4. Chemical bonding and molecular structure — Chemistry STD 11 Science — Question

So, for $\mathrm{KO}_{2}$ there would be 35 electrons $(19+8 \times 2)$

For $\mathrm{AlO}_{2}^{-}$ there would be 20 electrons $(13+8 \times 2+1)$ (for negative charge)

For $\mathrm{BaO}_{2}$ there would be 72 electrons $(56+16)$

For $\mathrm{NO}_{2}^{+}$ there would be 12 electron $\mathrm{s}(7+8 \times 2-1)$ (for positive charge)

So only $\mathrm{KO}_{2}$ is having odd number of electrons so it is having unpaired electrons.

$\mathrm{KO}_{2}$ has $\left(\mathrm{K}^{+}+\mathrm{O}_{2}^{-}\right)$ structure having one unpaired electron due to presence of superoxide ion

$\left(\mathrm{O}_{2}^{-}\right.$ ion $)$ in it, which is paramagnetic.

Hence option $\mathrm{C}$ is correct.