BF_3 on hydrolysis forms: | vedclass

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Question

(C) The hydrolysis of $BF_3$ occurs in two steps:$1$. $BF_3$ reacts with water to form boric acid $(H_3BO_3)$ and hydrogen fluoride $(HF)$:$BF_3 + 3 H

Vedclass · Accepted Answer

Both $(A)$ and $(B)$

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(C) The hydrolysis of $BF_3$ occurs in two steps:$1$. $BF_3$ reacts with water to form boric acid $(H_3BO_3)$ and hydrogen fluoride $(HF)$:$BF_3 + 3 H_2O \rightarrow B(OH)_3 + 3 HF$$2$. The $HF$ produced reacts with excess $BF_3$ to form fluoroboric acid $(HBF_4)$:$3 HF + 3 BF_3 \rightarrow 3 H[BF_4]$Combining these,the overall reaction is:$4 BF_3 + 3 H_2O \rightarrow B(OH)_3 + 3 H[BF_4]$Thus,both $H_3BO_3$ and $HBF_4$ are formed.

$BF_3$ on hydrolysis forms:

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