The p-Block Elements JEE 2018: Nitrogen on Heating

What is the correct answer to the p-block elements JEE 2018 question?

B and C are correct in the p-block elements JEE 2018 question: ammonium dichromate and barium azide release molecular nitrogen below 300 degrees Celsius. This multiple-correct question from JEE Advanced 2018, Paper 1, Chemistry, The p-Block Elements asks you to select every compound whose thermal decomposition meets both the product and temperature conditions.

Check each option independently:

A) & N H_{4} N O_{3} B) & (N H_{4})_{2} C r_{2} O_{7} C) & B a (N_{3})_{2} D) & M g_{3} N_{2}

The question bank tags it as medium difficulty, with a 90-second target. These are question-bank labels, not ratings issued by the exam authority.

Why is ammonium nitrate excluded?

First identify the nitrogen-containing gaseous product formed on heating below 300 degrees Celsius, then compare it with the required species: molecular nitrogen. Ammonium nitrate fails because the supplied solution gives nitrous oxide, not molecular nitrogen, under that condition.

N H_{4} N O_{3} \overset{Δ}{\to} N_{2} O + 2 H_{2} O

The equation balances:

Nitrogen: 2 atoms on each side.
Hydrogen: 4 atoms on each side.
Oxygen: 3 atoms on each side, with 1 in nitrous oxide and 2 in water on the right.

The nitrogen-containing product and the required gas are different substances:

N_{2} O (nitrous oxide)

N_{2} (molecular nitrogen)

According to the supplied solution, further decomposition of nitrous oxide into nitrogen and oxygen requires temperatures above the question’s limit. Option A is incorrect: containing nitrogen atoms does not make a gas molecular nitrogen.

Why do ammonium dichromate and barium azide qualify?

Both compounds produce molecular nitrogen directly in the decomposition reactions accepted under the given heating condition. They satisfy both tests: the product is molecular nitrogen, and its formation does not require heating beyond the stated limit.

For ammonium dichromate:

(N H_{4})_{2} C r_{2} O_{7} \overset{Δ}{\to} N_{2} + C r_{2} O_{3} + 4 H_{2} O

Check the atoms:

Nitrogen: 2 atoms on each side.
Hydrogen: 8 atoms on each side, requiring four water molecules.
Chromium: 2 atoms on each side.
Oxygen: 7 atoms on each side.

The product-side oxygen count is: $3 + 4 = 7$

Molecular nitrogen appears directly among the products under the given condition, so option B is correct. Barium azide also releases molecular nitrogen:

B a (N_{3})_{2} \overset{Δ}{\to} B a + 3 N_{2}

Two azide groups contain six nitrogen atoms, requiring three molecules of molecular nitrogen. One barium atom is also conserved.

2 \times 3 = 6, 3 \times 2 = 6

The azide liberates molecular nitrogen on heating under the stated condition, so option C is correct. Balancing verifies a recalled reaction; it does not establish the products or their temperature conditions.

Why is magnesium nitride rejected, and which options should you select?

Magnesium nitride does not release molecular nitrogen by thermal decomposition under the specified condition, so D is incorrect. Do not invent a decomposition equation because the compound contains nitrogen, or replace heating with a reaction involving water.

The four-option verdict is:

Final answer: B and C, ammonium dichromate and barium azide, respectively:

(N H_{4})_{2} C r_{2} O_{7} and B a (N_{3})_{2}

What is wrong with saying nitrous oxide can eventually form nitrogen?

That argument carries the reaction beyond the permitted temperature condition. The question asks what forms during the specified heating step, not what is obtainable after additional heating.

The incorrect chain of reasoning is:

Ammonium nitrate forms nitrous oxide.
Nitrous oxide can subsequently form molecular nitrogen.
Therefore, ammonium nitrate should be selected.

According to the supplied solution, the secondary decomposition below requires temperatures above 300 degrees Celsius. It cannot qualify option A.

2 N_{2} O \overset{Δ}{\to} 2 N_{2} + O_{2}

The error is replacing “formed under the stated conditions” with “obtainable after additional heating.” Conserving nitrogen atoms does not make nitrous oxide and molecular nitrogen interchangeable: nitrous oxide also contains oxygen and has a different chemical identity.

Use this check:

Underline the required gas and temperature limit.
Write the relevant decomposition reaction.
Reject any argument needing a later reaction outside that limit.

Which three practice questions check the same method?

Use these original follow-up practice questions to test product recall, balancing and mole ratios using the supplied reactions. They are not additional verified PYQs.

Which ammonium salt decomposes into chromium(III) oxide, molecular nitrogen and water?

The salt is ammonium dichromate. Its balanced decomposition is:

(N H_{4})_{2} C r_{2} O_{7} \overset{Δ}{\to} C r_{2} O_{3} + N_{2} + 4 H_{2} O

The two ammonium groups contain eight hydrogen atoms. Each water molecule contains two hydrogen atoms, so the water coefficient must be four.

How much nitrogen does 0.10 mol each of barium azide and ammonium dichromate produce?

Assuming complete decomposition, barium azide produces 0.30 mol of molecular nitrogen, while ammonium dichromate produces 0.10 mol. Their balanced equations give three moles and one mole of nitrogen per mole of compound, respectively.

For barium azide:

n (N_{2}) = 0.10 \times 3 = 0.30 m o l

For ammonium dichromate:

n (N_{2}) = 0.10 \times 1 = 0.10 m o l

The ratio, in the order barium azide to ammonium dichromate, is:

0.30 : 0.10 = 3 : 1

How should you correct a molecular nitrogen label when ammonium nitrate is heated below 300 degrees Celsius?

Replace molecular nitrogen with nitrous oxide. The correct decomposition equation is:

N H_{4} N O_{3} \overset{Δ}{\to} N_{2} O + 2 H_{2} O

The specified heating step does not justify adding secondary decomposition. Check the product’s formula against the required gas, then check that every reaction step stays within the temperature limit.

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Keep going with Orbital Motion of Satellites JEE 2018: Ratio Solution.

Frequently asked questions

What is the answer to the p-block elements JEE 2018 question?

The correct options are B and C: ammonium dichromate and barium azide. Both release molecular nitrogen by thermal decomposition below 300 degrees Celsius under the conditions accepted in the supplied solution.

Why is ammonium nitrate not a correct option?

Ammonium nitrate forms nitrous oxide, N₂O, rather than molecular nitrogen, N₂, under the specified heating condition. According to the supplied solution, further decomposition of nitrous oxide requires temperatures above the 300 degrees Celsius limit, so it cannot qualify option A.

Why is magnesium nitride rejected in this question?

Magnesium nitride does not release molecular nitrogen by thermal decomposition under the specified condition. A reaction involving water cannot replace the heating step stated in the question.

How much nitrogen does 0.10 mol of barium azide produce?

Assuming complete decomposition, 0.10 mol of barium azide produces 0.30 mol of molecular nitrogen. Its balanced decomposition equation gives three moles of nitrogen per mole of barium azide.