Question

Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd and Hg are soft metals.
(iii) ${{\text{E}}^{\text{o}}}$ value of the ${\text{M}}{{\text{n}}^{{\text{3 + }}}}{\text{/M}}{{\text{n}}^{{\text{2 + }}}}$couple is highly positive $\left( { + 1.57\,{\text{V}}} \right)$ as compared to ${\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{/C}}{{\text{r}}^{{\text{2 + }}}}$.

Answer 1

Hint: The valence orbitals of transition metals are not deep buried. The Zn, Cd and Hg have full-filled electronic configuration and if after the reduction a metal gain stable electronic configuration then the metal will have high reduction potential.

Complete step by step answer:
(i) The reason for the variable oxidation states of transition metals is as follows:
The valence electronic configuration of the transition metals is $\left( {{\text{n}} - 1} \right){{\text{d}}^{1 - 10}}\,{\text{n}}{{\text{s}}^2}$ .
The energy of $\left( {{\text{n - 1}}} \right){\text{d}}$ and ${\text{ns}}$ orbitals are the same so, the electrons can remove from both$\left( {{\text{n - 1}}} \right){\text{d}}$ and ${\text{ns}}$orbitals so, the transition metals show the variable oxidation state.
So, in presence of strong electron-withdrawing, a transition metal can donate all valence electrons.
Therefore, the transition metals show variable oxidation states due to almost the same energy of $\left( {{\text{n - 1}}} \right){\text{d}}$ and ${\text{ns}}$ orbitals.
(ii) The reason for the soft nature of Zn, Cd and Hg is as follows:
The valence electronic configuration of the Zn, Cd and Hg is $\left( {{\text{n}} - 1} \right){{\text{d}}^{10}}\,{\text{n}}{{\text{s}}^2}$ .
So, the valence shells of the Zn, Cd and Hg are fully-filled and hence very stable so they cannot gain or lose electrons easily so they form weak bonding hence they are soft metals.
Therefore, Zn, Cd and Hg are soft metals due to fully-filled electronic configuration.
Therefore, ${\text{M}}{{\text{n}}^{{\text{3 + }}}}$ ion is a strong oxidizing agent due to high reduction potential.
(iii) The reason for the highly positive ${{\text{E}}^{\text{o}}}$ value of the ${\text{M}}{{\text{n}}^{{\text{3 + }}}}{\text{/M}}{{\text{n}}^{{\text{2 + }}}}$ couple is as follows:
The valence electronic configuration of the ${\text{M}}{{\text{n}}^{{\text{3 + }}}}$ and ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ is,
${\text{M}}{{\text{n}}^{{\text{3 + }}}}\, = \,3{{\text{d}}^4}\,{\text{4}}{{\text{s}}^0}$ and ${\text{M}}{{\text{n}}^{{\text{2 + }}}}\, = \,3{{\text{d}}^5}\,{\text{4}}{{\text{s}}^0}$.
After the reduction ${\text{M}}{{\text{n}}^{{\text{3 + }}}}$ converts into ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$ which has half-filled electronic configuration so, the conversion of ${\text{M}}{{\text{n}}^{{\text{3 + }}}}{\text{/M}}{{\text{n}}^{{\text{2 + }}}}$ is favourable so, it has positive value of ${{\text{E}}^{\text{o}}}$.
The valence electronic configuration of the ${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ and ${\text{C}}{{\text{r}}^{{\text{2 + }}}}$ is,
${\text{C}}{{\text{r}}^{{\text{3 + }}}}\, = \,3{{\text{d}}^3}{\text{4}}{{\text{s}}^0}$ and ${\text{C}}{{\text{r}}^{{\text{2 + }}}}\, = \,3{{\text{d}}^4}\,{\text{4}}{{\text{s}}^0}$.
After the reduction${\text{C}}{{\text{r}}^{{\text{3 + }}}}$ converts into ${\text{C}}{{\text{r}}^{{\text{2 + }}}}$ this conversion does not lead to the stable electronic configuration.
Therefore, ${{\text{E}}^{\text{o}}}$ value of the ${\text{M}}{{\text{n}}^{{\text{3 + }}}}{\text{/M}}{{\text{n}}^{{\text{2 + }}}}$ couple is highly positive $\left( { + 1.57\,{\text{V}}} \right)$ as compared to ${\text{C}}{{\text{r}}^{{\text{3 + }}}}{\text{/C}}{{\text{r}}^{{\text{2 + }}}}$ due to stable electronic configuration.

Note: In transition metals, Mn shows the maximum $\left( {{\text{ + 3}}\,\,{\text{to}}\,\,{\text{ + 7}}} \right)$ and highest ($\,{\text{ + 7}}$) oxidation number. Soft metals show only + 2 oxidation states and have low melting and boiling point.