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I'm trying to build a molecular orbital diagram for BF3 and I'm running into problems with irreducible representations on the F side.

2s for B has an irreducible representation of A1. 2p for B has an irreducible representation of E' and A''2.

2s for F considered non bonding

2p (along the bond axis) for F has an irreducible representation of E' and A' 2p (z) for F Has an irreducible representation of E'' and A''2 2p (the remaining one) has an irreducible representation of E' and A'2

From this I built A1 bonding and antibonding orbitals, A'2 bonding and antibonding orbitals, 1 A'2 and 2 E''2 non bonding orbitals.

I'm now stuck because I have 4 E' on the F side and only 2 E' on the B side.

I found the following diagram for BF3 online but it doesn't generate the E' anti bonding and also doesn't generate enough molecular orbitals.

enter image description here

Does anyone know how to mix 4 E' with 2 E'

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I think you are most of the way to the answer, but I will start the process from scratch for the sake of a full explanation for future readers.

First, we determine the symmetry group of $\ce{BF3}$, which by quick inspection we can determine to be $D_{3h}$ (since it has a principle $C_3$ axis, $3\perp C_{2}$ axes, and a horizontal mirror plane of symmetry).

Point group flow chart

Knowing this, we can now make use of the $D_{3h}$ character table (thank you @orthocresol for compiling these in mathjax format) to determine the irreducible representation of the atomic orbitals (or their symmetry adapted linear combinations SALCS) within the group.

$$\begin{array}{|c|cccccc|c|c|} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & \text{linear/rotations} & quadratic\\ \hline \mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\ \mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \\ \mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy) \\ \mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\ \mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & \\ \mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) \\ \hline \end{array}$$

Using this, we can quickly determine the irreps of boron's orbitals using the 2nd column from the right.

$\ce{B}$: $2s=a_1'$, $2p_x,2p_y=e'$, and $2p_z=a_2'$.

Next, we will determine the irreps for the $3\ce{F}$ atoms, which will require that I consider combinations of $s,p_x,p_y, \text{and } p_z$ orbitals.

Below, I show how the symmetry operations of the group affect each set of orbitals and show the orbitals that result (image source).

$$\begin{array}{|c|cccccc|} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline \Gamma_s & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline \mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\ \mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 \\ \hline \end{array}$$

$\hspace{20ex}$s-orbitals

$$\begin{array}{|c|cccccc|} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline \Gamma_{p_x} & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline \mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\ \mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 \\ \hline \end{array}$$

$\hspace{12ex}$px orbitals

$$\begin{array}{|c|cccccc|} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline \Gamma_{p_y} & 3 & 0 & -1 & 3 & 0 & -1 \\ \hline \mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 \\ \mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 \\ \hline \end{array}$$

$\hspace{20ex}$py orbitals

$$\begin{array}{|c|cccccc|} \hline D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline \Gamma_{p_z} & 3 & 0 & -1 & -3 & 0 & 1 \\ \hline \mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 \\ \mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 \\ \hline \end{array}$$

$\hspace{21ex}$pz orbitals

These reductions to irreducible representations can be done via the reduction formula $$n_i=\frac{1}{h}\sum_c g_c \cdot \chi_{i_c} \cdot \chi_{r_c}$$ where $n_i$ is the number of times a particular irrep $i$ occurs in the reducible representation, $g_c$ is the number of symmetry operations $c$, $\chi_{i_c}$ is the character of the irrep $i$ for the symmetry operation $c$, and $\chi_{r_c}$ is the character of the reducible representation $r$ for the symmetry operation $c$.

This allows us to write the irreps of the three $\ce{F}$'s orbitals.

$3\ce{F}$: $a_1'+a_2'+a_2''+2e'+e''$

With all the formalities out of the way, we can construct the qualitative MO diagram for $\ce{BF3}$. Like much of this post, my source for the MO diagram is the pdf of lecture notes for a Dartmouth chemistry course located at http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

BF3 MOs

Working our way up through this diagram, we see that we start by assuming that the SALCs of the $s$ orbitals of the three $\ce{F}$'s are assumed to be low enough in energy where they do not interact with the orbitals of $\ce{B}$.

The next point of interest is the $e' (\sigma)$ MO, which we form from the $p_x$ SALC of the $\ce{F}$ atoms and the $p_x$ and $p_y$ of the $\ce{B}$. We could additionally consider the interaction of the $e'$ $p_y$ SALC to form a $\pi$ bonding interaction, but we will assume that, with the $\sigma$ interaction already present, this $\pi$ bonding interaction will be weak and thus can be neglected.

This previous point leads us to have a set of nonbonding orbitals $a_2'+e'(y)+e''$. We then also have antibonding orbitals for each of the bonding orbitals below.

As a check on our work, we can compare this to the somewhat more quantitative diagram given in Principles of Inorganic Chemistry by Brian PFennig using the Student Editon of Spartan:

semi-quantitative MO diagram

The only noticeable difference here is that the nonbonding MOs are now placed with respect to their computed energy rather than being lumped together. The classification of orbital as bonding, nonbonding, and antibonding seems to fit with the diagram we have above, with the caveat that this more quantitative diagram does suggest some $e'(y)$ bonding contribution, as its energy is somewhat lower than the p orbitals it came from.

Now that we have the correct diagram, we can then compute the bond order for each $\ce{B-F}$ using: $$\frac{\text{Bonding electrons } -\text{Antibonding electrons}}{2\cdot\text{number of bonds}}$$ which for this diagram gives a value of $\frac{4}{3}$ per bond. This gives a sense of why $\ce{BF_3}$ is a strong Lewis acid, since it can accept a pair of electrons into its antibonding LUMO and still leave a compound with a BO of $1$ per bond.

As a final point, it is worth noting that this method of classifying orbitals is far from perfect and in general we should really look at individual bonds to classify if the orbital is anti/bonding with respect to that bond. However this energy based approach is what is commonly taught in school and happens to work well in this simple case; just be aware that it can fail for even seemingly simpler molecules like $\ce{CO}$.

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  • 2
    $\begingroup$ That is a really good explanation. I personally dislike the classification of bonding, non-bonding, and anti-bonding according to energy, because it's not really that rigorous and sometimes leads to wrong conclusions (point in case CO), but it often works, as it's here the case. I personally would classify the 3e' orbitals as anti-bonding according to symmetry with respect to at least two bonds. The same would be the case for 4e', but that's a discussion for another time. $\endgroup$ – Martin - マーチン Aug 23 '17 at 12:02

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