Reaction of diborane with ammonia gives initially

Question

Reaction of diborane with ammonia gives initially (A) B2H6.NH3 (B) Borazole (C) B2H6.3NH3 (D) [BH2(NH3)2]+[BH4]-

Tardigrade Admin · Accepted Answer

B2H6 + NH3 xrightarrow[ textlow temperature] textexcess NH3 B2H6. 2NH3 Diborane with ammonia gives B 2 H 6 ⋅ 2 NH 3 that is formulated as [ BH 2( NH 3)2]+[ BH 4]-which when heated to 473 K, decomposes to give borazole.

Q. Reaction of diborane with ammonia gives initially

$B_{2} H_{6} . N H_{3}$

Borazole

$B_{2} H_{6} .3 N H_{3}$

$[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$

$B_{2} N_{4} H_{10}$

$B_{2} H_{6} + N H_{3} excess N H_{3} low temperature B_{2} H_{6} .2 N H_{3}$
Diborane with ammonia gives $B_{2} H_{6} \cdot 2 N H_{3}$ that is formulated as $[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$ which when heated to $473 K$ , decomposes to give borazole.

Q. Reaction of diborane with ammonia gives initially

B2​H6​.NH3​

Borazole

B2​H6​.3NH3​

[BH2​(NH3​)2​]+[BH4​]−

B2​N4​H10​

B2​H6​+NH3​excessNH3​low temperature​B2​H6​.2NH3​ Diborane with ammonia gives B2​H6​⋅2NH3​ that is formulated as [BH2​(NH3​)2​]+[BH4​]−which when heated to 473K, decomposes to give borazole.

$B_{2} H_{6} . N H_{3}$

$B_{2} H_{6} .3 N H_{3}$

$[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$

$B_{2} N_{4} H_{10}$

$B_{2} H_{6} + N H_{3} excess N H_{3} low temperature B_{2} H_{6} .2 N H_{3}$
Diborane with ammonia gives $B_{2} H_{6} \cdot 2 N H_{3}$ that is formulated as $[B H_{2} (N H_{3})_{2}]^{+} [B H_{4}]^{-}$ which when heated to $473 K$ , decomposes to give borazole.