Group 16 (the oxygen family or chalcogens) is O, S, Se, Te, Po with configuration ns2np4. Group 17 (the halogens) is F, Cl, Br, I, At with configuration ns2np5 — one electron short of a noble-gas shell, so they are the most reactive non-metals.
Group 16 trends
Oxidation states range from −2 to +6; oxygen is usually −2 (no d orbitals, so no +6) while S, Se and Te show +2, +4 and +6. Atomic radius and metallic character increase down the group; O and S are non-metals, Se and Te metalloids, Po a metal. Hydride stability falls H2O > H2S > H2Se > H2Te while acidic strength rises in the same order. Oxygen forms strong pπ–pπ bonds (O=O), so O2 is diatomic, whereas sulphur catenates into S8 rings.
Dioxygen and ozone
Ozone (O3) is an allotrope of oxygen formed by passing a silent electric discharge through O2: 3O2 → 2O3 (endothermic). O3 is bent (bond angle ~117°) with resonance, and is a powerful oxidising agent (it liberates I2 from KI). It shields the Earth from UV radiation.
Sulphur and sulphuric acid
Sulphur exists chiefly as the yellow rhombic (α) and monoclinic (β) allotropes, both built from puckered S8 crown rings. Sulphur dioxide (SO2, bent, acidic, reducing) is made by burning S or roasting sulphide ores. Sulphuric acid is manufactured by the Contact process: S/ore → SO2; 2SO2 + O2 ⇌ 2SO3 over a V2O5 catalyst (~720 K, ~2 bar); SO3 is absorbed in conc. H2SO4 to give oleum (H2S2O7), then diluted. H2SO4 is a strong dibasic acid, dehydrating agent and (when hot/concentrated) an oxidiser.
Group 17 trends
Halogens have the most negative electron-gain enthalpies and high electronegativity (F is the most electronegative element). They are coloured, exist as diatomic X2 molecules, and bond enthalpy is Cl2 > Br2 > F2 > I2 (the small F–F bond is weak due to lone-pair repulsion). Oxidising power decreases F2 > Cl2 > Br2 > I2, so a higher halogen displaces a lower one from its salt.
Interhalogens and oxoacids
Interhalogen compounds XX′n (n = 1, 3, 5, 7) such as ClF3 (T-shaped), BrF5 (square pyramidal) and IF7 (pentagonal bipyramidal) form between two halogens; they are more reactive than the parent halogens. Chlorine forms the oxoacids HOCl, HOClO, HOClO2 and HOClO3 (hypochlorous, chlorous, chloric, perchloric) — acid strength rises with oxidation state. Hydrogen chloride (HCl) is made by heating NaCl with conc. H2SO4; its aqueous solution is a strong monobasic acid.
Explain why the bond dissociation enthalpy of F2 is lower than that of Cl2, even though fluorine is the more reactive halogen.
Solution- In F2 the two fluorine atoms are very small, so the F–F internuclear distance is short.
- This brings the three lone pairs on each fluorine into close proximity, causing strong lone-pair–lone-pair repulsion.
- The repulsion weakens the F–F bond, lowering its dissociation enthalpy below that of Cl2.
- The weak F–F bond is actually part of why F2 is so reactive — it breaks easily.
Answer: Inter-electronic (lone-pair) repulsion in the small F2 molecule weakens the F–F bond, so its bond enthalpy is lower than Cl2.
Describe the Contact process for manufacturing sulphuric acid with all balanced equations and conditions.
Solution- Burn sulphur (or roast sulphide ore): S + O2 → SO2.
- Catalytically oxidise SO2: 2SO2 + O2 ⇌ 2SO3, ΔH < 0, using V2O5 at ~720 K and ~2 bar.
- Absorb SO3 in concentrated H2SO4 to form oleum: SO3 + H2SO4 → H2S2O7.
- Dilute oleum with water: H2S2O7 + H2O → 2H2SO4. (SO3 is not absorbed in water directly because it forms a corrosive mist.)
Answer: S → SO2 → SO3 (V2O5, ~720 K) → oleum (H2S2O7) → H2SO4 on dilution.
Ozone acts as a powerful oxidising agent. Write the reaction with potassium iodide solution and explain the oxidation-state changes.
Solution- Ozone oxidises iodide ion to iodine: 2KI + H2O + O3 → 2KOH + I2 + O2.
- Iodine changes from −1 (in I-) to 0 (in I2): it is oxidised.
- Ozone oxygen goes from 0 (effective) to −2 (in OH-): O3 is reduced, confirming it is the oxidiser.
- The liberated I2 turns starch blue — a standard test for ozone.
Answer: 2KI + H2O + O3 → 2KOH + I2 + O2; I- is oxidised to I2, so O3 is the oxidising agent.
Arrange HOCl, HClO2, HClO3 and HClO4 in order of increasing acid strength and justify the trend.
Solution- The oxidation state of chlorine rises: +1 (HOCl), +3 (HClO2), +5 (HClO3), +7 (HClO4).
- More terminal O atoms withdraw electron density, stabilising the conjugate base (ClOx-) by delocalising its negative charge.
- A more stable conjugate base means a stronger acid.
- Hence acid strength increases with oxidation state.
Answer: HOCl < HClO2 < HClO3 < HClO4; greater O-count stabilises the conjugate base, raising acidity.
Predict the shapes of ClF3 and BrF5 using VSEPR theory.
Solution- ClF3: Cl has 7 valence electrons; 3 are used in bonds, leaving 2 lone pairs → 5 electron pairs, sp3d.
- The two lone pairs occupy equatorial positions, giving a T-shaped molecule.
- BrF5: Br has 7 valence electrons; 5 bonds plus 1 lone pair → 6 electron pairs, sp3d2.
- One lone pair gives a square pyramidal shape.
Answer: ClF3 is T-shaped (sp3d); BrF5 is square pyramidal (sp3d2).
Why is concentrated H2SO4 described as a dehydrating agent? Give one illustrative reaction.
Solution- Concentrated H2SO4 has a very high affinity for water and can remove the elements of water from many compounds.
- It chars sugar by removing hydrogen and oxygen as water: C12H22O11 → 12C + 11H2O (absorbed by the acid).
- It also dehydrates formic acid: HCOOH → CO + H2O.
- Because it removes water rather than oxidising in these cases, it is acting as a dehydrating agent.
Answer: Its strong affinity for water lets it strip H and O as water, e.g. it chars sugar: C12H22O11 → 12C + 11H2O.