A) \[C{{H}_{3}}OH<{{H}_{2}}O~<O{{H}^{-}}~<C{{H}_{3}}{{O}^{-}}\]
B) \[{{H}_{2}}O~<C{{H}_{3}}OH<C{{H}_{3}}{{O}^{-}}<O{{H}^{-}}\]
C) \[{{H}_{2}}O~<C{{H}_{3}}OH<O{{H}^{-}}<C{{H}_{3}}{{O}^{-}}\]
D) \[C{{H}_{3}}OH<{{H}_{2}}O~<C{{H}_{3}}{{O}^{-}}<O{{H}^{-}}\]
Correct Answer: C
Solution :
Basicity depends upon the availability of lone pair of electrons for donation. Between\[{{H}_{2}}O\]and\[C{{H}_{3}}OH,\]the later is more basic because of the presence of electron donating\[-C{{H}_{3}}\]group. Thus, \[C{{H}_{3}}OH>HaO\] (basicity order) Conjugate base of weaker acid is strong. Thus, \[C{{H}_{3}}{{O}^{-}}>O{{H}^{-}}\](basicity order) Hence, the increasing order of basicity is \[C{{H}_{3}}{{O}^{-}}>O{{H}^{-}}>C{{H}_{3}}OH>{{H}_{2}}O\]
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