Question

Draw `BF_3` and assign a point group. How many degrees of vibrational freedom does the molecule have?

Answer 1

`D_(3h)` point group, 6 vibrational modes.

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LEWIS STRUCTURE

`"BF"_3` has:

• `7 xx 3` valence electrons from the fluorines
• `3` valence electrons from the boron

And thus, using `24` valence electrons, we assign `3 xx 6` nonbonding electrons to the fluorines and the remaining `6` as three single bonds.

POINT GROUP

Now, to get the point group, we consider the general symmetry operations:

`hatC_n`: Proper Rotation

Rotate the molecule `360^@/n` degrees, so that it returns to an orientation indistinguishable from the previous. For instance, `hatC_3` is a `120^@` rotation (`360^@/3 = 120^@`).

`hatsigma`: Reflection Plane

There exist `hatsigma_h`, `hatsigma_v`, and `hatsigma_d` (horizontal, vertical, dihedral).

• The `sigma_h` symmetry element is perpendicular to the `C_n` of the highest `n` (the principal rotation axis).
• The `sigma_v` symmetry element is colinear with the `C_n` of the highest `n`, but perpendicular to any present `sigma_h`
• The `sigma_d` symmetry element bisects and is perpendicular to two `C_2` axes that are coplanar with a `sigma_h`.

`hatS_n`: Improper Rotation

This is basically `hatC_nhatsigma_h` or `hatsigma_hhatC_n` in one. Rotate about the axis, then reflect through the plane perpendicular to that axis. If you know how to use `hatC_n` and `hatsigma_h`, you know how this works.

`hati`: Inversion

The easiest way I can describe this is, `(x,y,z) -> (-x,-y,-z)`. It can also be treated as `hatC_2sigma_h`, PROVIDED the `sigma_h` element is perpendicular to the `C_2` axis (and it may not be).

If you have a hard time with this, this website really helps with visualization.

Then, this flow chart may make it easier. However, I don't really use it, so finding point groups can be done without a flow chart except for really complicated molecules.

Here is my thought process for finding the point group of `"BF"_3`:

1. It has a single `C_3(z)` axis perpendicular to its plane (`120^@` rotation), and that is the highest-order rotation axis, so it is the principal rotation axis and we call it the `z` axis by convention.
2. We see that the `bb(sigma_h(xy))` perpendicular to the `C_3` axis is trivially present because the molecule is planar (it reflects onto itself and appears to do nothing).
3. There is a `bb(C_2"'")` axis perpendicular to the `bb(C_3(z))`, but coplanar with the `sigma_h (xy)`.

That is enough to conclude that we have a `bb(D_(3h))` point group, a dihedral group that has a `C_2` perpendicular to the `C_n` of the highest `n`.

VIBRATIONAL DEGREES OF FREEDOM

Lastly, the degrees of vibrational freedom for nonlinear polyatomic molecules are found as:

`"DOF"_(vib) = 3N - 6`

where `N` is the number of atoms. (For linear polyatomic molecules, like `"CO"_2`, it would have been `3N - 5`.)

Thus, `"BF"_3` has `bb6` vibrational modes (`A_1' + 2E' + A_2''`).