Question

Use the molecular orbital energy level diagram to show that \[\ce{N2}\] would be expected to have a triple bond, \[\ce{F2}\], a single bond and \[\ce{Ne2}\], no bond.

Answer 1

Formation of N₂ molecules.

Electrons configuration of N-tom .₇ N = `1s^2, 2s^2, 2p_x^1, 2p_y^1, 2p_z^1`,N₂ molecules = `sigma1s^2, sigma^∗1s^2, sigma2s^2, pi2p_x^2 = pi2p_y^2, pi2p_z^2`

Bond order = `1/2[N_b - N_a] = 1/2(10 - 4)` = 3.

Bond order value of 3 means that N₂ contains a triple bond.

Formation of F₂ molecule, _.9 F = `1s^2, 2s^2, 2p_x^2, sigma^∗2s^2, sigma2p_z^2, pi2p_x^2, pi2p_x^2 = pi2p_y^2, pi^∗2p_z^2 = pi^∗2p_y^2`

Bond order = `1/2[N_b - N_a] = 1/2(10 - 8)` = 1

Bond order value 1 means that \[\ce{F2}\] contains single bond.

Formation of Ne₂ molecule 10 Ne = `1s^2, 2p_x^2, 2p_y^2, 2p_z^2`

Ne molecule = `sigma1s^2, sigma^∗1s^2, sigma2s^2, sigma^∗2s^2, sigma2p_z^2, pi2p_z^2, = pi2p_y^2, pi^∗2p_x^2 = pi^∗2p_y^2, sigma^∗2p_z^2`

Bond order = `1/2[N_b - N_a] = 1/2(10 - 10)` = 0

Bond order value zero means that there is no formation of bond between two Ne-atoms. Hence, Ne₂ molecole does not exist.