Question

Using the molecular orbital filling order

5, using the molecular orbital filling order σ2s σ2s π2p Ơ2p π2p Ơ2p write the MO electron configurations, calculate the bond orders, and predict which is the most stable molecule for the following species: N2, N2, and N2.

Answer 1

N_2 Molecule: N (7) = 1s^2 2s^2 2p^2 2p^3 = 7e^(-) s N_2 Molecule has total = 14 e^(-) s M.O configuration sigma_1s^2 < sigma^*_1s^2 < sigma^*_2s^2 < sigma^*_2s^2 < sigma^2_2p_3 < pi 2p^2_x = pi 2p^2_y Bond order: = 1/2 [No. of bonding e^(-) s - No. of non bonding e^(-) s] B.O of N_2 = 1/2 [10 - 4] = 6/2 = 3 Bond order = 3 (exist & stable) And it is diamagnetic no unpaired e^(-) s If B.O = 0 not exist B.O = 1, 2, 3 ... integers exist v stable B.O = 0.5, 1.5, 2.5 ... exist but unstable N^-_2 Molecule: N (7) = 1s^2 2s^2 2p^3 N/-2 = 7 + 7 + 1 = 15 e^(-) M.O configuration using N_2 M.O diagram extra e^(-) enters into pi^*_2 px \r\n Fill 15e^(-) in M.O. configuration sigma_1s^2 < sigma^*_1s^2 < sigma_2s^2 < sigma^*_2s^2 < sigma^2_2p_3 < pi 2p^2_x = pi 2p^2_y < pi^*