Group 16 Elements (Chalcogens)

• Explain the allotropic forms of oxygen (O₂ and O₃) and the special properties of ozone
• Compare H₂O and H₂O₂ in terms of structure, oxidising power, and reactions
• Describe the allotropes of sulphur (rhombic, monoclinic, plastic) and their interconversion
• Write all steps of the Contact Process for manufacture of H₂SO₄
• Explain the oxidising and reducing behaviour of SO₂ and the formation of SO₃
• Name, draw, and classify the oxoacids of sulphur by oxidation state

Level 1 — Foundations

Group 16 — Overview

Anomalous high BP of H₂O: Extensive H-bonding (each O forms up to 4 H-bonds in ice).
Decreasing stability of hydrides: H₂O > H₂S > H₂Se > H₂Te (bond strength decreases).
Increasing acidic strength: H₂O < H₂S < H₂Se < H₂Te (bond weakens → easier to release H⁺).

Oxygen and Its Allotropes

O₂ (dioxygen):

• Bond order = 2 (O=O)
• Paramagnetic: 2 unpaired electrons in π* antibonding MOs (from MO theory)
• OS of O in O₂ = 0; in O₃ = 0; in H₂O = −2; in H₂O₂ = −1; in OF₂ = +2

O₃ (ozone):

• Bent molecule; bond angle = 117°; OS of central O = +1, terminal O = −½ (average)
• Resonance structure: O=O⁺–O⁻ ↔ ⁻O–O⁺=O
• Stronger oxidising agent than O₂: O₃ + 2KI + H₂O → I₂ + 2KOH + O₂ (liberates I₂ from KI — ozone test) O₃ + PbS → PbSO₄ (oxidises sulfides directly)
• O₃ in stratosphere absorbs UV-B and UV-C (240–320 nm)
• Depleted by CFCs: Cl• + O₃ → ClO• + O₂ (chain reaction)

Water (H₂O) and Hydrogen Peroxide (H₂O₂)

Comparison:

H₂O₂ as oxidising agent: H₂O₂ + 2KI → I₂ + 2KOH
H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O (Fenton's reagent context)

H₂O₂ as reducing agent: 2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → 2MnSO₄ + K₂SO₄ + 8H₂O + 5O₂
(MnO₄⁻ oxidises H₂O₂; H₂O₂ is the reducing agent here)

Bleaching action of H₂O₂: Nascent oxygen [O] released bleaches coloured materials:
H₂O₂ → H₂O + [O] (acidic conditions)

Volume strength of H₂O₂:
1 mole H₂O₂ → 1 mole O₂ at STP (22.4 L)
"10 volume" H₂O₂ = 1 L releases 10 L O₂ at STP
Molarity = Volume strength / 11.2

Sulphur Allotropes

Transition temperature (96°C): below → rhombic stable; above → monoclinic stable.

H₂SO₄ — Contact Process

Step 1: Burning sulphur or roasting sulphide ores
S + O₂ → SO₂ (or 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂)

Step 2: Catalytic oxidation of SO₂ to SO₃
2SO₂ + O₂ ⇌ 2SO₃ ΔH = −196 kJ/mol

Step 3: SO₃ absorbed in 98% H₂SO₄ (oleum formed), then diluted
SO₃ + H₂SO₄ → H₂S₂O₇ (oleum / fuming sulphuric acid)
H₂S₂O₇ + H₂O → 2H₂SO₄

(SO₃ is NOT absorbed directly in water — forms acid mist instead.)

Properties of H₂SO₄ (concentrated):

SO₂ and SO₃

SO₂:

• Bent molecule; S is sp³ hybridised (1 lone pair + 3 bond pairs if counting resonance) or sp² (delocalised)
• OS of S = +4
• Reducing agent: 2SO₂ + O₂ → 2SO₃; bleaches acidic dyes (temporary bleach — by reduction)
• Oxidising agent: SO₂ + 2H₂S → 3S + 2H₂O (S is reduced from +4 to 0)
• Acidic oxide: SO₂ + H₂O → H₂SO₃ (sulphurous acid)

SO₃:

• Planar trigonal; sp² hybridised; OS of S = +6
• Powerful oxidising and sulphonating agent
• Anhydride of H₂SO₄: SO₃ + H₂O → H₂SO₄

Oxoacids of Sulphur

Level 2 — JEE Depth

Why is O₂ paramagnetic? (MO theory — JEE favourite)
MO configuration of O₂:
σ1s² σ1s² σ2s² σ2s² σ2p² π2p⁴ π2p²
Two degenerate π orbitals each have 1 electron (Hund's rule) → 2 unpaired electrons → paramagnetic.
Bond order = (8−4)/2 = 2.

