How do you draw the Lewis structure of XeF₄?

XeF₄ has 36 valence electrons. Xenon forms 4 single bonds to fluorine, using 8 electrons; 28 remain. Each F gets 3 lone pairs (24 electrons). 4 electrons remain as 2 lone pairs on Xe. Total on Xe: 12 electrons.

Lewis Structure of XeF₄ (Xenon Tetrafluoride)

What is the Lewis structure of XeF₄?

XeF₄ has 36 valence electrons, 4 bonding electron pairs, and 14 lone pairs. XeF₄ has 36 valence electrons. Eight are in four Xe-F bonds, 4 form 2 lone pairs on xenon, and 24 form 3 lone pairs on each fluorine. Xenon has 12 electrons around it (expanded octet). The 2 lone pairs occupy axial positions, giving a square planar molecular geometry.

Lewis structure of XeF₄ (Xenon Tetrafluoride)

Formula	XeF₄
Name	Xenon Tetrafluoride
Total Valence Electrons	36
Lone Pairs	14
Has Resonance	No
Concept	Expanded Octet
Difficulty	Advanced

Step 1: Count Valence Electrons

Xe has 8, each F has 7: 8 + 4(7) = 36 total.

Step 2: Draw the Skeleton

Xenon forms 4 single bonds to fluorine, using 8 electrons; 28 remain.

Step 3: Distribute Lone Pairs

Each F gets 3 lone pairs (24 electrons). 4 electrons remain as 2 lone pairs on Xe. Total on Xe: 12 electrons.

Step 4: Form Multiple Bonds

No multiple bonds. Xe has 4 bonds + 2 lone pairs = 12 electrons around it.

Step 5: Check Formal Charges

Xe: FC = 8 - 4 - ½(8) = 0
F: FC = 7 - 6 - ½(2) = 0

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