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

Hint: 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.