0.7 Chemical bonding and molecular energy levels  (Page 5/7)

In ${\mathrm{He}}_2$ , both the bonding and the anti-bonding orbitals must be used in order to accommodate four electrons. The two electrons in the bonding orbital lower the energy of the molecule, but the two electrons in the anti-bonding orbital raise it. Since two He atoms will not bind together, then the net effect must be that the anti-bonding orbital more than offsets the bonding orbital.

We have now deduced an explanation for why the paired electrons in an atom do not contribute to bonding. Both bonding and anti-bonding orbitals are always formed when two atomic orbitals overlap. When the electrons are already paired in the atomic orbitals, then there are too many electrons for the bonding molecular orbital. The extra electrons must go into the anti-bonding orbital, which raises the energy of the molecule, preventing the bond from forming.

Returning to the ${\mathrm{Li}}_2$ example discussed above, we can develop a simple picture of the bonding. The two 1s electrons from each atom do not participate in the bonding, since the anti-bonding more than offsets the bonding. Thus, the paired “core” electrons remain in their atomic orbitals, unshared, and we can ignore them in describing the bond. The bond is formed due to overlap of the 2s orbitals and sharing of these electrons only. This is also consistent with our earlier view that the core electrons are closer to the nucleus, and thus unlikely to be shared by two atoms.

The model we have constructed seems to describe fairly well the bonding in the bound diatomic molecules listed above. For example, in a fluorine atom, the only unpaired electron is in a 2p orbital. Recall that a 2p orbital has two lobes, directed along one axis. If these lobes are assumed to lie along the axis between the two nuclei in $F_2$ , then we can overlap them to form a bonding orbital. Placing the two unpaired electrons into this orbital then results in a single shared pair of electrons and a stable molecular bond.

Observation 3: ionization energies of diatomic molecule

The energies of electrons in molecular orbitals can be observed directly by measuring the ionization energy. This is the energy required to remove an electron, in this case, from a molecule:

$H_2(g)\to H_2^{+}(g)+e^{-}(g)$

The measured ionization energy of $H_2$ is 1488 kJ/mol. This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is 1312 kJ/mol. Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom, so we can conclude that the electron has a lower energy in the molecule. If we attempt to pull the atoms apart, we must raise the energy of the electron. Hence, energy is required to break the bond, so the molecule is bound.

We conclude that a bond is formed when the energy of the electrons in the molecule is lower than the energy of the electrons in the separated atoms. This conclusion seems consistent with our previous view of shared electrons in bonding molecular orbitals.

As a second example, we consider the nitrogen molecule, $N_2$ . We find that the ionization energy of molecular nitrogen is 1503 kJ/mol, and that of atomic nitrogen is 1402 kJ/mol. Once again, we conclude that the energy of the electrons in molecular nitrogen is lower than that of the electrons in the separated atoms, so the molecule is bound.