Question

Arrange in decreasing order of basicity,

$H_{2}O,O{H}^{-},C{H}_{3}OH,C{H}_3{O}^{-}$

• A. $C{H}_{3}OH>C{H}_3{O}^{-}>H_{2}O>O{H}^{-}$
• B. $C{H}_3{O}^{-}>O{H}^{-}>H_{2}O>C{H}_{3}OH$
• C. $C{H}_3{O}^{-}>O{H}^{-}>C{H}_{3}OH>H_{2}O$
• D. $C{H}_{3}OH>C{H}_3{O}^{-}>O{H}^{-}>H_{2}O$

Answer 1

The correct option is C

$C{H}_3{O}^{-}>O{H}^{-}>C{H}_{3}OH>H_{2}O$

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 $C{H}_{3}OH$, 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 $C{H}_{3}OH$ 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 $O{H}^{-}$.

$More{e}^{-}density\Rightarrow e^{-}$ pair can be easily donated

Thus, the order will be

$C{H}_3{O}^{-}>O{H}^{-}>C{H}_{3}OH>H_{2}O$