Diborane undergoes cleavage and gives adduct with
(i) CO
(ii) $H_{2} O$
(iii) $N M e_{3}$
(iv) $N H_{3}$

i, ii

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i, iii

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ii, iii

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i, iv

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Solution

The correct option is B i, iii
(i) and (iii): Diborane undergoes cleavage reactions with Lewis bases (L) to give borane adducts, $B H_{3} . L$
$B_{2} H_{6} + 2 N M e_{3} \to 2 B H_{3} \cdot N M e_{3}$
$B_{2} H_{6} + 2 C O \to 2 B H_{3} \cdot C O$
(ii) diborane gets hydrolysed readily to give boric acid.
$B_{2} H_{6} Borane + 6 H_{2} O \to 2 H_{3} B O_{3} Orthoboric acid + 6 H_{2} ↑ Dihydrogen$
(iv) Diammoniate salt complex is formed
$B_{2} H_{6} + N H_{3} (excess) low temperature - -------- \to B_{2} H_{6} .2 N H_{3} o r [H_{2} B (N H_{3})_{2}]^{+} [B H_{4}]^{-}$
Hence correct option is b.

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