Molecular Orbitals - Part 2

Energy Level
Diagrams of N_{2} and O_{2}

**The N**_{2}**
Molecule**

In part 1 of this tutorial, you learned about
the various molecular orbitals in the nitrogen molecule. There
are eight MOs in the N_{2} 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 N_{2} 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, H_{2}, 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 s_{1s} 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 s_{1s}. 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 N_{2},
we must deal with many electrons. However, the method is the same
as that for the H_{2} molecule. The ground state electron
configuration for the N atom is 1s^{2} 2s^{2} 2p^{3}.
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 s_{2s} and s*_{2s}
orbitals, similar to the s_{1s}
and s*_{1s} MOs of H_{2}.
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 s_{2p}
MO, and the corresponding antibonding orbital, s*_{2p}.The other four p-orbitals
form four MOs, namely two p_{2p}
MOs and two p*_{2p} MOs. The
two p_{2p} 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 p_{2p} MOs. The
other bonding orbital, the s_{2p}
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 N_{2},
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 N_{2}. 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 O**_{2}**
Molecule**

The O_{2} molecule has two more valence
electrons that the N_{2} 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, O_{2}
has two unpaired electrons as shown below:

The major consequence of this is that O_{2} is
paramagnetic, due to the unpaired electron spins in the molecule.