Study of Compounds

The labelled diagram for the laboratory preparation of hydrogen chloride gas is shown below:

(a) Conc. H₂SO₄ is used for the preparation of hydrogen chloride gas in the laboratory.

As conc. H₂SO₄ is non-volatile and has a high boiling point, therefore, it displaces the volatile hydrogen chloride from the salt sodium chloride. Hence, conc. H₂SO₄ is used as a reactant in the laboratory preparation of HCl from sodium chloride.

(b) $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g.]}$

(c) Conc. H₂SO₄

Reason — Drying agent used for drying should only remove the moisture and not react with it, hence, conc. sulphuric acid is used as the drying agent.

(d) Phosphorus pentoxide and calcium oxide are good drying agents, but they cannot be used to dry hydrogen chloride gas since they react with it.

2P₂O₅ + 3HCl ⟶ POCl₃ + 3HPO₃

CaO + 2HCl ⟶ CaCl₂ + H₂O

(e) Direct absorption of hydrogen chloride gas in water is not feasible because it causes back suction in the delivery tube.

1. Hydrogen chloride gas is extremely soluble in water.
2. If a delivery tube through which HCl gas is passed is directly immersed in water, the rate of absorption of HCl gas is high and hence a partial vacuum is created in the tube.
3. The pressure outside being higher causes the water to be pushed up into the delivery tube and damages the apparatus. This is called back-suction.

(a) Below diagrams show the special funnel arrangement used for the absorption of HCl gas in water:

(b) The reasons are :

1. Prevents or minimizes back-suction of water.
2. Provides a large surface area for the absorption of the HCl gas.

(c) The equations are:

(A) $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

(B) $\text{2NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\gt 200 ^{\circ}\text{C}} \text{Na}_2\text{SO}_4 + \text{2HCl [g}]$ ↑

(a) Hydrogen chloride gas is not collected over water since it is highly soluble in water.

(b) Quicklime [CaO] is alkaline in nature and it reacts with HCl forming the respective chloride. Hence, it can't be used as a drying agent.

(c) When a glass rod dipped in ammonium hydroxide (NH₄OH) is brought near the mouth of a bottle full of HCl gas, then it leads to the formation of dense white fumes due to the production of ammonium chloride.

NH₄OH + HCl ⟶ NH₄Cl + H₂O

(d) Hydrogen chloride gas fumes in moist air because it dissolves in water vapour present in air to form tiny droplets of hydrochloric acid, which appear as white fumes.

(a) Reactants: Sodium chloride (common salt) and concentrated sulphuric acid.

Products: Hydrogen chloride gas and Sodium bisulphate (at low temperatures) or Sodium sulphate (at high temperatures).

(b) Conditions of reaction:

The reaction typically occurs in two stages depending on the temperature:
Below 200°C: The reaction yields sodium bisulphate.

$\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl} \uparrow$

Above 200°C: If the temperature is increased further, the sodium bisulphate reacts with more salt to form sodium sulphate. However, this is usually avoided in labs to protect the glass apparatus and prevent the formation of a hard crust.

(c) Identification of the gas :
Bring a glass rod dipped in ammonia solution near the jar mouth; dense white fumes of ammonium chloride (NH₄Cl) forms.
It also turns moist blue litmus paper red, showing its acidic nature.

(d) Precautions to be taken :

1. The lower end of the thistle funnel must be dipped in conc. sulphuric acid.
2. Delivery tube should be dipped in drying agent i.e., conc. H₂SO₄.
3. Avoid temperatures above 200°C to prevent glass apparatus from cracking, wasting fuel, and forming a hard sodium sulphate crust.

Polar covalent

Reason — Hydrogen chloride (HCl) is a polar covalent molecule because it is formed by sharing of electrons between hydrogen and chlorine atoms. However, chlorine is more electronegative than hydrogen, so it pulls the shared electrons closer to itself. This creates a partial negative charge on chlorine and a partial positive charge on hydrogen, making the molecule polar.

Conc. sulphuric acid.

Reason — Drying agent used for drying should only remove the moisture and not react with it, hence, conc. sulphuric acid is used as the drying agent.

HCl is highly soluble in water

Reason — Fountain experiment is performed to show the great solubility of HCl gas in water.

Only R

Reason — Toluene is an organic solvent and non-polar in nature. When HCl dissolves in toluene, it does not ionize to form hydronium [H₃O⁺] ions. Since there are no free hydrogen ions, the solution does not show acidic properties and also cannot react with sodium carbonate to release carbon dioxide gas. Therefore, only R is correct — the solution does not show any acidic property. b

Both A and R are true and R is the correct explanation of A.

Explanation— HCl is produced, when moist hydrogen gas combines with chlorine in presence of diffused sunlight. Hence, assertion (A) is true.

${\text{H}_2} + {\text{Cl}_2} \xrightarrow[\text{sunlight}]{\text{diffused}} {2\text{HCl}}$

The formation HCl is a explosive reaction in direct sunlight but it is negligible in dark. So, it is carried out in diffused sunlight. Hence, reason (R) is true.

Therefore, Reason (R) thus provides a valid explanation for why this reaction is occurring in diffused sunlight. Hence, Reason (R) is the correct explanation of assertion

A is true but R is false.

Explanation— The principle of fountain experiment is to show the great solubility of the HCl gas. Hence the assertion (A) is true.

The blue litmus solution turns red due to the acidic nature of hydrogen chloride gas in the Fountain experiment. Hence reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation— Hydrogen chloride gas is collected by the downward delivery (upward displacement of air). Hence, the assertion (A) is true.
HCl is collected by downward delivery because it is 1.28 times heavier than air. Hence, reason (R) is true and it is the correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation— Hydrogen chloride (HCl) gas fumes in moist air because it is highly soluble in water. In the presence of moisture, HCl gas dissolves in water vapour in the air to form tiny droplets of hydrochloric acid, which appear as white fumes. So, both the assertion and the reason are true, and the reason correctly explains the assertion.

Both A and R are true and R is the correct explanation of A.

Explanation— Hydrogen chloride (HCl) is a polar covalent compound, which means it can dissolve in polar solvents like water due to attraction between opposite charges. Interestingly, it can also physically dissolve in some non-polar organic solvents like toluene, but without ionizing. The reason HCl can dissolve in both is due to its polar covalent nature. Hence, both the Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

A is true but R is false.

Explanation— HCl gas dissociates into hydrogen and chlorine, when heated to 500°C and hydrogen chloride reacts with metals that come before hydrogen in the electrochemical series to form corresponding chlorides of those metals. Hence, assertion (A) is true.
When concentrated Hydrochloric acid is oxidised by strong oxidising agent like lead oxide lead chloride is formed. Hence, reason (R) is false.

Pb₃O₄ + 8HCl (conc.) ⟶ 3PbCl₂ + 4H₂O + Cl₂ ↑

A is true but R is false.

Explanation— When dilute hydrochloric acid (HCl) reacts with iron(II) sulphide (FeS), it undergoes a double displacement reaction to form iron(II) chloride and hydrogen sulphide gas (H₂S), which is a colourless gas with a rotten egg smell. Hence, assertion (A) is true.
Hydrogen sulphide gas (H₂S) does react with moist lead acetate paper. It turns the paper silvery black (or brownish-black) due to the formation of lead sulphide (PbS). Hence, reason (R) is false.

(a) Quicklime is not used to dry HCl gas because CaO is alkaline.

(b) When sodium chloride is heated with concentrated sulphuric acid below 200°C, one of the product formed is sodium hydrogen sulphate.

(a) Copper oxide (CuO)

(b) Hydrogen chloride and Ammonia

(c) Hydrogen and Oxygen

(d) Silver chloride

(e) Aqua Regia

(f) Fountain experiment

(g) Oxygen gas

(h) Dilute hydrochloric acid

(i) Hydrogen chloride

(j) Chlorine

Explanation

(a) Copper oxide (CuO) is black, reacts with HCl to form blue-coloured copper(II) chloride solution.
CuO + 2HCl ⟶ CuCl₂ + H₂O

(b) NH₃ (g) + HCl (g) ⟶ NH₄Cl (white solid)

(c) 2H₂ (g) + O₂ (g) ⟶ 2H₂O (liquid water)

(d) AgCl dissolves in excess NH₄OH due to formation of [Ag(NH₃)₂]⁺ complex.
AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂]⁺Cl^- + 2H₂O

(e) Aqua regia (1 part of conc. HNO₃ and 3 parts of conc. HCl) dissolves gold.

(f) Fountain experiment vividly shows HCl gas dissolving in water, creating a fountain effect due to pressure drop.

(g) When chlorine water is exposed to sunlight, chlorine reacts with water to form hypochlorous acid. This acid then breaks apart to give nascent oxygen (O) and hydrochloric acid (HCl). The nascent oxygen combines to form Oxygen gas:
Cl₂ + H₂ $\xrightarrow{\text{sunlight}}$ HClO + HCl
HClO $\xrightarrow{\text{sunlight}}$ HCl + [O] (nascent oxygen)
[O] + [O] ⟶ O₂

(h) AgNO₃ + HCl ⟶ AgCl (white ppt.) + HNO₃ which dissolves in excess NH₄OH.

(i) NH₃ [g] + HCl [g] ⟶ NH₄Cl [s]

(j) H₂ + Cl₂ ⟶ 2HCl; HCl is strongly acidic in water.

A → Silver nitrate
B → Hydrochloric acid
C → Silver chloride

Explanation — Solution A is silver nitrate which reacts with acid B i.e., Hydrochloric acid to give a white precipitate C of AgCl which is insoluble in nitric acid but soluble in ammonium hydroxide.

AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃

AgNO₃ + HNO₃ ⟶ no reaction

AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂Cl)] + 2H₂O

Aqua regia : 1 part of conc. HNO₃ and 3 parts of conc. HCl

Nitric acid present in aqua regia oxidizes HCl to chlorine.

It shows that hydrochloric acid (HCl) is oxidized to chlorine (Cl₂) by strong oxidizing agents such as MnO₂, PbO₂ and red lead.

(a) NH₄OH + HCl (dil.)⟶ NH₄Cl + H₂O

(b) NaHSO₃ + HCl (dil.) ⟶ NaCl + H₂O + SO₂

(c) Pb(NO₃)₂ + 2HCl (dil.) ⟶ PbCl₂+ 2HNO₃

(d) Pb₃O₄ + 8HCl (conc.) ⟶ 3PbCl₂ + 4H₂O + Cl₂ ↑

(a) When Sodium carbonate is treated with dil. HCl, odourless carbon dioxide gas is produced.

Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂ ↑

However, when sodium sulphite is treated with dil.HCl, sulphur dioxide gas with a suffocating odour (burning smell) is produced.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ ↑

(b) When sodium thiosulphate is treated with dil. HCl it produces sulphur dioxide gas and yellow sulphur precipitates.

Na₂S₂O₃ + 2HCl ⟶ 2NaCl + S + H₂O + SO₂ ↑

However, S is not produced when sodium sulphite is treated with dil.HCl.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ ↑

(a) Silver nitrate solution
AgNO₃ + HCl ⟶ AgCl (white ppt.) + HNO₃

(b) Magnesium foil
Mg + 2HCl ⟶ MgCl₂ + H₂

(c) Caustic soda solution
NaOH + HCl ⟶ NaCl + H₂O

(d) Zinc carbonate
ZnCO₃ + 2HCl [dil.] ⟶ ZnCl₂ + H₂O + CO₂ [g.]

(e) Manganese (IV) oxide
$\text{MnO}_2 + \underset{\text{Conc.}}{\text{4HCl}} \xrightarrow{\Delta} \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$ ↑

(f) Copper oxide
CuO + 2HCl ⟶ CuCl₂ + H₂O.

(a) Anhydrous HCl is a gas and does not contain any ions, hence it is a poor conductor of electricity. On the other hand, aqueous HCl (HCl dissolved in water) forms ions in solution, and thus becomes an excellent conductor of electricity.

HCl + H₂O ⟶ H₃O⁺ + Cl^-

(b) When hydrogen chloride gas is exposed to air, it gives white fumes, due to the formation of hydrochloric acid on dissolving in atmospheric water vapour.

(c) A solution of hydrogen chloride in water gives ions,

HCl + H₂O ⟶ H₃O⁺ + Cl^-

These ions are responsible for conducting electricity and turning blue litmus red, due to the presence of hydronium [H₃O⁺] ions which make aqueous solution of HCl acidic in nature.

Toluene is an organic solvent and HCl is a covalent compound. A solution of HCl in toluene contains only molecules and not ions due to which it does not conduct electricity. The absence of hydronium [H₃O⁺] ions make the solution neutral and it does not effect litmus paper.

(d) Dry hydrogen chloride gas does not contain any ions. Due to the absence of hydronium [H₃O⁺] ions, it is neutral and does not affect a dry strip of blue litmus paper. On the other hand, in the presence of a drop of water, HCl dissolves in it and dissociates into hydronium [H₃O⁺] ions and chloride ions [Cl^-]

HCl + H₂O ⟶ H₃O⁺ + Cl^-

The presence of hydronium [H₃O⁺] ions makes the aqueous solution acidic and blue litmus turns red.

(e) Hydrochloric acid forms a constant boiling mixture at 110°C, 22.2% by weight. On boiling further, the mixture evolves out the vapours of both acid and water in the same proportion as in the liquid.

Hence, dilute HCl cannot be concentrated beyond 22.2% by boiling.

(a) $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

(b) Fe + 2HCl (dil.) ⟶ FeCl₂ + H₂ [g]

(c) NH₄OH + HCl (dil.)⟶ NH₄Cl + H₂O

(d) $\text{PbO}_2 + 4\text{HCl} \xrightarrow{\Delta} \text{PbCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$ ↑

(a) Na₂S₂O₃ + 2HCl ⟶ 2NaCl + S + H₂O + SO₂ ↑

(b) Ca(HCO₃)₂ + 2HCl ⟶ CaCl₂ + 2H₂O + 2CO₂ ↑

(c) 2KMnO₄ + 16HCl $\xrightarrow{\Delta}$ 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂↑

(d) Na₂S + 2HCl ⟶ 2NaCl + H₂S [g.] ↑

(e) NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂ ↑

The balanced equation are :

(a) Iron
Fe + 2HCl (dil.) ⟶ FeCl₂ + H₂ ↑

(b) Sodium hydrogen carbonate
NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂ ↑

(c) Iron (II) sulphide
FeS + 2HCl ⟶ FeCl₂ + H₂S ↑

(d) Magnesium sulphite
MgSO₃ + 2HCl ⟶ MgCl₂ + H₂O + SO₂ ↑

When silver nitrate soln. is added to dil. HCl, curdy white ppt. of silver chloride is formed whereas, there is no reaction when silver nitrate soln. is added to dil. HNO₃.

AgNO₃ + HCl ⟶ AgCl (white ppt.) + HNO₃

AgNO₃ + HNO₃ ⟶ no reaction

(a) A colourless gas having smell of rotten eggs is given off. The gas evolved is H₂S
CuS + 2HCl ⟶ CuCl₂ + H₂S ↑

(b) Curdy white precipitate of silver chloride [AgCl] is obtained, which is soluble in excess of NH₄OH.
AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃
AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂Cl)] + 2H₂O

(c) White precipitate of PbCl₂ is formed which is soluble in hot water.
Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

(d) Hydrogen gas is evolved with bubbles and it burns with a pop sound
Zn + 2HCl ⟶ ZnCl₂ + H₂ ↑

(e) Effervescence of CO₂ seen which turns lime water milky.
Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂ ↑

(f) When concentrated hydrochloric acid is added to manganese dioxide (MnO₂), a greenish-yellow gas (chlorine gas) is evolved with effervescence.
MnO₂ + 4HCl (conc.) $\xrightarrow{\Delta}$ MnCl₂ + 2H₂O + Cl₂↑

(a) On reaction with HCl, lead nitrate solution gives white precipitate which can be dissolved in hot water.

Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

Whereas, silver nitrate solution gives thick curdy white precipitate which is insoluble in hot water.

AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃

(b) On reaction with HCl, potassium sulphite gives SO₂ gas with pungent odour

K₂SO₃ + 2HCl ⟶ 2KCl + H₂O + SO₂ ↑

However, potassium sulphide gives H₂S gas with rotten eggs odour.

K₂S + 2HCl ⟶ 2KCl + H₂S ↑

(a) Hydrochloric acid (HCl)

(b) The reactants used are sodium chloride and sulphuric acid.

(c) The empty flask acts as an Anti-Suction device. If back suction takes place then the water gets collected in it and will not reach the generating flask.

(d) The drying agent used is Concentrated Sulphuric acid. It is chosen as drying agent because it does not react with HCl.

(e) The role of the inverted funnel in the arrangement is -

1. It prevents or minimizes back suction of water.
2. It provides a large surface area for absorption of HCl gas.

(a) The equation for the laboratory preparation of hydrogen chloride gas :

$\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

(b) Concentrated sulphuric acid is used as the drying agent used in the above preparation.

Reason — Drying agent used for drying should only remove the moisture and not react with it, hence, conc. sulphuric acid is used as the drying agent. Other drying agents like phosphorous pentoxide (P₂O₅) and quick lime (CaO) cannot be used since they react with hydrogen chloride gas.

(c) Temperature is maintained at nearly 200°C as at higher temperatures above 200°C the glass apparatus may crack, fuel is wasted, sodium sulphate formed, forms a hard crust which sticks to the glass and is difficult to remove.

(i) Using an active metal below magnesium

Fe + 2HCl ⟶ FeCl₂ + H₂↑

(ii) Using an oxidising agent not containing lead.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂↑

(i) Action on metals — Hydrochloric acid reacts with metals above hydrogen in the activity series forming metallic chlorides and evolving hydrogen.

Ca + 2HCl ⟶ CaCl₂ + H₂ ↑

(ii) Action on oxides — Hydrochloric acid reacts with oxides to form salt and water only.

CuO + 2HCl ⟶ CuCl₂ + H₂O

(iii) With salts of weaker acids — Hydrochloric acid decomposes salts of weaker acids.

Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂ ↑

(a) Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ + 2HNO₃
Hg₂(NO₃)₂ + 2HCl ⟶ Hg₂Cl₂ + 2HNO₃

(b) FeO + 2HCl ⟶ FeCl₂ + H₂O

(c) CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂ ↑

(d) Na₂S + 2HCl ⟶ 2NaCl + H₂S ↑

The balanced chemical reactions are:

A : $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

B : $\text{Zn} + 2\text{HCl ( dil.)} \xrightarrow{\Delta} \text{ZnCl}_2 + \text{H}_2$ ↑

C : AgNO₃ + HCl ⟶ AgCl (white ppt.) + HNO₃

D : $\text{MnO}_2 + \underset{\text{Conc.}}{\text{4HCl}} \xrightarrow{\Delta} \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$ ↑

E : NaHCO₃ + HCl ⟶ NaCl + H₂O + CO₂ ↑

(i) Dry HCl (Hydrogen chloride) gas is the gas Y

(ii) High solubility of HCl in water

(iii) Ammonia (NH₃) gas also demonstrates high solubility in water.

(i) H₂S gas is evolved when dilute hydrochloric acid reacts with Iron [II] sulphide.

FeS + 2HCl ⟶ FeCl₂ + H₂S ↑

(ii) SO₂ is evolved when dilute hydrochloric acid reacts with sodium sulphite.

Na₂SO₃ + 2HCl ⟶ 2NaCl + H₂O + SO₂ [g] ↑

(a) When HCl gas is passed through lead nitrate, white precipitate of lead chloride is formed.

Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ ↓ + 2HNO₃

(b) Hydrochloric acid, when reacted with silver nitrate solution it gives a thick curdy white precipitate of silver chloride. This white precipitate is insoluble in nitric acid.

AgNO₃ + HCl ⟶ AgCl ↓ + HNO₃

(c) On addition of ammonium hydroxide, the resultant of part (b) i.e., thick curdy white precipitate of silver chloride dissolves and forms a complex salt called diammine silver (I) chloride.

AgCl + 2NH₄OH ⟶ [Ag(NH₃)₂Cl)] + 2H₂O

P — $\text{NaCl} + \text{H}_2\text{SO}_4 \xrightarrow{\lt 200 ^{\circ}\text{C}} \text{NaHSO}_4 + \text{HCl [g]}$ ↑

Q — Fe + 2HCl (dil.) ⟶ FeCl₂ + H₂ [g]

R — NH₄OH + HCl (dil.)⟶ NH₄Cl + H₂O

S — $\text{PbO}_2 + 4\text{HCl} \xrightarrow{\Delta} \text{PbCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$ ↑

(a) Covalent bonding is present in ammonia as shown below —

(b) The formula of liquid ammonia is NH₄OH.

The aqueous solution of ammonia [NH₄OH] is a weak base. It dissociates partially to give hydroxyl ions [OH^1-].

