Draw the Lewis structure of HXeSH and describe the electron and molecular geometry around the central atom according to VSEPR theory?
Answer:
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 \(\mathrm{H}_{2} \mathrm{XeS},\) but it wasn’t. That means this is the structural formula, and so you already have the lewis structure.
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 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.