Molecular Orbitals - Part 2
Energy Level Diagrams of N2 and O2
The N2 Molecule
In part 1 of this tutorial, you learned about the various molecular orbitals in the nitrogen molecule. There are eight MOs in the N2 molecule, capableof holding 16 electrons in total. (Only six were shown in part 1, but recall that two of them are degenerate, i.e. they have twins at the same energies.) The N2 molecule has a total of only 10 valence electrons, so the question is "which orbitals are used, and which are empty?"
Simply put, the lowest energy orbitals are filled first. When we run out of valence electrons, the remaining MOs are left empty. The lowest energy MOs are generally those formed from the lowest energy atomic orbitals. In the case of making a molecule of hydogen, H2, we have only two electrons - one from each H atom. Thus, we start by looking at the MO formed by the overlap of the 1s orbitals. We show the atomic orbitals of each H atom, each containing one electron, and both at the same energy level. In these diagrams, the energy level is the vertical scale.
When the two H atoms come close enought to interact, the 1s orbitals overlap to form two molecular orbitals. Remember - the number of MOs formed = the number of atomic orbitals used. The lowest energy (i.e. most stable) MO formed is via constructive overlap of the two 1s atomic orbitals. This results in a s1s orbital at a lower energy than the atomic orbitals. The other is an anti-bonding orbital at a somewhat higher energy level, and is giventhe term s*1s, the asterisk denoting "antibonding". The energy level diagram now looks like:
Each MO can hold up to two electrons, just like an atomic orbital. Thus, the two electrons move from the atomic orbitals into the lowest energy MO available, i.e. the s1s. Note that the electron spins in the MO must be opposite. Note also that the two atomic orbitals cease to exist, and that the s*1s orbital is unoccupied, and therefore does not exist either. These unoccupied orbitals are included on an MO diagram for reference.
In the case of a larger molecule such as N2, we must deal with many electrons. However, the method is the same as that for the H2 molecule. The ground state electron configuration for the N atom is 1s2 2s2 2p3. The initial (atomic) orbitals have an energy level diagram as shown below, with an N atom on each side:
The 2s atomic orbitals combine to make s2s and s*2s orbitals, similar to the s1s and s*1s MOs of H2. Note that all four of the 2s electrons move into the two new MOs:
The 2p orbitals also overlap to make MOs. Since we have six 2p orbitals, we expect to make six MOs. Here, you must recall that the three 2p orbitals are perpendicular to one another. Thus, only the p-orbitals lined up along the same axis can interact end-on, forming a s2p MO, and the corresponding antibonding orbital, s*2p.The other four p-orbitals form four MOs, namely two p2p MOs and two p*2p MOs. The two p2p MOs are degenerate (have the same energy level) and the p*2p MOs are degenerate with one another as well.
The most stable (lowest energy) MO of these six are the two p2p MOs. The other bonding orbital, the s2p is slightly less stable (i.e. at a higher energy), followed by the two p*2p MOs and the s*2p MO at the top. The overall MO diagram thus looks like:
There are several things of note in the above diagram:
1. Since there only 10 valence electrons in N2,
only the five MOs lowest in energy are occupied. The p*2p and the s*2p
remain unoccupied, i.e. they do not exist in this molecule.
2. There are no un-paired electrons in N2. We do not expect the molecule to be paramagnetic (and it isn't).
3. There are more electrons in bonding MOs (8) than antibonding (2). We thus expect the molecule to be stable (and it is). We can be more quantitative about this using the concept of bond order (see part 3 of this tutorial).
4. The molecule is held together by four electrons in s bonds, and four in p bonds.
The O2 Molecule
The O2 molecule has two more valence electrons that the N2 molecule. Since these go into the next available MO, they end up in the p*2p. Note that this is a degenerate MO, consisting of two MOs at the same energy level. Remember Hund's rule? We must place an electron in each of these two MOs before pairing them up. Thus, O2 has two unpaired electrons as shown below:
The major consequence of this is that O2 is paramagnetic, due to the unpaired electron spins in the molecule.
On to Part 3, Bond Order
Back to Part 1, Shapes of MOs