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Although S has a higher electronegativity leading to stronger dipole-dipole interactions, Se has a larger atomic mass, resulting in stronger London dispersion forces which are more stronger than the dipole-dipole interactions. Thus, H2Se has a higher boiling point.
You are also already told initially that H2Se has a higher boiling point than H2S, so you don't have to worry about determining which one has the higher boiling point. All you have to do is explain why H2Se is the one with the higher boiling point, and in this case the only possible way for H2Se to have the higher boiling point is if its London dispersion forces are stronger than the dipole-dipole interactions of H2S.
Both species have 2 hydrogens, so we want to focus on the S and Se. Se will have higher polarizability because it is the bigger atom. Thus the instantaneous dipoles (London Dispersion Forces) formed between several H2Se molecules will be greater in strength than in H2S. We can conclude that H2Se has a higher boiling point.
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