A) \[{{H}_{3}}\overset{+}{\mathop{O}}\,<NH_{4}^{+}<HF<O{{H}^{-}}<{{H}_{2}}O\]
B) \[NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}<{{H}_{2}}O<O{{H}^{-}}\]
C) \[H{{O}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\]
D) \[{{H}_{3}}{{O}^{+}}>HF>{{H}_{2}}O>NH_{4}^{+}>O{{H}^{-}}\]
Correct Answer: C
Solution :
Among the given species, correct order of increasing acidic strength is \[O{{H}^{-}}<{{H}_{2}}O<NH_{4}^{+}<HF<{{H}_{3}}{{O}^{+}}\] \[O{{H}^{-}}\] is basic (L e., it tends to gain a proton) and hence, is least acidic. \[{{H}_{2}}O\] is neutral species. \[{{H}_{3}}{{O}^{+}}\] is most acidic as it readily lose proton.
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