Chemistry - CHEMICAL BONDING Question with Solution | TestHub

${\mathrm{O}}_2^{+}$ ion - Total number of electrons $(16-1)=15$ E.C.: $$\begin{array}{r}\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\\ \pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y\\ \text{ Bond order }=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}=\frac{10-5}{2}=\frac{5}{2}=2\frac{1}{2}\end{array}$$ ${\mathrm{O}}_2^{-}$ (Super oxide ion): Total number of electrons $(16+1)=17$ E.C.: $$\begin{array}{l}{\text{ sigma1s }}^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\\ \pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^1\\ \text{ Bond order }=\frac{\left({\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}\right)}{2}=\frac{10-7}{2}=\frac{3}{2}=1\frac{1}{2}\end{array}$$ ${\mathrm{O}}_2^{2+}$ ion: Total number of electrons $=(16-2)=14$ E.C. $$\begin{array}{l}\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\sigma 2p_z^2\\ \text{ Bond order }=\frac{\left(N_b-N_a\right)}{2}=\frac{10-4}{2}=\frac{6}{2}=3\end{array}$$ So bond order: ${\mathrm{O}}_2^{-}<{\mathrm{O}}_2^{+}<{\mathrm{O}}_2^{2+}$