2.2.14Additional lessons
Expansion of the Octet (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 6 parts Lesson 38 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 6What Expansion Means
Some central atoms form compounds in which they surround themselves with more than four electron pairs.
Definition
- Definition A central atom accommodates more than eight electrons in its valence shell.
Period 3 and below
- Period 3 and below These elements have energetically accessible d orbitals in their valence shell, so more pairs can fit.
Period 2 limit
- Period 2 limit Carbon, nitrogen, oxygen and fluorine have no d orbitals in the second shell, so they never exceed eight.
Typical outer atoms
- Typical outer atoms Small, electronegative atoms such as F, Cl and O are usually attached to the expanded atom.
Section 2 of 6Common Examples
Common Examples
Phosphorus pentachloride
- Phosphorus pentachloride PCl₅ has five bonding pairs, giving ten electrons around phosphorus.
Sulfur hexafluoride
- Sulfur hexafluoride SF₆ has six bonding pairs, giving twelve electrons around sulfur.
Xenon tetrafluoride
- Xenon tetrafluoride XeF₄ has four bonding pairs and two lone pairs, so twelve electrons on xenon.
Sulfate ion
- Sulfate ion The S atom may be drawn with two double bonds to oxygen, giving twelve electrons.
| Species | Domains on central atom | Shape |
|---|---|---|
| PCl₅ | 5 | Trigonal bipyramidal |
| SF₆ | 6 | Octahedral |
| XeF₄ | 6 | Square planar |
Section 3 of 6Drawing the Formula
1
Count the total valence electrons of all atoms and subtract or add for the charge.
2
Connect the outer atoms to the central atom and give each outer atom a full octet.
3
Place all the remaining electrons on the central atom, even if it ends up with more than eight.
4
Check that the total number of electrons drawn equals the number counted.
5
Use the number of domains to predict the shape.
Section 4 of 6Why Period 2 Is Different
NF₃ exists
- NF₃ exists Nitrogen forms NF₃ with eight electrons, but NF₅ does not exist.
PF₅ exists
- PF₅ exists Phosphorus can use its vacant 3d orbitals, so PF₅ forms readily.
Interpretation
- Interpretation The d orbitals model is a useful simple explanation for the electron counts.
Section 5 of 6Worked Example
Question Determine the electron count and shape of XeF₄.
- Valence electrons8 + 4 × 7 = 36.
- Four Xe–F bonds use 8 electrons; each F takes 3 lone pairs (24 electrons), leaving 4 electrons.
- These 4 electrons form 2 lone pairs on Xe, giving 6 domains.
Answer
- Octahedral arrangement of domains;
- with two lone pairs opposite each other, XeF₄ is square planar.
Section 6 of 6Exam Checklist & Tips
2.2.14 – Key Points for the Exam
- Expanded octet A central atom with more than eight valence electrons.
- d orbitals Orbitals available from the third shell onwards that allow extra electron pairs.
- Domain A region of electron density around an atom.
- Hypervalent Describes a compound whose central atom has an expanded octet.
Common Mistakes
- Saying nitrogen can form NCl₅.
- Forgetting the lone pairs on xenon when counting electrons.
- Giving an octet to every atom in SF₆.
- Thinking all the electrons must be bonding pairs.
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