A) \[{{H}_{2}}Te>{{H}_{2}}O>{{H}_{2}}Se>{{H}_{2}}S\]
B) \[{{H}_{2}}O>{{H}_{2}}S>{{H}_{2}}Se>{{H}_{2}}Te\]
C) \[{{H}_{2}}Te>{{H}_{2}}Se>{{H}_{2}}S>{{H}_{2}}O\]
D) \[{{H}_{2}}O>{{H}_{2}}Te>{{H}_{2}}Se>{{H}_{2}}S\]
Correct Answer: D
Solution :
The correct order of boiling point of \[{{H}_{2}}O,\,\,{{H}_{2}}S,\,\,{{H}_{2}}Se\] and \[{{H}_{2}}Te\] is \[\underset{373\,\,K}{\mathop{{{H}_{2}}O}}\,>\underset{269\,\,K}{\mathop{{{H}_{2}}Te}}\,>\underset{232\,\,K}{\mathop{{{H}_{2}}Se}}\,>\underset{213\,\,K}{\mathop{{{H}_{2}}S}}\,\] The high boiling point of water is due to the association of \[{{H}_{2}}O\] molecules through hydrogen bonding. However, H-bonding is not present in other hydrides. The intermolecular forces between the hydrides (except \[{{H}_{2}}O\]) are van der Waals forces. These forces increase with increase in molecular size and therefore, boiling points increase on moving from \[{{H}_{2}}S\] to \[{{H}_{2}}\]Te.
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