Determing the electronic structure of ammonia will introduce the new ideas of degenerate orbitals and degenerate axes. It is important to understand these concepts because of the large number of molecules that have point groups such as C_{3v} or D_{3h}.
To determine the MO's of ammonia
The Point Group
Ammonia has the following symmetry elements: E, 2 C_{3}, 3 sigmav. This places it in the C_{3v} point group. The character table for C_{3v }is shown below.
C_{3v} 
E 
2 C_{3v} 
3 Sigmav 

A_{1} 
1 
1 
1 
z 
x^{2}+y^{2}, z^{2} 
A_{2} 
1 
1 
1 
R_{z} 

E 
2 
1 
0 
(x,y)(R_{x},R_{y}) 
(x^{2}y^{2}, xy)(xz,yz) 
Identification of the valence atomic orbitals
Ammonia has the following atomic orbitals. Of these, we select all but the nitrogen 1s. The energetic match of this corelevel orbital with the valence orbitals is very poor.
Atom  Atomic Orbital  Ionization Potential (eV) 
H 
1s 
13.6 
N 
1s 
409.9 
N 
2s 
20.3 
N 
2p 
14.5 
Transformation of the valence atomic orbitals in C_{3v }symmetry
The following table summarizes how the nitrogen and hydrogen atomic orbitals transform. Additional discussion follows regarding the assignment of characters for the 2px and 2py orbitals which transform together as a pair.
C_{3v} 
E 
C_{2} 
sigma(xz) 

N 2s 
1 
1 
1 
A_{1} 
N 2p_{z} 
1 
1 
1 
A_{1} 
N 2p_{x}, 2p_{y} 
2 
1 
0 
E 
3 H 1s 
3 
0 
1 
A_{1} + E 
In thinking about how the px and py orbitals transform in C_{3v} symmetry, it is important to remember that our pictures represent mathematical functions. The real questions is, how does the symmetry operation affect the function. In this case, the C3 rotation does not move the p_{x} or p_{y} orbitals back into their original positions on the x and y axes respectively, but it does leave a partial projection of the orbital on the original axis. Consider the case of a unit vector aligned along the yaxis as shown below.
If unit vector A is rotated by some angle alpha, the projection of unit vector A remain along the yaxis is equal to cos(alpha). To summarize:
alpha = 0  cos(alpha) = 1  Equivalent to orbital remaining in place 
alpha = 90  cos(alpha) = 0  Equivalent to orbital moving to a new position 
alpha = 180  cos(alpha) = 1  Equivalent to a phase inversion 
alpha = 120  cos(alpha) = 0.5  The amount of the orbital projected on the original position 
Having selected the valence orbitals and transformed them according to the symmetry of the molecule under consideration, we are now prepared to make the MO diagram for ammonia.
The initial step taken in this case was to mix the 2s and 2pz orbital, they are of the same symmetry. This is a convenient way of breaking energetic degeneracies early in the problem. Also, by mixing orbitals of like symmetry on the same atom first, you can then generally procede by mixing pairs of orbitals of like symmetry from the different atoms by starting with the lowest energy orbitals and work upwards in energy. For chemistry 302, you should be able to get the energetic ordering of the bonding and lonepair orbitals correct, we will be less concerned with the antibonding orbitals.
The Molecular Orbital Diagram for Ammonia
Now it is time to draw the molecular orbitals of ammmonia and calculate them using CAChe. As you can see, the drawings based on our linear combinations of the atomic orbitals look very much like those calculated by CAChe.
Finally, comparing the calculations to an experimental measure of the electronic structure such as photoelectron spectroscopy is important. The expected number of peaks, three, is observed, although our energy estimates are rather poor. The experimental numbers in eV are:
2 a_{1}  1 e  1 a_{1} 
10.2  15.0  27.0 