Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The molecular configuration is depicted below: Molecular orbital configuration of $$\begin{array}{rl}& {\mathrm{O}}_2\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1{\mathrm{s}}^2\to \sigma 2s^2\to {\sigma }^{\ast }\\ & 2s^2\to \sigma 2{\mathrm{p}}_z^2\pi 2p^2{}_x\to \pi 2p^2{}_y\\ & {\pi }^{\ast }2{\mathrm{p}}_x^1\to {\pi }^{\ast }2p^1{}_y\\ & \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{10-6}{2}=2\\ & {\mathrm{O}}_2^{+}\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\to \sigma 2s^2\to \\ & {\sigma }^{\ast }2s^2\to \sigma 2p^2{}_z\to \pi 2{\mathrm{p}}_x^2\\ & \pi 2p^2{}_y\to {\pi }^{\ast }2p^1{}_x\\ & \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{10-5}{2}=2.5\\ & {\mathrm{C}}_2\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\to \sigma 2s^2\to \\ & {\sigma }^{\ast }2s^2\to \pi 2p^2{}_x\to \pi 2p^2\\ & \Rightarrow \text{ Diamagnetic }\\ & \text{ Bond order }=\frac{8-4}{2}=2\\ & {\mathrm{C}}_2^{+}\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\to \sigma 2s^2\to \\ & {\sigma }^{\ast }2s^2\to \pi 2p^2{}_x\to \pi 2p^1{}_y\end{array}$$ $$\begin{array}{rl}& \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{7-4}{2}=15\\ & \mathrm{NO}\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\to \sigma 2s^2\to \\ & {\sigma }^{\ast }2s^2\to \sigma 2p^2=\pi 2p_x^2\to \\ & \pi 2p_y^2\to {\pi }^{\ast }2p_x^1\\ & \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{10-5}{2}=2.5\\ & {\mathrm{NO}}^{+}\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\sigma 2s^2\to {\sigma }^{\ast }2s^2\\ & \sigma 2p_z^2\to \pi 2p_x^2\to \pi 2p_y^2\\ & \Rightarrow \text{ Diamagnetic }\\ & \text{ Bond }\text{ order }=\frac{10-4}{2}=3\\ & {\mathrm{N}}_2\Rightarrow \sigma 1s^2\to {\sigma }^{\ast }1s^2\sigma 2s^2\to {\sigma }^{\ast }2s^2\\ & \to \pi 2p_x^2\to \pi 2p_y^2\to \sigma 2p^1=\\ & \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{10-4}{2}=3\\ & {\mathrm{N}}_2^{+}\Rightarrow \sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p^2{}_x\\ & \pi 2p^2\sigma 2p_z^1\\ & \Rightarrow \text{ Paramagnetic }\\ & \text{ Bond order }=\frac{9-4}{2}=2.5\\ & \end{array}$$