Diborane forms adduct with

$P F_{3}$

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$B F_{3}$

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$H^{+}$

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$N H_{4}^{+}$

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Solution

The correct option is A
$P F_{3}$

$P F_{3}$ acts as a Lewis base and forms the adduct $[F_{3} P \to B H_{3}]$ with $B_{2} H_{6}$ .

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C O

H_{2} O

N M e_{3}

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H_{2} O

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B F_{3}

N H_{3}

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