JEE Advanced 2022 · Paper 2 · Chemistry Q11

ChemistryMultiple correct+4 / -2Question 47 of 54

The compound(s) which react(s) with $\ce{NH3}$ to give boron nitride (BN) is(are)

\ce{B}

\ce{B2H6}

\ce{B2O3}

\ce{HBF4}

Answer and worked solution

Official answer: B, C

Routes to boron nitride.

Diborane: $\ce{B2H6 + 2NH3}$ gives $\ce{B2H6.2NH3}$ , which on heating gives borazine and then BN.
Boric oxide: $\ce{B2O3 + 2NH3 -> 2BN + 3H2O}$ at high temperature.
Elemental boron reacts with $\ce{NH3}$ only at very high temperature. The official key also accepts (A), (B), (C).
$\ce{HBF4}$ does not give BN.

(B), (C).

JEE Advanced 2022 · Paper 2, Chemistry question 11