Lewis structure of XeF4

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Step method to draw lewis structure of Xenon tetrafluoride

Step method to draw lewis structure for XeF4( This molecules is an example of expanded  octet)

Step 1: Find valence e- for all atoms. Add them together.

Xe:8
F:7x4=28

Total=36

Step2: Find octet e- for each atom and add them together.

Xe=12
F:8x4=32

Total=44

Bromine gets 12 electrons in order to make 5 bonds with surrounding atoms.

Step3: Gives you bonding e-. Subtract step 1 total from step 2

44-36=8e-

Step 4: Find number of bonds by diving the number in step 3 by 2(because each bond is made of 2 e-)

8e-/2= 4 bond pairs

Step 5: Find the number of nonbonding (lone pairs) e-. Subtract step 3 number from step 1.

36-8= 28e-=14 lone pairs

Use information from step 4 and 5 to draw the lewis structure.


XeF4 lewis structure

Lewis dot structure of XeF4

Alternatively a dot method can be used to draw the lewis structure.
Calculate the total valence electrons in the molecule.
Xe:8
F:7x4=28

Total=36


Put xenon in the center and four fluorine atoms on the sides.
Put a pair of electrons connecting the side atom with central atom.Put remaining electrons on the side atoms.Make sure each side atom get 8 electrons to get octet state.

XeF4 lewis dot structure