Question

$\text{M}{{\text{n}}^{2+}}$ can be converted to $\text{M}{{\text{n}}^{7+}}$ by reacting with:
A. $\text{Pb}{{\text{O}}_{2~}}$
B. $\text{SnC}{{\text{l}}_{2~}}$
C. $\text{S}{{\text{O}}_{2}}$
D. $\text{C}{{\text{l}}_{2~}}$

Answer 1

Hint: If we have to convert $\text{M}{{\text{n}}^{2+}}$ to $\text{M}{{\text{n}}^{7+}}$ we know that we have to oxidize $\text{M}{{\text{n}}^{2+}}$. For this we need a compound in which metal is readily reducible and is in high oxidation state. It should also have a stable lower oxidation state. In this case, the other metal used will be reduced and $\text{M}{{\text{n}}^{2+}}$ will be oxidized.

Complete answer:
Here a redox reaction will take place which can be defined as a chemical reaction in which electrons are transferred between two reactants participating in it. The transfer of electrons can be identified if we observe the changes in the oxidation states of the reacting species. Since, we need to use a compound having high oxidation state and also has a stable lower oxidation state, we can use $\text{Pb}{{\text{O}}_{2~}}$ for this purpose as in this compound Pb is in \[+4\] oxidation state and we know that $\text{P}{{\text{b}}^{2+}}$ is more stable than $\text{P}{{\text{b}}^{4+}}$ due to ‘Inert pair effect’.

$5Pb{{O}_{2}}+2Mn{{\left( N{{O}_{3}} \right)}_{2}}+6HN{{O}_{3}}\to 2HMn{{O}_{4}}+5Pb{{\left( N{{O}_{3}} \right)}_{2}}+2{{H}_{2}}O$

In this reaction, $\text{M}{{\text{n}}^{2+}}$ is oxidized to $\text{M}{{\text{n}}^{7+}}$ and $\text{P}{{\text{b}}^{4+}}$ is reduced to $\text{P}{{\text{b}}^{2+}}$. Strong, oxidising agents such as $\text{Pb}{{\text{O}}_{2~}}$ or sodium bismuthate \[(NaBi{{O}_{3}})\] oxidise , to. Hence, the correct answer is $\text{Pb}{{\text{O}}_{2~}}$. which is option A.

Note: The inert pair effect is the tendency of the two electrons in the outermost atomic s-orbital to remain unshared and do not participate in bonding in compounds of post-transition metals. As a result, the inert pair of ns electrons remains more tightly held by the nucleus and hence participates less in bond formation so as we move down a group in p-block it becomes hard for s-orbital electrons to leave and lower oxidation state becomes more stable than higher oxidation state.