Question

1. How many peaks would you expect to see in the Fluorine (19F) NMR of PF5 at room temperature and how many at 110 K? Support your answer.

2. The reaction of sodium carbonate, boron oxide, and silicon dioxide gives a borosilicate glass. Explain why the powder diffraction pattern of this product shows no diffraction maxima.

Answer 1

Consider a molecule such as PF5 . In principle, this molecule could adopt any of several structures. For example, the molecule could have D3h symmetry or alternatively it could have C4v symmetry or it might even be possible for the molecule to be planar and have D5h symmetry. On the other hand, the PF5 molecule could be pyramidal with a structure belonging to the C5v point group. Here’s a question we can ask concerning these various possible structures for the PF5 molecule: for a given structure, how many fluorine environments are present? In the first case, the D3h structure provides two different fluorine environments which are present in a 2 to 3 ratio. Actually, we can determine this through application of the operations of the D3h point group. For example, carrying out a C3 operation leads to exchange of fluorine positions in the equatorial plane, but does not affect the two axial fluorine atoms. On the other hand, we also see that if we carry out a C2 operation, this leads to pairwise exchange of equatorial fluorine atoms at the same time as it leads to pairwise exchange of the axial fluorine atoms. What we will find is that none of the group operations leads to exchange between the axial and the equatorial positions. This is why D3h symmetry for PF5 mandates the presence of two distinct fluorine environments in a 2 to 3 ratio. Think about it this way: an axial fluorine atom has three neighbors close by at 90◦ angles, whereas an equatorial fluorine has four neighbors (two at 90◦ angles and two at 120◦ angles). Carrying out a similar analysis, in the case of the C4v geometry, PF5 again would have two fluorine environments, but now these would be present in a 1 to 4 ratio. This can be seen by carrying out the various operations of the C4v point group. Carrying out a reflection operation leads to exchange of equatorial fluorine atoms, as does carrying out a fourfold rotation operation. None of the operations of the C4v point group leads to exchange of the axial fluorine with any of the equatorial fluorine atoms. The two other possible structures we can conceive of for PF5 , namely the D5h and the C5v geometries, share a common feature: they both have only a single fluorine environment. If we had access to some kind of spectroscopy that could tell us how many fluorine environments are actually present, then we could potentially distinguish between the above possibilities with the exception of the D5h–C5v case. Fortunately, we do have access to a type of spectroscopy that will tell us how many fluorine environments there are! The spectroscopy we generally use for this is called nuclear magnetic resonance, or NMR, spectroscopy. This takes advantage of the fact that when atomic nuclei are in different environments, they will resonate at different frequencies when placed inside a strong magnetic field and. The most common applications of NMR spectroscopy are two problems involving nuclei possessing a magnetic dipole moment. Examples of such nuclei are 1H and 13C. The former conveniently has a high natural abundance, whereas the latter has a low natural abundance. Both fluorine and phosphorus have a high natural abundance isotope that is very useful for NMR spectroscopy: 31P and 19F are the nuclei in question. Therefore, we ought to be able to just take the 19F NMR spectrum of PF5 to find out the number of fluorine environments. In most cases, this approach works very well to determine the number of environments that are present for a particular type of NMR active nucleus in the molecule. However, it turns out that in the case of PF5 , and other molecules in this category termed “fluxional”, this does not work. The reason it does not work is that NMR is a relatively slow type of spectroscopy. Unlike in the case of an infrared spectrum, wherein the atomic nuclei are effectively frozen on the time scale of measurement, with NMR spectroscopy is sometimes possible for the atoms in a molecular structure to rearrange themselves rapidly relative to the time it takes to acquire the measurement; in this case the environments become averaged. So indeed, although we do find only a single 19F NMR signal for PF5 at room temperature, the infrared spectrum is consistent only with the D3h structure. Now the question becomes, what is the mechanism by which axial and equatorial fluorine atoms can exchange with one another so rapidly? The accepted explanation is called the Berry pseudorotation, or BPR.