NEET Chemistry: The p-Block Elements (Oxygen Family) – Detailed Notes and Practice Questions

Chapter 7B: The p-Block Elements (Oxygen Family – Group 16)

1. Introduction to Group 16 Elements

• Group 16 elements (Oxygen Family or Chalcogens) include Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), and Polonium (Po).
• The general electronic configuration is ns2np4.
• The name “Chalcogens” means ‘ore forming elements’, as many metal ores are oxides or sulphides.
• Oxygen, Sulphur, and Selenium are non-metals. Tellurium is a metalloid, and Polonium is a metal (and radioactive).

2. General Characteristics of Group 16 Elements

A. Electronic Configuration:

• ns2np4.
• O: [He]2s22p4
• S: [Ne]3s23p4
• Se: [Ar]3d104s24p4
• Te: [Kr]4d105s25p4
• Po: [Xe]4f145d106s26p4

B. Atomic and Ionic Radii:

• Atomic and ionic radii generally increase down the group due to the addition of new electron shells.

C. Ionization Enthalpy:

• Generally decreases down the group due to increasing atomic size.
• Group 16 elements have lower ionization enthalpies compared to Group 15 elements (due to less stable np4 configuration vs np3 half-filled configuration) and higher than Group 13 elements in corresponding periods.

D. Electronegativity:

• Electronegativity generally decreases down the group.
• Oxygen is the second most electronegative element in the periodic table (after Fluorine).

E. Electron Gain Enthalpy:

• Oxygen has a less negative electron gain enthalpy than Sulfur. This is due to the small size of oxygen, leading to significant inter-electronic repulsions in the relatively compact 2p subshell when an electron is added.
• Electron gain enthalpy becomes less negative (or more positive) down the group from S onwards.

F. Oxidation States:

• Common oxidation states are −2,+2,+4, and +6.
• −2 Oxidation State:
  • Most common for Oxygen (e.g., in oxides), except in peroxides (−1) and compounds with fluorine (+2 in OF2​, +1 in O2​F2​).
  • Tendency to exhibit −2 oxidation state decreases down the group as metallic character increases.
• +2,+4,+6 Oxidation States:
  • Sulphur, Selenium, and Tellurium commonly show +2,+4, and +6 oxidation states.
  • The stability of the +6 oxidation state decreases down the group, while the stability of the +4 oxidation state increases down the group. This is due to the inert pair effect.
  • Polonium mainly shows +2 and +4 oxidation states. The +6 state is very rare for Polonium.

G. Allotropy:

• All elements of Group 16 exhibit allotropy.
• Oxygen: Dioxygen (O2​), Ozone (O3​).
• Sulphur: Rhombic (alpha-sulphur), Monoclinic (beta-sulphur), Plastic sulphur, and others. Rhombic sulphur is the most stable form.
• Selenium: Red (amorphous), Grey (metallic/crystalline), Black.
• Tellurium: Crystalline, Amorphous.
• Polonium: Alpha and Beta forms.

H. Metallic Character:

• Non-metallic character decreases down the group.
• Oxygen, Sulphur: Non-metals.
• Selenium, Tellurium: Metalloids.
• Polonium: Metal.

3. Anomalous Behaviour of Oxygen

Oxygen differs from other members of its group due to:

1. Small size.
2. High electronegativity.
3. Absence of d-orbitals in its valence shell. (This is the most significant reason for many differences).

• Diatomic molecule (O2​): Oxygen exists as a diatomic molecule with a double bond. Other elements exist as polyatomic molecules (e.g., S8​).
• Maximum covalency: Oxygen can exhibit a maximum covalency of 2 (e.g., in H2​O). Other elements can extend their covalency beyond 2 due to the presence of vacant d-orbitals (e.g., SF6​ where S shows covalency of 6).
• Hydrogen Bonding: Oxygen forms strong hydrogen bonds (e.g., in H2​O, alcohols) due to its high electronegativity, which is not prominent in hydrides of other elements.
• Physical state of hydrides: H2​O is a liquid, while H2​S is a gas (at room temperature) due to hydrogen bonding in water.
• Acidic Nature of Hydrides: The acidic nature of hydrides increases down the group (H2​O<H2​S<H2​Se<H2​Te).

