# Question 2013d

Jan 23, 2015

In order to determine rotation, let's start by determining the molecular geometry of the sulphur hexafluoride (SF"_6#) molecule.

The molecule's Lewis structure must account for 48 valence electrons, 6 from $\text{S}$ and 7 from each $\text{F}$ atom. Here's what the Lewis structure looks like:

Indeed, all the 48 valence electrons are accounted for - 2 for each of the six sigma bonds, and 6 for each flourine atom. According to VSEPR Theory, the ${\text{SF}}_{6}$ molecule has an octahedral molecular geometry, the central atom being ${\text{sp"^3"d}}^{2}$ hydridized.

Notice that the molecule has no double or triple bonds, which means that is has no pi bonds. The absence of pi bonds ensures free rotation about all the six sigma bonds. Free rotation about sigma bonds takes place because the bond's electron distribution exists along an imaginary line drawn between the two atoms' nuclei - this is caused by the head-on overlapping of the hybrid orbitals.

Rotation about this imaginary axis would not cause changes in the electron distribution, which means that bond strength will not be affected by rotation.

As a result, sulphur hexafluoride has free rotation about all its sigma bonds.