Question

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

Answer 1

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.