London dispersion forces.
Xenon tetrafluoride is "XeF"_4XeF4, a covalent compound with a square planar geometry. When you draw out the Lewis structure using the electron-counting method:
- "Xe"Xe =>⇒ "8 valence"8 valence
- 4xx"F"4×F =>⇒ 4xx"7 valence"4×7 valence
With 3636 valence electrons, and fluorine tending to make single bonds, we have bb88 bonding valence electrons surrounding "Xe"Xe for the four single bonds total, and the bb(4xx6) nonbonding valence electrons surrounding the fluorine atoms total.
That leaves 4 to be distributed onto xenon, which symmetrically distribute themselves to minimize repulsions.
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Since "XeF"_4 is symmetrical, any polarities in its bonds cancel out, leaving it nonpolar. Therefore, it has london dispersion, and that's pretty much it.