Question
What happens when diborane reacts with Lewis bases?
A.
It forms boron trihydride $$\left( {B{H_3}} \right)$$ due to cleavage.
B.
It undergoes cleavage to give borane adduct $$B{H_3} \cdot L$$ ( where, $$L =$$ Lewis base ).
C.
It oxidises to give $${B_2}{O_3}.$$
D.
It does not react with Lewis bases.
Answer :
It undergoes cleavage to give borane adduct $$B{H_3} \cdot L$$ ( where, $$L =$$ Lewis base ).
Solution :
$${B_2}{H_6} + \mathop {2L}\limits_{\left( {{\text{Lewis base}}} \right)} \to \mathop {2B{H_3} \cdot L}\limits_{{\text{Adduct}}} $$