IB DP Chemistry - S2.1.1 Formation of ions- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.1.1 Formation of ions – 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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Structure 2.1.1 — Formation of Ions
Structure 2.1.1 — Formation of Ions
What Happens When Atoms Gain or Lose Electrons?
- Metals lose electrons → form positive ions (cations).
- Non-metals gain electrons → form negative ions (anions).
How to Predict the Charge on an Ion
You can predict the charge based on the number of valence electrons and the element’s position in the periodic table:
- Group 1 → lose 1 electron → charge = +1 (e.g., Na⁺)
- Group 2 → lose 2 electrons → charge = +2 (e.g., Mg²⁺)
- Group 13 → lose 3 electrons → charge = +3 (e.g., Al³⁺)
- Group 15 → gain 3 electrons → charge = -3 (e.g., N³⁻)
- Group 16 → gain 2 electrons → charge = -2 (e.g., O²⁻)
- Group 17 → gain 1 electron → charge = -1 (e.g., Cl⁻)
- Group 18 → full outer shell → no ions formed under normal conditions.
Example:
Sodium (Na)
Electron configuration: \( 1s^2\,2s^2\,2p^6\,3s^1 \)
It loses the 3s¹ electron → becomes \( \text{Na}^+ \) with configuration \( 1s^2\,2s^2\,2p^6 \)
Chlorine (Cl)
Electron configuration: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \)
Gains one electron → becomes \( \text{Cl}^- \) with full outer shell: \( 3p^6 \)
Transition Metals: Variable Charges
Transition metals can form ions with multiple possible charges. This is because:
- They can lose different numbers of electrons from both the outer 4s and inner 3d subshells.
- The energy levels of the 4s and 3d orbitals are close together.
Examples:
- Iron (Fe): Can form \( \text{Fe}^{2+} \) (loses 2 electrons) or \( \text{Fe}^{3+} \) (loses 3 electrons)
- Copper (Cu): Can form \( \text{Cu}^+ \) or \( \text{Cu}^{2+} \)
Example
What is the likely ion formed by calcium? Use its electron configuration to explain your answer.
▶️Answer/Explanation
Electron configuration of Ca: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2 \)
Calcium is in Group 2. It loses its two 4s electrons to achieve a noble gas configuration.
Ion formed: \( \text{Ca}^{2+} \)
Ion configuration: \( [\text{Ar}] \)
Example
Use the electron configuration of chlorine to predict the ion it forms.
▶️Answer/Explanation
Electron configuration of Cl: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^5 \)
Chlorine is in Group 17. It gains one electron to complete the 3p subshell.
Ion formed: \( \text{Cl}^- \)
Ion configuration: \( 1s^2\,2s^2\,2p^6\,3s^2\,3p^6 \)
Example
Iron commonly forms two different ions. Predict both using its electron configuration.
▶️Answer/Explanation
Electron configuration of Fe: \( [\text{Ar}]\,4s^2\,3d^6 \)
- Fe²⁺: loses 2 electrons → configuration = \( [\text{Ar}]\,3d^6 \)
- Fe³⁺: loses 3 electrons (2 from 4s, 1 from 3d) → configuration = \( [\text{Ar}]\,3d^5 \)
Fe³⁺ is more stable due to half-filled 3d subshell (d⁵).
Example
Copper forms both +1 and +2 ions. Predict both ions using its electron configuration and justify the stability.
▶️Answer/Explanation
Electron configuration of Cu: \( [\text{Ar}]\,4s^1\,3d^{10} \)
- Cu⁺: loses 1 electron (from 4s) → configuration = \( [\text{Ar}]\,3d^{10} \)
- Cu²⁺: loses 2 electrons (1 from 4s, 1 from 3d) → configuration = \( [\text{Ar}]\,3d^9 \)
Cu⁺ has a full d-subshell (d¹⁰) and is quite stable, but Cu²⁺ is more common in aqueous solution due to hydration energy and lattice enthalpy considerations.