Question

Which is the best oxidizing agent?
A. $G{e^{4 + }}$
B. \[P{b^{4 + }}\]
C. $S{n^{4 + }}$
D. $S{n^{2 + }}$

Answer 1

Hint: To solve this question, we need to refer to the carbon family. Then, we need to arrange these options with respect to the carbon family. Carbon is the top most element in the family, after that there is Silicon, Germanium, Strontium, and Lead on the other end of the family.

Complete step by step answer:
Firstly we must know that an oxidizing agent is a reactant that eliminates electrons from the other reactants through a redox reaction.
As we know that lead is present on the end of the carbon family and before that there is strontium. So, from here we get to know that $S{n^{4 + }}$ is more stable than $S{n^{2 + }}$ because to achieve more stable state the element needs to lose more electrons whereas the other elements need to gain electrons. Here, $S{n^{2 + }}$ acts as a reducing agent and $P{b^{4 + }}$ as an oxidizing agent. So, $P{b^{^{2 + }}}$ will be more stable than $P{b^{4 + }}$.
When we refer to the carbon family, the stability of the oxidation state of +4 decreases as we move from top to bottom due to the inert pair effect. Thus, it means that the oxidation state of +4 of Pb is less stable than the +4 oxidation state of Sn.
Here, $P{b^{4 + }}$ will perform reduction and the other ions or elements will perform oxidation. Therefore, $P{b^{4 + }}$ is known to be one of the strongest oxidizing agents from all the above options.
$\therefore $Option B is the correct answer.

Note:
We must know that the fluorine is considered to be one of the strongest oxidizing agents and Lithium tends to be the weakest oxidizing agent as we go down the list. Oxidizing agents are electron acceptors. And the best oxidizing agents are one which have large affinity for electrons.