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

##### 1 Answer
Aug 23, 2017

${D}_{3 h}$ point group, 6 vibrational modes.

LEWIS STRUCTURE

${\text{BF}}_{3}$ has:

And thus, using $24$ valence electrons, we assign $3 \times 6$ 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:

${\hat{C}}_{n}$: Proper Rotation

Rotate the molecule ${360}^{\circ} / n$ degrees, so that it returns to an orientation indistinguishable from the previous. For instance, ${\hat{C}}_{3}$ is a ${120}^{\circ}$ rotation (${360}^{\circ} / 3 = {120}^{\circ}$).

$\hat{\sigma}$: Reflection Plane

There exist ${\hat{\sigma}}_{h}$, ${\hat{\sigma}}_{v}$, and ${\hat{\sigma}}_{d}$ (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 ${\sigma}_{h}$.

${\hat{S}}_{n}$: Improper Rotation

This is basically ${\hat{C}}_{n} {\hat{\sigma}}_{h}$ or ${\hat{\sigma}}_{h} {\hat{C}}_{n}$ in one. Rotate about the axis, then reflect through the plane perpendicular to that axis. If you know how to use ${\hat{C}}_{n}$ and ${\hat{\sigma}}_{h}$, you know how this works.

$\hat{i}$: Inversion

The easiest way I can describe this is, $\left(x , y , z\right) \to \left(- x , - y , - z\right)$. It can also be treated as ${\hat{C}}_{2} {\sigma}_{h}$, 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 ${\text{BF}}_{3}$:

1. It has a single ${C}_{3} \left(z\right)$ axis perpendicular to its plane (${120}^{\circ}$ 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 $\boldsymbol{{\sigma}_{h} \left(x y\right)}$ 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 $\boldsymbol{{C}_{2} \text{'}}$ axis perpendicular to the $\boldsymbol{{C}_{3} \left(z\right)}$, but coplanar with the ${\sigma}_{h} \left(x y\right)$.

That is enough to conclude that we have a $\boldsymbol{{D}_{3 h}}$ 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:

${\text{DOF}}_{v i b} = 3 N - 6$

where $N$ is the number of atoms. (For linear polyatomic molecules, like ${\text{CO}}_{2}$, it would have been $3 N - 5$.)

Thus, ${\text{BF}}_{3}$ has $\boldsymbol{6}$ vibrational modes (${A}_{1} ' + 2 E ' + {A}_{2} ' '$).