
Be able to explain the concept of bonding and antibonding molecular orbitals.
The addition of atomic orbitals increases the probability of electrons lying between atom centres.
Bonding molecular orbitals are lower in energy than the two parent atomic orbitals, they are labelled
as sigma (σ). The subtraction of overlapping
orbitals decreases the probability of electrons
lying between the atom centres. These are
antibonding molecular orbitals and are higher in
energy than the parent atomic orbitals; they are
labelled as sigma star (σ*). The probability of
finding an electron at a node in an anti-bonding
molecular orbital is zero. Bonding molecular
orbitals are considered to be constructive
overlapping whereas,
anti-bonding molecular
orbitals are considered
to be destructive.
In the example of hydrogen (H2), the two atomic orbitals (one from each
atom) form two molecular orbitals. One is bonding and is lower in energy
then the second anti-bonding. The H2 configuration is written as
(σ1s)2(σ*1s)0 or simply just (σ1s)2. The two electrons reside in the bonding
MO because of its lower energy and each have an opposite spin (Hund’s
Rule).

Be able to draw molecular orbital energy-level diagrams and use them to obtain bond orders
and electron configurations of homonuclear diatomic molecules.
Bond order is defined as half of the difference between the number of electrons in the bonding MOs
and the number of electrons in the antibonding MOs. If there are more bonding electrons than
antibonding electrons molecular stability is usually greater and a bond order >0 indicates stability. In
the case of H2, the molecule has 2 bonding and no antibonding electrons so; bond order = ½ (2-0) =1
and is therefore predicted to be a stable molecule. A bond
order of 1 also defines a single bond.
In the case of He2, it would have a total of 4 electrons; two
would have to go in the bonding MO and two in the antibonding. Therefore, the bond order = ½ (2-2) = 0. This
predicts the molecule to be unstable and thus it is very
unlikely that it exists naturally.
In the case of Li2; the 1s and 2s orbital pairs each make two σ MOs. It has six
electrons, 4 in the 1s and 2 in the 2s orbitals. The bond order = ½ (4-2) = 1;
indicating a single bond between the lithium atoms.
P-orbitals interact to form 2 π and 2 π* molecular orbitals. The two π and
the two π* are of the same energy and therefore, each pair is said to be
degenerate however, * orbitals always remain higher in energy than the
bonding molecular orbitals. These π bond can change their sign as they are
rotated around the axis which is important in spectroscopy.
As the molecular orbitals diagrams become
more complex with more and more orbitals,
the 1s orbitals aren’t really taken into account
since they are very low in energy and don’t
really contribute. In the case of N2, the 2s
orbitals make bonding and anti-bonding σ
MOs, the 2p orbitals make bonding and
antibonding σ MOs as well as 2 bonding and 2 antibonding π MOs and
the ten valence electrons fill them up from the lowest energy level. The
result is 8 electrons in bonding MOs and 2 electrons in an anti-bonding
MO – the resultant bond order = 3 (indicating a triple bond).
If electrons are not spin paired when there aren’t enough to fill the orbitals, the atom will have
magnetic properties. Unpaired electron spins result in a non-zero net spin which leads the atom to be
attracted to a magnetic field; therefore, paramagnetic. If the electrons are paired, there is a zero net
spin and therefore no attraction leading to a diamagnetic atom. Molecular orbital predicts that there
are 2 unpaired electrons per molecule and therefore O2 would be considered to be paramagnetic.
Whether a molecule is paramagnetic or diamagnetic depends on the pairing of spin in the orbitals –
which is a direct result of the number of electrons.

Understand the relationships between bond order, bond length and bond strength.
The higher the bond order, the shorter the bond length and the stronger the bond.