Step 1 — Electronic Structure of Boron: Boron has atomic number 5, configuration 1$s^{2}$ 2$s^{2}$ 2$p^{1}$. In compounds, it typically uses $sp^{2}$ hybridization, contributing 3 electrons to 3 bonds.

Step 2 — Electron Count in $H_{3}BO_{3}$: In $H_{3}BO_{3}$, boron forms 3 bonds to oxygen ($sp^{2}$ hybridized). This gives B only 6 electrons around it (3 bonding pairs × 2 electrons = 6), NOT 8. Boron is therefore electron-deficient — it lacks a complete octet.

Step 3 — What Does an Electron-Deficient Atom Want? An electron-deficient atom is an electron-pair acceptor by definition. This is precisely the definition of a Lewis acid.

Step 4 — How Does $H_{3}BO_{3}$ React with Water? Water ($H_{2}O$) has lone pairs on oxygen. When $H_{3}BO_{3}$ encounters water, the electron-deficient B accepts a lone pair from one O of $H_{2}O$ (acting as Lewis base), forming a new B-O coordinate bond: $H_{3}BO_{3}$ + $H_{2}O$ → [B(OH)_{4}]^{-} + $H^{+}$

Step 5 — Why Is the Proton Released? When B accepts $OH^{-}$ from water, the water molecule loses $OH^{-}$ and releases $H^{+}$ into solution. The $H^{+}$ comes from WATER, not from $H_{3}BO_{3}$.

Step 6 — Why Monobasic, Not Tribasic? Despite three OH groups in $H_{3}BO_{3}$, only ONE $H^{+}$ is released per molecule (only one B-OH addition occurs per B centre). Boron becomes tetrahedral [B(OH)_{4}]^{-} after accepting one $OH^{-}$ — it is now satisfied (8 electrons) and no further reaction occurs.

Step 7 — Conclusion: $H_{3}BO_{3}$ is a LEWIS acid (electron-pair acceptor), NOT a Bronsted acid ($H^{+}$ donor). Its acidity arises from boron's electron deficiency, not from O-H bond polarity. It is monobasic because only one $H^{+}$ is produced per molecule.