A) \[{{H}_{2}}O\]
B) \[{{H}_{2}}S\]
C) \[{{H}_{2}}Se\]
D) \[{{H}_{2}}Te\]
Correct Answer: B
Solution :
\[{{H}_{2}}O\] \[2NaOH+C{{l}_{2}}\,\to \,NaClO+NaCl+{{H}_{2}}O\] \[{{H}_{2}}Se\] \[{{H}_{2}}Te\] 373K 213K 269K 232K \[{{H}_{2}}S\] has lowest boiling point and \[{{H}_{2}}O\] has highest boiling point because if any compound has hydrogen bond. Its boiling point is high.
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