JEE Main 2021 · Chemistry · s-Block, p-Block Elements and Hydrogen · medium

Diborane reacts with excess ammonia at high temperature (about 450 K, sealed system) to give a cyclic compound Y in which B and N alternate. Y is

Single correct option

Diborane reacts with excess ammonia at high temperature (about

450\ \mathrm{K}

, sealed system) to give a cyclic compound

\mathrm{Y}

in which

\mathrm{B}

and

\mathrm{N}

alternate.

\mathrm{Y}

A. borazine, $\mathrm{B_3N_3H_6}$ , a planar ring with $sp^2$ boron and nitrogen
B. boron nitride, $\mathrm{(BN)_n}$ , a diamond-like network
C. $\mathrm{B_2H_6\cdot2NH_3}$ , a molecular adduct with pyramidal boron
D. ammonium borate, $\mathrm{(NH_4)_3BO_3}$

Show answer and solution

Answer: (A) borazine, $\mathrm{B_3N_3H_6}$ , a planar ring with $sp^2$ boron and nitrogen

At low temperature, diborane is cleaved unsymmetrically by

\mathrm{NH_3}

to give the salt

[\mathrm{BH_2(NH_3)_2}]^+[\mathrm{BH_4}]^-

. Heating

\mathrm{B_2H_6}

with excess

\mathrm{NH_3}

at about

450\ \mathrm{K}

drives off hydrogen:

3\mathrm{B_2H_6}+6\mathrm{NH_3}\to2\mathrm{B_3N_3H_6}+12\mathrm{H_2}

. The product is borazine (inorganic benzene), a six-membered planar ring of alternating B and N atoms, each

sp^2

hybridised with delocalised

\pi

electrons; B–N bond length is intermediate between single and double. BN network forms only at far higher temperature, and

\mathrm{B_2H_6\cdot 2NH_3}

is a salt, not a molecular adduct.

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake