Chemistry - CHEMICAL BONDING Question with Solution | TestHub

The reaction shown, $B_2{H}_6+2X\to 2(X\to B{H}_3)$, represents the symmetrical cleavage of diborane. This specific pathway occurs when diborane reacts with bulky Lewis bases (like tertiary or secondary amines) or soft bases (like CO).

Symmetrical cleavage (X): $(C{H}_3{)}_{2}NH$, CO, $P{F}_3$, and Pyridine (the cyclic structure) all result in the formation of $B{H}_3$ adducts.

Asymmetrical cleavage: Small, hard bases like $N{H}_3$ and $C{H}_{3}N{H}_2$ lead to ionic products like $[B{H}_2(N{H}_3{)}_2{]}^{⁺}[B{H}_4{]}^{⁻}$.

Excluded: $H^{⁺}$ is an electrophile/acid and cannot act as a Lewis base to cleave diborane.

The total number of species 'X' is 4.