there are no electrons present in the d-orbital, the d-d transition will not occur.
4. The oxygen atom in ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{-2}$that act as a ligand and due to charge transfer transfers its one electron to chromium and due to which the colored compound is formed.
5. Similarly, in ${{\mathrm{MnO}}_4}^{-}$the oxidation number of Manganese$\Rightarrow \mathrm{x}+4(-2)=-1,\mathrm{x}=+7$. The electronic configuration of Manganese is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^5$ so it will lose seven electrons to form ${\mathrm{Mn}}^{+7}$
6. Electronic configuration of${\mathrm{Mn}}^{+7}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^6$, here also no d-orbital is present so no d-d transition will take place.
7. The oxygen atom in ${{\mathrm{MnO}}_4}^{-}$acts as a ligand and due to charge transfer transfers its one electron to manganese due to which a colored compound is formed.

Therefore, ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ and ${\mathrm{KMnO}}_4$ are colored compounds due to charge transfer or ligand to metal transfer.