Question

What is the n-factor of $H_{3}B{O}_3$?

• A. 3
• B. 1 or 2 or 3
• C. 1
• D. 1 or 2

Answer 1

The correct option is C

1

In $H_{3}B{O}_3$, although all 3H are linked to oxygen, yet all 3H are not replaceable. Here, boron atom is electron deficient, so it acts as a lewis acid. When $H_{3}B{O}_3$ is added to water, then oxygen atom of $H_{2}O$ through its lone pair attack the Boron atom, as follows

Net reaction is $H_{3}B{O}_3$ + 2$H_2$O $\to$ ${\left[B\right(OH{)}_4]}^{-}$+ ${H_{3}O}^{+}$

^{So, we get only one $H^{+}$ per molecule of} $H_{3}B{O}_3$ ^{which its n-factor = 1}