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Lewis structure

Lewis structure of XeOF₄

42 valence electrons. One Xe=O double bond and four Xe-F single bonds, with one lone pair on Xe, so Xe has 14 electrons around it (an expanded octet).

FFXeOFF
  • Xe
  • O
  • F
  • lone pair

XeOF₄Xenon oxytetrafluoride

Square pyramidal

AB₅E: 5 bond pairs, 1 lone pair

  • Hybridisationsp³d²
  • Bond angleabout 90°
  • Polaritypolar

Lewis structure

42 valence electrons
XeOFFFF

Formal charges: none

Why this shape

  1. Xe has 8 valence electrons. It uses six in one double bond to O and four single bonds to F, so two are left as one lone pair.
  2. Steric number = five sigma bonds + one lone pair = 6, so Xe is sp³d² hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
  3. Six electron pairs point to the corners of an octahedron (octahedral electron geometry). With one lone pair in place of atoms, the atoms make a square pyramidal shape.
  4. The lone pair takes any one of the six equal positions, and the five bonds form a square pyramid.
  5. The O sits on top, opposite the lone pair, and the four F atoms make the square. The angles are close to 90 degrees.
  6. The bond dipoles do not cancel in the square pyramidal shape, so XeOF₄ is polar.
Working

How to draw it

  1. Count the valence electrons: Xe 8 + O 6 + 4 × F 7 = 42 valence electrons, 21 pairs.
  2. Put Xe in the middle and join it to the five outer atoms with single bonds.
  3. 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.
  4. 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.
  5. Result: one Xe=O double bond and four Xe-F single bonds, with one lone pair on Xe, so Xe has 14 electrons around it (an expanded octet).
  6. Formal charge = valence electrons − lone pair electrons − number of bonds. Here: none.

Class 11, Chapter 4 Chemical Bonding and Molecular Structure.

Questions

Lewis structure of XeOF4: quick answers

How many valence electrons are in XeOF4?

Xe 8 + O 6 + 4 × F 7 = 42 valence electrons, 21 pairs.

What are the formal charges in the Lewis structure of XeOF4?

Every atom has a formal charge of zero.

How many lone pairs are in the Lewis structure of XeOF4?

15 in all, 1 of them on the central Xe.

Try any molecule