XeF4 (Xenon tetrafluoride) Lewis Structure

In XeF4 (Xenon tetrafluoride) lewis structure, there are four sigma bonds and two lone pairs around xenon atom. Each fluorine atom has three lone pairs. In this tutorial, we will learn how to draw lewis structure of XeF4 step by step.

Lewis structure of XeF4

xenon tetrafluoride XeF4 lewis structure

Xenon atom is the center atom and each fluorine atom has made a single bond with xenon atom. There are two lone pairs on xenon atom. This is a rare example of a noble gas forming a chemical compound.

Steps of drawing lewis structure of XeF4

There are several steps to draw the lewis structure of XeF4. Each step is explained in detail in next sections. If you are a beginner to lewis structure drawing, follow these sections slowly and properly to understand it's method completely. Look the figures to understand each step.

  1. Find total number of electrons of the valance shells of xenon and fluorine atoms
  2. Total electrons pairs
  3. Center atom selection
  4. Put lone pairs on atoms
  5. Check the stability and minimize charges on atoms by converting lone pairs to bonds until most stable structure is obtained.

Find total number of electrons of the valance shells of XeF4

There are only two elements in XeF4; xenon and fluorine. Xenon is a group IA element and has 8 electrons in its last shell (valence shell). Fluorine is a group VIIA element in the periodic table and contains 7 electrons in their last shell. Now we know how many electrons includes in valence shells of xenon and fluorine atom.

  • valence electrons given by fluorine atoms = 7 * 4 = 28
  • valence electrons given by xenon atom = 8 * 1 = 8

  • Total valence electrons = 28 + 8 = 36

Determine total valence electrons pairs

Total valance electrons pairs = σ bonds + π bonds + lone pairs at valence shells

Total electron pairs are determined by dividing the number total valence electrons by two. For, XeF4, total pairs of electrons are 18 (=36/2) in their valence shells.

Selection of center atom and sketch of XeF4 molecule

To be the center atom, ability of having greater valance and being a electropositive element are important facts. However, from our experience we know that there is a very low possibility that fluorine atom cannot be a center atom because fluorine's maximum valence is 1 (fluorine cannot make two or more bonds).

Therefore, xenon becomes the center atom and each fluorine atom is joint with xenon atom. The basic sketch is given below.

XeF4 sketch

Mark lone pairs on xenon and fluorine atoms

After deciding the center atom and basic sketch of XeF4, we can start to mark lone pairs on atoms. Remember that, there are total of 18 electron pairs to mark on atoms as lone pairs and bonds.

  • There are already 4 sigma bonds in the above drawn basic sketch. Now only 14 (18-4) electron pairs are remaining to mark on atoms.
  • Usually as a theory, those remaining electron pairs should be first marked on outside atoms. In XeF4, fluorine atoms are the outside atoms. Each fluorine atom will take three lone pairs. Then all four fluorine atoms will take 12 lone pairs.
  • Now, only 2 (14-12) lone pairs are remaining. Therefore, then mark those two electrons pairs on xenon atom. Then all remained lone pairs are finished and there are no more lone pairs to mark.
mark lone pairs on xenon and fluorine atoms in XeF4

Mark charges on atoms and check the stability and minimize charges on atoms by converting lone pairs to bonds<

There are no charges on atoms in above sketch and do not need to worry about reducing charges to obtain best stable structure. That means, we have already got the lewis structure of XeF4.


Lewis Structures of Molecules