4. Important Compounds of Oxygen

A. Dioxygen (O2​):

• Preparation:
  • Laboratory:
    • By heating KClO3​ with MnO2​ catalyst: 2KClO3​MnO2​,Heat​2KCl+3O2​
    • By decomposition of hydrogen peroxide: 2H2​O2​MnO2​ or Heat​2H2​O+O2​
  • Industrial: From air by fractional distillation of liquid air or by electrolysis of water.
• Properties: Colourless, odourless gas. Paramagnetic (due to two unpaired electrons in antibonding π∗ molecular orbitals). Supports combustion.
• Uses: Respiration, combustion, welding (oxy-acetylene flame), in steel manufacturing.

B. Ozone (O3​):

• Preparation: When a slow dry stream of oxygen is passed through a silent electrical discharge, ozone is formed. 3O2​(g)⇌2O3​(g) ($ \Delta H^\circ = +142 , kJ/mol$).
  • The reaction is endothermic, and silent electrical discharge is used to prevent its decomposition.
• Properties: Pale blue gas (pure ozone is dark blue liquid/violet-black solid). Pungent smell. Unstable and decomposes to oxygen. Powerful oxidizing agent.
  • Oxidizes KI to I2​: 2KI+H2​O+O3​→2KOH+I2​+O2​
• Structure: Angular molecule (bent shape, sp2 hybridized central oxygen with one lone pair). Resonance structures.
• Uses: Germicide, disinfectant, sterilizing water, bleaching oils, protecting earth from UV radiation (ozone layer).

5. Important Compounds of Sulfur

A. Allotropic Forms of Sulfur:

1. Rhombic Sulphur (α-sulphur):
   • Structure: Yellow solid, forms S8​ rings (puckered eight-membered rings).
   • Properties: Most stable allotropic form at room temperature (<369K). Insoluble in water, soluble in CS2​.
2. Monoclinic Sulphur (β-sulphur):
   • Structure: Long needle-like crystals, forms S8​ rings.
   • Properties: Stable above 369K. Slowly changes to rhombic sulphur below 369K. Soluble in CS2​.

B. Sulfur Dioxide (SO2​):

• Preparation:
  • Laboratory: By reacting sodium sulphite with dilute H2​SO4​. Na2​SO3​+H2​SO4​→Na2​SO4​+H2​O+SO2​
  • Industrial: By burning sulphur or iron pyrites. S+O2​→SO2​ 4FeS2​+11O2​→2Fe2​O3​+8SO2​
• Properties: Colourless gas with a pungent smell. Readily liquefiable. Acidic oxide (forms sulphurous acid, H2​SO3​, with water). Reducing agent (e.g., decolorizes acidified KMnO4​).
• Uses: Bleaching wool/silk/straw, disinfectant, preservative, refining petroleum and sugar.

C. Sulphuric Acid (H2​SO4​):

• Preparation:
  • Contact Process (Industrial):
    1. Production of sulphur dioxide: S+O2​→SO2​
    2. Catalytic oxidation of sulphur dioxide to sulphur trioxide (SO3​): 2SO2​(g)+O2​(g)⇌2SO3​(g) ($ \Delta H^\circ = -196.6 , kJ/mol$) Conditions: V2​O5​ catalyst, optimal temperature (720K), 2 bar pressure.
    3. Absorption of SO3​ in H2​SO4​ to form Oleum (H2​S2​O7​): SO3​+H2​SO4​→H2​S2​O7​ (Oleum)
    4. Dilution of Oleum with water to produce concentrated H2​SO4​: H2​S2​O7​+H2​O→2H2​SO4​
• Properties: Strong acid, oxidizing agent, dehydrating agent, non-volatile.
• Uses: Manufacturing fertilizers, detergents, chemicals, in petroleum refining, metallurgy.

D. Oxoacids of Sulfur: (Important to remember oxidation state and structure)

• Sulphurous acid (H2​SO3​): +4 oxidation state.
• Sulphuric acid (H2​SO4​): +6 oxidation state.
• Peroxomonosulphuric acid (Caro’s acid), H2​SO5​: +6 oxidation state (contains one peroxo linkage, -O-O-).
• Peroxodisulphuric acid (Marshall’s acid), H2​S2​O8​: +6 oxidation state (contains one peroxo linkage, -O-O-).
• Pyrosulphuric acid (Oleum), H2​S2​O7​: +6 oxidation state (contains one bridging oxygen atom, -O-).

