Diborane reacts with ammonia under different | vedclass

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(B) Diborane $(B_2H_6)$ reacts with ammonia $(NH_3)$ under different conditions to yield various products:$1$. At low temperature, it forms an ionic a

Vedclass · Accepted Answer

$B_{12}H_{12}$

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(B) Diborane $(B_2H_6)$ reacts with ammonia $(NH_3)$ under different conditions to yield various products:$1$. At low temperature, it forms an ionic addition product: $B_2H_6 + 2NH_3 \rightarrow [BH_2(NH_3)_2]^+ [BH_4]^-$ (often written as $B_2H_6 \cdot 2NH_3$).$2$. At high temperatures, it forms borazine $(B_3N_3H_6)$, which is known as inorganic benzene: $3B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 12H_2$.$3$. At very high temperatures, it forms boron nitride $((BN)_n)$, which is isoelectronic with graphite: $B_2H_6 + 2NH_3 \rightarrow 2BN + 6H_2$.Therefore, $B_{12}H_{12}$ is not formed in the reaction between diborane and ammonia.

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 under different conditions to give a variety of products. Which one among the following is not formed in these reactions?

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