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

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

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

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(B N)_{x}

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Borazine

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Solution

The correct option is B $B_{2} H_{6} .2 N H_{3}$
$B_{2} H_{6} + N H_{3} (e x c e s s) l o w - -- \to t e m p B_{2} H_{6} . 2 N H_{3}$

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