# Why does symmetry have to be maintained in molecular orbitals?

Using the example of $\ce{XeF4}$: What is the physical explanation enforcing the symmetry of the $\ce{1b_{1g}}$ orbital on the fluorine atoms? Why isn't the symmetry of a nonbonding orbital arbitrary? If it's going to be nonbonding anyways, why can't we, for example, have a Fluorine p-orbital arrangement facing towards Xenon with three positive p-orbitals and one negative p-orbital?

To elaborate:

If I imagine a free Xenon atom in space, and the approach of four individual fluorine atoms, I would expect the bond formation to be randomized with respect to the orientation of the fluorine p-orbitals, and therefore for some arrangements to not be perfectly symmetrical, such as 3 positive p-orbitals, 1 negative, facing inward. I understand bonds can't be made without symmetry between the Xenon and Fluorine orbitals; that makes physical sense because we can argue it by looking at orbital overlap that dictates bonds can only occur with appropriate symmetry. But in a nonbonding case, such as $\ce{1b_{1g}}$ above, I don't understand why symmetry is also required.

Source for pictures:

The orbital and geometrical symmetry are closely related. You know that $\ce{XeF_4}$ is square planar, therefore $\ce{D_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.

So if you perform some manipulation of the molecule, e.g. rotation by $\ce{90^{\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.

One could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.

• Hi, thanks for the prompt answer. To clarify: "So if you perform some manipulation of the molecule, e.g. rotation by 90∘, you must end up with the same picture" Why must you end up with the same picture? I understand there's a pattern of symmetry, but I don't understand any physical explanation why. If I were a Fluorine orbital, why do I care how the Fluorine orbital opposite me looks, in particular if neither of us are involved in bonding? Jan 20, 2015 at 11:55
• You as the fluorine play hide-and-seek with poor observer and you should be indistinguishable from other 3 fluorines, as symmetry dictates. Would the molecule be non-symmetrical, this request is lifted and you can do whatever you wish. But you cannot. Jan 20, 2015 at 13:30

It's not so much that these are arbitrary arrangements or that symmetry is "conserved."

We take the 4 fluorine atoms as a set and consider the reproducible representation of all fluorine 2p orbitals.

Under the $D_{4h}$ point group of $\ce{XeF4}$, the 4 fluorine orbitals consist of an $a_{1g}$, $b_{1g}$ and $e_u$ representations. (There are other representations of the out-of-plane 2p orbitals too.) Four atomic orbitals, 2x1D 1x2D representation.

Now we try combining the representations of the fluorine orbitals.

It's a matter of math... the reason $b_1g$ is non-bonding is because there is no Xe atomic orbital with the same symmetry. If you consider the direct product of the fluorine $b_{1g}$ with any Xe orbital, you get zero - they are orthogonal and have no overlap.

Now, you ask why can't you have 3:1.. well, you have 4 identical fluorines. If you had $\ce{XeF3Cl}$ or something like this, you'd then have different symmetry and different orbitals.

• Hi, thanks for taking the time to respond. I actually understand why the orbital is nonbonding. I guess my sticking point is that all the explanations so far approach from the angle of symmetry. Of course if we define a problem as symmetry, and then explain it using symmetry, it seems self-consistent. But nature itself isn't aware of a concept like symmetry. Nature is composed of fundamental forces dictating its behavior. In non-bonding, my confusion is what fundamental property of nature requires certain symmetries? In bonding, it's orbital overlap so that probabilities add, but nonbonding... Jan 21, 2015 at 11:02
• but non-bonding orbitals don't require orbital overlap to exist. Why is a non-bonding orbital forced then to be in a non-arbitrary arrangement? Jan 21, 2015 at 11:03
• There's nothing that "requires" symmetry. You asked a question about $\ce{XeF4}$ which has a symmetric property. Because the molecule is symmetric, the orbitals will be symmetric. There's nothing arbitrary about it - it's the base property of the molecule itself. If I look out my window, I see snowflakes right now. The ice lattice dictates the symmetry of the crystal, much like the symmetry of the molecule (i.e., four equivalent fluorines) dictate the symmetry of the orbitals. Jan 21, 2015 at 15:07
• A 3:1 mixture will be possible under a different less-symmetric molecule. Under $D_{4h}$ such a mixture is a reducible representation, not an irreducible one. Jan 21, 2015 at 15:08
• If you haven't studied it, perhaps you should read about using projection operators to create the MOs. Jan 21, 2015 at 15:08

Symmetry reduces the level of energy of any thing. Most of the things of nature has inbuilt symmetry with them. It is the property of nature to have different kinds of symmetry. So that it reduces the energy of the orbital or molecule