Orbital energies and nodal planes

Computer programs that calculate molecular orbitals and their energy tell you of course which energy level belongs to which orbital. With the qualitative approach that we are using in this course things are not so clear. For example, if we use just the 1s orbitals of atomic hydrogen to form the MO's of molecular hydrogen, we get a bonding and an antibonding orbital. We know that they are 1s_A+1s_B and 1s_A-1s_B, but we don't know which one is the bonding and which one is the antibonding. The orbital energies can be written as (a+b)/(1+S) and (a-b)/(1-S), respectively. This means that 1s_A+1s_B is the bonding if (a+b)/(1+S) < (a-b)/(1-S). With 0 < S < 1 this is true if and only if aS > b. However, showing that this last relation holds is not at all simple. Fortunately there is another way to see that 1s_A+1s_B is the bonding orbital. It is not as rigorous, but much easier. It is just a matter of comparing the number of nodal planes.

Let's start with atomic hydrogen. We know that the energy of the s orbitals increases in the order 1s, 2s, 3s, ... We also know that the number of spherical nodal planes is 0, 1, 2, ..., respectively. The s orbitals have no other nodal planes. So the Ns orbital has N-1 nodal planes. We see that the energy increases with the number of nodal planes. If we look at a 2p orbital we see a flat nodal plane: for 2p_z the nodal plane is z = 0. A 2p orbital is indeed higher in energy than the 1s orbital, which has no nodal planes. Going to 3p, 4p, 5p, ... we get additional spherical nodal planes, and an increase in energy. A 3d orbital like 3d_xy has two flat nodal planes (x = 0 and y = 0). Also 3d_z² has two nodal planes, but they may be a bit difficult to visualize. They are both cones with the apex in the origin. So having two nodal planes the 3d orbitals are higher in energy than the 1s, 2s, and 2p orbitals.

For other atoms we find the same relation. More nodal planes means higher energy. The reverse is not always true, however. A 2p orbital is higher in energy than the 2s orbital for all atoms except hydrogen, but both orbitals have the same number of nodal planes.

Also for molecular orbitals more nodal planes generally means higher energy. For molecular hydrogen 1s_A+1s_B has no nodal plane, and 1s_A-1s_B has one (the plane perpendicular to and bisecting the molecular axis). So 1s_A-1s_B is higher in energy and the antibonding orbital.

If we want to compare molecular orbitals that are constructed from different atomic orbitals we have to be very careful, however. If we look at molecular nitrogen we find the following. There are two MO's that are approximately 2s_A+2s_B and 2s_A-2s _B. The latter has one extra nodal plane, and so it has a higher energy. Two other MO's are approximately 2p_zA+2p_zB and 2p_zA-2p_zB. The former has one extra nodal plane and has a higher energy (take care: the z axis is chosen parallel to the molecular axis, and the signs of the lobes of the 2p_z orbitals pointing to each other are different). So far it is as to be expected. The situation becomes a bit messy if we compare all four MO's with each other. Whatever calculation one does 2s_A+2s_B is always to one with the lowest energy and 2p_zA-2p_zB is never the one with the higher energy. Other statements cannot be made. Things are extra complicated, because there are substantial interactions between 2s_A+2s_B and 2p_zA-2p_zB, and between 2s_A-2s_B and 2p_zA+2p_zB. These interactions decrease the energy of 2s_A+2s_B and increase the energy of 2p_zA-2p_zB. They also increase the energy difference between 2s_A-2s_B and 2p_{z A}+2p_zB, but it depends on details of the calculations and of the form of the atomic orbitals which molecular orbital goes up and which one goes down in energy.

There is a good reason for the relation between the number of nodal planes and the energy of an orbital. This is the form of the operator for the kinetic energy of the electrons. This operator takes second derivatives of an orbital. The value of an orbital with a nodal plane changes more than the value of an orbital without nodal planes. This means that the (second) derivatives of an orbital with a nodal plane are higher (in absolute sense). As a consequence the kinetic energy of an electron in an orbital with a nodal plane is higher. There may be some compensation from potential energy contributions, but in general the energy of an orbital is higher when it has more nodal planes.