NEET Chemistry: The p-Block Elements (Oxygen Family) – Practice Questions

I. Multiple Choice Questions (MCQs)

1. Which of the following elements belongs to the oxygen family?
   a) Nitrogen
   b) Carbon
   c) Sulphur
   d) Fluorine
2. What is the general electronic configuration of Group 16 elements?
   a) ns²np²
   b) ns²np³
   c) ns²np⁴
   d) ns²np⁵
3. Which of the following is a metalloid in Group 16?
   a) Oxygen
   b) Sulphur
   c) Tellurium
   d) Polonium
4. Oxygen exhibits anomalous behavior among Group 16 elements due to:
   a) Its high atomic mass
   b) Presence of vacant d-orbitals
   c) Its ability to form pπ–pπ multiple bonds
   d) Its low electronegativity
5. What is the correct order of electron gain enthalpy (magnitude only) in Group 16 elements?
   a) O > S > Se > Te
   b) S > O > Se > Te
   c) O > Se > S > Te
   d) Te > Se > S > O
6. Which allotropic form of sulphur is stable below 369 K (room temperature)?
   a) Monoclinic sulphur
   b) Plastic sulphur
   c) Rhombic sulphur
   d) Liquid sulphur
7. Which catalyst is used in the contact process for manufacturing sulphuric acid?
   a) Fe₂O₃
   b) MnO₂
   c) V₂O₅
   d) Ni
8. Ozone (O₃) is prepared from oxygen (O₂) using:
   a) High temperature
   b) High pressure
   c) Silent electrical discharge
   d) Sunlight
9. Which oxide of sulphur contains sulfur in the +4 oxidation state?
   a) H₂SO₄
   b) H₂S₂O₇
   c) SO₂
   d) H₂SO₅
10. What is the molecular geometry of ozone (O₃)?
    a) Linear
    b) Trigonal planar
    c) Bent (Angular)
    d) Tetrahedral
11. Which property of sulphur dioxide (SO₂) is responsible for its bleaching action on wool and silk?
    a) Acidic nature
    b) Reducing nature
    c) Oxidizing nature
    d) Pungent smell
12. Which of the following compounds contains an S–S bond?
    a) H₂SO₄
    b) H₂SO₅
    c) H₂S₂O₈
    d) H₂S₂O₇
13. Why is water (H₂O) a liquid while hydrogen sulphide (H₂S) is a gas at room temperature?
    a) Higher molecular mass of H₂O
    b) Hydrogen bonding in H₂O
    c) Smaller size of oxygen
    d) Greater polarity of H₂S
14. Which is the most stable oxidation state of sulphur?
    a) –2
    b) +2
    c) +4
    d) +6
15. Which of the following statements about oxygen is correct?
    a) It exhibits +4 and +6 oxidation states.
    b) It forms hydrogen bonds more readily than nitrogen.
    c) It is paramagnetic.
    d) It has d-orbitals in its valence shell.

II. Assertion-Reason Type Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is NOT the correct explanation of A. c) A is true but R is false. d) A is false but R is true.

16. Assertion (A): Oxygen has a less negative electron gain enthalpy than sulfur. Reason (R): Oxygen atom is smaller than sulfur atom.

17. Assertion (A): Ozone is a powerful oxidizing agent. Reason (R): Ozone is highly unstable and readily decomposes to form nascent oxygen.

18. Assertion (A): H2​SO4​ is a strong dehydrating agent. Reason (R): It has a strong affinity for water.

19. Assertion (A): The stability of +6 oxidation state decreases down Group 16. Reason (R): The inert pair effect becomes more prominent down the group.

20. Assertion (A): Oxygen exists as a diatomic gas while sulfur exists as an S8​ polyatomic molecule. Reason (R): Oxygen forms pπ−pπ multiple bonds readily due to its small size and high electronegativity.

III. Short Answer / Conceptual Questions

21. Question: Explain why oxygen exhibits only −2,−1,+1,+2 oxidation states, but not higher positive states like +4 or +6.