The basic nature of NH₄OH is due to the presence of hydroxyl ions [OH^1-].

NH₃ + H₂O ⟶ NH₄OH

NH₄OH ⇌ NH₄⁺ + OH^-

(a) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃ [g]

(b) In order to get dry ammonia, the gas is passed through a drying tower containing lumps of quicklime [CaO]. Ammonia gas is collected in inverted gas jars by the downward displacement of air.

(c) As ammonia gas is highly soluble in water, therefore, it is not collected over water.

(a) The diagram below shows the preparation of aqueous ammonia:

Procedure : Water is taken in a container and only a small portion of the mouth of the funnel is dipped in water.

As ammonia dissolves in water at a higher rate than it's production in the flask, the pressure in the funnel above water level decreases for a moment and water rushes into the funnel.

As a result, the rim of the funnel loses it's contact with water. Since, ammonia produced pushes the water down, the funnel comes in contact with water again. In this way, ammonia dissolves in water without back suction of water.

(b) Ammonia being basic in nature reacts chemically with P₂O₅ and CaCl₂

6NH₃ + P₂O₅ + 3H₂O ⟶ 2(NH₄)₃PO₄

4NH₃ + CaCl₂ ⟶ CaCl₂.4NH₃

Hence, P₂O₅ and CaCl₂ are not used to dry NH₃

Substance A is Ammonium chloride (NH₄Cl) and gas B is Ammonia

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃ [g]

(a) Nitrogen reacts with hydrogen under specific conditions liberating ammonia

N₂ + 3H₂ ⇌ 2NH₃ + Δ

Conditions :

Temperature : 450-500°C [Optimum temperature]

Pressure : 200 to 900 atmospheres [Optimum pressure]

Catalyst : Finely divided iron [Fe]

Promotor : Molybdenum [Mo]

[Catalyst - iron [III] oxide [Fe₂O₃] may also be used containing promoters about 1% K₂O and 3% Al₂O₃]

(b) According to the above equation, nitrogen and hydrogen combines in 1:3 ratio by volume.

(c) Nitrogen is obtained by fractional distillation of liquid air. Hydrogen is obtained from the water gas (Bosch process) or from natural gas.

(d) The formation of ammonia is promoted by the use of high pressure as it favours the forward direction.

(e) Two possible ways by which ammonia produced is removed from unchanged gases are:

1. By Liquefaction — ammonia is liquefied easily as compared to nitrogen and hydrogen.
2. By Absorbing in water — because ammonia is highly soluble in water, as hydrogen and nitrogen are very slightly soluble.

(f) Function of finely divided iron and molybdenum are:

1. Finely divided iron increases the rate of reaction.
2. Molybdenum acts as a promoter to increase the efficiency of the catalyst.

(g) 15%

(h) The unchanged nitrogen and hydrogen are recirculated through the plant to get more ammonia, by this way we can achieve 98% of ammonia.

Due to the high solubility of Ammonium compounds in water it does not occur in minerals.

As ammonium nitrate is explosive in nature and may itself decompose forming nitrous oxide and water vapour, hence it is not used in the preparation of ammonia.

NH₄NO₃ $\xrightarrow{\Delta}$ N₂O + 2H₂O

As sulphuric acid reacts chemically with ammonia to form ammonium sulphate hence, it is not used as a drying agent for drying ammonia.

2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₄

1. In the lab preparation of ammonia, ammonium chloride is used. Ammonium chloride is sublime, during the reaction, the heat will cause ammonium chloride to vapourise. To prevent this, calcium hydroxide is used in excess. It absorbs excess heat and prevents the loss of ammonium chloride as vapours.
2. The flask is fitted in slanting position so that the water formed in the reaction does not trickle back into the heated flask and thus break it.

Conditions :

(a) Temperature : 450-500°C [Optimum temperature]

(b) Pressure : 200 to 900 atmospheres [Optimum pressure]

(c) Catalyst : Finely divided iron [Fe]

(a) Ammonium chloride is a soluble salt prepared by neutralization.

NH₃ (gas) + HCl (gas) ⟶ NH₄Cl (solid)

(b) When ammonium chloride is heated, it undergoes thermal dissociation.

NH₄Cl ⇌ NH₃ + HCl

(c) Heating ammonium chloride with sodium hydroxide produces ammonia.

NH₄Cl + NaOH ⟶ NaCl+ H₂O + NH₃

An element has 2 electrons in it's N shell, hence, it has

1. 2 (electrons in K shell) +
2. 8 (electrons in L shell) +
3. 8 (electrons in M shell) +
4. 2 (electrons in N shell) = 20 electrons total i.e. Calcium

Non metal (atomic number 7) is Nitrogen

Calcium and Nitrogen react to form calcium nitride.

3Ca + N₂ ⟶ Ca₃N₂

Ca₃N₂ reacts with water to give calcium hydroxide and ammonia.

Ca₃N₂ + 6H₂O ⟶ 3Ca(OH)₂ + 2NH₃ [g]

(a) When metal for A is magnesium, then, compound X is Mg₃N₂

N₂ + 3Mg ⟶ Mg₃N₂

(b) Magnesium nitride

Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃ [g]

(c) Ammonia is a reducing agent and hence it reduces the less reactive metal oxide to the respective metal.

2NH₃ + 3MgO ⟶ 3Mg + 3H₂O + N₂

(a) A dirty green precipitate is obtained when ammonium hydroxide is added to ferrous sulphate.

(b) Aqueous ammonia is a solution of NH₃ in water.

(c) Finely divided iron is used in Haber's Process.

(d) Quicklime, is a drying agent for NH₃,

(e) Ammonium salts, on heating with caustic alkali, decompose to give ammonia.

Magnesium nitride

Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃ [g]

Iron

Reason — The catalyst used in haber's process of manufacture of ammonia is Finely divided iron.

Ammonium nitrite

Reason — When ammonium nitrate is heated it decomposes into nitrogen gas and water.

NH₄NO₂ $\xrightarrow\Delta$ 2H₂O + N₂

Mg₃N₂

Reason — Ammonia can also be obtained by the action of warm water on nitride of metal like magnesium.

Mg₂N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃ ↑

NH₃

Reason — Ammonia (NH₃) gas is collected in inverted gas jars by downward displacement of air because it is lighter than air.

1 : 3

Reason — In the Haber's process, the ratio of reactants nitrogen and hydrogen is 1 : 3. One volume of nitrogen (from liquid air) and three volume of hydrogen (water gas from Bosch process).

NCl₃

Reason — When ammonia reacts with an excess of chlorine, products formed are hydrogen chloride and yellow coloured highly explosive liquid nitrogen trichloride.
NH₃ + 3Cl₂ ⟶ 3HCl + NCl₃

Ostwald process

Reason — Ammonia is used as a starting material for manufacturing nitric acid using ostwald process.

Only R

Reason — Ammonia burns in oxygen with greenish yellow fumes. This reaction is explosive and dangerous. Hence, by keeping ammonia tube higher than oxygen tube, limited oxygen will react with ammonia.

A is true but R is false.

Explanation — Pure ammonia (NH₃) in its gaseous or liquid form is covalent and does not contain free ions to conduct electricity. Hence, the assertion (A) is true.
Ammonia consist of molecule, it doesn't have free ions to conduct electricity. Hence, the reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation— Ammonia is dried by passing it through a drying tower containing lumps of quicklime (CaO). Hence, the assertion (A) is true.
Other drying agents like conc. sulphuric acid, phosphorous pentoxide and anhydrous calcium chloride are not used, as ammonia is basic in nature, reacts with other drying agents. Hence, the reason (R) is true and it is the correct explanation of assertion (A).

A is true but R is false.

Explanation— Ammonia and ammonium compounds do not occur as minerals because they are highly soluble in water. Hence, the assertion (A) is true. Ammonia is highly soluble in water. Hence, the reason (R) is false.

A is true but R is false.

Explanation — In the Haber's process of manufacturing ammonia, nitrogen and hydrogen are used as reactants. Where, hydrogen is obtained as water gas from Bosch process. Hence, the assertion (A) is true.
The ratio of reactants nitrogen and hydrogen in the Haber's process is 1 : 3. One volume of nitrogen (from liquid air) and three volume of hydrogen (water gas from Bosch process). Hence, reason (R) is false.

A is true but R is false.

Explanation — Haber's process is used to manufacture ammonia. Hence, the assertion (A) is true.
The catalyst used in Haber's process is finely divided iron. Hence, the reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation — Ammonia obtained from Haber's process is separated from the unreacted nitrogen and hydrogen by liquification process. Hence, the assertion (A) is true.
Ammonia is recovered by liquification process because ammonia is liquified easily as compared to Nitrogen and Hydrogen. Hence, the reason (R) is true and it is the correct explanation of assertion (A).

Both A and R are true but R is not the correct explanation of A.

Explanation — Liquid ammonia is used as a refrigerant in ice plants because anhydrous ammonia is a clear, colourless liquid under pressure. It evaporates rapidly and produces cooling effect. This makes ammonia a good refrigerant. Hence, the assertion (A) is true. CFC, the main refrigerant, decompose by ultraviolet rays to form free chlorine radicals. These free radicals contribute to global warming and ozone depletion. Hence, the reason (R) is true.

Since, reason (R) doesn't explain why liquid ammonia is used as refrigerant in ice plants, it is not the correct explanation of assertion (A).

(a) Ammonia gas is collected by downward displacement of air.

(b) Ammonia reacts with excess chlorine to form nitrogen trichloride
NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(i) Ammonia

(ii) Alkaline

(iii) Ammonium

(iv) Hydroxyl

(v) Dirty green

(a) Ammonia

(b) Hydrogen chloride and chlorine gas

(c)
    (i) dry cell — ammonium chloride
    (ii) explosives — ammonium nitrate
    (iii) medicine — ammonium carbonate

(d) Acidic gas — Cl₂ ; Basic gas — Ammonia; Neutral gas — N₂
     2NH₃ + 3Cl₂ ⟶ N₂ + 6HCl

(e) Silver chloride

(f) Nitrogen

(g) Magnesium nitride

(h) Lead oxide

(i) Ammonium chloride

(j) Ammonia (NH₃)

(k) Nitrogen
    8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(l) Ammonium salts like ammonium chloride, ammonium sulphate

(m) Ammonia gas
NH₃ + HCl ⟶ NH₄Cl

(a) Ammonia gas

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

(b) Nitrogen gas

NH₄Cl + NaNO₂ ⟶ NaCl + NH₄NO₂

NH₄NO₂ ⟶ 2H₂O + N₂

Ammonia

Reducing nature of ammonia:
2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

Gas 'P' is Ammonia and its formula is NH₃.

