Chemical Bonding and Molecular Structure Part 2

(1) Bond order $=\;\frac{1}{2}[N_b - N_a]$$N_b=$ No of electrons in bonding molecular orbital$N_a=$ No of electrons in anti bonding molecular orbital(2) upto 14 electrons, molecular orbital configuration isHere $N_a=$ Anti bonding electron $=4$ and $N_b=10$(3) After 14 electrons to 20 electrons molecular orbital configuration is -Here $N_a=10$and $N_b=10$In $\mathrm{O}$ atom 8 electrons present, so in $\mathrm{O}_2, 8 \times 2 = 16$ electrons present.in $\mathrm{O}_2^{2-}$ no of electrons $=18$(A) Molecular orbital configuration of $\mathrm{O}_2^{2-}(18$ electrons) is$\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \sigma_{2p_z^2} \pi_{2p_x^2} = \pi_{2p_y^2} \pi_{2p_x^2}^* = \pi_{2p_y^2}^*$$\therefore \; N_b = 10$$N_a = 8$$\therefore \; BO = \frac{1}{2}[10-8] = 1$ (B) $\mathrm{C}_2^{2-}$ has 14 electrons.Moleculer orbital configuration of $\mathrm{C}_2^{2-}$ is$\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2} = \pi_{2p_y^2} \sigma_{2p_z^2}$$\therefore N_a = 4$$\therefore N_b = 10$$\therefore \; BO = \frac{1}{2}[10-4] = 3$(C) $\mathrm{N}_2^{2-}$ has 16 electrons.Moleculer orbital configuration of $\mathrm{N}_2^{2-}$ is$\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2} = \pi_{2p_y^2} \sigma_{2p_z^2} \pi_{2p_x^1}^* = \pi_{2p_y^1}^*$$\therefore N_a = 6$ $\therefore N_b = 10$$\therefore \; BO = \frac{1}{2}[10-6] = 2$The correct order of bond orders of $\mathrm{C}_2^{2-}, \mathrm{N}_2^{2-}$ and $\mathrm{O}_2^{2-}$$\mathrm{O}_2^{2-} < \mathrm{N}_2^{2-} < \mathrm{C}_2^{2-}$