22. Question: Describe the preparation of sulphur dioxide on a large scale. State two of its uses.

23. Question: Draw the structure of ozone (O3​) and state its hybridization.

24. Question: How does the acidic character of hydrides of Group 16 elements change down the group? Give a reason.

25. Question: Why is a silent electrical discharge used during the preparation of ozone from oxygen?

26. Question: Write the balanced chemical equations for the following reactions: a) Oxidation of sulfur dioxide to sulfur trioxide in Contact Process. b) Action of ozone on potassium iodide solution.

27. Question: Name two oxoacids of sulfur which contain a peroxo linkage. Write their formulas.

28. Question: Differentiate between Rhombic sulfur and Monoclinic sulfur based on their stability and crystalline form.

29. Question: Explain why oxygen is paramagnetic, despite having an even number of electrons.

30. Question: What is Oleum? How is it formed in the Contact Process?

Answers and Explanations

I. Multiple Choice Questions (MCQs) – Answers

1. Answer: c) Sulphur Explanation: Group 16 is known as the Oxygen family or Chalcogens, which includes Oxygen, Sulphur, Selenium, Tellurium, and Polonium.

2. Answer: c) ns2np4 Explanation: Group 16 elements have 6 valence electrons, with 2 in the s-orbital and 4 in the p-orbital.

3. Answer: c) Tellurium Explanation: In Group 16, Oxygen and Sulphur are non-metals. Selenium and Tellurium are metalloids. Polonium is a metal.

4. Answer: c) Its ability to form pπ−pπ multiple bonds Explanation: Oxygen’s small size and high electronegativity allow it to form stable pπ−pπ multiple bonds (e.g., O=O). Other elements cannot do this effectively due to their larger size and diffused p-orbitals. Oxygen also uniquely has no d-orbitals for expanded valency.

5. Answer: b) S > O > Se > Te Explanation: Electron gain enthalpy generally becomes less negative (or more positive) down a group. However, Oxygen has a less negative (or less exothermic) electron gain enthalpy than Sulfur. This is due to the small size of the oxygen atom, which leads to significant inter-electronic repulsions when an extra electron is added to its compact 2p subshell. Thus, sulfur has the most negative electron gain enthalpy in the group.

6. Answer: c) Rhombic sulphur Explanation: Rhombic sulphur (or α-sulphur) is the most stable allotropic form of sulphur at room temperature (below 369K). Monoclinic sulphur (β-sulphur) is stable above 369K.

7. Answer: c) V2​O5​ Explanation: Vanadium pentoxide (V2​O5​) is used as a catalyst in the contact process for the industrial manufacture of sulphuric acid, specifically for the oxidation of sulfur dioxide to sulfur trioxide.

8. Question: c) Silent electrical discharge Explanation: Ozone is formed from oxygen by passing a slow, dry stream of oxygen through a silent electrical discharge. This method provides the necessary energy for the endothermic reaction (3O2​⇌2O3​) without causing the decomposition of ozone due to excessive heat.

9. Answer: c) SO2​ Explanation: a) H2​SO4​ (Sulphuric acid): S is in +6 oxidation state. b) H2​S2​O7​ (Oleum): S is in +6 oxidation state. c) SO2​ (Sulphur dioxide): S is in +4 oxidation state (x+2(−2)=0⇒x=+4). d) H2​SO5​ (Caro’s acid): S is in +6 oxidation state (one peroxo oxygen is -1).

10. Answer: c) Bent (Angular) Explanation: The ozone (O3​) molecule has a bent (angular) shape. The central oxygen atom is sp$^2$ hybridized and has one lone pair of electrons, which causes repulsion and reduces the bond angle from the ideal 120∘ of a trigonal planar geometry. It exhibits resonance.

11. Answer: b) Reducing nature Explanation: Sulphur dioxide (SO2​) acts as a reducing agent (due to S in +4 oxidation state, which can be oxidized to +6). It bleaches wool and silk by reduction. The bleaching effect is temporary and reversed by exposure to air.

12. Answer: c) H2​S2​O8​ Explanation: H2​S2​O8​ (Peroxodisulphuric acid or Marshall’s acid) contains a peroxo linkage (-O-O-) between two sulphur atoms, leading to an S-O-O-S structure. The other options (H2​SO4​, H2​SO5​, H2​S2​O7​) do not have S-S bonds. H2​SO5​ has a P-O-O-H linkage, not S-S.

