Q4. Short questions

i) How does nitrogen differ from other elements of its group?

• Answer: Nitrogen differs from other Group 15 elements due to its small size, high electronegativity, and ability to form multiple bonds (N≡N). It shows maximum covalency of four, while heavier elements like phosphorus and arsenic can show higher covalency. Nitrogen also forms a stable diatomic molecule (N₂), whereas other elements do not. Additionally, nitrogen does not have vacant d-orbitals, limiting its ability to expand its octet, unlike other Group 15 elements.

ii) Why does aqua regia dissolve gold and platinum?

• Answer: Aqua regia, a mixture of concentrated hydrochloric acid (HCl) and concentrated nitric acid (HNO₃), dissolves gold and platinum due to the formation of chloroauric acid (HAuCl₄) or chloroplatinic acid (H₂PtCl₆). Nitric acid oxidizes the metal to form metal ions, while hydrochloric acid provides chloride ions to stabilize these metal ions in solution as complex ions. The overall reactions for gold and platinum are:
  Au + 3HNO₃ + 4HCl → HAuCl₄ + 3NO₂ + 2H₂O
  Pt + 4HNO₃ + 6HCl → H₂PtCl₆ + 4NO₂ + 2H₂O

iii) Why do the elements of Group VIA other than oxygen show more than two oxidation states?

• Answer: Group VIA elements like sulfur, selenium, and tellurium have access to vacant d-orbitals, allowing them to show higher oxidation states (+4, +6) in addition to the common -2 oxidation state. Oxygen, due to its small size and absence of d-orbitals, is limited to -2 and rarely shows other oxidation states.

iv) Write down a comparison of the properties of oxygen and sulphur.

• Answer:
• Oxygen is a diatomic gas (O₂) at room temperature, while sulfur exists as a solid (S₈) with puckered rings.
• Oxygen is more electronegative (3.44) than sulfur (2.58).
• Oxygen forms strong hydrogen bonds in water, making water a liquid, while sulfur does not form hydrogen bonds.
• Oxidation States: Oxygen generally shows -2 oxidation states, while sulfur shows a range of oxidation states (-2, +4, +6).
• Chemical Reactivity: Oxygen is more reactive than sulfur, forming oxides with almost all elements, while sulfur is less reactive and primarily reacts at higher temperatures.

v) Write down the equation for the reaction between conc. H₂SO₄ and copper and explain what type of reaction it is.

• Answer:
  The reaction between concentrated sulfuric acid (H₂SO₄) and copper (Cu) is a redox reaction where sulfuric acid acts as an oxidizing agent. The equation is:
  Cu + 2H₂SO₄ (conc) → CuSO₄ + SO₂ + 2H₂O
  Copper is oxidized to Cu²⁺, and sulfur in H₂SO₄ is reduced from +6 in H₂SO₄ to +4 in SO₂.

Q5.

(a) Explain the Birkeland and Eyde’s process for the manufacture of nitric acid.

• Answer: The Birkeland and Eyde process is an older method for producing nitric acid (HNO₃) by oxidizing nitrogen gas (N₂) from the atmosphere to nitric oxide (NO) using an electric arc. The reaction proceeds as follows:
  N₂ + O₂ → 2NO
  The nitric oxide is further oxidized to nitrogen dioxide (NO₂), which dissolves in water to produce nitric acid:
  2NO + O₂ → 2NO₂
  3NO₂ + H₂O → 2HNO₃ + NO
  This process was replaced by the Ostwald process, which is more efficient.

(b) Which metals evolve hydrogen upon reaction with nitric acid? Illustrate along with chemical equations.

• Answer: Most metals do not evolve hydrogen when reacting with nitric acid because nitric acid is a strong oxidizing agent, and it reduces to nitrogen oxides instead. However, metals like magnesium (Mg) and manganese (Mn) can release hydrogen when reacting with dilute nitric acid under specific conditions. For example:
  Mg + 2HNO₃ (dil) → Mg(NO₃)₂ + H₂
  Mn + 2HNO₃ (dil) → Mn(NO₃)₂ + H₂

(c) What is meant by fuming nitric acid?

• Answer: Fuming nitric acid refers to concentrated nitric acid that contains dissolved nitrogen dioxide (NO₂), giving it a red-brown color and producing fumes. It is more corrosive and reactive than regular concentrated nitric acid and is typically used in nitration reactions.

Q6.

(a) Sulphuric acid is said to act as an acid, an oxidizing agent, and a dehydrating agent. Describe two reactions in each case to illustrate the truth of this statement.

