Chemistry - CHEMICAL BONDING Question with Solution | TestHub

According to molecular orbital theory, the energy level of the given molecules are $\begin{array}{rl}& {\mathrm{C}}_2^{2-}\to \sigma 1s^2{\sigma }^{\star }1s^2\sigma 2s^2{\sigma }^{\star }2s^2\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2\\ & \text{ B.O. }=1/2[10-4]=3\\ & {\mathrm{He}}_2^{+}\to \sigma 1s^2{\sigma }^{\star }1s^1\\ & \text{ B.O. }=1/2[2-1]=1/2=0.5\\ & \mathrm{NO}\to \sigma 1s^2{\sigma }^{\star }1s^2\sigma 2s^2{\sigma }^{\star }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2\\ & {\pi }^{\star }2p_x^1\end{array}$ $\begin{array}{rl}& \text{ B.O. }=1/2[10-5]=2.5\\ & {\mathrm{O}}_2^{-}\to \sigma 1s^2{\sigma }^{\star }1s^2\sigma 2s^2{\sigma }^{\star }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\star }2p_x^2\\ & \text{ B.O. }=1/2[10-7]=1.5\\ & {\pi }^{\star }2p_y^1\\ & \end{array}$ So, the correct order of their increasing bond order is ${\mathrm{He}}_2^{+}<{\mathrm{O}}_2^{-}<\mathrm{NO}<{\mathrm{C}}_2^{2-}$