13. Answer: b) Hydrogen bonding in H2​O Explanation: Water (H2​O) is a liquid at room temperature because oxygen’s high electronegativity and small size allow it to form extensive intermolecular hydrogen bonds. Hydrogen sulphide (H2​S) does not form significant hydrogen bonds (due to lower electronegativity of S), so its molecules are held by weaker dipole-dipole forces, making it a gas.

14. Answer: d) +6 Explanation: For sulfur, the +6 oxidation state is the most stable and common among its positive oxidation states, typically seen in sulphuric acid and sulphates. This is because sulfur can utilize all its valence electrons (3s23p4) for bonding.

15. Answer: c) It is paramagnetic. Explanation: Molecular orbital theory explains that the oxygen molecule (O2​) has two unpaired electrons in its anti-bonding π∗ molecular orbitals. This presence of unpaired electrons makes O2​ paramagnetic. a) Oxygen generally exhibits −2,−1,+1,+2 oxidation states, not +4 or +6 (due to absence of d-orbitals). b) Oxygen is more electronegative than nitrogen, and hydrogen bonds are strongest with F, O, N. Both form H-bonds, but it’s not a reason for “more readily than N” in all contexts. d) Oxygen is in the 2nd period and does not have d-orbitals in its valence shell.

II. Assertion-Reason Type Questions – Answers

16. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Oxygen indeed has a less negative electron gain enthalpy (meaning it’s less exothermic) than sulfur. This is directly due to its very small atomic size (Reason R). The small size of the oxygen atom leads to strong inter-electronic repulsions when an incoming electron is added to its already compact 2p subshell, thus counteracting the nuclear attraction.

17. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Ozone (O3​) is a powerful oxidizing agent. This is because it is thermodynamically unstable and readily decomposes to form oxygen (O2​) and highly reactive nascent oxygen (O), which is a very strong oxidizing species. O3​→O2​+[O]

18. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Concentrated sulphuric acid is a very strong dehydrating agent. This property stems from its exceptionally high affinity for water, with which it forms hydrates. It can remove water from various substances, including organic compounds, often causing charring.

19. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: The stability of the highest oxidation state (in this case, +6) generally decreases down the group for p-block elements. This is due to the inert pair effect, where the ns2 valence electrons become increasingly reluctant to participate in bonding in heavier elements, favoring lower oxidation states like +4.

20. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Oxygen exists as a diatomic gas (O2​) with a double bond (O=O), while sulfur exists as polyatomic S8​ rings or chains. This difference is because oxygen’s small size and high electronegativity enable it to form strong pπ−pπ multiple bonds, whereas larger sulfur atoms prefer single S-S bonds and extended structures to maximize bond energy.

III. Short Answer / Conceptual Questions – Answers

21. Answer: Oxygen primarily exhibits oxidation states of −2 (most common in oxides), −1 (in peroxides), +1 (in O2​F2​), and +2 (in OF2​). It does not show higher positive oxidation states like +4 or +6, which are common for other elements in Group 16 (S, Se, Te). The main reason for this is the absence of d-orbitals in the valence shell of oxygen. Unlike sulfur and other heavier elements, oxygen cannot expand its octet and therefore cannot accommodate more than two bonds or a formal positive charge higher than +2 by involving d-orbital participation.

22. Answer: Preparation of Sulphur Dioxide (SO2​) on a large scale: Sulphur dioxide is primarily produced on a large scale by the combustion of elemental sulphur or iron pyrites in air.

1. By burning sulphur: S(s)+O2​(g)Heat​SO2​(g)
2. By roasting iron pyrites: 4FeS2​(s)+11O2​(g)Heat​2Fe2​O3​(s)+8SO2​(g) Two uses of Sulphur Dioxide:
3. Bleaching Agent: It is used as a bleaching agent for delicate materials like wool, silk, straw, and paper, which are damaged by chlorine-based bleaches.
4. Preservative: It acts as a preservative in food and beverages due to its antiseptic and germicidal properties.

23. Answer: Structure of Ozone (O3​): Ozone has an angular (bent) molecular geometry. The central oxygen atom is sp$^2$ hybridized. It forms one double bond and one single bond with the other two oxygen atoms, and it also possesses one lone pair of electrons. The bond angle is approximately 116.7∘. The molecule exhibits resonance, meaning the double bond character is delocalized over both O-O bonds.

