Reaction of ammonia with diborane gives initially B₂H₆.2NH₃ which can also be written as

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${[{BH}_{2} {({NH}_{3})}_{2}]}^{+} {[{BH}_{4}]}^{-}$

${[{BH}_{4}]}^{+} {[{BH}_{2} {({NH}_{3})}_{2}]}^{-}$

${[{BH}_{3} {NH}_{3}]}^{+} {[{BH}_{4}]}^{-}$

${[B_{2} N_{2} H_{6}]}^{+} {[H_{3}]}^{-}$

answer is A.

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Detailed Solution

Boron hydride reacts with excess ammonia to given B₂H₆.2NH₃ which is white ionic solid and consists of ${(H_{3} N ⟶ {BH}_{2} ⟵ {NH}_{3})}^{+} and {BH}_{4}^{-}$

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