# What are the order of melting points for water, hydrogen sulfide, hydrogen telluride, and hydrogen selenide?

Apr 8, 2017

I would guess, ${H}_{2} T e$ $>$ ${H}_{2} S e$ $>$ ${H}_{2} S$ $\text{<<}$ ${H}_{2} O$ with respect to melting point.
I would guess, ${H}_{2} T e$ $>$ ${H}_{2} S e$ $>$ ${H}_{2} S$ $\text{<<}$ ${H}_{2} O$ with respect to melting point.
But as a chemist, as a physical scientist, you are required to interpret data, not remember them. As a matter of fact I do remember the normal boiling points (I will tell you why in a minute), ${H}_{2} T e , - 2.2$ ""^@C, ${H}_{2} S e , - 41.3$ ""^@C, and ${H}_{2} S$, $- 60$ ""^@C. Clearly, the dominant intermolecular force is hydrogen bonding, which is expressed most strongly in $\text{water}$. Boiling points of the lower hydrides follow the order we would expect for dispersion forces.
When I was much younger, I worked in an inorganic lab whose specialty was inorganic sulfides and selenides. Now you have probably gotten a whiff of ${H}_{2} S$, which is pretty nasty. The lower group hydrides, ${H}_{2} T e$, and ${H}_{2} S e$, are even worse: they smell like dead dogs. Very early in my career I disposed of some selenide residues without proper precautions (i.e. bleaching the residues out). I managed to clear an entire building of students and ancillary staff, and since I did the right thing and put my hand up and copped the blame, my name was mud for the rest of my time in that department. I also found out that ${H}_{2} T e$, and ${H}_{2} S e$ were significantly more toxic than $H C \equiv N$. At least I didn't kill anyone (including myself).