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XeF4

Xe has 8 valance electrons.

F has 7 valance electrons * 4 = 28 valance electrons

8 + 28 = 36 valance electron total.

Now, there are 4 bonds between Xe and the 4 F's, so that is a total of 8 electrons shared.

36 - 8 = 28 valance electrons left over.

That means that 6 each go around the fluorine atoms as three lone pair per atom and one electron for the exon atom, unless this is a charged molecule.

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12y ago

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Q: How do you count lone pair in XeF4?
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