A student claims that since aluminium (atomic number 13) is a reactive metal, it should form an ionic compound with fluorine (atomic number 9) in which one aluminium ion bonds with one fluoride ion. Identify the error in this claim and show, using electronic configurations, the correct formula of the ionic compound formed.
Generated by claude-sonnet-4-6 · 2026-06-26 01:06 · grounding rag
Model Answer
Error in the student's claim:
Aluminium has atomic number 13, so its electronic configuration is 2, 8, 3. It has 3 valence electrons and loses all three to form Al³⁺ (not Al⁺). The student incorrectly assumed aluminium forms a 1+ ion.
Electronic configurations:
- Al (2, 8, 3) → Al³⁺ (2, 8) + 3e⁻
- F (2, 7) + 1e⁻ → F⁻ (2, 8)
Correct formula:
Each Al³⁺ ion loses 3 electrons. Each F⁻ ion gains only 1 electron. Therefore, 3 fluoride ions are needed to accept the 3 electrons lost by one aluminium ion.
$$\text{Al}^{3+} + 3\text{F}^- \rightarrow \text{AlF}_3$$
The correct formula of the ionic compound formed is AlF₃ (aluminium fluoride).
Source: Chapter 3, Section 3.3 — How do Metals and Non-metals React?
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Explanation
- Examiners award marks for: (1) identifying the specific error (Al forms Al³⁺, not Al⁺), (2) showing both electronic configurations with electron transfer, and (3) deriving the correct formula using charge balance.
- Always write electronic configurations in shell notation (2, 8, 3) and show the ion formed after electron loss/gain.
- The formula follows from charge neutrality: one Al³⁺ needs exactly three F⁻ ions → AlF₃. This mirrors the MgCl₂ example in the textbook.