Question

Using the molecular orbital energy ordering for second-row homonuclear diatomic molecules in which the π2p orbitals lie at lower energy than the σ2p, predict the bond order in a molecule or ion with each of the following numbers of total valence electrons.(Use the drawing MO energy diagrams) Will the molecule or ion be diamagnetic or paramagnetic? Part A Determine the bond order in a molecule or ion with 4 valence electrons. Part B Will this molecule or ion be diamagnetic or paramagnetic? Part C Determine the bond order in a molecule or ion with 6 valence electrons. Part D Will this molecule or ion be diamagnetic or paramagnetic? Part E Determine the bond order in a molecule or ion with 8 valence electrons. Part F Will this molecule or ion be diamagnetic or paramagnetic? Part G Determine the bond order in a molecule or ion with 9 valence electrons. Part H Will this molecule or ion be diamagnetic or paramagnetic?

Answer 1

(A) Molecule with 4 valence electron is Be_2 its molecular orbital configuration is sigma 1 s^2 sigma 1 s^2 sigma 2 s^2 sigma 2 s^2 Bond order = 1/2 (Bonding electron - Antibonding electron) = 1/2 (2 - 2) = 0 (B) Since, the bond order is zero, the molecule will not exist. (c) Molecule with 6 valence electrons is B_2. It's molecular orbital configuration is sigma 1 s^2 sigma 1 s^2 sigma 2 s^2 sigma 2 s^2 pi 2 P_x^1 pi 2 Py^1 Bond order = 1/2 (4 - 2) = 1 (d) Since, B_2 has unpaired electrons it is paramagnetic. (E) Molecule with 8 valence electrons is C_2. Its molecular orbital configuration is sigma 1 s^2 sigma 1 s^2 sigma 2 s^2 sigma 2 s^2 pi 2 P_x^2 pi 2 P_y^2 Bond order = 1/2 (6 - 2) = 4/2 = 2\r\n (F) Since, the molecule has no unpaired electrons, it is diamagnetic.