Arrange in decreasing order of basicity,

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

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

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

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

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

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Solution

The correct option is C
$C H_{3} O^{-} > O H^{-} > C H_{3} O H > H_{2} O$

One of the definitions of basicity is availability of electron pair and how easy is it to donate it to the proton.

In $H_{2} O$ and $C H_{3} O H$ , there are electron pairs on oxygen.

But they are complete molecules, and hence quite stable.

In organic chemistry, you'll learn that $C H_{3} -$ has an electron donating effect. Thus, O in $C H_{3} O H$ has more electron density as compared to that of $H_{2} O$ .

Using the same logic, we can see that there is more electron density on the Oxygen atom of $C H_{3} O^{-}$ than that of the Oxygen atom in $O H^{-}$ .

$M o r e e^{-} d e n s i t y \Rightarrow e^{-}$ pair can be easily donated

Thus, the order will be

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

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Similar questions

Arrange in decreasing order of basicity,

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

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

C H_{3} N H_{2}

C_{6} H_{5} C H_{2} N H_{2}

p - C H_{3} O - C_{6} H_{4} - N H_{2}

m - C H_{3} O - C_{6} H_{4} - N H_{2}

C H_{3} C | | O - O^{-}

C H_{3} O^{-}

C N^{-}

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