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

Apr 5, 2016

I don't know why you have that image, but anyways, $\text{HXeSH}$...

• $\text{Xe}$ is a noble gas which contributes $8$ valence electrons to the lewis structure.
• $\text{S}$ is group 16, which contributes $6$ electrons to the lewis structure.
• Each $\text{H}$ contributes $1$ electron.

So, we have a total of $16$ electrons to distribute. Note that the formula could have been written $\text{H"_2"XeS}$, but it wasn't. That means this is the structural formula, and so you already have the lewis structure.

$\text{H"-"Xe"-"S"-"H}$

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 $\setminus m a t h b f \left(14\right)$ 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 $\text{Xe}$.

Thus, the structure looks like this: Since $\text{Xe}$ 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 $\text{S}$ 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:

• $\text{Xe}$ owns $8$ electrons (three lone pairs and one from each single bond) and needs $8$, thus its formal charge is $8 - 8 = 0$.
• $\text{S}$ owns $6$ electrons (two lone pairs and one from each single bond) and needs $6$, thus its formal charge is $8 - 8 = 0$.
• $\text{H}$ 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.