BCl₃ + H₂O longrightarrow Product, here product

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p-Block Elements - I

hard

$BCl_3 + H_2O \longrightarrow$ Product, here product

A

$H_3BO_3 + HCl$

B

$B_2O_3 + HOCl$

C

$B_2H_6 + HCl$

D

$B_2O_3 + HCl$

Solution

$BCl _3+3 H _2 O \rightarrow H _3 BO _3+3 HCl$

The products formed in the reaction are $H _3 BO _3+3 HCl$.

Standard 11

Chemistry

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