Chemistry - Complete d-block Question with Solution | TestHub

1. The deep blue colour 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 $$\begin{gathered} {\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}\Rightarrow 2\mathrm{x}+7\times \left(-2\right) \\ \mathrm{x}=+6 \end{gathered}$$. 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 coloured compound is formed.
5. Similarly, in ${\mathrm{MnO}}_4^{-}$the oxidation number of Manganese$$\begin{gathered} \Rightarrow \mathrm{x}+4(-2)=-1 \\ \mathrm{x}=+7 \end{gathered}$$. 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 coloured compound is formed.

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