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Question

(A) Boric acid ($H_3BO_3$ or $B(OH)_3$) is a weak monobasic Lewis acid.It does not act as a proton donor (Brønsted acid) by releasing $H^{+}$ ions dir

Vedclass · Accepted Answer

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Vedclass · Answer

(A) Boric acid ($H_3BO_3$ or $B(OH)_3$) is a weak monobasic Lewis acid.It does not act as a proton donor (Brønsted acid) by releasing $H^{+}$ ions directly.Instead,it acts as a Lewis acid by accepting a lone pair of electrons from the $OH^{-}$ ion of a water molecule.The reaction is: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.This process releases $H^{+}$ ions (as $H_3O^+$) into the solution,which is why it exhibits acidic properties.Therefore,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation of $A$.

Similar Questions

What is the decreasing order of Lewis acid strength of boron halides?

In the reactions $I$ and $II$,the covalencies of $Be$ and $Al$ in $X$ and $Y$ are respectively:
$I$ $Be(OH)_2 + NaOH \text{ (excess)} \rightarrow X$
$II$ $Al(OH)_3 + NaOH \text{ (excess)} \rightarrow Y$

The incorrect statement regarding the above reactions is:
$Al \xrightarrow{HCl(aq.)} 'X' + \text{Gas } 'P'$
$Al \xrightarrow{NaOH(aq.) + H_2O} 'Y' + \text{Gas } 'Q'$

In the following reactions:
$3B_2H_6 + 6NH_3 \text{ (excess)} \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2$
$6NaH + 2BF_3 \xrightarrow{450 \ K} B_2H_6 + 6NaF$
$B_2H_6 + 6H_2O \longrightarrow 2H_3BO_3 + 6H_2$
$X, Y$ and $Z$ are respectively:

Aluminium reacts with $NaOH$ and forms compound '$X$'. If the coordination number of aluminium in '$X$' is $6$,the correct formula of $X$ is:

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