Video: Permanent Dipole Moment of Phosphorus Trifluoride

Which of the following molecules has a permanent dipole moment? [A] BF₃ [B] AlF₃ [C] CF₄ [D] PF₃ [E] PF₅

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Video Transcript

Which of the following molecules has a permanent dipole moment? A) BF₃, B) AlF₃, C) CF₄, D) PF₃, or E) PF₅.

To begin to answer this question, let’s review what we know about permanent dipole moments. The first keyword in this question is permanent. This means we’re not talking about induced or instantaneous dipoles. Now, let’s consider what we mean by a dipole moment. Sometimes, given the letter 𝜇, a dipole moment is essentially a measure of polarity. Polarity in a molecule arises from partial charges on atoms. And these partial charges come from electronegativity differences between elements and features of the bonding in a molecule. Let’s take a simple example of a diatomic, like H–F.

The fluorine is a more electronegative element than the hydrogen. This can result in the molecule being polar. This gives a 𝛿+ charge on the hydrogen and a 𝛿- partial charge on the fluorine. We would denote the direction of the dipole moment with an arrow, like this. The arrow points towards the more negatively charged element. You can remember this because the other end looks a little bit like a plus sign. But what other features determine whether a molecule has a permanent dipole moment?

In the case of polyatomic molecules, it turns out that symmetry of the molecule is hugely important in determining whether it has an overall dipole moment. In fact, molecular symmetry is even more important than which elements are in your molecule. Let’s look at the examples of ozone, O₃, and carbon dioxide, CO₂.

Ozone is homonuclear and triatomic, and it has a permanent dipole. Whereas CO₂, also triatomic, but this time heteronuclear, does not have a dipole. Why is this the case? The answer is the symmetry of the molecule. We can see that in the case of ozone, the molecule is nonlinear, whereas in CO₂, the molecule is linear.

Let’s look first at CO₂. Both oxygen atoms are slightly more electronegative than the central carbon atom. This places a partial negative charge on both oxygens and a partial positive charge on the carbon. Using this information, we can draw in the dipole moment arrows. As you can see, each arrow is the same. This is because each C=O bond is equivalent. Moreover, the two dipole arrows point in opposite directions, effectively balancing each other out. This means that overall, CO₂ does not have a permanent dipole moment.

Now let’s look at ozone. You may also see ozone drawn like this, with a formal positive and negative charge and one double bond. But I prefer the delocalized form. In either depiction, you can clearly see that the three oxygen atoms are not equivalent. We actually end up with a positive central oxygen and the two other oxygens having a partial negative charge. Again, we can use this to draw in our dipole moment arrows. Although the dipole moment on the left is equivalent to the dipole moment on the right, there is an overall dipole. This is because, due to the bend in the molecule, the two dipoles are not in opposing directions. So, they don’t cancel themselves out, like was in the case of CO₂. Here, we actually end up with an overall dipole moment a bit like this.

So, let’s use these techniques to work out the answer to our question. Here, we have BF₃. The important thing about this molecule is that it is trigonal planar and symmetrical. Each of the fluorines is more electronegative than the central boron. And this means, we can draw in our dipole moments. Each of the B–F bonds is equivalent due to its symmetry. And this means that each of the dipole moments has the same magnitude. Moreover, they each pull equally in opposing directions, which means that overall, there’s no dipole moment. So, this is not a correct answer to our question.

If you’re struggling to think about how these dipole moments pull in opposing directions and whether they balance each other out, you could simply rotate the molecule with its dipole moments and realize that it would be the same each time you rotate.

Let’s move on to look at AlF₃. As it turns out, aluminum trifluoride is actually very similar. It has the same trigonal planar shape and the same level of symmetry. Aluminum, of course, is just below boron on the periodic table and therefore in the same group. So, for the same reasons as BF₃, AlF₃ also does not have a dipole moment.

Once again, we can draw on our partial charges, fluorine being more electronegative than carbon. Again, thanks to the symmetry of this molecule, each carbon fluorine bond is equivalent. Again, if we draw on our dipole moment arrows, we can see that each of them is the same. Moreover, due to the symmetry, each one is pulling in an opposing direction to another. This means they all cancel each other out. And we’re left with no permanent dipole moment. So, this is not a correct answer.

If you favor a more mathematical explanation for the dipole moments canceling each other out, this can be done using vector addition. This is also something which can be done computationally.

Let’s look at the final two molecules side by side. The question here is how do we arrange the fluorine atoms around the phosphorus. In the case of PF₅, the best way to arrange five fluorine atoms around the central phosphorus is in a trigonal bipyramidal shape. The problem comes with PF₃. You might think that this should be trigonal planar just like BF₃ and AlF₃. But it’s not quite that simple. Phosphorus has five valence electrons. If it forms three covalent bonds, that counts with three of the five valence electrons. So, what happens to the other two?

The answer, of course, is that they form a lone pair. And we must take this into account when looking at symmetry. This means that PF₃ actually forms a trigonal pyramidal structure. This is very similar to the structure in CF₄, the tetrahedral structure, except that instead of having a fourth P–F bond, we have a lone pair.

So, now let’s look at our partial charges. In the case of PF₅, we have P–F equivalent bonds all the way around in a very symmetrical structure. This means that each of the dipole moments is equivalent and each of them cancels the others out. So, we’re left with no overall dipole moment. Let’s look closer at PF₃. Each of the phosphorus fluorine bonds is equivalent to one another. But this lone pair on the top means that the overall molecule isn’t entirely symmetrical. We have three dipole moments along the P–F bonds. But there is no opposing dipole moment in the vertical direction. This means that not all of the dipole moments are balanced out. This, in fact, gives us an overall permanent dipole moment in the direction shown.

So, of the five molecules we’re given, PF₃ is the only one with a permanent dipole moment. So, this is our correct answer.

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