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

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

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''#).