Diborane reacts with ammonia to form X, which | vedclass

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(A) The reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ at low temperature results in the formation of an ionic adduct $X$, which is diammoniate

Vedclass · Accepted Answer

$[BH_2(NH_3)_2]^+[BH_4]^-$

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(A) The reaction of diborane $(B_2H_6)$ with ammonia $(NH_3)$ at low temperature results in the formation of an ionic adduct $X$, which is diammoniate of diborane, $[BH_2(NH_3)_2]^+[BH_4]^-$. The chemical equation is: $B_2H_6 + 2NH_3 \longrightarrow [BH_2(NH_3)_2]^+[BH_4]^-$. Upon heating, this compound $X$ undergoes decomposition to form borazine $(B_3N_3H_6)$ and hydrogen gas $(H_2)$: $3[BH_2(NH_3)_2]^+[BH_4]^- \xrightarrow{\Delta} 2B_3N_3H_6 + 12H_2$.

Column-$1$ (Reaction)	Column-$2$ (Main product)
$A$. $B_2H_6 + 2CO \longrightarrow$	$I$. $B_2O_3$
$B$. $B_2H_6 + 3O_2 \longrightarrow$	$II$. $2BH_3 \cdot CO$
$C$. $B_2H_6 + 6H_2O \longrightarrow$	$III$. $2H_3BO_3$

Diborane reacts with ammonia to form $X$, which on heating gives $H_2$ and borazine. $X$ is

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