# Which of these molecules is polar? "SiF"_4, "XeF"_4, "BF"_3, "SF"_4

Jan 5, 2017

${\text{SF}}_{4}$, which is a see-saw molecular geometry, a not totally-symmetrical structure.

A polar molecule has both an uneven distribution of electrons and no total symmetry to counteract it. This image sums it up:

When a molecule has a great deal of symmetry, you can rotate it in certain ways that return it to an indistinguishable position, meaning that any polarity it has will cancel out.

(By an indistinguishable position, I mean that you can turn your back, someone else can rotate the molecule in a particular way, and when you turn around, you can't tell anything happened.)

Assuming you aren't very familiar with vectors...

For your purposes, if the molecule has any lone pairs of electrons, you can consider it not totally symmetric. Therefore, it will be polar by virtue of its shape, even if its individual bonds are not.

Using the valence electron-counting method, draw each of these Lewis structures.

## 1)${\text{SiF}}_{4}$

$\text{Si}$: $4$
$\text{F}$: $7 \times 4$
$\text{Total} : 32$

Each $\text{F}$ holds $6$ nonbonding electrons, so $6 \times 4 = 24$ surrounding the $\text{F}$ atoms in total. That leaves $8$. For each bond, we need $2$ electrons, so $\text{Si}$ makes one bond with each $\text{F}$.

Therefore, we have four electron groups (of those four, all are bonding groups), giving a tetrahedral structure:

This is totally symmetric, so this is a nonpolar molecule, even without checking the polarity of each individual bond.

(If we did, $\Delta \text{EN" = "EN"_"F" - "EN"_"Si} = 4.0 - 1.8 = 2.2$, but again, all four dipoles cancel out, giving no net dipole, and thus net nonpolarity.)

## 2)${\text{XeF}}_{4}$

$\text{Xe}$: $8$
$\text{F}$: $7 \times 4$
$\text{Total} : 36$

Each $\text{F}$ holds $6$ nonbonding electrons, so $6 \times 4 = 24$ surrounding the $\text{F}$ atoms in total. That leaves $8$. For each bond, we need $2$ electrons, so $\text{Si}$ makes one bond with each $\text{F}$. Fluorine tends to make only one bond as an outer atom, so we have two lone pairs.

Therefore, we have six electron groups, two of which are lone pairs, giving a square planar structure (take two atoms off of the same axis in an octahedral molecular geometry to derive it):

This is totally symmetric, because each $\text{F}$ atom is counteracted by the one directly across from it, and each lone pair is ${180}^{\circ}$ from the other.

So, this is a nonpolar molecule, even without checking the polarity of each individual bond.

## 3)${\text{BF}}_{3}$

I'll let you follow the previous parts of the answer to derive the molecular geometry, which is trigonal planar (recall that boron is OK with six valence electrons).

This is also totally symmetric, so this is nonpolar.

## 4)${\text{SF}}_{4}$

${\text{SF}}_{4}$ is rather tricky, but I think you can derive that it has a see-saw molecular geometry (from the trigonal bipyramidal electron geometry; remember to take the atom off the equatorial plane).

The $\text{F}$ on the same axis counteract each other, but the two on the central plane (along the vertical in the image above) that are about ${90}^{\circ}$ from each other do not. The lone pair crunches the two coaxial $\text{F}$ atoms together, so that some of the bond angles are altered from the expected ${90}^{\circ}$.

This is not totally symmetric, so this is a polar molecule.

In summary, for your purposes, if the molecule has any lone pairs of electrons, you can consider it not totally symmetric. Therefore, it will be polar by virtue of its shape, even if its individual bonds are not.