How many lone pairs of electrons are on the Xe atom in XeF_6? A) 0 B) 3 C) 2 D) 1

May 25, 2018

D) $1$ lone pair.

Explanation:

First, let's draw the Lewis Structure for $X e {F}_{6}$:

1. The total number of electrons in $X e {F}_{6}$ is $8 + \left(7 \times 6\right) = 50$ electrons.
2. $X e$ will be the central atom. So, let's just connect $X e$ and $F$ atoms with a single bond each.
3. We have $50 - 12 = 38$ electrons left to draw on.
This means that we have enough electrons to not need double bonds and complete the octet for $F$ using lone pairs.
4. However, we still have $38 - 36 = 2$ electrons left to use.
These electrons will need to be placed on $X e$, since $X e$, having its valence electrons in the $5 p$ orbital, have empty $5 d$, $4 f$, and $5 f$ orbitals—allowing it have more electrons than an octet.
So, there are $2$ electrons on $X e$—which is $1$ pair.