With chlorine, ammonia gas gives dense white fumes of NH₄Cl.

8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

Reaction of copper (II) hydroxide with aqueous solution of Ammonia (NH₄OH) is given below:

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

(a) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃↑

(b) 2NH₃ + 3PbO ⟶ 3Pb + 3H₂O + N₂ [g]

(c) NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(d) 2NH₃ + CO₂ ⟶ NH₂CONH₂ + H₂O

(i) Ammonia acts as a reducing agent as it loses hydrogen in equation (c).

(ii) Urea is prepared from equation (d). Ammonia reacts with carbon dioxide at 150°C and 150 atm, pressure to give urea.

(a) Ammonia is basic in nature and rest are acidic.

(b) Copper oxide is less reactive and can be reduced by C, CO or hydrogen whereas aluminium oxide, sodium oxide, magnesium oxide are reduced by electrolysis.

White gelatinous precipitate of zinc hydroxide is formed which dissolves in excess of NH₄OH solution.

Zn(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Zn(OH)₂ ↓

(i) Dirty green ppt. of ferrous hydroxide is formed which is insoluble in excess of NH₄OH.

(ii) Reddish brown ppt. of ferric hydroxide is formed which is insoluble in excess of NH₄OH.

(iii) White ppt. of lead hydroxide is formed which is insoluble in excess of NH₄OH.

(iv) White gelatinous ppt. of Zinc hydroxide is formed which is soluble in excess of NH₄OH.

(b) The balanced chemical equations are given as :

(i) FeSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Fe(OH)₂ ↓

(ii) FeCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Fe(OH)₃ ↓

(iii) Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂ ↓

(iv) Zn(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Zn(OH)₂ ↓

Solution B is Copper Sulphate and cation present is (Cu²⁺). The colour of solution B is Blue.

Ammonium hydroxide gives a blue precipitate when it combines with a solution of copper salt, due to the formation of Cu(OH)₂

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper [II] sulphate, an azure blue soluble complex salt.

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

(a) NH₄Cl ⇌ NH₃ + HCl

This reaction is an example of thermal dissociation.

(b) NH₄NO₃ ⇌ N₂O + 2H₂O

This reaction is an example of thermal dissociation.

Hydrogen chloride gas (HCl) and Ammonia (NH₃) are the two gases which can be used to study the fountain experiment.

The common property demonstrated by the fountain experiment is the solubility of gases in water.

Hydroxyl ion [OH^-1] and ammonium ions [NH₄⁺] are formed.

NH₃ + H₂O ⟶ NH₄OH

NH₄OH ⇌ NH₄⁺ + OH^-1

The alkaline NH₄OH, due to the presence of hydroxyl ions [OH^-] turns red litmus blue and phenolphthalein soln. pink

Conditions for catalytic oxidation of ammonia to nitric oxide are :

1. Temperature : 800°C
2. Catalyst : Platinum (Pt)

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}} \text{4NO} + \text{6H}_\text{2}\text{O} + \Delta$

(a) Liquid ammonia is a clear, colourless liquid under pressure. It evaporates rapidly and produces cooling effect. Moreover, Ammonia is environmentally compatible, has superior thermodynamic properties and its leaks are easily detectable due to its recognizable odour. All these reasons make liquid ammonia a good refrigerant.

(b) As aqueous solution of ammonia emulsifies or dissolves fats, grease etc., hence it is used for removing grease stains from woollen clothes.

(c) Pungent smell of aqueous solution of ammonia is due to the presence of ammonia, which has strong, pungent choking smell.

(d) In aqueous solution, the ammonia molecule combines with a hydrogen atom H⁺ by sharing the lone pair of electrons of nitrogen atom to form ammonium ion (NH₄⁺). Thus, in water, ammonium hydroxide (NH₄OH) dissociates into NH₄⁺ (ammonium ion) and OH^- (hydroxide ion) as follows:

NH₄OH ⟶ NH₄⁺ + OH^-

Due to this ionisation, aqueous solution of ammonia (NH₄OH) conducts electricity.

(a) FeSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Fe(OH)₂ ↓

(b) Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃↑

(c) NH₄Cl + NaOH ⟶ NaCl + H₂O + NH₃

(d) 8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(e) 4NH₃ + 5O₂ $\xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}}$ 6H₂O + 4NO↑ + Δ

(f) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

(g) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

(h) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃↑

(i) NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(j) 2NH₃ + H₂SO₄ (dil.) ⟶ (NH₄)₂SO₄

(k) 8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(a) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ ↑

(b) NH₃ + 3Cl₂ ⟶ 3HCl + NCl₃ ↑

Ammonia gas is a strong reducing agent. Below equation shows reduction of Black Copper [II] oxide to brown Copper by Ammonia.

2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

Magnesium nitride reacts with warm water to liberate ammonia along with magnesium hydroxide.

(a) When we add ammonium hydroxide solution to both the given salt solutions, Zinc Chloride reacts to form a gelatinous white precipitate of Zinc Hydroxide (ZnOH), whereas, no such observations i.e no precipitate will form in case of Calcium Chloride solution because ammonium hydroxide is a weak base and it cannot react with calcium salts to precipitate the hydroxide of calcium.

ZnCl₂ + 2NH₄OH ⟶ Zn(OH)₂ + 2NH₄Cl

(b) On reaction with ammonium hydroxide, Iron [II] sulphate (ferrous salt) forms a dirty green ppt whereas iron [III] sulphate (ferric salt) forms a reddish brown ppt. Hence, the two can be distinguished easily.

FeSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Fe(OH)₂ ↓

Fe₂(SO₄)₃ + 6NH₄OH ⟶ 3(NH₄)₂SO₄ + 2Fe(OH)₃ ↓

(c) When zinc nitrate is treated with ammonium hydroxide, white gelatinous ppt. of Zinc hydroxide is formed which is soluble in excess of NH₄OH.

Zn(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Zn(OH)₂ ↓

Whereas, when lead nitrate is treated with ammonium hydroxide, white ppt. of lead hydroxide is formed which is insoluble in excess of NH₄OH

Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂ ↓

(a) Ammonia to nitrogen using an acidic gas :

8NH₃ [excess] + 3Cl₂ ⟶ 6NH₄Cl + N₂

(b) Ammonia to brown gas:

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}} \text{4NO} + \text{6H}_\text{2}\text{O} + \Delta$

2NO + O₂ ⟶ 2NO₂ [brown gas]

(c) Ammonia to nitrogen trichloride

NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(d) Ammonia solution to an amphoteric hydroxide :

AlCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Al(OH)₃

(e) A nitride of a trivalent metal to ammonia:

AlN + 3H₂O ⟶ Al(OH)₃ + NH₃ ↑

(f) Lead oxide to lead

2NH₃ + 3PbO ⟶ 3Pb + 3H₂O + N₂ [g]

(a) On reaction with ammonium hydroxide, Iron [II] sulphate forms a dirty green ppt whereas iron [III] sulphate forms a reddish brown ppt. Hence, the two can be distinguished easily.

FeSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Fe(OH)₂ ↓

Fe₂(SO₄)₃ + 6NH₄OH ⟶ 3(NH₄)₂SO₄ + 2Fe(OH)₃ ↓

(b) A lead salt gives a chalky white ppt. on reaction with ammonium hydroxide that is insoluble in excess of ammonium hydroxide.

Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂ ↓

On the other hand, zinc salt forms a white gelatinous ppt. which dissolves when excess of ammonium hydroxide is added. Hence, the two can be distinguished.

ZnSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Zn(OH)₂ ↓

[Zn(OH)₂] + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Zn(NH₃)₄]SO₄ + 4H₂O

(a) Black copper [II] oxide is reduced to brown copper.

2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ [g]

(b) Green or greenish yellow flame is seen when ammonia gas is burnt in an atmosphere of oxygen. The reaction of combustion of Ammonia is:

4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O

(c) Ammonium hydroxide if first added in small quantity and then in excess to a solution of copper sulphate, a pale blue ppt. of copper hydroxide is formed which dissolves in excess of ammonium hydroxide forming a soluble complex salt.

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

(d) When Al metal is burnt in a jar of nitrogen gas, its nitride i.e., AlN is formed. When warm water is added to AlN, it undergoes hydrolysis and is decomposed by warm water to give pungent smelling ammonia gas. The corresponding insoluble metal hydroxide (i.e., Al(OH)₃) is precipitated out. The reactions are shown below:

2Al + N₂ ⟶ 2AlN

AlN + 3H₂O ⟶ Al(OH)₃ + NH₃ [g]

(e) Colourless ammonia gas reacts with greenish yellow excess chlorine giving a yellow explosive liquid (Nitrogen trichloride).

NH₃ + 3Cl₂ [excess] ⟶ 3HCl + NCl₃

(f) Pungent smelling gas (ammonia) is given out.

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

(a) Ammonia is less dense than air because vapour density of ammonia is 8.5 and that of air is 14.4.

(b) Fountain Experiment demonstrates the high solubility of ammonia gas in water.

(c) 2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₄

Ammonia solution in water gives a blue precipitate when it combines with a solution of copper salt, due to the formation of Cu(OH)₂

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper [II] sulphate, an azure blue soluble complex salt.

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

Ammonia dissociates into nitrogen and hydrogen at high temperature or by electric sparks

2NH₃ ⇌ N₂ + 3H₂

When an ammonium salt is warmed with sodium hydroxide solution, ammonia gas is evolved.

NH₄Cl + NaOH ⟶ NaCl + H₂O + NH₃

Three ways in which ammonia gas can be identified is:

1. It has a sharp characteristic odour.
2. It turns:
   1. moist red litmus blue,
   2. moist turmeric paper brown,
   3. phenolphthalein solution pink.
3. It gives dense white fumes with conc. hydrochloric acid.
   NH₃ + HCl ⟶ NH₄Cl

As the 'A' turns red litmus blue it is a base. Now the gas 'A' combines with 'B' in presence of Catalyst to give colourless gas Nitrogen monoxide. It reacts with oxygen to give brown gas which is Nitrogen dioxide. The gases are given as :

A = NH₃
B = O₂
C = NO
D = NO₂

4NH₃ + 5O₂ $\xrightarrow[800 ^{\circ} \text{C}]{\text{Pt.}}$ 4NO↑ + 6H₂O + Δ

2NO + O₂ ⟶ 2NO₂ [brown gas]

(a) The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer and contribute to global warming. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form.

