Question #e6edb

1 Answer
Oct 22, 2016

#"15 lone pairs"#


Your strategy here will be to draw the Lewis structure of xenon oxytetrafluoride, #"XeOF"_4#, and count the number of lone pairs of electrons present in a molecule.

Start by calculating the total number of valence electrons you have in xenon oxytetrafluoride

  • xenon contributes eight valence electrons
  • oxygen contributes six valence electrons
  • each of the four fluoride atoms contributes seven valence electrons

The total number of valence electrons will be

#"8 e"^(-) + "6 e"^(-) + 4 xx "7 e"^(-) = "42 e"^(-)#

Now, the xenon atom will be the central atom of the molecule. It will form a double bond with the oxygen atom and a single bond with each of the four fluorine atoms.

These bonds will account for

#overbrace(1 xx "4 e"^(-))^(color(red)("one double bond")) + overbrace(4 xx "2 e"^(-))^(color(blue)("four single bonds")) = "12 e"^(-)#

These electrons act as bonding electrons because they are used by the atoms to form covalent bonds.

The remaining number of electrons will be distributed as lone pairs

#overbrace("42 e"^(-))^(color(purple)("total no. of valence electrons")) - overbrace("12 e"^(-))^(color(brown)("bonding electrons")) = overbrace("30 e"^(-))^(color(darkgreen)("in lone pairs"))#

Since each lone pair is composed of two electrons, the total number of lone pairs present in a xenon oxytetrafluoride molecule will be

#color(green)(bar(ul(|color(white)(a/a)color(black)("no. of lone pairs" = "30 e"^(-)/2 = "15 lone pairs")color(white)(a/a)|)))#

Now, these lone pairs will be distributed as follows

  • three lone pairs on each of the four fluorine atoms
  • two lone pairs on the oxygen atom
  • one lone pair on the xenon atom

The Lewis structure of xenon oxytetrafluoride looks like this

As you can see, the #15# lone pairs of electrons that we came up with are confirmed by the structure.

Notice that xenon has a total of #14# valence electrons in this structure; that is the case because it can expand its octet and accommodate the extra #6# valence electrons.