Question

A compound of boron X reacts at `200^(@)C` temperature with `NH_(3)` to give another compound Y which is called as inorganic benzene. The compound Y is a colourless liquid and is highly light sensitive. Its melting point is `-57^(@)C`. The compound X with excess of `NH_(3)` and at a still higher temperature gives boron nitride `(BN)_(n)`. The compounds X and Y are respectively:
A. `B_(2)H_(6),B_(3)N_(3)H_(6)`
B. `B_(2)O_(3),B_(3)N_(3)H_(6)`
C. `BF_(3),B_(3)N_(3)H_(6)`
D. `B_(3)N_(3)H_(6),B_(2)H_(6)`

Answer 1

Correct Answer - A
(i) Reaction of ammonia with diborane gives initially `B_(2)H_(6).2NH_(3)` which is fermulated as `[BH_(2)(NH_(3))_(2)]+[NH_(4)]` further heating gives borazine, `B_(3)N_(3)H_(6)` also called borazole.
`underset(underset((X))("Diborane"))(3B_(2)H_(6))+6NH_(3)overset(473K)(rarr)underset(underset((Y))("Borazole"))(2B_(3)N_(3)H_(6))+12H_(2)`
Borazole has cyclic structure and is isoelectronic and isosteric with benzene and thus called inorganic benzene to triborine or borazine.
(ii) Diborane can be prepared by the reduction of `BF_(3)` with lithium aluminium hydride in diethyl ether
`4BF_(3)+3LiAlH_(4)rarr underset((X))(2B_(2)H_(6))+2AlF_(3)+3LiF`.