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Question: \({{B}_{2}}{{H}_{6}}+N{{H}_{3}}\to \)Addition compound (X) \((X)\xrightarrow{450k}Y+Z(g)\) In t...

B2H6+NH3{{B}_{2}}{{H}_{6}}+N{{H}_{3}}\to Addition compound (X)
(X)450kY+Z(g)(X)\xrightarrow{450k}Y+Z(g)
In the above sequence Y and Z are respectively:
(A)- borazine, H2O{{H}_{2}}O
(B)- boron, H2{{H}_{2}}
(C)- boron nitride, H2{{H}_{2}}
(D)- borazine and hydrogen

Explanation

Solution

The addition product formed by the action of ammonia (NH3N{{H}_{3}}) on diborane (B2H6{{B}_{2}}{{H}_{6}}) diammoniate of diborane., i.e. B2H6.2NH3{{B}_{2}}{{H}_{6}}.2N{{H}_{3}}. It is ionic in nature. Reaction of B2H6{{B}_{2}}{{H}_{6}} with NH3N{{H}_{3}} is one of the methods used for the synthesis of borazine.

Complete step by step answer:
The products of the reaction of diborane (B2H6{{B}_{2}}{{H}_{6}}) with ammonia (NH3N{{H}_{3}}) depends on the reaction conditions.
At low temperature, B2H6{{B}_{2}}{{H}_{6}} reacts with NH3N{{H}_{3}} to form an addition product. The chemical reaction is given below:
B2H6+2NH3B2H6.2NH3{{B}_{2}}{{H}_{6}}+2N{{H}_{3}}\to {{B}_{2}}{{H}_{6}}.2N{{H}_{3}}

B2H6.2NH3{{B}_{2}}{{H}_{6}}.2N{{H}_{3}} is ionic in nature and exists as [H3NBH2NH3]+[BH4]{{\left[ {{H}_{3}}N\to B{{H}_{2}}\leftarrow N{{H}_{3}} \right]}^{+}}{{\left[ B{{H}_{4}} \right]}^{-}} .
When the reaction between B2H6{{B}_{2}}{{H}_{6}} and NH3N{{H}_{3}} is carried out around 450 K temperature, the addition product formed gets converted into borazine and hydrogen. The sequence of chemical reactions taking place is given below:
B2H6+2NH3B2H6.2NH3 3B2H6.2NH3450K2B3N3H6+12H2(g) \begin{aligned} & {{B}_{2}}{{H}_{6}}+2N{{H}_{3}}\to {{B}_{2}}{{H}_{6}}.2N{{H}_{3}} \\\ & 3{{B}_{2}}{{H}_{6}}.2N{{H}_{3}}\xrightarrow{450K}2{{B}_{3}}{{N}_{3}}{{H}_{6}}+12{{H}_{2}}(g) \\\ \end{aligned}

Since B2H6.2NH3{{B}_{2}}{{H}_{6}}.2N{{H}_{3}} is unstable at 450 K and only appears as an intermediate in the reaction, we can also write the overall reaction as
3B2H6+6NH32B3N3H6+12H23{{B}_{2}}{{H}_{6}}+6N{{H}_{3}}\to 2{{B}_{3}}{{N}_{3}}{{H}_{6}}+12{{H}_{2}}

We can identify the compound Y and Z from the above sequence of reactions. Compound Y is borazine (B3N3H6{{B}_{3}}{{N}_{3}}{{H}_{6}}) and compound Z is hydrogen (H2{{H}_{2}}) gas.
So, the correct answer is “Option D”.

Additional Information:
Borazine is known as inorganic benzene because its structure is similar to that of benzene.

Note: We might get confused between the options. So remember that diborane only forms boron nitride if it is treated with excess of ammonia at high temperature. Thus, here option C cannot be correct. And since there is no oxygen involved in the reaction, option A is also not correct.