16.2 Writing formulas using oxidation numbers | @everexams
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16.2 Writing chemical formulas of ionic compounds by applying oxidation numbers

cross‑multiply charges · periodic table groups · sum to zero

We can write the chemical formula of a compound if we know the oxidation numbers of its constituent elements. The periodic table and the electronic structure of the atoms can help us in predicting the principal oxidation number of the element. The formula of the compound may then be written using the oxidation number of the atoms present in that compound. For example, we shall write the formula of calcium chloride in the following way.

The element calcium is present in group II of the periodic table and its principal oxidation number is +2. Chlorine is present in Group 17 and we know that when it is attached with less electronegative element in binary compounds its oxidation number is -1. The formula for calcium chloride will, therefore, be CaCl₂ because only with this formula the sum of the oxidation numbers of both the elements will add upto zero.

Ca 2+ Cl⁻ → CaCl₂
📌 16.4 Example – Aluminum oxide

Write the chemical formula of aluminum oxide with the help of oxidation numbers of the elements involved.

Solution: Aluminum is present in Group 13 so its principal oxidation number shall be +3. Oxygen being more electronegative atom normally shows –2 oxidation state in normal oxides. The formula of aluminum oxide will then be Al₂O₃ because the sum of oxidation numbers of both elements will be equal to zero with this formula.

Al³⁺ O²⁻ → Al₂O₃
⚡ cross‑over method: Al O → Al₂O₃ (3 and 2 swap)
✨ Quick steps to write formula:
1. Write symbols with oxidation numbers (e.g., Ca²⁺ Cl⁻).
2. Swap the numbers (without sign) as subscripts → Ca₁Cl₂ → CaCl₂.
3. Reduce subscripts if possible (e.g., Mg²⁺ O²⁻ → Mg₂O₂ → MgO).
4. Check sum of oxidation numbers = 0.

⭐ Note: This method works for ionic compounds. For polyatomic ions, treat the ion as a unit (e.g., NH₄⁺, SO₄²⁻).

⏳ 35‑min lesson · formulas from oxidation numbers

0–5 minEngage: Show formulas CaCl₂, Al₂O₃ – ask “why 2 and 3?”. Recall oxidation number rules.
5–12 minDirect: Explain cross‑over method using CaCl₂. Show how group number gives oxidation state (Group 2 → +2, Group 17 → –1).
12–18 minExample 16.4: Work Al₂O₃ step‑by‑step (Al³⁺, O²⁻ → swap → Al₂O₃). Emphasise sum zero: 2×(+3)+3×(–2)=0.
18–25 minGuided practice: Pairs write formulas: Na⁺ & S²⁻ → Na₂S ; Mg²⁺ & N³⁻ → Mg₃N₂ ; Al³⁺ & SO₄²⁻ → Al₂(SO₄)₃.
25–32 minQuiz mix: 7 MCQs from below (identify correct formula, find oxidation state). Discuss.
32–35 minExit slip: Write formula of calcium phosphate (Ca²⁺, PO₄³⁻) → Ca₃(PO₄)₂.

📝 Step‑by‑step solutions (attached examples + extra)

🔹 Calcium chloride (CaCl₂): Ca (group 2) → +2 ; Cl (group 17, binary) → –1. To get total zero, need two Cl⁻: (+2) + 2×(–1) = 0. Formula CaCl₂.
🔹 Aluminum oxide (Al₂O₃): Al (group13) → +3 ; O (usual) → –2. LCM of 3 and 2 is 6. 2 Al (+6) and 3 O (–6) → Al₂O₃. (swap: 3→Al, 2→O).
🔹 Sodium sulfide: Na (group1) → +1 ; S (group16) → –2 (more electronegative). Swap: Na₂S. Check: 2×(+1)+(–2)=0.
🔹 Magnesium nitride: Mg (group2) → +2 ; N (group15) → –3 (in nitride). Swap: Mg₃N₂. 3×(+2)+2×(–3)=0.
🔹 Aluminum sulfate: Al³⁺ and SO₄²⁻ → swap: Al₂(SO₄)₃. Total: 2×(+3) + 3×(–2)=0 (sulfate ion has –2).

✅ Always verify sum of oxidation numbers = 0 for neutral compound.

📋 50 MCQs · formulas & oxidation numbers