# How do you write the lewis structure for XeF_4?

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anor277 Share
Dec 23, 2017

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Well, there are 8 valence electrons around xenon.....

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...and 7 valence electrons around each fluorine atom..

VESPER predicts an $A {X}_{4} {E}_{2}$ structure, that is octahedral with respect to electronic geometry, and SQUARE planar with respect to molecular geometry. $\angle F - X e - F \equiv {90}^{\circ}$ to a first approx.

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