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

1 Answer
Apr 5, 2016

I don't know why you have that image, but anyways, #"HXeSH"#...

  • #"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 #16# electrons to distribute. Note that the formula could have been written #"H"_2"XeS"#, but it wasn't. That means this is the structural formula, and so you already have the lewis structure.

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

is the skeleton structure. We've accounted for #6# valence electrons. We would be able to add four more valence electrons onto sulfur and four more onto xenon, accounting for a total of #\mathbf(14)# so far.

Since it wouldn't make sense for sulfur to get the remaining #2# electrons (xenon is larger and can hold more electrons around it more easily), we put the remaining #2# on #"Xe"#.

Thus, the structure looks like this:

Since #"Xe"# has 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 #"S"# has 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.