Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The molecular orbital configurations of ${\mathrm{O}}_2^{+},{\mathrm{O}}_2^{-},{\mathrm{O}}_2^{2-}$ and ${\mathrm{O}}_2$ are $$\begin{array}{rl}& {\mathrm{O}}_2^{+}=\sigma 1s^2,\sigma 1s^2,\sigma 2s^2,\sigma 2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\\ & \pi 2p_x^1\approx \pi 2p_y^0\\ & {\mathrm{O}}_2^{-}=\sigma 1s^2,\sigma \cdot \%1s^2,\sigma 2s^2,\sigma 2s^2,\sigma 2p_z^2,\pi 2p_x^2\\ & \approx \pi 2p_y^2,\pi 2p_x^2\approx \pi 2p_y^1\\ & {\mathrm{O}}_2^{2-}=\sigma 1s^2,\sigma \cdot s^2,\sigma 2s^2,\sigma 2s^2,\sigma 2p_z^2,\pi 2p_x^2\\ & \approx \pi 2p_y^2,\pi 2p_x^2\approx \pi 2p_y^2.\\ & {\mathrm{O}}_2=\sigma 1s^2,\sigma 1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\\ & \approx \pi 2p_y^2,\pi 2p_x^1\approx \pi 2p_y^1\end{array}$$ And the electronic configuration of $\mathrm{O}$ and ${\mathrm{O}}^{+}$ are $$\begin{array}{r}\mathrm{O}=1s^2,2s^2,2p_x^2,2p_y^1,2p_z^1\\ {\mathrm{O}}^{+}=1s^2,2s^2,2p_x^1,2p_y^1,2p_z^1\end{array}$$ As ${\mathrm{O}}_2^{+},{\mathrm{O}}_2,{\mathrm{O}}_2^{-},\mathrm{O}$ and ${\mathrm{O}}^{+}$have unpaired electrons, hence are paramagnetic.