Hydrolysis of Boron Halides

Reaction Equations:

$$\text{BF}_3 + 3\text{H}_2\text{O} \to \text{B(OH)}_3 + 3\text{HF}$$

$$4\text{BF}_3 + 3\text{H}_2\text{O} \to \text{B(OH)}_3 + 3\text{H}[\text{BF}_4] \quad \text{(Partial Hydrolysis due to } [\text{BF}_4]^- \text{ formation)}$$

$$\text{BCl}_3 + 3\text{H}_2\text{O} \to \text{B(OH)}_3 + 3\text{HCl} \quad \text{(Complete Hydrolysis)}$$

Concept: $$\text{BF}_3$$ does not undergo complete hydrolysis because the generated $$\text{HF}$$ reacts with unreacted $$\text{BF}_3$$ to yield the stable fluoroborate ion $$[\text{BF}_4]^-$$. Other halides ($$\text{BCl}_3, \text{BBr}_3, \text{BI}_3$$) undergo complete hydrolysis.