8
$\begingroup$

What will the $\ce{^129Xe}$ NMR spectrum of $\ce{XeOF4}$ look like?

I know that its electron pair geometry is octahedral, and the molecular pair geometry is square pyramidal.

I believe that the lone pair on $\ce{Xe}$ will take the equatorial position (less sterically crowded), alongside the oxygen and 2 other equatorial fluorines in the same plane, to minimize electron pair repulsion.

So there will be 2 equatorial $\ce{F}$ and 2 axial $\ce{F}$. Therefore, am I right to say that the $\ce{^129Xe}$ NMR will be a triplet of triplets?

| improve this question | |
$\endgroup$
8
$\begingroup$

The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, quintuplet of peaks (reference, p. 445). As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (reference), but the lone pair occupies an axial position.

enter image description here

Therefore, the 4 equatorial fluorines are equivalent and since they are spin 1/2 nuclei they will couple with the $\ce{^{129}Xe}$ nucleus (also spin 1/2) producing the observed quintuplet. $\ce{XeOF4}$ is an IUPAC suggested standard for $\ce{^{129}Xe}$ nmr spectroscopy (but xenon gas is more commonly used because $\ce{XeOF4}$ can decompose in the presence of moisture to produce $\ce{HF}$).

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hi, I am curious as to why the lone pair chooses to adopt an axial position. Doesn't this lead to the lone pair being at right angles to 4 other substituents vs being at the equatorial position? And also why is the oxygen at the axial position for the same reason? $\endgroup$ – kane9530 May 25 '15 at 20:22
  • 2
    $\begingroup$ You might want to ask that as a separate question, the answer requires more than a few sentences. If this answer was helpful please consider accepting and\or up-voting it, thanks. $\endgroup$ – ron May 25 '15 at 21:19

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.