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, p. 445).  As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (reference), but the lone pair occupies an axial position.



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 $\ce{HF}$).