# General methods of Analysis for FTIR Results

What can we say about a molecule from the position and number of bands in FTIR spectrum. I know that the more symmetric a molecule is, the less number of bands it has. What else can we say, for example about the position of the band? Does it show the frequency of vibrations and therefore the strength of the bonds? For example, the spectrum of MoF4 has one band at 674 cm -1. XeF4 has 3 bands at 586, 291, and 161. How can we compare the structure of these materials from the above information?

• I understand that this link has almost everything in detail for XeF4 : users.csbsju.edu/~frioux/group/XeF4.pdf ; My question is more on how to differentiate between the two compounds by checking the IR bands, since one may expect the structures to be close. The MoF4 is a bit shaky though, I am not sure if the compound exist in stable form. – Kinformationist Mar 12 '17 at 17:22
• IR is not a good spectroscopy for differentiation. What I mean is that, if you have two samples of which you know nothing at all, the experiment you should run is not IR. IR gives you some additional information once you already know a lot about your powder. However, in your case, knowing the formulas and the structures, you can expect MoF4 to show higher frequencies in comparison to XeF4 due to the mass effect I discussed shortly in my answer. – Anon Mar 13 '17 at 9:18

Let me start by saying that the XeF$_4$ is a standard example of group theory applied to IR, which is generally treated in Uni courses, for which you can find plenty of notes on the web. For these reasons, I will not explain it, but you can rather put a bit of effort and read it yourself.
$$\omega = \sqrt{\frac{k}{\mu}} \\ \mu = \frac{m_1 m_2}{m_1 + m_2}$$
where $k$ is the force constant of the bond, and $\mu$ is its reduced mass. So, what this tells you is that stronger bonds vibrate at a higher frequency. It also tells you that bonds involving lighter atoms vibrate at a higher frequency. Another thing that you can deduce is that bending frequencies are lower than stretching frequencies. So, to give you a hint, what you can say immediately by looking at those frequencies you reported is that the 586 cm$^{-1}$ vibration is a stretching.