Chemical Bonding and Molecular Structure Part 1

(A) Moleculer orbital configuration of $\mathrm{C}_2$ (12 electrons) $=\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2}=\pi_{2p_y^2}$$\therefore N_a = 4$$N_b = 8$$\therefore \; \; \text{BO} = \; \frac{1}{2}[8-4]=2$ Here no unpaired electron present, so it is diamagnetic. Moleculer orbital configuration of $\mathrm{C}_2^{+}$(11 electrons) $\; = \sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2} = \pi_{2p_y^1}$$\; \therefore \; \; N_a = 4$$\; N_b = 7$$\; \therefore \; \; \text{BO} = \; \frac{1}{2}[7-4]=1.5$Here 1 unpaired electron present, so it is paramagnetic. (B) Moleculer orbital configuration of $\mathrm{N}_2$ (14 electrons)$\; = \sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2} = \pi_{2p_y^2} \sigma_{2p_z^2}$$\; \therefore N_a = 4$$\; N_b = 10$$\; \therefore \; \; \text{BO} = \; \frac{1}{2}[10-4]=3$ Here no unpaired electron present, so it is diamagnetic.Moleculer orbital configuration of $\mathrm{N}_2^{+}$(13 electrons) $=\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \pi_{2p_x^2} = \pi_{2p_y^2} \sigma_{2p_z^1}$$\therefore N_a = 4$$N_b = 9$$\therefore \; \; \text{BO} = \; \frac{1}{2}[9-4]=2.5$ Here 1 unpaired electron present, so it is paramagnetic. (C) Moleculer orbital configuration of NO (15 electrons) is $\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \sigma_{2p_z^2} \pi_{2p_z^2} = \pi_{2p_y^2} \pi_{2p_x^1}^* = \pi_{2p_y^0}^*$ $\therefore N_a = 5$ $N_b = 10$ $\therefore \; \; B O = \; \frac{1}{2}[10-5]=2.5$ Here is 1 unpaired electron, So it is paramagnetic. Moleculer orbital configuration of $\mathrm{NO}^{+}(14$ electrons) is $\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \sigma_{2p_z^2} \pi_{2p_z^2} = \pi_{2p_y^2} \pi_{2p_x^0}^* = \pi_{2p_y^0}^*$ $\therefore N_a = 4$$N_b = 10$$\therefore \; \; B O = \; \frac{1}{2}[10-4]=3$ Here is no unpaired electron, So it is diamagnetic. (D) Molecular orbital configuration of $\mathrm{O}_2$ ( 16 electrons) is $\sigma_{1s^2} \sigma_{1s^2}^* \sigma_{2s^2} \sigma_{2s^2}^* \sigma_{2p_z^2} \pi_{2p_x^2} = \pi_{2p_y^2} \pi_{2p_x^1}^* = \pi_{2p_y^1}^*$ $\; \therefore N_a = 6$$\; N_b = 10$$\; \therefore \; \; B O = \; \frac{1}{2}[10-6]=2$Here 2 unpaired electrons present, so it is paramagnetic.Molecular orbital configuration of $\mathrm{O}_2^{+}$(15 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^1}^* = \pi_{2p_y^o}^*$$\; \therefore \; \; N_b = 10$$\; N_a = 5$$\; \therefore \; \; \text{BO} = \; \frac{1}{2}[10-5]=2.5$Here 1 unpaired electrons present, so it is also paramagnetic.