# Question e6edb

Oct 22, 2016

$\text{15 lone pairs}$

#### Explanation:

Your strategy here will be to draw the Lewis structure of xenon oxytetrafluoride, ${\text{XeOF}}_{4}$, 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

${\text{8 e"^(-) + "6 e"^(-) + 4 xx "7 e"^(-) = "42 e}}^{-}$

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

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{no. of lone pairs" = "30 e"^(-)/2 = "15 lone pairs}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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.