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Question

The correct order of boiling point is:

A
H2O<H2S<H2Se<H2Te
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B
H2O>H2Se>H2Te>H2S
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C
H2O>H2S>H2Se>H2Te
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D
H2O>H2Te>H2Se>H2S
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Solution

The correct option is D H2O>H2Te>H2Se>H2S
Generally large molecules have more electrons and nuclei that create van der Waal's attractive force. So, the heavier compounds generally have higher boiling points than similar compounds made up of smaller molecules. But here in H2O, due to the high electronegativity of O atoms, intermolecular H bonding is present. Due to the presence of H bonding H2O has the highest boiling point.
Among H2S, H2Se and H2Te the correct order is H2Te>H2Se>H2S, because on moving down the group the size of the atoms increases.
Hence the correct order is:
H2O>H2Te>H2Se>H2S

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