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  • Correct order of boiling point of group 16 hydrides:Option: 1 <img alt="\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{S

Correct order of boiling point of group 16 hydrides:

Option: 1

\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{Te}


Option: 2

\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}>\mathrm{H}_2 \mathrm{O}


Option: 3

\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{O}


Option: 4

\mathrm{H}_2 \mathrm{O}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}


Answers (1)

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\mathrm{H}_2 \mathrm{O} has maximum boiling point because it exhibits hydrogen bonding. On moving down the group size of atom increases and hence magnitude of van der Waals forces increases. Therefore, the correct order of boiling point is \mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{O}.

Among hydrides of group 16 elements, with the increase in molecular weight, van der Waal forces increase so boiling point also increases. However,\mathrm{H}_2 \mathrm{O}unexpectedly has highest boiling point due to presence of hydrogen bonding.

The rest of the order can be explained on the account of increasing strength of van der Waals forces due to increasing size & molar masses.

 

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