Diborane on hydrolysis gives a compound X. Th | vedclass

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(B) The hydrolysis of diborane $(B_2H_6)$ is represented by the reaction: $B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$. Thus,the compound $X$ is ortho

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$II$ & $III$

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(B) The hydrolysis of diborane $(B_2H_6)$ is represented by the reaction: $B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2$. Thus,the compound $X$ is orthoboric acid $(H_3BO_3)$. Regarding the properties of $H_3BO_3$: $I$. It is a weak monobasic acid,not a tribasic acid,because it acts as a Lewis acid by accepting an $OH^-$ ion from water: $B(OH)_3 + 2H_2O \rightarrow [B(OH)_4]^- + H_3O^+$. $II$. This statement is correct. $III$. It has a layer structure in the solid state where planar $BO_3$ units are linked by hydrogen bonds. This statement is correct. $IV$. It is sparingly soluble in cold water but its solubility increases with temperature. Thus,it is not considered 'highly' soluble. Therefore,the correct statements are $II$ and $III$.

Diborane on hydrolysis gives a compound $X$. The correct statements about $X$ are
$I$. It is a tribasic acid
$II$. It is a weak monobasic acid
$III$. It has a layer structure
$IV$. It is highly soluble in water

Similar Questions

In $B_2H_6$:

Assertion $(A)$: $[B(OH_2)_6]^{3+}$ and $[B(OH)_4]^{-}$ form octahedral and tetrahedral structures.
Reason $(R)$: Being electron deficient,boron readily reacts with Lewis bases like $H_2O$ and $OH^{-}$.
The correct option among the following is

The bonding in diborane $(B_2H_6)$ can be described by

Which of the following properties does not support the anomalous behavior of $Boron$?

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Diborane on hydrolysis gives a compound $X$. The correct statements about $X$ are$I$. It is a tribasic acid$II$. It is a weak monobasic acid$III$. It has a layer structure$IV$. It is highly soluble in water