$\text{CF}_2\text{Cl}_2 \xrightarrow{\text{Ultraviolet}} \text{CF}_2\text{Cl} + \text{Cl} [\text{free radical}]$

The free radical [Cl] reacts with ozone and chlorine monoxide is formed.

Cl + O₃ [ozone] ⟶ ClO + O₂

This causes depletion of ozone. Chlorine monoxide further reacts with atomic oxygen and produces more free radicals.

ClO + O ⟶ Cl + O₂

Again this free radical [Cl] destroys ozone, and the process continues thereby giving rise to ozone depletion.

(b) Liquid ammonia

(c) Advantages of ammonia as refrigerant:

1. Ammonia is environmentally compatible. It does not deplete ozone layer and does not contribute towards global warming.
2. It has superior thermodynamic qualities as result ammonia refrigeration systems use less electricity.
3. Ammonia has a recognizable odour and so leaks are not likely to escape. It being lighter than air goes up in the atmosphere not affecting the life too much on earth.

Disadvantages of ammonia as a refrigerant are as follows:

1. It is not compatible with copper, so it cannot be used in any system with copper pipes.
2. It is poisonous in high concentration.

(a) Ammonia molecule has a lone pair of electrons on it's nitrogen atom.

Due to this lone pair of electrons, in aqueous solutions it forms ammonium ion.

NH₃ + H₂O ⟶ NH₄OH

NH₄OH ⇌ NH₄⁺ + OH^-

(b) Hydroxyl ion [OH^-].

(c) The red litmus paper turns blue due to the presence of hydroxyl ion in the solution.

(a) HCl gas is denser as it's vapour density = 18.25, and that of ammonia is 8.5. HCl gas is collected by the upward displacement of air.

(b) NH₃ + HCl ⟶ NH₄Cl

A : Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃ [g]

B:
$\underset{\text{[aq.]}}{\text{NH}_3} + \underset{\text{[vapour]}}{\text{HCl}} \longrightarrow \underset{\text{[dense white fumes]}}{\text{NH}_4\text{Cl}}$

C: 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃↑

(a) Haber's Process

(b) Nitrogen (N₂) and hydrogen (H₂) in the ratio 1:3 by volume

(c) Finely divided iron (Fe)

(d) N₂ + 3H₂ ⇌ 2NH₃ + Δ

(a) As ammonium nitrate is explosive in nature and may itself decompose forming nitrous oxide and water vapour, hence it is not used in the preparation of ammonia.

$\text{NH}_4\text{NO}_3 \xrightarrow\Delta \text{N}_2\text{O} + \text{2H}_2\text{O}$

(b) Quick lime [CaO] is used as a drying agent during the process.

(c) Downward displacement of air is the method used for the collection of the ammonia gas. As ammonia gas is highly soluble in water, therefore, it is not collected over water.

(a) Ammonia (NH₃) gas is collected.

(b) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

(c) Downward displacement of air is the method used to collect the gas.

(d) Quicklime (CaO) is the drying agent.

(e) A glass rod dipped in conc. HCl acid is brought near the mouth of the jar. If the jar is full-dense white fumes of ammonium chloride are formed.

(a) Gas P — Nitrogen [N₂] and Gas Q — Hydrogen [H₂]

(b) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂ (g)
Ammonia acts as a reducing agent reducing copper oxide to copper and itself is oxidised to nitrogen gas.

(c) Fertiliser S is Ammonium sulphate
2NH₃ + H₂SO₄ ⟶ (NH₄)₂ SO₄

(d) Process T is Ostwald's process Conditions that enable conversion of ammonia to nitric oxide are 800°C temperature and platinum as catalyst.

(e) 4NO₂ (g) + H₂O (l) + O₂ (g) ⟶ 4HNO₃ (aq.)

(f) Fertiliser A is Ammonium nitrate.

(g) Process R is the Haber's process.

(a) Haber's process

(b)
    A — Nitrogen
    B — Hydrogen
    C — Liquid ammonia

(c) Ratio of A and B is 1:3

(d)
Reaction:
N₂ + 3H₂ ⇌ 2NH₃ + heat

Favorable Conditions:

(e) Product is separated from unreacted reactants by Liquification. Ammonia is liquified easily as compared to nitrogen and hydrogen.

Cold dilute nitric acid reacts with copper to form nitric oxide.

Nitric acid (HNO₃) was formerly known aqua fortis, meaning strong water. It is so called because it reacts with nearly all metals. It can even dissolve silver which does not dissolve in other acids.

Conc. nitric acid (1 part by volume) when mixed with conc. hydrochloric acid (3 parts by volume) gives the mixture called Aqua Regia (meaning royal water).

The conversion of free atmospheric nitrogen (N₂) into useful nitrogenous compounds in the soil is known as fixation of nitrogen.

During thunderstorms, the nitrogen present in the atmosphere reacts with oxygen to form nitric oxide.

N₂ + O₂ ⟶ 2NO

Nitric oxide is further oxidized to nitrogen dioxide.

2NO + O₂ ⟶ 2NO₂

The nitrogen dioxide dissolves in atmospheric moisture or rain water in the presence of oxygen of the air and forms nitric acid which is washed down during rain and combines with the salt present on the surface of the earth.

4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

(a) Dry air (free from carbon dioxide and dust particles) and dry ammonia produced in Haber’s process.

(b) Platinum gauze

(c) Oxygen

(d) 1:10 i.e., 1 volume of ammonia and 10 volume of air.

(e) Quartz is acid resistant and when packed in layers, it slows down the movement of the gaseous NO₂ so that it can dissolve in water uniformly.

Below is the equation for the lab. preparation of HNO₃ from potassium nitrate and conc. H₂SO₄:

$\underset{[conc.]}{\text{KNO}_3} + \underset{[conc.]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[acid \space salt]}{\text{KHSO}_4} + \text{HNO}_3$

Conc. HCl is not used in place of conc. Sulphuric Acid (H₂SO₄) because HCl is volatile and hence nitric acid vapous will carry HCl vapours.

Concentrated nitric acid prepared in the laboratory is yellow in color due to dissolution of reddish brown nitrogen dioxide gas in the acid. This gas is produced due to the thermal decomposition of a portion of nitric acid.

4HNO₃ ⟶ 2H₂O + 4NO₂ + O₂

The yellow color of the acid is removed :

1. If dry air or CO₂ is bubbled through the yellow acid, the latter turns colourless because it drives out NO₂ gas from warm acid, which is further oxidised to nitric acid.
2. By adding excess water, nitrogen dioxide gas dissolves in water and thus yellow color of acid is removed.

The temperature of the reaction should not increase above 200°C, because sodium sulphate formed at higher temperature, forms a hard crust which sticks to the walls of the retort and is difficult to remove, although the yield of nitric acid is higher.

2NaNO₃ + H₂SO₄ $\xrightarrow{\gt 200 ^{\circ}\text{C}}$ Na₂SO₄ + 2HNO₃

The higher temperature

• may damage the glass apparatus.
• decomposition of nitric acid can also occur.
• wastage of fuel.

(a) An aqueous solution of nitric acid (68% concentration) forms a constant boiling mixture at 121°C. A constant boiling mixture is one which boils without change in composition.

Hence on boiling further, the mixture evolves out the vapours of both acid and water in the same proportion as in the liquid.

Therefore, dilute HNO₃ cannot be concentrated beyond 68% by boiling.

(b) Iron becomes passive or inert when treated with pure concentrated nitric acid due to the formation of the extremely thin layer of insoluble iron oxide (passivity) which stops the reaction.

Fe + 6HNO₃ ⟶ Fe(NO₃)₃ + 3H₂O + 3NO₂

This inert iron is known as passive iron.

Passivity can be removed by rubbing the surface layer with sandpaper or by treating with strong reducing agent.

(a) The products formed when carbon and conc. nitric acid is heated are carbon dioxide, nitrogen dioxide and water

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(b) The products formed when dilute HNO₃ is added to copper are copper nitrate, nitric oxide and water

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) Action on non-metals:

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Reaction with alkalis:

K₂O + 2HNO₃ ⟶ 2KNO₃ + H₂O

CuO + 2HNO₃ ⟶ Cu(NO₃)₂ + H₂O

(c) Nitric acid acting as oxidising agent:

C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

3Cu + 8HNO₃ [cold and dil.] ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) Sodium hydrogen carbonate when treated with nitric acid forms sodium nitrate, carbon dioxide and water.

NaHCO₃ + HNO₃ ⟶ NaNO₃ + H₂O + CO₂ ↑

(b) Cupric oxide reacts with dilute nitric acid, it forms copper nitrate and water.

CuO + 2HNO₃ ⟶ Cu(NO₃)₂ + H₂O

(c) Zinc reacts with dil. nitric acid to form zinc nitrate, nitric oxide and water

3Zn + 8HNO₃ ⟶ 3Zn(NO₃)₂ + 4H₂O + 2NO

(d) When concentrated nitric acid is heated it forms nitrogen dioxide gas, water and oxygen gas

4HNO₃ $\xrightarrow{\Delta}$ 4NO₂ + 2H₂O + O₂

(a) NaOH + HNO₃ [dil.] ⟶ NaNO₃ + H₂O

(b) CuO + 2HNO₃ [dil.] ⟶ Cu(NO₃)₂ + H₂O

(c) Pb + 4HNO₃ [conc.] ⟶ Pb(NO₃)₂ + 2H₂O + 2NO₂

(d) Mg(OH)₂ + 2HNO₃ [dil.] ⟶ Mg(NO₃)₂ + 2H₂O

(e) Fe + 6HNO₃ [conc.] ⟶ Fe(NO₃)₃ + 3H₂O + 3NO₂

(f) Aqua regia is a freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid mixed in the ratio of 3:1 by volume.

3HCl + HNO₃ ⟶ NOCl + 2H₂O + 2[Cl]

A : 3Cu + 8HNO₃ [cold and dil.] ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

B : 2Cu(NO₃)₂ $\xrightarrow{\Delta}$ 2CuO + 4NO₂ + O₂

C : 2Cu + O₂ ⟶ 2CuO

D : CuO + H₂ $\xrightarrow{\Delta}$ Cu + H₂O

(a) HNO₃ is a strong oxidizing agent

(b) NaNO₃ gives NaNO₂ and O₂ on heating.

(c) Constant boiling nitric acid contains 68% nitric acid by weight

(d) Nitric acid turns yellow when exposed to light.