Drawing:

      O
     // \\
    O   O

(This is a simplified representation showing the bent shape and delocalized bonding. The actual structure is a resonance hybrid.)

24. Answer: The acidic character of hydrides of Group 16 elements increases down the group. Order: H2​O<H2​S<H2​Se<H2​Te. Reason: As we move down the group from Oxygen to Tellurium, the atomic size of the central atom (O, S, Se, Te) increases. This leads to:

1. Increase in bond length: The M−H bond length (where M is the Group 16 element) increases down the group.
2. Decrease in bond dissociation enthalpy: A longer bond is weaker, so the bond dissociation enthalpy of the M−H bond decreases down the group.
3. Ease of Proton Release: A weaker M−H bond means that hydrogen can be released more easily as a proton (H+). The ease of releasing H+ defines the acidic character. Therefore, H2​Te is the strongest acid among these hydrides, and H2​O is the weakest.

25. Answer: A silent electrical discharge is used during the preparation of ozone (O3​) from oxygen (O2​) for two main reasons:

1. Endothermic Reaction: The conversion of oxygen to ozone (3O2​(g)⇌2O3​(g)) is a highly endothermic reaction (ΔH∘=+142kJ/mol). Electrical energy is required to drive this reaction.
2. Prevention of Decomposition: Ozone is thermodynamically unstable and has a tendency to decompose back into oxygen. Using a silent (non-sparking) electrical discharge ensures that the temperature does not rise significantly. High temperatures would cause the newly formed ozone to immediately decompose back into oxygen, reducing the yield. The silent discharge provides the necessary energy in a controlled manner without excessive heat.

26. Answer: a) Oxidation of sulfur dioxide to sulfur trioxide in Contact Process: 2SO2​(g)+O2​(g)V2​O5​ catalyst,720K,2bar​2SO3​(g) b) Action of ozone on potassium iodide solution: 2KI(aq)+H2​O(l)+O3​(g)→2KOH(aq)+I2​(s)+O2​(g)

27. Answer: Two oxoacids of sulfur which contain a peroxo linkage (-O-O-) are:

1. Peroxomonosulphuric acid (Caro’s acid): H2​SO5​
2. Peroxodisulphuric acid (Marshall’s acid): H2​S2​O8​

28. Answer: | Feature | Rhombic Sulphur (α-sulphur) | Monoclinic Sulphur (β-sulphur) | | :————— | :————————————————- | :————————————————– | | Stability | Stable below 369K (room temperature) | Stable above 369K (transition temperature) | | Crystalline Form | Octahedral crystals | Long needle-like prismatic crystals | | Colour | Yellow | Pale yellow | | Conversion | Converts to monoclinic sulphur above 369K | Converts to rhombic sulphur below 369K |

29. Answer: Oxygen (O2​) is paramagnetic, despite having an even number of electrons (16 electrons in total). This behavior cannot be explained by simple Lewis structures or Valence Bond Theory, but it is successfully explained by Molecular Orbital Theory (MOT). According to MOT, the molecular orbital electronic configuration of O2​ is: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz​)2(π2px​)2(π2py​)2(π∗2px​)1(π∗2py​)1 In this configuration, there are two unpaired electrons present in the anti-bonding π∗2px​ and π∗2py​ molecular orbitals. The presence of these two unpaired electrons gives O2​ its paramagnetic property.

30. Answer:

• Oleum: Oleum is the common name for pyrosulphuric acid (H2​S2​O7​). It can be thought of as a solution of sulfur trioxide (SO3​) in concentrated sulphuric acid (H2​SO4​).
• Formation in Contact Process: In the Contact Process for the manufacture of sulphuric acid, after the catalytic oxidation of sulfur dioxide (SO2​) to sulfur trioxide (SO3​), the SO3​ gas is not directly absorbed in water. Instead, it is absorbed in concentrated sulphuric acid to form oleum. This step is crucial because direct absorption of SO3​ in water forms a mist of sulphuric acid, which is difficult to condense. SO3​(g)+H2​SO4​(l)→H2​S2​O7​(l) (Oleum) Oleum is then diluted with water to obtain concentrated sulphuric acid of the desired strength. H2​S2​O7​(l)+H2​O(l)→2H2​SO4​(aq)