Bleaching by SO₂ vs Cl₂:

H₂O₂ as both oxidising and reducing agent — when to predict which:

• With reducing agents (KI, FeSO₄): H₂O₂ is the oxidiser (O goes from −1 to −2)
• With oxidising agents (KMnO₄, K₂Cr₂O₇, O₃): H₂O₂ is the reducer (O goes from −1 to 0 as O₂)

Contact process — why NOT use very high pressure:

• Equilibrium yield is already ~99.5% at 450°C even at 1–2 atm
• High pressure equipment is costly and S compounds cause catalyst sintering
• Unlike Haber process, high P not industrially justified here

Comparison of Group 16 oxides:

Trend: acidic → amphoteric → basic going down Group 16 (metallic character increases).

Oleum (H₂S₂O₇) structure:
Two SO₄ tetrahedra sharing one O: O₃S–O–SO₃
On adding water: H₂S₂O₇ + H₂O → 2H₂SO₄

Dihedral angle in H₂O₂:
Gas phase: 111.5°; solid: 90.2°; solution: variable.
Non-planar structure due to lone pair–lone pair repulsion on adjacent O atoms.

Example 1: Calculate the molarity of "20 volume" H₂O₂ solution.

Step 1: "20 volume" means 1 L of this solution releases 20 L of O₂ at STP.

Step 2: Moles of O₂ released = 20 / 22.4 = 0.893 mol

Step 3: From: 2H₂O₂ → 2H₂O + O₂
        Moles of H₂O₂ = 2 × moles of O₂ = 2 × 0.893 = 1.786 mol

Step 4: Volume of solution = 1 L
        Molarity = 1.786 mol / 1 L = 1.786 M ≈ 1.79 M

Shortcut: Molarity of H₂O₂ = Volume strength / 11.2
          = 20 / 11.2 = 1.786 M ✓

Answer: Molarity = 1.79 M (approximately 1.8 M)

Example 2: In the Contact Process, why is SO₃ absorbed in concentrated H₂SO₄ and not in water?

Step 1: Reaction with water:
        SO₃ + H₂O → H₂SO₄

Step 2: The reaction is highly exothermic and fast. When SO₃ gas meets
        water vapour or dilute H₂SO₄, it forms a dense acid mist
        (fine droplets of H₂SO₄ that do not settle easily and
        are hazardous to handle).

Step 3: Absorption in 98% H₂SO₄ (oleum formation) is much cleaner:
        SO₃ + H₂SO₄ → H₂S₂O₇  (oleum)
        Then: H₂S₂O₇ + H₂O → 2H₂SO₄ (controlled dilution)

Step 4: This avoids mist formation and gives pure H₂SO₄ with
        quantitative absorption efficiency.

Answer: Direct absorption in water forms acid mist; conc. H₂SO₄ gives
        clean oleum which is then safely diluted to pure H₂SO₄.

• Q1: State two chemical tests to distinguish O₂ from O₃.
• Q2: A sample of H₂O₂ is labelled "15 volume". Calculate its molarity.
• Q3: Why does rhombic sulphur convert to monoclinic sulphur above 96°C?
• Q4: Write the balanced equation for the reaction of hot concentrated H₂SO₄ with copper. Identify the oxidising and reducing agents.
• Stretch: Q5: Using MO theory, predict whether O₂⁺ (dioxygenyl cation) is paramagnetic or diamagnetic, and whether its bond order is higher or lower than O₂. Relate this to whether the O–O bond is shorter in O₂ or O₂⁺.

NCERT Chapter 7 link: The p-Block Elements, NCERT Chemistry Part 1, Class 12 (Group 16)

Exam connections: JEE Mains tests volume strength of H₂O₂, Contact Process conditions, and SO₂ bleaching mechanism. JEE Advanced has asked for MO treatment of O₂ and complete electrode-half-reaction balancing for reactions of H₂O₂ with KMnO₄ in acid medium.

Study strategy: Build a single comparison table: O₂ vs O₃ vs H₂O vs H₂O₂ (OS of O, structure, oxidising power). Draw the Contact Process as a flowchart with temperatures. For oxoacids of S, group by OS (+4: H₂SO₃; +6: H₂SO₄, H₂S₂O₇, Caro's, Marshall's; mixed: H₂S₂O₃).

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• Voice or text reflection with AI Mentor: explain the concept to a younger student or family member.

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• One hands-on project or measurement using the Drishti kit or household items that makes the concept physical.
• Direct link to at least one Future Skill track (Money Management, Green Tech, Cyber Defenders, Micro-Entrepreneurship, AI Mastery, Sustainable Living, Personality Development).
• Coding extension where relevant (simple script, simulation, or data logging).

This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.

Portfolio Evidence Idea: Your photo/table/reflection/project + one sentence on "How this helps me in real life or a possible future path."

Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.

See curriculum for cross-links and the full future-skills/robotics chapters.