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How do you count lone pair in XeF4? - Answers

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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How do you count lone pair in XeF4? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_count_lone_pair_in_XeF4

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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https://math.answers.com/math-and-arithmetic/How_do_you_count_lone_pair_in_XeF4

How do you count lone pair in XeF4? - Answers

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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      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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