Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\begin{array}{rl}& \text{(a) }\text{MO configuration of }{\mathrm{O}}_2^{+}(8+8-1=15)\\ & =\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2,\sigma 2p_z^2,\\ & \pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^1\approx \pi 2p_y^0\\ & \mathrm{BO}=\frac{N_b-N_a}{2}\end{array}$ (where, $N_b=$ number of electrons in bonding molecular orbital $N_a=$ number of electrons in anti-bonding molecular orbital $\therefore \mathrm{BO}=\frac{10-5}{2}=2.5$ Similarly, (b) ${\mathrm{O}}_2^{-}(8+8+1=17)$ so $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-7}{2}=1.5$ (c) ${\mathrm{O}}_2^{2-}(8+8+2=18)$ $\mathrm{BO}=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1$ (d) ${\mathrm{O}}_2(8+8=16)$ $\mathrm{BO}=\frac{10-6}{2}=2$ Thus, ${\mathrm{O}}_2^{-}$shows the bond order 1.5.