A) \[O{{H}^{-}}>NH_{2}^{-}>{{H}_{2}}O>N{{H}_{3}}\]
B) \[NH_{2}^{-}>O{{H}^{-}}>N{{H}_{3}}>{{H}_{2}}O\]
C) \[N{{H}_{3}}>{{H}_{2}}O>NH_{2}^{-}>O{{H}^{-}}\]
D) \[{{H}_{2}}O>N{{H}_{3}}>O{{H}^{-}}>NH_{2}^{-}\]
Correct Answer: B
Solution :
\[N{{H}_{3}}\]is more basic than\[{{H}_{2}}O,\]therefore,\[NH_{2}^{-}\]is a stronger base than\[O{{H}^{-}}\]. Thus, the decreasing order of basic strength is \[NH_{2}^{-}>O{{H}^{-}}>N{{H}_{3}}>{{H}_{2}}O\]
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