Question

Assertion :$PbC{l}_2$ is more stable than $PbC{l}_4$ Reason: $PbC{l}_4$ is a powerful oxidising agent

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion: $PbC{l}_2$ is more stable than $PbC{l}_4$

The inert pair effect is the tendency of the two electrons in the outermost atomic s orbital to remain unionized or unshared in compounds of post-transition metals., out of which (Pb) lead is one. The term inert pair effect is often used in relation to the increasing stability of oxidation states, that are two less than the group valency for the heavier elements of groups 13, 14, 15 and 16.

The s electrons are more tightly bound to the nucleus and therefore more difficult to ionize.

Due to inert pair effect, the higher oxidation state of an element is not stable as we move down the group.

In other words, $P{b}^{+4}$ is less stable than $P{b}^{+2}$, hence $PbC{l}_4$ is also less stable.

Reason: $PbC{l}_4$ is a powerful oxidizing agent.

As we go down in a group 14 elements of periodic table, we observe that +4 oxidation state decreases while that of +2 oxidation state increases due to inert pair effect. So, the outermost s-orbital electrons show reluctance to take part in the reaction. This leads to greater stability of +2 oxidation state in group 14 elements.
Instead of losing 4 electrons[2 electrons from S and 2 from P orbitals}, they tend to lose the p orbitals electrons, hence $P{b}^{+4}$ can easily gain 2 electrons to form more stable $P{b}^{+2}$ ions.
Pb(lead) show +2 oxidation state and easily gets reduced and therefore oxides others
Hence $PbC{l}_4$ is the stronger oxidizing agent.