A) \[{{H}_{2}}S\]
B) \[N{{H}_{3}}\]
C) \[P{{H}_{3}}\]
D) \[{{H}_{2}}O\]
Correct Answer: B
Solution :
\[N{{H}_{3}}\]is more basic due to more tendency of N to donate its lone pair (having lower electronegativity) than\[{{H}_{2}}O\]. Hence, it has the highest proton affinity.
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