Question

Elements of group $14$.
This question has multiple answers.
A) Exhibit oxidation state of $+ 4$ only
B) Exhibit oxidation state of $+ 2$ and $+ 4$
C) Form ${M^{2 + }}$ and ${M^{4 + }}$ ions
D) Form ${M^{2 + }}$ and ${M^{4 + }}$ ions

Answer 1

We have to know that all the elements in group $14$ have $4$ electrons in the outermost shell, the valency of group $14$ elements is $4$. They use these electrons in the bond formation in order to obtain octet configuration. We need to remember that the radii of group $14$ elements are lesser than that of group $13$ elements. This can be explained by the increase in the effective nuclear charge.

Complete answer:
We will look at options one by one:
Option A) this option is incorrect as group $14$ elements exhibit oxidation state of $+ 4$ but with that they also show $+ 2$ oxidation state.
Option B) this option is correct as group $14$ elements exhibit oxidation states of $+ 2$ and $+ 4$.
Option C) this option is incorrect as group $14$ elements do not form ${M^{2 + }}$ ions.
Option D) this option is correct as group $14$ elements form ${M^{2 + }}$and ${M^{4 + }}$oxidation states. We need to remember that the elements of group $14$, have an outer electronic configuration of $n{s^2}n{p^4}$ this shows $+ 2$ and $+ 4$ oxidation states. As we know that on moving down the group; the stability of $+ 2$ oxidation state increases and the stability of the $+ 4$ oxidation state decreases.

Option B is the correct answer.

Note:
We know that the general oxidation states exhibited by the group $14$ elements are $+ 4$, and $+ 2$. As we go down the group, the tendency to form $+ 2$ ion increases. This is due to the inert pair effect. This effect is exhibited by p-block elements. This can be explained using the inert pair effect. It is the non-participation of the s-orbital during bonding due to the poor shielding of the intervening electrons.