Final answer:
Among the hydrogen chalcogenides H₂Se, H₂O, H₂S, and H₂Te, H₂S is expected to have the lowest boiling point due to its lower molecular weight and the weaker van der Waals forces compared to the others.
Step-by-step explanation:
The student asked which of the given hydrogen chalcogenides would have the lowest boiling point, considering the compounds H₂Se, H₂O, H₂S, and H₂Te. To determine the boiling points of different molecules, we consider both molecular weight and the type of intermolecular forces present in the substance. H₂O has the highest boiling point among the options due to hydrogen bonding, a strong intermolecular force.
H₂S, H₂Se, and H₂Te primarily experience van der Waals forces, with somewhat weaker dipole-dipole interactions for H₂S and H₂Se since they are polar molecules. As the molecular mass increases from H₂S to H₂Se to H₂Te, the van der Waals forces generally increase, leading to higher boiling points. Hence, H₂S, with the lowest molecular weight and the weakest intermolecular forces among these three, is expected to have the lowest boiling point.