Lewis structure of XeO₄
32 valence electrons. Four Xe=O double bonds, with no lone pair on Xe, so Xe has 16 electrons around it (an expanded octet).
Lewis structure
32 valence electronsFormal charges: none
Why this shape
- Xe has 8 valence electrons. It uses eight in four double bonds to O, so none are left over.
- Steric number = four sigma bonds + no lone pairs = 4, so Xe is sp³ hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
- Four electron pairs point to the corners of a tetrahedron (tetrahedral electron geometry). With no lone pairs the atoms take the same shape: tetrahedral.
- All the bond pairs repel each other equally, so the angle is the ideal 109.5°.
- The bond dipoles cancel in the symmetric tetrahedral shape, so XeO₄ is non-polar.
How to draw it
- Count the valence electrons: Xe 8 + 4 × O 6 = 32 valence electrons, 16 pairs.
- Put Xe in the middle and join it to the four outer atoms with single bonds.
- Give each outer atom lone pairs until it has an octet (H needs only two electrons), then put any electrons left over on the central atom.
- Where an atom is still short of an octet, turn a lone pair of its neighbour into a shared pair: that makes a double or triple bond.
- Result: four Xe=O double bonds, with no lone pair on Xe, so Xe has 16 electrons around it (an expanded octet).
- Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Lewis structure of XeO4: quick answers
How many valence electrons are in XeO4?
Xe 8 + 4 × O 6 = 32 valence electrons, 16 pairs.
What are the formal charges in the Lewis structure of XeO4?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of XeO4?
8 in all, none on the central Xe.