(e) Magnesium reacts with very dilute (about 1%) nitric acid to liberate hydrogen gas.

KNO₃

Reason — The alkali metal nitrates such as sodium or potassium nitrates when heated melts into colourless liquids which decompose on heating to give oxygen gas.

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$ ↑

FeSO₄

Reason — In the brown ring test the conc. sulphuric acid being heavier settles down and the iron [II] sulphate layer remains above it resulting in formation of the brown ring at the junction.

6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil. ) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO

[Nitroso Ferrous sulphate, a brown compound]

NO₂

Reason — Lead nitrate decomposes on heating to give lead oxide, nitrogen dioxide and oxygen.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

conc. H₂SO₄

Reason — In the laboratory preparation of Nitric acid by distillation process, potassium nitrate (nitre) is heated with conc. sulphuric acid (H₂SO₄).

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

Ostwald process

Reason — Nitric acid is manufactured by Ostwald process developed by a german chemist Ostwald, in 1914.

Pt

Reason — The catalyst used in the manufacture of HNO₃ by Ostwald's process is Platinum.

NO

Reason — Cold and dilute HNO₃ reacts with Copper and oxidises it to give Nitric oxide (NO)

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

NO₃^-

Reason — The brown ring test is a qualitative test used to detect the presence of nitrate ions (NO₃^-) in a solution. A brown ring forms at the junction of the two liquids if nitrate ions are present.

H₂SO₄

Reason — Conc. Nitric acid (HNO₃) reacts with non-metals like sulphur to give sulphuric acid and nitrogen dioxide.

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

Freshly prepared FeSO₄

Reason — The solvent for NO is freshly prepared FeSO₄, because upon exposure to the atmosphere, it is oxidised to ferric sulphate which will not give brown ring.

HNO₃

Reason — Nitric acid (HNO₃) is used to etch designs on brassware because it acts as solvent for a large number of metals except noble metals.

decompose in the presence of light

Reason — Nitric acid (HNO₃) decomposes in presence of sunlight even at room temperature and turn into yellow colour solution. To avoid the decomposition, nitric acid is normally stored in reagent bottles.

NH₄NO₃

Reason — Ammonium nitrate (NH₄NO₃) decomposes explosively leaving behind no residue.

Only R

Reason — On heating zinc nitrate, the products formed are zinc oxide, nitrogen dioxide and oxygen. Zinc oxide is yellow when hot, white when cold.

2Zn(NO₃)₂ $\xrightarrow{\Delta}$ 2ZnO + 4NO₂ + O₂

Both A and R are true and R is the correct explanation of A.

Explanation— In the laboratory preparation of Nitric acid, nitric acid is obtained by distilling the reactants, conc. sulphuric acid (H₂SO₄) and potassium nitrate (KNO₃). Hence, the assertion (A) is true.
Sulphuric acid is a non-volatile acid and produces volatile nitric acid on reacting with potassium nitrate or sodium nitrate. Hence, the reason (R) is true.
Reason (R) explains why conc. sulphuric acid (H₂SO₄) and potassium nitrate (KNO₃) are used as reactants for the preparation of nitric acid. Hence, reason (R) is a correct explanation of assertion (A).

A is true but R is false.

Explanation— Pure nitric acid is colourless but the acid obtained in the laboratory is slightly yellow. The yellow colour is due to dissolution of reddish brown coloured nitrogen dioxide gas in the acid. This gas is produced due to the thermal decomposition of a portion of nitric acid. Hence, the assertion (A) is true.

Brown gas is not used in nitric acid preparation but brown coloured nitrogen dioxide gas is produced due to the thermal decomposition of portion of nitric acid. Hence, the reason (R) is false.

A is true but R is false.

Explanation— Catalytic oxidation of Ammonia by Ostwald's process does not require any external heat because the reaction itself is exothermic and the heat evolved maintains the temperature of the catalytic chamber. Hence, the assertion (A) is true.

Catalytic oxidation of ammonia is an exothermic process. Hence, the reason (R) is false.

Both A and R are true and R is the correct explanation of A.

Explanation — The yellow colour of nitric acid is removed by bubbling carbon dioxide gas or dry air into it. Hence, the assertion (A) is true.
The yellow colour will disappear because carbon dioxide gas or dry air drive out NO₂ gas from the warm nitric acid. Hence, the reason (R) is true.

Reason (R) explains how the use of carbon dioxide and dry air turn yellow colour nitric acid into a colourless compound. Hence, reason (R) is a correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation — Nitric acid reacts with alkalies like NaOH/KOH and neutralises it to form salt and water. Hence, the assertion (R) is true.

NaOH + HNO₃ ⟶ NaNO₃ + 2H₂O

Alkalies react with acids to form salt and water. Such type of reactions are called as neutralisation reaction. Hence, the reason (R) is true.

Since, reason (R) depicts the neutralisation reaction that takes place between Nitric acid and NaOH/KOH, it is the correct explanation of assertion (A).

Both A and R are true but R is not the correct explanation of A.

Explanation — Cold and dilute nitric acid oxidises metals to their nitrates and nitric oxide. Hence, the assertion (A) is true.

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

Hot dilute nitric acid or concentrated nitric acid liberates nitrogen dioxide. Hence, the reason (R) is true.

Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

However, reason (R) doesn't explain the action of Cold and dilute nitric acid on metals. So, it is not the correct explanation of assertion (A).

(a) A yellow non-metal, sulphur reacts with nitric acid to liberate NO₂ (Nitrogen dioxide).

(b) Brown ring test is used to identify the nitrate radical.

(c) Dilute HNO₃ reacts with copper evolving NO (Nitric oxide)) gas.

(d) Nitric acid is obtained on a large scale by the Ostwald process and the catalyst used is platinum.

(e) In Brown ring test, freshly prepared Iron (II) sulphate solution is used.

(f) Nitric acid in a very dilute form reacts with magnesium to form hydrogen.

(a) Sodium nitrate
2NaNO₃ $\xrightarrow{\Delta}$ 2NaNO₂ + O₂↑

(b) Ammonium nitrate
NH₄NO₃ $\xrightarrow{\Delta}$ N₂O + 2H₂O

(c) Calcium nitrate
2Ca(NO₃)₂ $\xrightarrow{\Delta}$ 2CaO + 4NO₂ + O₂

(d) Silver nitrate
2AgNO₃ $\xrightarrow{\Delta}$ 2Ag + 2NO₂ + O₂

(e) Freshly prepared ferrous sulphate
6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil.) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO

(f) Nitric oxide
2NO + O₂ ⟶ 2NO₂ (brown gas)

(g) Nitrogen dioxide
Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Nitrogen dioxide gas

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Oxygen gas

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$

(a) Nitric acid

S + 6HNO₃ ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Nitric acid

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[700-800 ^{\circ}\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + 21.5 \text{K cals}$

$2\text{NO} + \text{O}_2 \xrightarrow{50 ^{\circ}\text{C}} \text{2NO}_\text{2}$

4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

Complete table is as follows:

When nitric acid undergoes decomposition it yields nascent oxygen. This nascent oxygen oxidises copper allowing it to react with nitric acid. Hence, the oxidising nature of nitric acid helps it to react with copper.

(a) All glass apparatus is used in the laboratory preparation of nitric acid since the vapours of nitric acid being highly corrosive attack rubber, cork, etc.

(b) When nitric acid is kept in a reagent bottle for a long time, it turns dark yellowish brown in colour.
This is because nitric acid is unstable hence, it decomposes slightly even at ordinary temperatures and in the presence of sunlight. When it is kept for a long time, the decomposition is complete resulting in the formation of reddish brown nitrogen dioxide [NO₂] that dissolves in the acid giving it a darker yellowish brown colour.

4HNO₃ ⟶ O₂ + 2H₂O + 2NO₂

(c) Iron or aluminium are rendered passive on reaction with fuming HNO₃ due to formation of a thin oxide coating on the surface of the metal which prevents further reaction

(d) Dilute nitric acid is generally considered a typical acid except for it's reaction with metals since it generally does not liberate hydrogen on reaction with metals. Nitric oxide on decomposition forms nascent oxygen which oxidizes the hydrogen to water.

(e) Pure nitric acid [HNO₃] is colourless and unstable and decomposes slightly even at ordinary temperatures and in the presence of sunlight.
The decomposition results in formation of reddish brown nitrogen dioxide [NO₂] which remains dissolved in the acid thus imparting a slight yellowish brown colour.

HNO₃ (Nitric acid)

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(a) When concentrated nitric acid is reacted with sulphur, dense brown fumes of nitrogen dioxide are observed.

S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

(b) Lead nitrate decomposes on heating to give nitrogen dioxide, oxygen gas and yellow residue of lead oxide which fuses with the glass of the test tube.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

(c) Reddish brown nitrogen dioxide gas is evolved on heating zinc nitrate crystals.

$2\text{Zn(NO}_3)_2 \xrightarrow{\Delta} 2\text{ZnO} + \text{O}_2 + 4\text{NO}_2$

(d) Reddish brown nitrogen dioxide gas is evolved.

Black coloured residue of CuO is obtained in the test tube.

$2\text{Cu(NO}_3)_2 \xrightarrow{\Delta} 2\text{CuO} + \text{O}_2 + 4\text{NO}_2$

(a) CuCO₃ + 2HNO₃ [dil.] ⟶ Cu(NO₃)₂ + H₂O + CO₂

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(c) Balanced equation for laboratory preparation of nitric acid:

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

(d) Balanced equation for action of heat on a mixture of copper and concentrated nitric acid:

$\text{Cu} + 4\underset{[\text{conc.}]}{\text{HNO}_3} \longrightarrow \text{Cu(NO}_3 \text{)}_2 + 2\text{H}_2 \text{O} + 2\text{NO}_2$

(a) The common anion is the nitrate ion.

(b) The cation in X is either sodium or potassium

(c) The metal ion present in Y is the lead ion.

(d) Z contains the ammonium ion

(e) The balanced equations are:

1. Reaction between Potassium nitrate and concentrated sulphuric acid (below 200° C):

$\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

2. 2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

3. Cu + 4HNO₃ (conc.) ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂.

Explanation

(a) On adding copper to acidified nitrates and heating, dense reddish brown fumes of nitrogen dioxide [NO₂] are evolved.

(b) The alkali metal nitrates such as sodium or potassium nitrates when heated melts into colourless liquids which decompose on heating to give oxygen gas.

