The product obtained in the reaction of diborane with excess of ammonia at low temperature is:

3 B_{2} H_{6} \cdot 2 N H_{3}

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(B N)_{2}

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B_{3} N_{3} H_{6}

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(B N)_{4}

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Solution

The correct option is A $3 B_{2} H_{6} \cdot 2 N H_{3}$
The product obtained in the reaction of diborane with excess of ammonia at low temperature is $3 B_{2} H_{6} \cdot 2 N H_{3}$ $B_{2} H_{6} + N H_{3} (excess) low temperature - -------- \to 3 B_{2} H_{6} .2 N H_{3} o r [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-}$
Hence correct option is a.

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N H_{3}

Diborane reacts with ammonia under different condition to give a variety of products. Which of the following are formed in these conditions.

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