• Answer:
• As an Acid:
  1. Reaction with metals:
     Zn + H₂SO₄ → ZnSO₄ + H₂
     Sulphuric acid reacts with metals like zinc, liberating hydrogen gas.
  2. Reaction with bases:
     H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
     Neutralization reaction with sodium hydroxide (NaOH).
• As an Oxidizing Agent:
  1. Reaction with copper:
     Cu + 2H₂SO₄ (conc) → CuSO₄ + SO₂ + 2H₂O
     Concentrated H₂SO₄ oxidizes copper, releasing sulfur dioxide (SO₂).
  2. Reaction with carbon:
     C + 2H₂SO₄ (conc) → CO₂ + 2SO₂ + 2H₂O
     Sulfuric acid oxidizes carbon to carbon dioxide (CO₂) and sulfur dioxide (SO₂).
• As a Dehydrating Agent:
  1. Dehydration of sugar:
     C₁₂H₂₂O₁₁ (sugar) + H₂SO₄ → 12C + 11H₂O
     Sulfuric acid removes water from sugar, leaving behind carbon.
  2. Dehydration of ethanol:
     C₂H₅OH → C₂H₄ + H₂O
     Concentrated sulfuric acid dehydrates ethanol to form ethene (C₂H₄).

(b) Give the advantages of the contact process for the manufacture of sulphuric acid.

• Answer:

1. Higher Efficiency: The contact process is highly efficient and capable of producing large quantities of sulfuric acid.
2. Purity: The acid produced by the contact process is highly pure (around 98% concentration).
3. Economic Viability: The process is cost-effective due to the recycling of raw materials like sulfur dioxide.
4. Environmental Benefits: The contact process emits fewer pollutants compared to older methods.

Q.7 (a)

The industrial preparation of sulfuric acid involves the Contact Process. The steps are as follows:

1. Sulfur is burned in the air to produce sulfur dioxide (SO₂).

2. The sulfur dioxide is then oxidized to sulfur trioxide (SO₃) using a vanadium oxide (V₂O₅) catalyst in the presence of excess oxygen.

3. The sulfur trioxide is dissolved in concentrated sulfuric acid to produce oleum (H₂S₂O₇).

4. Finally, oleum is diluted with water to form sulfuric acid (H₂SO₄).

Q.7 (b)

SO₃ is dissolved in H₂SO₄ instead of water because it reacts violently with water, producing a mist of sulfuric acid that is difficult to handle. The reaction with sulfuric acid is more controlled, producing oleum which is later diluted with water.

Q.7 (c)

Sulfuric acid is a strong acid and acts as both an oxidizing agent and a dehydrating agent. It reacts with metals to form metal sulfate and hydrogen gas. For example:

Zn + H₂SO₄ → ZnSO₄ + H₂

Q.8

NO₂ can be prepared by heating concentrated nitric acid with copper:

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

NO₂ is a reddish-brown gas and is a major component of air pollution.

Q.9

PCl₃ and PCl₅ are used as chlorinating agents in organic chemistry. They are used to convert alcohols to alkyl chlorides and carboxylic acids to acyl chlorides.

For example:

CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

Q.10 (i)

The ‘Ring test’ for nitrates involves adding concentrated sulfuric acid and iron(II) sulfate to a nitrate solution. A brown ring forms at the interface, indicating the presence of nitrate ions.

Q.10 (ii)

NO₂ is a strong oxidizing agent. It can oxidize metals such as copper and non-metals such as carbon:

C + 2NO₂ → CO₂ + 2NO

Q.10 (iii)

When HNO₃ reacts with arsenic and antimony, it forms arsenic and antimony oxides and nitrogen dioxide gas.

For example:

As + 5HNO₃ → H₃AsO₄ + NO₂ + H₂O

Q.10 (iv)

Phosphorus trichloride (PCl₃) is prepared by reacting phosphorus with chlorine:

P₄ + 6Cl₂ → 4PCl₃

Q.10 (v)

Phosphorus pentoxide (P₂O₅) is a powerful dehydrating agent. It can be used to dehydrate acetic acid to ketene:

CH₃COOH → CH₂=C=O + H₂O

Q.11

Complete and balance the following chemical equations:

i) P + NO → P₂O₃

ii) NO + Cl₂ → NOCl

iii) H₂S + NO → S + H₂O + N₂

iv) Pb(NO₃)₂ → PbO + NO₂ + O₂

v) NO₂ + H₂O → HNO₃ + NO

vi) NO + H₂SO₄ → NO₂ + H₂O

vii) HNO₃ + HI → I₂ + H₂O + NO₂

viii) HNO₃ + (COOH)₂ → CO₂ + H₂O + NO₂

ix) KNO₃ + H₂SO₄ → HNO₃ + K₂SO₄

Q.12

Phosphorus pentoxide is prepared by burning white phosphorus in excess oxygen:

P₄ + 5O₂ → P₄O₁₀

Phosphorus pentoxide is a powerful dehydrating agent and is used in organic synthesis for removing water molecules.

Q.13

The Group VIA elements (chalcogens) show the following trends in physical properties:

1. Atomic size increases down the group as the number of electron shells increases.

2. Ionization energy decreases down the group due to increased atomic size and shielding effect.

3. Electronegativity decreases as atomic size increases.

4. Melting and boiling points increase down the group, except for oxygen.