A) \[{{H}_{3}}{{O}^{+}}<NH_{4}^{+}<HF<O{{H}^{-}}<{{H}_{2}}O\]
B) \[NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}<{{H}_{2}}O<O{{H}^{-}}\]
C) \[O{{H}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\]
D) \[{{H}_{3}}{{O}^{+}}>HF>{{H}_{2}}O>NH_{4}^{+}>O{{H}^{-}}\]
Correct Answer: C
Solution :
The order of acidic nature of \[NH_{4}^{+},\,{{H}_{2}}O,{{H}_{3}}{{O}^{+}},HF\] and \[O{{H}^{-}}\] is as \[O{{H}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\] \[O{{H}^{-}}\] is basic (ie, it tends to gain a proton). So, it is least acidic. \[{{H}_{2}}O\] is a neutral species. HF is a weak acid because the bond dissociation energy of HF is very high. \[{{H}_{3}}{{O}^{+}}\] is most acidic because it readily loose proton. Note - Presence of negative charge on the atom holding the electron pair increases the basicity. While presence of positive charge on the atom; holding the electron pair deceases the basicity (or Increases acidity).
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