JEE Main · 2023 · Shift-ImediumPB12-039

"A" obtained by Ostwald's method involving air oxidation of NH3, upon further air oxidation produces "B". "B" on…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

"A" obtained by Ostwald's method involving air oxidation of NH3\mathrm{NH_3}, upon further air oxidation produces "B". "B" on hydration forms an oxoacid of Nitrogen along with evolution of "A". The oxoacid also produces "A" and gives positive brown ring test. A and B respectively are:

Options
  1. a

    NO2, N2O5\mathrm{NO_2,\ N_2O_5}

  2. b

    NO2, N2O4\mathrm{NO_2,\ N_2O_4}

  3. c

    NO, NO2\mathrm{NO,\ NO_2}

    ✓
  4. d

    N2O3, NO2\mathrm{N_2O_3,\ NO_2}

Correct Answerc

NO, NO2\mathrm{NO,\ NO_2}

Detailed Solution

Strategy: Trace the reaction sequence of the Ostwald process for nitric acid synthesis.

Step 1: Formation of A Ammonia (\ceNH3)(\ce{NH3}) is oxidized by air in the presence of a catalyst (\cePt/Rh)(\ce{Pt/Rh}) to give Nitric Oxide (\ceNO)(\ce{NO}). \ce4NH3+5O2→4NO+6H2O\ce{4NH3 + 5O2 \to 4NO + 6H2O} Thus, A = NO.

Step 2: Formation of B and oxoacid Nitric oxide (\ceNO)(\ce{NO}) quickly reacts with more oxygen (\ceO2)(\ce{O2}) to form Nitrogen Dioxide (\ceNO2)(\ce{NO2}). \ce2NO+O2→2NO2\ce{2NO + O2 \to 2NO2} Thus, B = NO2. \ceNO2\ce{NO2} on hydration (reaction with water) gives Nitric acid (\ceHNO3)(\ce{HNO3}) and evolves \ceNO\ce{NO} (A). \ceHNO3\ce{HNO3} is the oxoacid that gives a positive brown ring test and can produce \ceNO\ce{NO} upon reduction.

Answer: (Option) [c] \ceNO,NO2\boxed{\text{Answer: (Option) [c] $\ce{NO, NO2}$}}

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