IB DP Chemistry - S2.2.13 Expanded Octet of Electrons- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.2.13 Expanded Octet of Electrons- 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
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Structure 2.2.13 – Expanded Octet and VSEPR Geometry (5 & 6 Domains)
Structure 2.2.13 – Expanded Octet and VSEPR Geometry (5 & 6 Domains)
Atoms from Period 3 and beyond (e.g. phosphorus, sulfur, chlorine, etc.) have access to empty d orbitals and can therefore accommodate more than 8 electrons in their valence shell. This leads to the formation of molecules with expanded octets. These molecules often involve five or six electron domains around the central atom.
Key Principle (VSEPR Model):
Electron pairs (bonding and non-bonding) repel each other and arrange themselves around the central atom to minimize repulsion. The resulting 3D shape depends on the number of domains and whether they are bonding or lone pairs.
Five Electron Domains:
- Electron Domain Geometry: Trigonal bipyramidal
- Bond Angles: 90° and 120°
Six Electron Domains:
- Electron Domain Geometry: Octahedral
- Bond Angles: 90°
Visualizing Expanded Octet Structures:
Use Lewis structures showing all bonding and lone pairs. Then apply the VSEPR model to determine the geometry by minimizing repulsion between domains.
Important Notes:
- Only atoms in Period 3 and beyond can expand the octet.
- Central atoms in these species often have more than four bonded atoms or lone pairs.
- Lone pairs take up more space than bonding pairs and distort bond angles.
Example
Predict the shape and bond angles of \( \text{PCl}_5 \).
▶️Answer/Explanation
Step 1: Count valence electrons on phosphorus → 5
Step 2: Each chlorine contributes 1 bonding pair
Step 3: Total = 5 bonding pairs, 0 lone pairs
Electron domain geometry: Trigonal bipyramidal
Molecular geometry: Trigonal bipyramidal
Bond angles: 90° and 120°
Expanded octet: Phosphorus holds 10 valence electrons
Example
Predict the geometry and identify any lone pairs in \( \text{BrF}_5 \).
▶️Answer/Explanation
Step 1: Bromine has 7 valence electrons
Step 2: 5 fluorine atoms form bonds → 5 bonding pairs
Step 3: 7 – 5 = 2 electrons → 1 lone pair
Electron domain geometry: Octahedral (6 regions)
Molecular geometry: Square pyramidal (5 bonded, 1 lone pair)
Bond angles: ~85° (slightly less due to lone pair repulsion)
Expanded octet: Bromine holds 12 electrons
Example
Determine the shape and bonding in \( \text{XeF}_4 \).
▶️Answer/Explanation
Step 1: Xenon has 8 valence electrons
Step 2: Forms 4 bonds with fluorine → 4 bonding pairs
Step 3: 8 – 4 = 4 electrons → 2 lone pairs
Electron domain geometry: Octahedral (6 regions)
Molecular geometry: Square planar (lone pairs opposite each other)
Bond angles: 90°
Expanded octet: Xenon holds 12 electrons