Question

Arrive at the Lewis structure of XeF4

Answer 1

Step-1: Adding the valance electrons of Xenon and Fluoride

Xenon: 8

Fluoride: 7×4 = 28

So, total: 8+28 = 36

Step-2: Adding the octet electrons of Xenon and Fluoride

Xenon: 12

Fluoride: 8×4= 32

So, total: 12+32 = 44

Step-3: Finding the bonding electrons by subtracting valance electrons from octet electrons

44-36 = 8 electrons

Step-4: Now finding the pairs by dividing the bonding electrons by 2

8/2= 4 pairs

Step-5: Now finding the non-bonding electrons and pairs by subtracting bonding electrons from valence electrons

36-8 = 28 nonbonding electrons

28/2 = 14 nonbonding pairs

Now, use these details to draw the Lewis Structure of XeF4 (Xenon Tetrafluoride):

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We can also use the dot method to draw the Lewis Structure of Xenon Tetrafluoride with the help of the total 36 valence electrons.


Xenon is in the center, and four Fluoride atoms are in the all four directions of it. Each atom on the side gets 8 electrons to has the state of an octet.

Answer 2

Answer:

Step-1: Adding the valance electrons of Xenon and Fluoride

Xenon: 8

Fluoride: 7×4 = 28

So, total: 8+28 = 36

Step-2: Adding the octet electrons of Xenon and Fluoride

Xenon: 12

Fluoride: 8×4= 32

So, total: 12+32 = 44

Step-3: Finding the bonding electrons by subtracting valance electrons from octet electrons

44-36 = 8 electrons

Step-4: Now finding the pairs by dividing the bonding electrons by 2

8/2= 4 pairs

Step-5: Now finding the non-bonding electrons and pairs by subtracting bonding electrons from valence electrons

36-8 = 28 nonbonding electrons

28/2 = 14 nonbonding pairs

Now, use these details to draw the Lewis Structure of XeF4 (Xenon Tetrafluoride):

We can also use the dot method to draw the Lewis Structure of Xenon Tetrafluoride with the help of the total 36 valence electrons.

Xenon is in the center, and four Fluoride atoms are in the all four directions of it. Each atom on the side gets 8 electrons to has the state of an octet.