JEE Advanced 2018 · Paper 1 · Chemistry Q1

ChemistryMultiple correct+4 / -2Question 19 of 54

The compound(s) which generate(s) $\ce{N2}$ gas upon thermal decomposition below 300°C is (are)

\ce{NH4NO3}

\ce{(NH4)2Cr2O7}

\ce{Ba(N3)2}

\ce{Mg3N2}

Answer and worked solution

Official answer: B, C

(A) $\ce{NH4NO3 ->[\Delta] N2O + 2H2O}$ . Below 300 °C it gives nitrous oxide, not $\ce{N2}$ . No.
(B) $\ce{(NH4)2Cr2O7 ->[\Delta] N2 + Cr2O3 + 4H2O}$ . Yes.
(C) $\ce{Ba(N3)2 ->[\Delta] Ba + 3N2}$ . This is a standard way to make very pure $\ce{N2}$ . Yes.
(D) $\ce{Mg3N2}$ is thermally stable. It releases nitrogen only as $\ce{NH3}$ on hydrolysis. No.

(B), (C)

Topics:Group 15

JEE Advanced 2018 · Paper 1, Chemistry question 1