Question

Stability of ions of ${\text{Ge, Sn and Pb}}$ will be in the order:
A) ${\text{G}}{{\text{e}}^{2 + }} < {\text{S}}{{\text{n}}^{2 + }} < {\text{P}}{{\text{b}}^{2 + }}$
B) ${\text{G}}{{\text{e}}^{4 + }} > {\text{S}}{{\text{n}}^{4 + }} > {\text{P}}{{\text{b}}^{4 + }}$
C) ${\text{S}}{{\text{n}}^{4 + }} > {\text{S}}{{\text{n}}^{2 + }}$
D) All are correct.

Answer 1

We know that ${\text{Ge}}$, ${\text{Sn}}$ and ${\text{Pb}}$ are the elements of group 14 of the periodic table. To solve this we must know the stability of $+ 2$ and $+ 4$ oxidation state of the elements of the group 14 of the periodic table. The stability of oxidation states is affected by the inert pair effect.
Complete step by step answer:
We know that ${\text{Ge}}$, ${\text{Sn}}$ and ${\text{Pb}}$ are the elements of group 14 of the periodic table. The electronic configuration of group 14 is $n{s^2}n{p^2}$.
-The elements of group 14 in the order are carbon $\left( {\text{C}} \right)$, silicon $\left( {{\text{Si}}} \right)$, germanium $\left( {{\text{Ge}}} \right)$, tin $\left( {{\text{Sn}}} \right)$, lead $\left( {{\text{Pb}}} \right)$ and flerovium $\left( {{\text{Fl}}} \right)$.
-We know that oxidation state is the charge acquired by the species may be positive, negative or zero by gain or loss of electrons.
-The maximum number of electrons the p-orbital can occupy are 6. From the electronic configuration, -we can say that the most common oxidation state exhibited by the elements of group 14 must be $+ 4$.
-In group 14 as we move down the group the $+ 2$ oxidation state becomes more and more common. On moving down the group, the $+ 4$ oxidation state becomes less stable.
-The $+ 4$ oxidation state becomes less stable due to the inert pair effect. The tendency of the two electrons of the valence atomic orbital to remain unshared in compounds is known as inert pair effect.
-The elements ${\text{Ge}}$, ${\text{Sn}}$ and ${\text{Pb}}$ show both $+ 2$ and $+ 4$ oxidation state. But the stability of $+ 2$ oxidation state increases down the group. Thus,
-${\text{G}}{{\text{e}}^{2 + }} < {\text{S}}{{\text{n}}^{2 + }} < {\text{P}}{{\text{b}}^{2 + }}$
And the stability of $+ 4$ oxidation state decreases down the group. Thus,
${\text{G}}{{\text{e}}^{4 + }} > {\text{S}}{{\text{n}}^{4 + }} > {\text{P}}{{\text{b}}^{4 + }}$
The ${\text{S}}{{\text{n}}^{4 + }}$ is more stable than ${\text{S}}{{\text{n}}^{2 + }}$. Thus,
${\text{S}}{{\text{n}}^{4 + }} > {\text{S}}{{\text{n}}^{2 + }}$
Thus, stability of ions of ${\text{Ge, Sn and Pb}}$ will be in the order ${\text{G}}{{\text{e}}^{2 + }} < {\text{S}}{{\text{n}}^{2 + }} < {\text{P}}{{\text{b}}^{2 + }}$, ${\text{G}}{{\text{e}}^{4 + }} > {\text{S}}{{\text{n}}^{4 + }} > {\text{P}}{{\text{b}}^{4 + }}$ and ${\text{S}}{{\text{n}}^{4 + }} > {\text{S}}{{\text{n}}^{2 + }}$.
Thus, the correct option is (D) all are correct.

Note: The tendency of the two electrons of the valence atomic orbital to remain unshared in compounds is known as inert pair effect. As a result, the inert pair of the ns electrons remains more tightly held by the nucleus. Thus, the inert pair of the ns electrons participates less in the formation of bond.