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I would guess, $H_2Te$ $>$ $H_2Se$ $>$ $H_2S$ $"<<"$ $H_2O$ 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), $H_2Te, -2.2$ $""^@C$, $H_2Se, -41.3$ $""^@C$, and $H_2S$, $-60$ $""^@C$. Clearly, the dominant intermolecular force is hydrogen , which is expressed most strongly in $"water"$. 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 $H_2S$, which is pretty nasty. The lower group hydrides, $H_2Te$, and $H_2Se$, 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 $H_2Te$, and $H_2Se$ were significantly more toxic than $HC-=N$. At least I didn't kill anyone (including myself).

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