How to Draw the Lewis Structure for XeO4

To draw the Lewis structure for XeO₄ (Xenon tetroxide), follow these steps:

1. Determine the total number of valence electrons:
   • Xenon (Xe) is in Group 18, so it has 8 valence electrons.
   • Oxygen (O) is in Group 16, so each oxygen atom has 6 valence electrons. Since there are 4 oxygen atoms, the total number of valence electrons from oxygen is 4 × 6 = 24.
   • Add the valence electrons from xenon and oxygen: 8 (Xe) + 24 (O) = 32 valence electrons.
2. Place the least electronegative atom in the center:
   • Xenon is less electronegative than oxygen, so it will be the central atom.
3. Connect the central atom to the surrounding atoms with single bonds:
   • Draw single bonds from xenon to each of the four oxygen atoms. This uses 4 × 2 = 8 electrons.
4. Distribute the remaining electrons:
   • Subtract the electrons used in the bonds from the total valence electrons: 32 – 8 = 24 electrons.
   • Place the remaining electrons around the oxygen atoms to satisfy the octet rule. Each oxygen atom needs 6 more electrons to complete its octet, so 4 × 6 = 24 electrons are used.
5. Check the octet rule for the central atom:
   • Xenon has 8 electrons around it (4 from the single bonds), satisfying the octet rule.
6. Finalize the Lewis structure:
   • Xenon is surrounded by four oxygen atoms, each with a single bond and three lone pairs of electrons.

The final Lewis structure for XeO₄ shows xenon in the center with four single bonds to oxygen atoms, and each oxygen atom has three lone pairs of electrons.