Question

Assertion: $PbC{l_2} $ is more stable than $PbC{l_4} $
Reason: $PbC{l_4} $ is a powerful oxidizing agent.
(A) Both assertion and reason are correct and the reason is the correct explanation for assertion
(B) Both assertion and reason are correct but the reason is not the correct explanation for assertion
(C) the assertion is correct but the reason is incorrect
(D) the assertion is incorrect but the reason is correct

Answer 1

Hint: Pb has electronic configuration: $\left[{Xe}\right]4{f^{14}}5{d^{10}}6{s^2}6{p^2}$. Its s electrons are more strongly held that p electrons. Thus, it loses two p-electrons to form stable $PbC{l_2} $ and can also lose 2 s-electrons to form less stable $PbC{l_4} $. $PbC{l_4}$ being less stable easily gets converted to $PbC{l_2}$ while another reactant gets oxidized.

Complete step by step solution:
We have been given that: ​ $PbC{l_2} $is more stable than $PbC{l_4} $,

The inert pair effect is the tendency of 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.
Now, $PbC{l_4} $​ is a powerful oxidizing agent.

As we go down in a group 14 elements of the 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 Pb+4 can easily gain 2 electrons to form more stable Pb+2 ions.
Pb(lead) show +2 oxidation state and easily gets reduced and therefore oxides others
Hence PbCl4​ is the stronger oxidizing agent.
So, we can say that both assertion and reason are correct but the reason is not the correct explanation for the assertion.

Therefore, option (b) is correct.

Note: Due to the poor shielding effect of d block and f block (it means due to poor contraction of d block) electrons in the s orbital are more strongly held in the upper period of the same group. So, s electrons are inert and cannot be removed and give a valency of 4. Hence, Pb forms 2 ions instead of 4 and $PbC{l_2}$ becomes stable.