What does the 129Xe NMR spectrum of XeOF4 look like?

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?




The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, **quintuplet** of peaks ([reference](https://books.google.com/books?id=tUmcAQAAQBAJ&pg=PA445&lpg=PA445&dq=xeof4%20nmr&source=bl&ots=i19TFzAU6f&sig=WJEgg7V-64lsYqhPbvdcFjVQqYI&hl=en&sa=X&ei=h3VjVaeKA4mLsAW5-YCQAw&ved=0CEYQ6AEwBg#v=onepage&q=xeof4%20nmr&f=false), p. 445). As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry ([reference](http://en.wikipedia.org/wiki/Xenon_oxytetrafluoride)), but the lone pair occupies an axial position.


![enter image description here](https://i.stack.imgur.com/vOE4B.png)


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](http://chem.ch.huji.ac.il/nmr/techniques/1d/row5/xe.html#ref) 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 $\ce{HF}$).


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