A) \[{{H}_{2}}O\]
B) \[{{H}_{2}}S\]
C) \[{{H}_{2}}Se\]
D) \[{{H}_{2}}Te\]
Correct Answer: A
Solution :
\[\underset{\text{(aq)}}{\mathop{2NaOH}}\,+\underset{\text{(}g\text{)}}{\mathop{C{{l}_{2}}}}\,+\underset{(g)}{\mathop{{{H}_{2}}}}\,\] consist of highest boiling point than other hydride (Due to presence of the hydrogen bonding).
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