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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 ad

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$B_{12}H_{12}$

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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.

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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