IB DP Chemistry – S2.1.1 Formation of ions- Study Notes

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

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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.

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