16.1 Oxidation number · 35‑min lesson + examples | @everexams
18px

16.1 Oxidation number of atoms in compounds and ions

apparent charge · redox · formula writing

The assignment of oxidation number to atoms in compounds and ions helps us to identify whether the reaction involves oxidation and reduction or not. It also helps to identify oxidizing and reducing agents in a given oxidation reduction reaction. Oxidation numbers are also used to write chemical formulas and names of the chemical compounds. This concept also helps to keep track of the new distribution of electrons during an oxidation-reduction reaction.

The oxidation number is an apparent charge on an atom in a compound or an ion. It may have a positive or a negative sign.

The oxidation number of atoms can be found out by applying the following rules.

1 · Free element

Atom in free element → oxidation number zero. Na (metal) = 0; O in O₃ = 0.

💡 “alone and free — zero is the key”
2 · Monoatomic ions

Oxidation number = charge on ion. Ca²⁺ = +2, Al³⁺ = +3.

💡 sign stays, number after element = oxidation state
3 · Group 1,2,3 (except H)

Group1→+1, Group2→+2, Group3→+3.

💡 “Group 1,2,3 → plus 1,2,3”
4 · Binary compounds (electronegativity)

More electronegative atom gets negative ON. Fluorine always –1. In NH₃, N = –3 ; H₂S, S = –2.

💡 “EN wins negative, other positive.”
5 · Group 17 (halogens)

Usually –1 with less electronegative element. Cl in HCl, PCl₃, CHCl₃ = –1.

6 · Hydrogen

+1 with more electronegative atoms (HCl); –1 with less electronegative (NaH, sodium hydride).

💡 “Hydride hates (–1); others plus one”
7 · Oxygen

Usually –2 in most compounds.

8 · Sum of oxidation numbers

Neutral compound = 0 ; polyatomic ion = charge on ion.

16.1 Example: sulphur in SO₄²⁻ → S + 4(–2) = –2 → S = +6

16.2 Example: manganese in KMnO₄ → (+1)+Mn+4(–2)=0 → Mn = +7

16.3 Example: Fe in Fe₂O₃ → 2Fe+3(–2)=0 → Fe = +3

⚡ QUICK CHECK! K₂Cr₂O₇ (Cr = +6), (NH₄)₂SO₄ (S = +6)

⭐ Point to note: Oxidation numbers are not real charges except in simple ions. Oxidation state = oxidation number.

📅 35‑minute lesson planner · oxidation numbers

0–5 minStarter: Show a metal (Zn) and rust (Fe₂O₃) – ask why charges appear. Introduce “apparent charge”.
5–12 minDirect instruction: Teach rules 1–4 using visuals (periodic table). Emphasise fluorine, oxygen, hydrogen.
12–18 minGuided practice: Solve examples 16.1 (SO₄²⁻), 16.2 (KMnO₄), 16.3 (Fe₂O₃) step‑by‑step on board.
18–25 minTeam challenge: In pairs, find ON of Cr in K₂Cr₂O₇ and S in (NH₄)₂SO₄. Share answers.
25–32 minQuiz (7 MCQs): Use 7 questions from below – students show fingers 1‑4. Clarify misconceptions.
32–35 minExit ticket: “What is the oxidation number of Mn in MnO₂?” (+4) and one doubt.

🔍 Detailed solutions (examples from text)

📌 16.1 SO₄²⁻ : Total charge = –2. Oxygen ON = –2 (rule 7). Let S = x. x + 4×(–2) = –2 → x –8 = –2 → x = +6. ✓
📌 16.2 KMnO₄ : Neutral compound → sum = 0. K = +1 (group1), O = –2. +1 + Mn + 4×(–2) = 0 → Mn –7 = 0 → Mn = +7. ✓
📌 16.3 Fe₂O₃ : Neutral, O = –2. 2×Fe + 3×(–2) = 0 → 2Fe –6 = 0 → Fe = +3. ✓
📌 K₂Cr₂O₇ (quick check): K = +1, O = –2. 2(+1) + 2Cr + 7(–2) = 0 → 2 + 2Cr –14 = 0 → 2Cr = 12 → Cr = +6.
📌 (NH₄)₂SO₄ (quick check): NH₄⁺ : N + 4(+1) = +1 → N = –3. SO₄²⁻: S + 4(–2)= –2 → S = +6.

✨ All examples follow the 8 rules. Note: In polyatomic ions, sum = charge.

📋 50 MCQs · Oxidation number quiz