Question

Which of the following ions is the most stable?
A. ${\text{S}}{{\text{n}}^{2 + }}$
B. ${\text{G}}{{\text{e}}^{2 + }}$
C. ${\text{S}}{{\text{i}}^{2 + }}$
D. ${\text{P}}{{\text{b}}^{2 + }}$

Answer 1

Hint:The stability here represents the chemical stability of the ions. These ions are from the same non - metal series, they all belong to the carbon family. Recall some general trends of the periodic table, like atomic size and oxidation state, and try to predict the most stable amongst them.


Complete solution:
When we go down the group we see the increase in d, f orbitals, and these groups are poor in shielding effect due to which the effective nuclear charge does not decrease significantly down the group and gives rise to what is known as inert pair effect.First, let’s identify the ions, in the given option we have Tin, Germanium, Silicon, and lead. As mentioned they all belong to the carbon family or are called group 14 elements
Look at the oxidation states of all these metal ions, they all are in the same +2 oxidation state, but they are placed at different hierarchies in group 15. If we arrange them in the increasing order of their atomic size we can write;
Si < Ge < Sn < Pb
The most stable oxidation state of group 14 is +4, but this is not uniform throughout the group as we go down the group the +4 stability decrease due to the inert pair effect, and the +2 oxidation state becomes predominant, therefore we can say that ${\text{P}}{{\text{b}}^{2 + }}$ is the most stable.
Therefore, from the above explanation we can say that the correct option is (D).


Note:We can say that the Inert pair effect is an exceptional effect because as we go down the group the increase in the atomic radius should increase the element's ability to donate the electrons present in the outermost orbital, but this is not observed due to this effect. The poor shielding effect of d and f orbitals results in the outermost electron experiencing relatively more nuclear energy.