$2\text{KNO}_3 \xrightarrow{\Delta} 2\text{KNO}_2 + \text{O}_2$

$2\text{NaNO}_3 \xrightarrow{\Delta} 2\text{NaNO}_2 + \text{O}_2$

(c) Lead nitrate decomposes on heating to give nitrogen dioxide, oxygen gas and yellow residue of lead oxide which fuses with the glass of the test tube.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + 4NO₂ + O₂

(d) Ammonium nitrate decomposes explosively on heating leaving behind no residue.

NH₄NO₃ $\xrightarrow{\Delta}$ N₂O + 2H₂O

Ammonium nitrate reacts with sodium hydroxide to give sodium nitrate, ammonia, and water. Ammonia gas turns moist red litmus paper blue.

NH₄NO₃ + NaOH ⟶ NaNO₃ + NH₃ + H₂O

(a) In the catalytic oxidation of ammonia by Ostwald's process, the temperature should be 800° C and catalyst used is platinum.

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 ^{\circ}\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + \text{Heat}$

(b) 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃

(a)

1. To etch designs on copper and brassware.
   Property : Acts as solvent for a large number of metals except noble metals
2. To purify gold.
   Property : Gold may contain Cu, Ag, Zn, Pb etc. as impurities which dissolve in nitric acid.
3. To prepare aqua regia.
   Property : It dissolves noble metals

(b) Brown ring test for nitric acid

To the aq. solution of a nitrate or nitric acid :

1. Add freshly prepared saturated solution of iron [II] sulphate.
2. Now add conc. sulphuric acid carefully from the sides of the test tube, so that it should not fall drop-wise in the test tube.
3. Cool the test tube in water.
4. A brown ring appears at the junction of the two liquids.

Equations for brown ring test:

6FeSO₄ + 3H₂SO₄ + 2HNO₃ (dil. ) ⟶ 3Fe₂(SO₄)₃ + 4H₂O +2NO

FeSO₄ + NO ⟶ FeSO₄.NO
[Nitroso Ferrous sulphate, a brown compound]

(c) A freshly prepared ferrous sulphate solution is used, because on exposure to the atmosphere, it is oxidised to ferric sulphate, which will not give the brown ring test.

(a) X is copper nitrate [Cu(NO₃)₂],
     Y is nitrogen dioxide [NO₂] and
     Z is hydrogen sulphide [H₂S].

(b) 2Cu(NO₃)₂ $\xrightarrow{\Delta}$ 2CuO + 4NO₂ + O₂

(c) When a H₂S gas, which has a rotten egg smell, is passed through a solution of Cu(NO₃)₂, a black ppt. of CuS is formed.
Cu(NO₃)₂ + H₂S ⟶ CuS + 2HNO₃

(a) Dilute nitric acid is considered a typical acid except for it's reaction with metals since it does not liberate hydrogen. It is a powerful oxidising agent and the nascent oxygen formed oxidises the hydrogen to water.

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

3Zn + 8HNO₃ ⟶ 3Zn(NO₃)₂ + 4H₂O + 2NO

3Fe + 8HNO₃ ⟶ 3Fe(NO3)2 + 4 H₂O + 2NO

(b) Reaction with dil. HNO₃:

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

Reaction with conc. HNO₃:

Cu + 4HNO₃ (conc.) ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂.

(a) A — concentrated sulphuric acid,

B — sodium nitrate or potassium nitrate and

C — is nitric acid.

(b) Nitric acid decomposes to liberate nitrogen dioxide along with oxygen and form water.

4HNO₃ $\xrightarrow{\Delta}$ 4NO₂ + 2H₂O + O₂

(c) Copper metal is oxidised by concentrated nitric acid to copper nitrate.

Cu + 4HNO₃ $\xrightarrow{\Delta}$ Cu(NO₃)₂ + 2H₂O + 2NO₂

(a) Freshly prepared FeSO₄ is used in the brown ring test, because upon exposure to the atmosphere, it is oxidised to ferric sulphate which will not give brown ring.

(b) The brown ring of nitroso ferrous sulphate decomposes on disturbing the test tube. The heat evolved decomposes the unstable brown ring.

(c) Lead nitrate (Pb(NO₃)₂) contains Pb²⁺ ions, which can form an insoluble precipitate of PbSO₄ when H₂SO₄ (sulphuric acid) is added.

This precipitate interferes with the formation of the necessary layers and prevents proper diffusion and interaction with Fe²⁺, thus suppressing the brown ring formation.

(d) Nitroso ferrous sulphate - FeSO₄.NO

(a) Sulphuric acid is referred as the King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries. It has been known for a long time.

(b) Sulphuric acid is referred to as Oil of vitriol because in the later middle ages, it was obtained as an oily viscous liquid by heating crystals of green vitriol.

(a) 4FeS₂ (iron pyrite) + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂ ↑

S + O ⟶ SO₂ ↑

(b) The conditions for the oxidation of SO₂ are:

1. The temperature should be as low as possible. The yield has been found to be maximum at about 410 ⁰C – 450 ⁰C.
2. High pressure favours the reaction because the product formed has less volume than reactant. Pressure : 1 to 2 atmosphere.
3. Excess of oxygen increases the production of sulphur trioxide.

(c) Vanadium pentoxide [V₂O₅]

(d) Sulphuric acid is not obtained by directly reacting SO₃ with water since sulphur trioxide does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid.

(e) The chemical used to dissolve SO₃ is concentrated sulphuric acid and the product formed is oleum.

Main reactions of this process are:

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

Impurity of arsenic oxide must be removed before passing the mixture of SO₂ and air through the catalytic chamber otherwise the catalyst loses it's efficiency.

(a) Vanadium pentoxide

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons [450-500\text{^{\circ} C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \text{45 K cals}$

(b) Main reactions of this process are:

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

(c) Dilute Sulphuric acid

Na₂SO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + SO₂[g] ↑

(d) 2NaOH + SO₂ ⟶ Na₂SO₃ + H₂O

Hydrogen sulphide

Reason —

FeS + H₂SO₄ [dil.] ⟶ FeSO₄ + H₂S

Oxidising agent

Reason — The oxidising property of concentrated sulphuric acid leads to its thermal decomposition, and yields nascent oxygen [O]. In the above reaction, Hydrogen sulphide is oxidised by this nascent oxygen released from sulphuric acid.

Lead nitrate

Reason —

Pb(NO₃)₂ + H₂SO₄ (dil) ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed]

White ppt. of PbSO₄ is visible.

H₂SO₄

Reason — On passing SO₂ through chlorine water, oxidation of aqueous solution of sulphur dioxide takes place and H₂SO₄ is formed.

SO₂ + H₂O + Cl₂ ⟶ H₂SO₄ + 2HCl

Non volatile acid

Reason — Concentrated sulphuric acid is used for preparing volatile acid hydrochloric acid from their salts by double decomposition, because sulphuric acid is considered to be a non-volatile acid with high boiling point (338 °C).

V₂O₅

Reason — The catalyst preferred in contact process is vanadium pentoxide (V₂O₅). It is highly efficient at converting SO₂ to SO₃.

H₂SO₄

Reason — The substance added to SO₃ to manufacture H₂SO₄ by contact process is H₂SO₄. Sulphur trioxide on dissolving in water can produce sulphuric acid, but it does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid so it is not directly absorbed by water. It is first absorbed by sulphuric acid to produce oleum. oleum is further diluted to produce H₂SO₄.

Oxidising agent

Reason — The oxidising property of concentrated sulphuric acid leads to its thermal decomposition, and yields nascent oxygen [O]. In the above reaction, sulphur is oxidised by this nascent oxygen released from sulphuric acid.

Copper

Reason — Dilute H₂SO₄ and conc. H₂SO₄ can be distinguished by Copper. Copper does not react with dilute H₂SO₄, because dilute H₂SO₄ cannot act as an oxidising agent. Whereas, conc. sulphuric acid on heating with copper evolves sulphur dioxide.

black

Reason — On adding conc. H₂SO₄ to cane sugar (carbohydrate), they react immediately to give a black spongy mass of carbon.

Conc. H₂SO₄

Reason — All carbohydrates react with Conc. H₂SO₄ immediately to give a black spongy mass of carbon. Carbon obtained in this reaction is very pure and is called as sugar charcoal.

Conc. H₂SO₄

Reason — Conc. H₂SO₄ has a great affinity for water. It readily removes chemically combined water molecules i.e., elements of water from other compounds. Hence, it acts as drying as well as dehydrating agent.

Both P and Q

Reason — The compounds formed by the oxidation of sulphur by nitric acid are Sulphuric acid (H₂SO₄) and Nitrogen dioxide (NO₂).

S + 6HNO₃ ⟶ H₂SO₄ + 6NO₂ + 2H₂O

Both A and R are true and R is the correct explanation of A.

Explanation— Sulphuric acid is referred to as Oil of vitriol because in the later middle ages, it was obtained as an oily viscous liquid by heating crystals of green vitriol. Hence, the assertion (A) is true.
Sulphuric acid was first obtained by heating crystals of green vitriol. Hence, reason (R) is true.

Reason (R) explains why sulphuric acid is known as the oil of vitriol. Hence, reason (R) is the correct explanation of assertion (A).

Both A and R are true but R is not the correct explanation of A.

Explanation— In the catalytic oxidation of sulphur dioxide, vanadium pentoxide (V₂O₅) or platinum (Pt) is used as a catalyst. Hence, the assertion (A) is true.
The catalytic oxidation of sulphur dioxide is highly exothermic, the catalyst is heated only in the beginning to about 450°C. This temperature is maintained by the heat evolved during the reaction. Hence, the reason (R) is true.

Reason (R) doesn't explain why V₂O₅ or Pt is used as the catalyst. Hence, reason (R) is not the correct explanation of assertion (A).

A is false but R is true.

Explanation— Sulphuric acid when dissolved in water forms hydronium ion and shows acidic properties. Hence, the assertion (A) is false.

Sulphuric acid ionises in two stages and is called dibasic acid. Hence, the reason (R) is true.

H₂SO₄ + 2H₂O $\rightleftharpoons$ H₃O⁺ + HSO₄^-

HSO₄^- + 2H₂O $\rightleftharpoons$ H₃O⁺ + SO₄^2-

Both A and R are true but R is not the correct explanation of A.

Explanation— Pure (concentrated) H₂SO₄ is not ionized significantly because no water is present to dissociate it into ions. Therefore, it does not conduct electricity. Hence, the assertion (A) is true.
When H₂SO₄ is diluted with water, it ionizes completely to give H^- and SO₄^2- ions, which are free to move. These ions carry electric current, making it a strong electrolyte. Hence, the reason (R) is true.

