Acidic Behavior of Orthoboric Acid in Water

To determine the correct option, let us analyze both statements:

1. Analysis of STATEMENT-1:
   Orthoboric acid, represented as $\text{H}_3\text{BO}_3$ or $\text{B(OH)}_3$, has a central boron atom surrounded by six valence electrons, which leaves its octet incomplete. In an aqueous solution, it acts as a Lewis acid by accepting a lone pair of electrons from the oxygen atom of a hydroxide ion ($\text{OH}^-$) derived from a water molecule:
   $\text{B(OH)}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^- + \text{H}_3\text{O}^+$
   Because each mole of $\text{B(OH)}_3$ results in the liberation of one mole of hydronium ions ($\text{H}_3\text{O}^+$), and the equilibrium constant is small ($K_a \approx 5.8 \times 10^{-10}$), it behaves as a weak monobasic acid.
   Therefore, STATEMENT-1 is True.

2. Analysis of STATEMENT-2:
   A Brønsted-Lowry acid acts directly as a proton donor by ionizing to release $\text{H}^+$ from its own molecular structure. Orthoboric acid does not donate its own protons; instead, it behaves as a Lewis acid by accepting an $\text{OH}^-$ ion from water, causing the water molecule to release a proton ($\text{H}^+$).
   Therefore, orthoboric acid acts as a Lewis acid (electron-pair acceptor) rather than a proton donor.
   Therefore, STATEMENT-2 is False.

Since Statement-1 is True and Statement-2 is False, the correct choice is (C).