Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Bond order $=\frac{N_b-N_a}{2}$ In NO, total electrons $=7+8=15$ $\therefore$ Configuration of $\mathrm{NO}$ $$\begin{array}{rl}& =KK,\sigma \left(2s^2\right),\stackrel{\ast }{\sigma }\left(2s^2\right),\sigma \left(2p_z^2\right),\pi \left(2p_x^2\right)\\ & \approx \pi \left(2p_y^2\right)\stackrel{\ast }{\pi }\left(2p_x^1\right)\\ & \end{array}$$ $\therefore$ Bond order $=\frac{8-3}{2}=\frac{5}{2}=2.5$ In ${\mathrm{O}}_2^{-}$, total electrons $=16+1=17$ $$\begin{array}{rl}& \therefore \text{ Configuration of }{\mathrm{O}}_2^{-}\\ & =KK,\sigma \left(2s^2\right),\stackrel{\ast }{\sigma }\left(2s^2\right),\sigma \left(2p_z^2\right),\pi \left(2p_x^2\right)\\ & \approx \pi \left(2p_y^2\right),\pi \left(2p_x^2\right)\approx \pi \left(2p_y^1\right)\\ & \therefore \text{ Bond order }=\frac{8-5}{2}=\frac{3}{2}=1.5\\ & \text{ In }{\mathrm{C}}_2^{2-}\text{, total electrons }=12+2=14\\ & \therefore \text{ Configuration of }{\mathrm{C}}_2^{2-}\\ & =KK,\sigma \left(2s^2\right),\stackrel{\ast }{\sigma }\left(2s^2\right),\sigma \left(2p_z^2\right),\approx \pi \left(2p_x^2\right),\pi \left(2p_y^2\right)\\ & \therefore \text{ Bond order }=\frac{8-2}{2}\\ & =\frac{6}{2}=3\\ & \text{ In }{\mathrm{He}}_2^{+}\text{, total electrons }=4-1=3\\ & \therefore \text{ Configuration of }{\mathrm{He}}_2^{+}=\sigma \left(1s^2\right),\stackrel{\ast }{\sigma }\left(1s^1\right)\\ & \therefore \text{ Bond order }=\frac{2-1}{2}\\ & =\frac{1}{2}=0.5\\ & \end{array}$$ Hence, correct order of bond order is $${\mathrm{He}}_2^{+}<{\mathrm{O}}_2^{-}<\mathrm{NO}<{\mathrm{C}}_2^{2-}$$