Which one of the following has no unpaired electrons?

(a) In $\mathrm{O}_{2}^{-}$total number of electrons $$ =16+1=17 $$ $\therefore$ Orbital arrangement $$ \begin{aligned} =K K^{\prime}, \sigma 2 s^{2}, \sigma 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2}, \\ \pi 2 p_{x}^{2} &=\pi 2 p_{y}^{1} \end{aligned} $$ (b) In $\mathrm{O}_{2}^{+}$, total number of electrons $$ =16-1=15 $$ $\therefore$ Orbital arrangement $=\mathrm{KK}^{\prime}, \sigma 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}$, $$ \sigma 2 \mathrm{p}_{\mathrm{z}}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi 2 \mathrm{p}_{\mathrm{y}}^{2}, \stackrel{\approx}{\pi} 2 \mathrm{p}_{\mathrm{x}}^{1} $$ (c) In $\mathrm{O}_{2}^{2-}$, total number of electrons $$ =16+2=18 $$ $\therefore$ Orbital arrangement $=K_{K}^{\prime}, \sigma 2 \mathrm{~s}^{2}, \sigma 2 \mathrm{~s}^{2}$, $\sigma 2 \mathrm{p}_{\mathrm{z}}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi 2 \mathrm{p}_{\mathrm{y}}^{2} \pi 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi 2 \mathrm{p}_{\mathrm{y}}^{2}$ (d) In $\mathrm{O}_{2}$, total number of electrons $=16$ $\therefore$ Orbital arrangement $=K K K^{\prime}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}$, $\sigma 2 \mathrm{p}_{\mathrm{z}}^{2}, \pi 2 \mathrm{p}_{\mathrm{x}}^{2} \approx \pi 2 \mathrm{p}_{\mathrm{y}}^{2}, \stackrel{\approx}{\pi} 2 \mathrm{p}_{\mathrm{x}}^{1} \approx \pi / 2 \mathrm{p}_{\mathrm{y}}^{1}$ Hence, only $\mathrm{O}_{2}^{2-}$ has no unpaired electrons.