Lewis structure of XeF₆
50 valence electrons. Six Xe-F single bonds, with one lone pair on Xe, so Xe has 14 electrons around it (an expanded octet).
Lewis structure
50 valence electronsFormal charges: none
Why this shape
- Xe has 8 valence electrons. It uses six in six single bonds to F, so two are left as one lone pair.
- Steric number = six sigma bonds + one lone pair = 7, so Xe is sp³d³ hybridised.
- Seven electron pairs point to the corners of a pentagonal bipyramid (pentagonal bipyramidal electron geometry). With one lone pair in place of atoms, the atoms make a distorted octahedral shape.
- The lone pair pushes the six bonds out of a perfect octahedron.
- Repulsion falls in the order lone pair to lone pair > lone pair to bond pair > bond pair to bond pair. The lone pair pushes the bond pairs closer, so the angle is no single angle.
- XeF₆ is slightly polar (distorted octahedral, the dipole is very small).
How to draw it
- Count the valence electrons: Xe 8 + 6 × F 7 = 50 valence electrons, 25 pairs.
- Put Xe in the middle and join it to the six 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.
- Result: six Xe-F single bonds, with one lone pair on Xe, so Xe has 14 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 XeF6: quick answers
How many valence electrons are in XeF6?
Xe 8 + 6 × F 7 = 50 valence electrons, 25 pairs.
What are the formal charges in the Lewis structure of XeF6?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of XeF6?
19 in all, 1 of them on the central Xe.