c
Generally, the elements possess odd number of electrons will have unpaired electrons.

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.