Q.Boric acid, $\text{B(OH)}_3$, is described as a weak monobasic acid, yet it does not release a proton like a typical Brønsted acid. Explain how $\text{B(OH)}_3$ behaves as a Lewis acid in water, and write the equilibrium that produces the acidic solution.

[!TLDR] Boron in $\text{B(OH)}_3$ is electron-deficient (an empty $2p$ orbital) and acts as a Lewis acid, accepting $\text{OH}^-$ from water rather than donating a proton itself. [!ANSWER] $\text{B(OH)}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^{-} + \text{H}_3\text{O}^{+}$: boron accepts a hydroxide ion's electron pair …

Boron in orthoboric acid, $\text{B(OH)}_3$, is $sp^2$ hybridised and three-coordinate, surrounded by only six bonding electrons (three $\sigma$ bonds to the three $-\text{OH}$ groups) and no lone pair; this leaves an empty $2p$ orbital on boron, making the whole molecule electron-deficient and an effective Lewis acid — an electron-pair acceptor. When $\text{B(OH)}_3$ is dissolved in water, it does not ionise as a normal Brønsted acid would (by donating one of its own $-\text{OH}$ protons); instead, the electron-deficient boron atom accepts a lone pair of electrons from an incoming water molecule's oxygen, effectively capturing a hydroxide ion from that water molecule and releasing a proton from it. The overall equilibrium is:\n$\text{B(OH)}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^{-} + \text{H}_3\text{O}^{+}$\nIn the product, boron is now four-coordinate, $sp^3$ hybridised, in the tetrahydroxyborate ion $[\text{B(OH)}_4]^{-}$, its octet now complete. Because a proton ($\text{H}_3\text{O}^+$) does appear in solution as a net result, the solution measurably behaves as a weak acid, but the mechanism is fundamentally different from a classical Brønsted acid: boric acid works by accepting a hydroxide ion (indirectly generating $\text{H}_3\text{O}^+$ from the water …

Identify boron's electron deficiency (empty 2p orbital, 6 electrons only) in B(OH)3, explain Lewis-acid behaviour as accepting OH- fr …

Writing a simple Brønsted ionisation $\text{B(OH)}_3 \to \text{B(OH)}_2\text{O}^- + \text{H}^+$, which misrepresents the mechanism; forgetting that the p …