JEE Chemistry 2023 Question with Solution

Given: The compounds are $\mathrm{Li_2O}$, $\mathrm{Na_2O_2}$, and $\mathrm{KO_2}$.

Find: Their magnetic behavior in the given order.

Magnetic behavior depends on whether the oxygen-containing ion has all electrons paired or contains an unpaired electron.

1. In $\mathrm{Li_2O}$, the anion is $\mathrm{O^{2-}}$. This ion has a completely filled electronic configuration, so it is diamagnetic.

2. In $\mathrm{Na_2O_2}$, the anion is $\mathrm{O_2^{2-}}$. The peroxide ion has all electrons paired, so it is diamagnetic.

3. In $\mathrm{KO_2}$, the anion is $\mathrm{O_2^-}$. The superoxide ion has one unpaired electron, so it is paramagnetic.

Therefore, the order is diamagnetic, diamagnetic, and paramagnetic. Hence, the correct option is D.

The solution also contains a conflicting page label saying "The Correct Option is B", but the worked explanation and final answer clearly state $\left(4\right)$ diamagnetic, diamagnetic, paramagnetic, which corresponds to option D.

Given: $\mathrm{Li_2O}$ contains $\mathrm{O^{2-}}$, $\mathrm{Na_2O_2}$ contains $\mathrm{O_2^{2-}}$, and $\mathrm{KO_2}$ contains $\mathrm{O_2^-}$.

Find: Which sequence of magnetic properties matches these ions.

Use the standard facts:

• $\mathrm{O^{2-}}$ is diamagnetic.
• $\mathrm{O_2^{2-}}$ is diamagnetic.
• $\mathrm{O_2^-}$ is paramagnetic.

So,

• $\mathrm{Li_2O} \rightarrow$ diamagnetic
• $\mathrm{Na_2O_2} \rightarrow$ diamagnetic
• $\mathrm{KO_2} \rightarrow$ paramagnetic

Thus the required sequence is diamagnetic, diamagnetic, and paramagnetic.

Therefore, the correct option is D.