Question

Consider the molecular orbital diagram shown below, a diagram that is often applied to the first row diatomic molecules. Note that this diagram comprises orbitals derived from the n = 2 valence shell of orbitals (2s and 2p). Use this diagram to determine (1) if the following diatomic molecules or ions will be diamagnetic (all electrons paired) or paramagnetic (one or more unpaired electrons), and (2) for paramagnetic species, the number of unpaired electrons.

O2, N2, CO, NO, NO-, NO+, CN-, FO-

Answer 1

1. O₂: Formed by combination of atomic orbitals on two O atoms having electronic configuration [He]2s²2p⁴

The σ_2s orbital has 2 electrons and the σ_2s* has 2. There are a total of (4 + 4) = 8 p electrons which go into the σ_2p, π and π^* orbitals. The σ_2p has 2 electrons, π_x has 2 and π_y has two. The remaining two electrons go into the π_x^* and π_y^* orbitals, each orbital containing 1 electron. Thus, O₂ has 2 unpaired electrons and is paramagnetic.

N₂: Formed by combination of two N atomic orbitals, each having configuration [He]2s²2p³.

We follow the same approach as before and note that the σ_2s and σ_2s* orbitals are completely filled. The remaining 6 electrons go into σ_2p, π_x and π_y. Each of these orbitals houses 2 electrons each; there are no unpaired electrons and hence N₂ is diamagnetic.

CO: Numbers of electrons in the outermost orbitals of C and O atoms are 4 (C:[He]2s²2p²) and 6 (O:[He]2s²2p⁴) – total number of electrons is 10. 4 electrons go into the σ_2s and σ_2s* orbitals (just like O₂ in 1 above). There are 6 electrons remaining and they fill up the σ_2p, π_x and π_y orbitals (like N₂ in 2 above). Hence there are no unpaired electrons and thus CO is diamagnetic.

NO: Total number of electrons in the outermost molecular orbitals = 5 (from N atom) + 6 (from O atom) = 11. Just like the three examples above, 4 electrons go into the σ_2s and σ_2s* orbitals. There are 7 electrons remaining, out of which 6 go into σ_2p, π_x and π_y orbitals (like N₂ above). The remaining 1 electron go into either π_x^* or π_y^* orbital; thus NO is paramagnetic due to 1 unpaired electron.

NO^-: Same as NO, expect that there are 12 electrons now (1 extra due to negative charge). The extra electron occupies π_x^* or π_y^* orbitals (both the non-bonding π_x^* and π_y^* orbitals must be singly occupied as per Hund’s rule). There are 2 unpaired electrons and hence NO^- is paramagnetic.

NO⁺: Same as NO, except that there are 10 electrons (1 electron short due to positive charge). Total is 10 electrons that fill up σ_2s, σ_2s*, σ_2p, π_x and π_y orbitals completely. There is no unpaired electron and hence NO⁺ is diamagnetic.