The correct order of increasing basicity of the following is $H_{2} O, O H^{-}, C H_{3} O H, C H_{3} O^{-}$ :

C H_{3} O H < H_{2} O < O H^{-} < C H_{3} O^{-}

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H_{2} O < C H_{3} O H < C H_{3} O^{-} < O H^{-}

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H_{2} O < C H_{3} O H < O H^{-} < C H_{3} O^{-}

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C H_{3} O H < H_{2} O < C H_{3} O^{-} < O H^{-}

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Solution

The correct option is C $H_{2} O < C H_{3} O H < O H^{-} < C H_{3} O^{-}$
Basicity depends upon the availability of lone pair of electrons for donation. Between $H_{2} O$ and $C H_{3} O H$ , the later is more basic because of the presence of electron donating $- C H_{3}$ group.
Thus, $C H_{3} O H > H_{2} O$ (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
$C H_{3} O^{-} > O H^{-} > C H_{3} O H > H_{2} O$ .

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Arrange in increasing order of basicity,

$H_{2} O, O H^{-}, C H_{3} O H, C H_{3} O^{-}$

Arrange in decreasing order of basicity,

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