Question

Which of the following statements about $N{a_2}{O_2}$ is not correct?
$(a)$ $N{a_2}{O_2}$ oxidises $C{r^{3 + }}{\text{ to Cr}}{{\text{O}}_4}^{2 - }$ in acid medium
$(b)$ It is a derivative of ${H_2}{O_2}$
$(c)$ It is superoxide of sodium
$(d)$It is diamagnetic in nature

Answer 1

Hint – In this question use the concept that $N{a_2}{O_2}$ is a peroxide of sodium and is considered as derivative of hydrogen peroxide ${H_2}{O_2}$ as it contains the monoxide ion of${O^{2 - }}$. This will help getting the option that does not resemble about $N{a_2}{O_2}$.
As we know $N{a_2}{O_2}$ is peroxide of sodium.

Complete answer:

As it contains peroxide ion ${O_2}^{2 - }$ which can be represented as ${}^ - O - {O^ - }$.
Hence it is considered as derivative of hydrogen peroxide ${H_2}{O_2}$.
As we know monoxide ion is ${O^{2 - }}$ and superoxide ion is ${O_2}^ -$.
And usually sodium forms the peroxide ion i.e. $N{a_2}{O_2}$
And it does not contain any unpaired electrons, (as sodium and oxygen paired their electrons so the last shell of sodium and oxygen is complete with 8 electrons) thus it is diamagnetic in nature.
So from the given options option (C) is wrong.
So this is the required answer.
Hence option (C) is the correct answer.

Note – Those elements that have paired electrons in the molecular orbit are referred to as diamagnetic. Thus $N{a_2}{O_2}$ does not have an unpaired electron in the molecular orbit thus it is diamagnetic. Now when $N{a_2}{O_2}$ reacts with $C{r_2}{O_3}$ then the equation follows as $3N{a_2}{O_2} + C{r_2}{O_3} + {H_2}O \to 6N{a^ + } + 2{\left( {Cr{O_4}} \right)^{2 - }} + 2O{H^ - }$. It is clear that $N{a_2}{O_2}$ is an oxidizing agent and $C{r_2}{O_3}$is a reducing agent. The mentioned reaction is an oxidation-reduction that is a redox reaction.