Lewis structure of XeF₄
36 valence electrons. Four Xe-F single bonds, with two lone pairs on Xe, so Xe has 12 electrons around it (an expanded octet).
Lewis structure
36 valence electronsFormal charges: none
Why this shape
- Xe has 8 valence electrons. It uses four in four single bonds to F, so four are left as two lone pairs.
- Steric number = four sigma bonds + two lone pairs = 6, so Xe is sp³d² hybridised.
- Six electron pairs point to the corners of an octahedron (octahedral electron geometry). With two lone pairs in place of atoms, the atoms make a square planar shape.
- The two lone pairs sit opposite each other, 180 degrees apart, so their repulsion is least and the four bonds lie in a square.
- The lone pairs balance each other, so the bonds are not pushed out of place and the angle stays at 90°.
- The bond dipoles cancel in the symmetric square planar shape, so XeF₄ is non-polar.
How to draw it
- Count the valence electrons: Xe 8 + 4 × F 7 = 36 valence electrons, 18 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.
- Result: four Xe-F single bonds, with two lone pairs on Xe, so Xe has 12 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 XeF4: quick answers
How many valence electrons are in XeF4?
Xe 8 + 4 × F 7 = 36 valence electrons, 18 pairs.
What are the formal charges in the Lewis structure of XeF4?
Every atom has a formal charge of zero.
How many lone pairs are in the Lewis structure of XeF4?
14 in all, 2 of them on the central Xe.