Question

Statement -1 : $Pb{O}_2$​ is an oxidising agent and reduced to $PbO$ .
Statement - 2 : Stability of $Pb(II)$$>Pb(IV)$ on account of inert pair effect.
A.Statement – I is True, statement- 2 is True; Statement 2 is a correct explanation of statement – I.
B.Statement – I is True, statement- 2 is True; Statement 2 is NOT a correct explanation of statement – I.
C.Statement – I is True, statement- 2 is False
D.Statement – I is False, statement- 2 is True

Answer 1

Hint: The atomic number of lead $(Pb)$is $82$ . The electrons in $6s^2$ appears to participate in the chemical bonding formation. This is why lead occurs in two states of oxidation: $+2$ and $+4$.

Complete step by step answer:
Before analysing the question, let's first learn about inert pair effects.
Inert pair effect is the reluctance of $s$ orbitals electrons to take part in chemical reaction.
As we know lead has two oxidation states. Among them, the $+4$ state is very unstable because the $6s$ and $6p$ electrons have a huge energy difference. So promoting the $6s$ electrons to the $6p$ subshell for the formation of $+4$ oxidation state by lead removing it's $4$ electrons is not an easy process because it requires a lot of energy. This is the effect of inert electron pairs.
In lead dioxide, lead has an oxidation state of $+4$ as a result lead donates two oxygen atoms to form lead monoxide in which lead has an oxidation state of $+2$ to achieve stability. That is why lead dioxide is reduced and other compounds are oxidised and we recognise that oxygen elimination is an example of reduction. That's why it acts as a good oxidising agent.
From the above explanation, it is clear that statement $1$ is True, and statement $2$ is a correct explanation of statement $2$ .

So, the correct answer is Option A.

Note:
An oxidising agent is a substance that undergoes reduction and a reducing agent is a substance that actually undergoes oxidation. Reduction is the process of gaining electrons whereas oxidation is the process of losing electrons