Question

In which of the following the inert pair effect is most prominent?
A.${\text{C}}$
B.${\text{Si}}$
C.${\text{Ge}}$
D.${\text{Pb}}$

Answer 1

All the elements given in the option belong to the same group. In a group as we move down the group inert pair effect increases. So the element at the bottom of the group will have maximum inert pair effect.

Complete step by step answer:
Chemical symbol for carbon is ${\text{C}}$
Chemical symbol for silicon is ${\text{Si}}$
Chemical symbol for germanium is ${\text{Ge}}$
Chemical symbol for Lead is ${\text{Pb}}$
Inert pair effect is defined as the reluctance on non participation of the s electrons in bonding because of the high intermolecular force of attraction is known as inert pair effect.
This happens due to the involvement of the f electrons which increase the intermolecular forces of attraction and hence the valence shell s orbital tends to remain unbounded.
Since as we move down the group inert pair effect increases all elements given in the option belong to group number 14 that is carbon family and lead is the bottom most member of this group. Hence lead has the maximum inert pair effect.
One of the major applications of the inert pair effect is the stability of the lower Oxidation State as we move down the group. Normally the stable oxidation state of group 14 elements is ${\text{ + 4}}$ but the most stable oxidation state for lead is ${\text{ + 2}}$. In the same way nitrogen and oxygen families also show inert pair effects.

Hence the correct option is D.

Note:
Since the s orbital contains a maximum of 2 electrons that is why the difference in oxidation state is always 2 as we have seen in case of Lead. In Case of nitrogen groups both plus 5 and plus 3 oxidation states exist. There are also differences between the two. Similarly in the oxygen family the oxidation states shown are 6 and 4.