Chem 109  General Chemistry  Spring 2015
Lecture Notes 33: 20 April
Return Exam 3
Molecular Orbital Model of Bonding
As with atoms, we will begin with the simplest system, in this case the dihydrogen molecule, H_{2}. (Strictly speaking, the simplest molecule is the dihydrogen molecular ion, H_{2}^{+}, with a single electron.)
As I noted in the beginning of our discussion of modern bonding, orbitals are conserved, so if we add two hydrogen atoms, H_{a} & H_{b} together, the two 1s orbitals should give us two molecular orbitals, MO_{1} and MO_{2}:
MO_{1} = 1s_{a} + 1s_{b}
MO_{2} = 1s_{a}  1s_{b}
Note that one orbital will have a lower energy and the second a higher energy as expected from the approximate conservation of orbital energies we noted earlier. [text Fig 9.27]And when we add and subtract the two atomic orbitals they give molecular orbitals of quite different shapes. [text Fig 9.25]
dihydrogen MO diagram
illustration from Wikimedia Commons
The molecular orbitals resulting from this combination are symmetrical along the atomic axis between the bonded atoms, and as before are referred to as sigma (sigma) molecular orbitals. The two orbitals, however have much different properties.
 The ground level (lower energy orbital) is a bonding orbital and called simply a sigma orbital. The electron density for this orbital is largely distributed between the atoms.
 The high energy orbital actually has most of the electron density not between the nuclei, so the nuclei and electrons will repel each other, and no bond is formed. This orbital is referred to as an antibonding orbital and given the designation sigma star (sigma*).
 Note that if electrons occupy both the bonding and antibonding orbitals there will be no net bond formed!
Bond Order = (#bonding electrons  # antibonding electrons)/2. Divide by two to get "classical" two electron bond. Bond order gives a measure of bond strength in units of an electronpair bond.
 If we look at hydrogen, H_{2}, both electrons go into the ground state (lowest energy) MO, giving a bond order of one, so H_{2} has a single bond. [text Fig 9.28]
 If we look at the next possible homonuclear diatomic molecule, He_{2}, the four electrons will first fill the lowest energy MO, but the next two go into the higher energy, antibonding MO. The bond order is then 0, and theory predicts no bonding and no He_{2} molecule. [text Fig 9.30, 9.32]
So far we've looked only at atoms with selectrons and sorbitals. What happens when we have pelectrons? The first element with pelectrons is boron, with a valence electronic configuration of 2s^{2}2p^{1}. So what happens if we combine two boron atoms and calculate the new energy levels for the potential molecule?
 First let's look at a simple calculation assuming the s and p orbitals do not interact with each other.
 Because the s and p orbitals are of significantly different energies we'll get two distinct sets of MO's.
 As expected the 2s orbitals will combine to give sigma MO's with a splitting just like we saw for helium with both the sigma_{2s} and sigma_{2s}* filled and thus cancelling each other.
 The porbitals will be a bit more complex. [text Fig 9.33a]
 One set, call them the p_{x} orbitals will overlap with cylindrical symmetry about the axis connecting the nuclei [text Fig 9.33d] giving a set of sigma and sigma* orbitals, one bonding and one antibonding. [text Fig p 432]
 The other two sets have planar symmetry and give pi orbitals. [text Fig p433]
 With this calculation the energy diagram shows, starting at the lowest energy, a sigma_{2s}, a sigma_{2s}*, a sigma_{2p}, two pi orbitals of equal energy (y, and z), two pi* orbitals of equal energy, and a sigma_{2p}* orbital. [text Fig 9.35]
 Filling from the bottom with the six electrons of the two boron atoms we should see two bonds and one antibond giving a total of one bond, which is what we observe. However, diboron is paramagnetic, which is not at all expected from our filling diagram. What's wrong? Our model is too simple.
 Recall from our earlier discussion that when we look at a molecule the electrons of that molecule are now just that  they belong to the molecule. In our first calculation above we assumed we could treat the electron energy levels the same as we did for the atoms. But when we put the two atoms together as a molecule we shouldn't be surprised that the energy levels are more complex  the valence s and p orbitals of the atoms are all considered together and a whole new set is calculated. Your author refers to this as "mixing" the s and p orbitals, but really there are no s or p valence orbitals in the molecule.)
 So when we calculate the orbitals fresh, assuming the molecule will of course still get the same number of orbitals, and even the same types, but the energy levels differ. [text Fig 9.37]
 The sigma_{2s}, and sigma_{2s}* orbitals remain separate from the other orbitals, but sigma_{2s}* energy is lowered.
 The order of the sigma_{2p} and pi_{2p} orbitals is reversed, with the two pi_{2p} at a lower energy. The order of the excited (*) orbitals remains the same.
 As a result, filling from the lowest energy, we see a sigma_{2p} bond, a sigma_{2p}* antibond, and a pi_{2p} bond made up of two unpaired electrons in different pi_{2p} orbitals. Thus the diboron molecule is predicted to be paramagnetic, as observed.
Lets now go back and and look at the bond orders and bonding of the homonuclear diatomic molecules of the second period. [text Fig 9.38] As we can see in each case the bonding is as predicted from molecular orbital theory. ( used clickers to predict bond orders and paramagnetism)
Molecular Orbital Energy Levels and Bonding in Diatomic Homonuclear Molecules, Li_{2}Ne_{2}

Li_{2} 
Be_{2} 
B_{2} 
C_{2} 
N_{2} 
O_{2} 
F_{2} 
Ne_{2} 
sigma_{2p}* 








pi_{2p}* 








sigma_{2p} 








pi_{2p} 








sigma_{2s}* 








sigma_{2s} 








Bonds 
1 
0 
1 
2 
3 
2 
1 
0 
Paramagnetic 
N 
N 
Y 
N 
N 
Y 
N 
N 
© R A Paselk
Last modified 20 Apr 2015