# Draw the Lewis structure of #"HXeSH"# and describe the electron and molecular geometry around the central atom according to VSEPR theory?

##### 1 Answer

I don't know why you have that image, but anyways,

#"Xe"# is a noble gas which contributes#8# valence electrons to the lewis structure.#"S"# is group 16, which contributes#6# electrons to the lewis structure.- Each
#"H"# contributes#1# electron.

So, we have a total of **you already have the lewis structure**.

#"H"-"Xe"-"S"-"H"#

is the skeleton structure. We've accounted for **four** more valence electrons onto sulfur and **four** more onto xenon, accounting for a total of

Since it wouldn't make sense for sulfur to get the remaining

Thus, the structure looks like this:

Since **five** electron groups around it, if we treat it as a central atom, it has a **trigonal bipyramidal** *electron geometry*, and a **linear** *molecular geometry* because it only has two bonding groups.

Since **four** electron groups around it, if we treat it as a central atom, it has a **tetrahedral** *electron geometry*, and a **bent** *molecular geometry* because it only has two bonding groups.

As for the formal charges, we have:

#"Xe"# owns#8# electrons (three lone pairs and one from each single bond) and needs#8# , thus its formal charge is#8 - 8 = 0# .#"S"# owns#6# electrons (two lone pairs and one from each single bond) and needs#6# , thus its formal charge is#8 - 8 = 0# .#"H"# owns#1# electron (from its single bond) and needs#1# , thus its formal charge is#1 - 1 = 0# .

*Therefore, all formal charges have been minimized.*