IB DP Chemistry -S2.1.2 Ionic bonds and binary compounds - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.1.2 Ionic bonds and binary compounds – Study Notes – New Syllabus
IITian Academy excellent Study Notes and effective strategies will help you prepare for your IB DP Chemistry exam.
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
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- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Structure 2.1.2 — Ionic Bonding and Naming Ionic Compounds
Structure 2.1.2 — Ionic Bonding and Naming Ionic Compounds
An ionic bond is formed due to the strong electrostatic attraction between oppositely charged ions (cations and anions).
How it works:
- Metals lose electrons → form cations (positive ions)
- Non-metals gain electrons → form anions (negative ions)
- Electrostatic forces hold the ions together in a crystal lattice
Forming formulas of ionic compounds:
- Write the correct formula and charge of each ion
- Balance the total positive and negative charges
- Use subscripts to show the number of each ion
Example: Formula of Magnesium Bromide
Identify electron configurations of each element
- Magnesium (Mg): Atomic number = 12 → Electron configuration: \( 1s^2\, 2s^2\, 2p^6\, 3s^2 \)
- Bromine (Br): Atomic number = 35 → Electron configuration: \( 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^5 \)
Determine ions formed by each atom
- Magnesium (Group 2): Loses 2 electrons → forms \( \text{Mg}^{2+} \)
- Bromine (Group 17): Gains 1 electron → forms \( \text{Br}^- \)
Magnesium ion electron configuration: \( 1s^2\, 2s^2\, 2p^6 \)
Bromide ion electron configuration: \( 1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6 \)
Combine ions to form a neutral compound
- \( \text{Mg}^{2+} \) has 2+ charge
- Each \( \text{Br}^- \) has 1− charge
- To balance: Need 2 Br⁻ ions for every Mg²⁺
The formula of magnesium bromide is: \( \text{MgBr}_2 \)
Naming binary ionic compounds:
- Cation is named first (use Roman numerals if it’s a transition metal)
- Anion is named second with the suffix “-ide”
- Polyatomic ions retain their names (e.g., sulfate, nitrate, hydroxide)
Common Polyatomic Ions (IBDP)
| Name | Formula | Charge |
|---|---|---|
| Ammonium | \( \text{NH}_4^+ \) | +1 |
| Hydroxide | \( \text{OH}^- \) | -1 |
| Nitrate | \( \text{NO}_3^- \) | -1 |
| Hydrogencarbonate | \( \text{HCO}_3^- \) | -1 |
| Carbonate | \( \text{CO}_3^{2-} \) | -2 |
| Sulfate | \( \text{SO}_4^{2-} \) | -2 |
| Phosphate | \( \text{PO}_4^{3-} \) | -3 |
Notes: These ions commonly appear in formulas and reactions.
You must:
- Recognize them by both name and formula
- Use brackets in formulas if there’s more than one of the polyatomic ion (e.g. \( \text{Ca(NO}_3)_2 \))
Example :
Deduce the formula of sodium chloride from its ions.
▶️Answer/Explanation
- Sodium ion: \( \text{Na}^+ \)
- Chloride ion: \( \text{Cl}^- \)
- Charges are balanced 1:1 → no subscripts needed
Formula: NaCl
Example :
Write the formula for calcium fluoride.
▶️Answer/Explanation
- Calcium ion: \( \text{Ca}^{2+} \)
- Fluoride ion: \( \text{F}^- \)
- Need 2 fluoride ions to balance the 2+ charge
Formula: \( \text{CaF}_2 \)
Example :
Write the formula for iron(III) oxide.
▶️Answer/Explanation
- Iron(III): \( \text{Fe}^{3+} \)
- Oxide: \( \text{O}^{2-} \)
- Least common multiple of charges = 6
- Need 2 Fe³⁺ and 3 O²⁻ ions
Formula: \( \text{Fe}_2\text{O}_3 \)
Example :
What is the name of the compound with formula \( \text{Al}_2(\text{SO}_4)_3 \)?
▶️Answer/Explanation
- Aluminum ion: \( \text{Al}^{3+} \)
- Sulfate ion: \( \text{SO}_4^{2-} \)
- To balance charges: 2 Al³⁺ and 3 SO₄²⁻ → total charges = +6 and -6
Name: Aluminum sulfate