# How to predict the 31P NMR spectra?

What can I predict about the $$\ce{^{31}P}$$ spectra of $$\ce{TiP2O7}$$?

To get the required information about the crystal structure of the compound

Here is a screenshot from the ICSD website: Since there is only one oxidation state (+5) for $$\ce{P}$$, there will be only one peak. And from the Wyckoff symbol, 4c, the intensity of the signal is 4, i.e. 4 $$\ce{P}$$ atoms in the unit cell.

I can't go any further to complete the task and extract all the relevant information. I'm using SIMPSON software to simulate the spectra. Can somebody walk me step by step through this?

• No, you cannot derive that there is only one peak because phosphorus only has one oxidation state. If there are four atoms in a unit cell, there can be up to four peaks of phosphorous out there, depending on whether they translate to each other or not. – Jan Sep 21 '15 at 22:31

If the question you were asked to answer was literally "what can you predict about the $\ce{^31P}$ NMR spectrum of $\ce{TiP2O7}$ (from the crystal structure), then I believe you have vastly overcomplicated the problem.
This is a question about solid-state NMR (as $\ce{TiP2O7}$ is a solid).
From the screenshot, it says that "formula units per cell" = 4. As the formula is $\ce{TiP2O7}$, this indicates there are 4*2 = 8 different phosphorus atoms in the unit cell.
So the answer is: You expect to observe 8 phosphorus signals in the $\ce{^31P}$ NMR spectrum.