# Molecular orbital diagram and irreducible representations for dinitrogen

I'm trying to understand how to draw molecular orbital diagrams, since I'd like to use the molecular orbitals to determine the total state of the molecule (i.e. $A_1$, $B_1$ etc.), so I prefer the notation with irreducible representations of orbitals like it's shown in this video for $\ce{H2O}$.

The problem is that I'm not sure how to make molecular orbitals for $\ce{N2}$ and $\ce{N2^+}$.

I know that $\ce{N}$ has atomic orbitals $\ce{1s^{2} 2s^{2} 2p^{3}}$. $\ce{N2}$ is a linear molecule, so it belongs to the $D_{\mathrm{\infty h}}$ symmetry point group, but we can descend in symmetry, so we'll use $D_{\mathrm{2h}}$.

But now I have no idea how to proceed. Should I work with both nitrogen atoms together, i.e. with group orbitals, similarly to the case with two hydrogen atoms in $\ce{H2O}$? Or am I supposed to work with them separately?

So could you please write down one of the diagrams using irreducible representations and explain it step-by-step? I think it would be really useful here, as I couldn't find any source with some general explanation.

Because this is a diatomic molecule, there are no group orbitals. Put another way, the group orbitals are the molecular orbitals. Knowing the nitrogen atomic orbitals (AOs) and their irreducible representation (irrep) labels is enough.

Since we'll work in the $D_{\mathrm{2h}}$ point group, we need its character table:

$$\begin{array}{c|cccccccc|cc} \hline D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline \mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\ \mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\ \mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\ \mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\ \mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\ \mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\ \mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\ \mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline \end{array}$$

Using the nitrogen atom's electron configuration $\ce{1s^{2} 2s^{2} 2p^{3}}$ as our minimal basis, no d-orbitals will be present, so we can assign irreps to each AO right away, in part because we assume the principal rotation axis (the one of highest order) is aligned along the z-axis:

$$\begin{array}{cccc} \hline \mathrm{s} & \mathrm{p}_x & \mathrm{p}_y & \mathrm{p}_z \\ \hline \mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\ \hline \end{array}$$

Then, proceed by starting to form the standard MO diagram for a diatomic, but add the irrep labels to each AO:

Note that I haven't spaced the energy levels properly; the $\ce{1s}$ should be much lower than it is relative to the $\ce{2s}$. More on this later. Regardless, because this is a homodiatomic, all AOs will mix at each energy level, even the core $\ce{1s}$ AOs. Then, attempt to form the MOs:

This is probably wrong, but it's a starting point. Again, energy levels will be discussed later. Finally, add symmetry labels to each MO, remembering that

1. MO symmetry is derived from AO symmetry,
2. numbering is consecutive within each irrep, not with respect to the set of all MOs, and
3. lowercase is used to signify that these are MOs; uppercase is for the irreps themselves.

This is our final MO diagram for $\ce{N2}$. To form the diagram for $\ce{N2^+}$, remove an electron from $\mathrm{4a_g}$. This ignores orbital relaxation effects, but for the purposes of working this out on paper, it should be fine.

Now for the matter of relative and absolute energy levels. It is probably possible to get the correct relative ordering of the MO energy levels. Here, I assume that since AO mixes with an identical partner on the other atom, the splitting for $\ce{1s}$ would be the same as $\ce{2s},~\ce{2p_z}$, etc. Since I drew the $\ce{1s}$ too high, the $\mathrm{3a_g}$ is almost certainly too high, and perhaps should even go below what is labeled as $\mathrm{2a_g}$. The way to confirm this, and the only way to get absolute energy levels, is to perform a quantum chemical calculation. Since we've used a minimal basis for the drawing, we'll stick a minimal basis in the calculation. Here is a Psi4 input file:

molecule {
N 0.0 0.0 0.0
N 0.0 0.0 1.0975
}

set {
basis sto-3g
scf_type direct
df_scf_guess false
cubeprop_orbitals [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
}

e, wfn = energy('hf', return_wfn=True)
cubeprop(wfn)


and from its output:

Orbital Energies (a.u.)
-----------------------

Doubly Occupied:

1Ag   -15.518067     1B1u  -15.516124     2Ag    -1.442840
2B1u   -0.722491     1B2u   -0.573123     1B3u   -0.573123
3Ag    -0.539495

Virtual:

1B2g    0.281319     1B3g    0.281319     3B1u    1.123476


It looks like I might have messed up the diagram, because the ordering isn't what's expected (why isn't $\mathrm{B_{1u}}$ degenerate with the other two p-orbitals?), and the virtual degenerate p-orbitals have gerade symmetry. Time to plot!

In my haste, I forgot an important point. When you make the MOs from AOs, you're making two linear combinations:

\begin{align} \psi_{s} &= \frac{1}{\sqrt{2}} (\chi_{l} + \chi_{r}) \\ \psi_{a} &= \frac{1}{\sqrt{2}} (\chi_{l} - \chi_{r}), \end{align}

which means that an antisymmetric combination of two s orbitals will look like a $\mathrm{p}_z$ orbital. This doesn't fully explain everything, but is a good starting point. The lesson up to now is that while drawing the diagrams by hand is a useful exercise, it is not enough to do only that when the goal is to perform a correlated ab initio electronic structure calculation.

• Please forgive me for the ugly diagrams, I can try and redraw them on a tablet if desired. I am working on an answer for your other (Molpro) question, but it will take me a little more time. – pentavalentcarbon Feb 26 '18 at 4:54
• Why do the cubes look so strange? – pH13 - Yet another Philipp Feb 26 '18 at 9:31
• Thank you very much! And don't bother with other new diagrams, these are certainly good enough ;) – Eenoku Feb 26 '18 at 12:52
• @Eenoku Group orbitals are used when one of your bases for the MO diagram is just that: a group. For example, in ammonia, the nitrogen atom may be on one side, but the $\ce{H3}$ is taken together as a group on the other side. Same thing would go for a $\ce{ML6}$. The way I think about it is that you can't have more than two sides on an MO diagram...what do you do when you have more than two atoms? Combine ones that have a characteristic symmetry. – pentavalentcarbon Feb 26 '18 at 19:02
• What about a trigonal bipyramidal complex, where there may be 3 distinct groups ($\ce{AX3Y2}$)? The whole thing is $D_{\mathrm{3h}}$. Form $\ce{Y2}$ in $D_{\mathrm{3h}}$ (MO diagram not necessary), form $\ce{X3}$ in $D_{\mathrm{3h}}$ (MO diagram not necessary), then have each group on a side of an MO diagram to help form $\ce{X3Y2}$, then finally make an MO diagram with $\ce{X3Y2}$ on one side and the central atom on the other. The $\ce{X3}$, $\ce{Y2}$, and $\ce{X3Y2}$ are all group (intermediate) orbitals. – pentavalentcarbon Feb 26 '18 at 19:07

This is a correctly ordered molecular orbital diagram with $$D_{2h}$$ irreducible representations and energies included for every orbital.

The energy values and orbital ordering were get with a help of Psi4 script from pentavalentcarbon's answer.