However, reason (R) does not explain why pure sulphuric acid is a poor conductor of electricity. Hence, reason (R) is not the correct explanation of assertion (A).

Both A and R are true and R is the correct explanation of A.

Explanation— In the manufacture of sulphuric acid, purification of gases is done by removing the impurity of arsenic oxide before passing the mixture of SO₂ and air through the catalytic chamber. Hence, the assertion (A) is true.
Purification of gases is done because arsenic oxide poisons the catalyst and the catalyst loses its efficiency. Hence, the reason (R) is true and it is the correct explanation of assertion (A).

A is true but R is false.

Explanation— As concentrated sulphuric acid is hygroscopic and has a great affinity for water hence, it absorbs moisture from the atmosphere. Therefore, it is kept in air tight bottles. Hence, the assertion (A) is true.
The main concern to keep sulphuric acid in air tight bottles is to avoid moisture from entering, not impurities. Impurities in air do not significantly affect concentrated sulphuric acid. Hence, the reason (R) is false.

Both A and R are true but R is not the correct explanation of A.

Explanation— Sulphuric acid acts as an oxidising agent due to the fact that on thermal decomposition, it yields nascent oxygen [O]. This nascent oxygen oxidises non-metals, metals and inorganic compounds. Hence, the assertion (A) is true.

H₂SO₄ ⟶ 2H₂O + SO₂ + [O]

Dilute sulphuric acid reacts with metals like zinc, that is present above hydrogen in the activity series to form metallic sulphate and hydrogen at ordinary temperature. Hence, the reason (R) is true.

Zn + H₂SO₄ ⟶ ZnSO₄ + H₂ ↑

But, reason (R) doesn't explain the reason behind oxidising property of sulphuric acid. Hence, reason (R) is not the correct explanation of assertion (A).

(i) Dehydrating agent

(ii) Non-volatile acid

(iii) Typical acid

(iv) Typical acid

(a) B: Dehydrating agent

(b) D: Oxidizing agent

(c) C: Non-volatile acid

(d) A: Dilute acid

(e) D: Oxidising agent

(a) Sulphur can be oxidised to sulphur dioxide by conc. sulphuric acid.

(b) Conc. sulphuric acid convert ethanol to ethene.

(c) Sulphate ion can be tested by barium chloride solution or lead nitrate.

(d) Sulphur trioxide reacts with conc. sulphuric acid to form oleum.

(e) The basicity of sulphuric acid is two. It can form acid salt and normal salt.

(f) Metal sulphate solution gives white precipitate with lead nitrate solution.

(a) Active metal + acid ⟶ metallic salt + hydrogen gas

(b) Base + Acid ⟶ salt + water

(c) Carbonate/hydrogen carbonate + Acid ⟶ salt + water + carbon dioxide

(d) Sulphite/hydrogen sulphite + Acid ⟶ salt + water + sulphur dioxide

(e) Sulphide + Acid ⟶ salt + hydrogen sulphide

(a) Sulphurous acid

(b) Carbon dioxide gas

(c) Liquid E is Ethanol [C₂H₅OH]
     C₂H₅OH + H₂SO₄ (conc.) ⟶ C₂H₄ + H₂O

(d) Hydrogen sulphide gas (H₂S)
     H₂S + H₂SO₄ [conc.] ⟶ S + SO₂ + 2H₂O

(e) Dilute sulphuric acid.

(f) Hydrogen sulphide (H₂S)

(g) Dilute sulphuric acid.
     BaCl₂ + H₂SO₄ (dil) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

(h) Conc. sulphuric acid

(i) Dilute sulphuric acid

(a) Sulphite

(b) Sulphate

(a) Dehydrating agent.
C₁₂H₂₂O₁₁ + nH₂SO₄ ⟶ 12C + 11H₂O

(b) Conc. H₂SO₄ is a non-volatile acid.

(a) Sulphuric acid behaves as a non-volatile acid.

$\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 ^{\circ}\text{C}} \text{KHSO}_\text{4} + \text{HNO}_\bold{3}$

(b) Sulphuric acid behaves as an oxidizing agent

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂

(c) Dehydrating property

C₂H₅OH $\xrightarrow[170^{\circ}\text{C}]{\text{Conc. H}_2\text{SO}_4}$ C₂H₄↑ + H₂O

(a) Sulphuric acid forms two types of salts i.e., acid salt and normal salt with an NaOH (alkali) since it's basicity is two.

NaOH [insufficient] + H₂SO₄ ⟶ H₂O + NaHSO₄ [acid salt]

2NaOH [excess] + H₂SO₄ ⟶ 2H₂O + Na₂SO₄ [normal salt]

(b) A piece of wood becomes black because of the dehydrating property of sulphuric acid. It readily removes elements of water from other compounds.

$[\text{C}_{6}\text{H}_{10}\text{O}_{5}]\text{n}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6[C]}_\text{n} + \text{5[H}_2\text{O}]_\text{n}$

(c) When oil vitriol is added to sodium carbonate brisk effervescence are seen due to the evolution of Carbon dioxide.

Na₂CO₃ + H₂SO₄ ⟶ H₂O + Na₂SO₄ + CO₂ ↑

Water is never poured on concentrated H₂SO₄ to dilute it as large amount of heat is evolved which changes poured water to steam. The steam so formed causes spurting of acid which can cause burn injuries, so dilution is done by pouring acid on a given amount of water in a controlled manner by continuous stirring, else acid being heavier will settle down. The evolved heat is dissipated in the water itself and hence the spurting of the acid is minimized.

(a) As concentrated sulphuric acid is hygroscopic and has a great affinity for water hence, it absorbs moisture from the atmosphere. Therefore, it is kept in air tight bottles.

(b) Sulphuric acid is not a drying agent for H₂S because it reacts with H₂S to form sulphur.
H₂S + H₂SO₄ ⟶ S + 2H₂O + SO₂ ↑

(c) Concentrated sulphuric acid has a high boiling point (338°C) and so it is considered to be a non-volatile acid. It is, therefore, used for preparing volatile acids like hydrochloric acid and nitric acid from their salts by double decomposition.

NaCl + H₂SO₄ [conc.] $\xrightarrow{\lt 200 ^{\circ}\text{C}}$ NaHSO₄ + HCl

NaNO₃ + H₂SO₄ [conc.] $\xrightarrow{\lt 200 ^{\circ}\text{C}}$ NaHSO₄ + HNO₃

(a) Steam is evolved from the test tube and it becomes very hot. Black spongy mass of carbon is formed.

$\underset{\text{Cane Sugar}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} \text{ (s)} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{Sugar Charcoal}}{12\text{C (s)}} + 11\text{H}_2\text{O}$

(b) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

(a) Na₂SO₃ + H₂SO₄ (dil.) ⟶ Na₂SO₄ + H₂O + SO₂

(b) Contact process:

Sulphur or Pyrite Burner:
S + O₂ ⟶ SO₂
4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

Contact Tower :

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{^{\circ} C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + 45 \text{ K cals}$

Absorption Tower :
SO₃ + H₂SO₄ ⟶ H₂S₂O₇

Dilution Tank:
H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

(c) S + 2H₂SO₄ [conc.] ⟶ 3SO₂ + 2H₂O

(d) $\underset{\text{Cane Sugar}}{\text{C}_{12}\text{H}_{22}\text{O}_{11}} \text{ (s)} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{Sugar Charcoal}}{12\text{C (s)}} + 11\text{H}_2\text{O}$

(e) C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

(a) BaCl₂ + H₂SO₄ (dil) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

(b) Na₂S + H₂SO₄ ⟶ Na₂SO₄ + H₂S

(c) ZnS + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂S

Conc. sulphuric acid acts as an oxidising agent and oxidises metals like copper and Zinc.

Cu + 2H₂SO₄ (conc.) ⟶ CuSO₄ + H₂O + 2SO₂ ↑

Zn + 2H₂SO₄ (conc.) ⟶ ZnSO₄ + H₂O + 2SO₂ ↑

Whereas, dil. sulphuric acid is not an oxidising agent and it cannot oxidise copper and zinc.

Hence, Conc. sulphuric acid and dil. sulphuric acid can be distinguished by considering the oxidising property of sulphuric acid.

(a) Acidic nature

1. Mg + H₂SO₄ (dil.) ⟶ MgSO₄ + H₂ ↑
2. Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂ ↑

(b) Oxidising agent

1. C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑
2. Zn + 2H₂SO₄ ⟶ ZnSO₄ + 2H₂O + SO₂ ↑

(c) Dehydrating Nature

1. $\text{HCOOH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O}$
2. $\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170^{\circ}\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} ↑ + \text{H}_2\text{O}$

(d) Non-volatile Nature

1. NaCl + H₂SO₄ ⟶ NaHSO₄ + HCl
2. KCl + H₂SO₄ ⟶ KHSO₄ + HCl

(a) When dil sulphuric acid is reacted with lead nitrate it produces a white ppt. of lead sulphate which remains insoluble even on heating.

Pb(NO₃)₂ + H₂SO₄ (dil) ⟶ 2HNO₃ + PbSO₄ ↓ [white ppt. formed]

When dil hydrochloric acid is reacted with lead nitrate, a white ppt. of lead chloride is formed which dissolves on heating.

Pb(NO₃)₂ + 2HCl ⟶ 2HNO₃ + PbCl₂ ↓ [white ppt. formed]

(b) Dilute sulphuric acid when reacted with zinc produces Hydrogen gas which burns with a pop sound.

Zn + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂ ↑

Concentrated sulphuric acid when reacted with zinc produces SO₂ gas which turns acidified potassium dichromate paper green.

Zn + 2H₂SO₄ [conc.] ⟶ ZnSO₄ + 2H₂O + SO₂ ↑

(c) Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO₄) but with dilute hydrochloric acid no white ppt. is produced since barium chloride is soluble in dil. sulphuric acid.

H₂SO₄ (dil.) + BaCl₂ (aq.) ⟶ 2HCl + BaSO₄ ↓ [white ppt. formed]

HCl (dil.) + BaCl₂ (aq.) ⟶ No reaction

The conditions for the conversion of sulphur dioxide to sulphur trioxide are:

1. Temperature — 450 to 500°C
2. Pressure — 1 to 2 atmosphere
3. Presence of Catalyst — Vanadium pentoxide (V₂O₅)
4. Excess of Oxygen

(a) Zn + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂

(b) Na₂SO₃ + H₂SO₄ [dil.] ⟶ Na₂SO₄ + H₂O + SO₂