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IB Chemistry: VSEPR & Molecular Shapes — FAQs
Answered by RevisionPrep's IB Educators
VSEPR trips students up not because it's conceptually hard, but because IB examiners want the exact shape name, the exact bond angle, and a reasoned explanation together. Below, an IB Chemistry educator answers the questions students and parents actually ask about this topic, from how it's tested to where HL goes further than SL.
Understanding VSEPR & Molecular Shapes
What is VSEPR theory in IB Chemistry?
VSEPR (Valence Shell Electron Pair Repulsion) theory predicts a molecule's 3D shape by assuming electron domains — bonding pairs and lone pairs — around a central atom arrange themselves to minimise repulsion. In IB Chemistry, it's examined as a model you apply to explain shape, bond angle and polarity together, not shape in isolation.
According to the IB Chemistry guide (first exams 2025), this content sits under Structure 2.2, "the covalent model" — so expect it linked to bonding questions, not treated as a standalone topic.
How is VSEPR & molecular shapes tested in IB Chemistry?
VSEPR shows up in Paper 1 as multiple-choice questions identifying a shape or bond angle from a formula, and in Paper 2 as structured questions asking you to draw a Lewis structure, name the shape, state the bond angle, and explain the reasoning — usually worth 3-4 marks across the sub-parts.
A typical Paper 2 command sequence looks like: "Deduce the Lewis structure of SF4. State its shape and bond angle. Explain, in terms of electron domain repulsion, why the bond angle deviates from the ideal." Marks are awarded separately for the structure, the shape name, the angle, and the explanation — miss one and you lose that mark even if the rest is correct. Both SL and HL sit this style of question; HL simply draws on a longer list of possible shapes.
What's the difference between electron domain geometry and molecular geometry?
Electron domain geometry counts every electron domain — bonding pairs and lone pairs — around the central atom. Molecular geometry describes only where the atoms sit, ignoring lone pairs visually. Water's electron domain geometry is tetrahedral, but its molecular shape is bent — and examiners mark the atom-shape name, not the domain shape.
Quick tip: if the question says "shape of the molecule", give the molecular geometry (bent, trigonal pyramidal, see-saw). If it says "electron domain geometry" or "arrangement of electron pairs", give the parent geometry (tetrahedral, trigonal bipyramidal) instead.
Why do lone pairs affect bond angles in VSEPR shapes?
Lone pairs occupy more space than bonding pairs because they're held by only one nucleus rather than shared between two atoms, so they repel neighbouring electron domains more strongly. Each lone pair squeezes bonding pairs closer together, which is why NH3 measures 107° and H2O measures 104.5°, both below the ideal tetrahedral 109.5°.
The repulsion strength ranking examiners expect you to quote: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. State this ranking explicitly in "explain" questions — a shape name alone without this reasoning typically scores zero on the explanation mark.
SL vs HL Chemistry — What's Different
Do SL and HL chemistry study the same molecular shapes?
No — SL Chemistry covers molecules with two to four electron domains (linear, trigonal planar, bent, tetrahedral, trigonal pyramidal), while HL adds five- and six-domain shapes involving expanded octets. Both levels use identical VSEPR repulsion logic; HL students just apply it to a longer, more complex shape list.
If your child is choosing between SL and HL Chemistry, know that VSEPR isn't the deciding factor on difficulty — the extra HL shapes follow the same rules SL students already master, so it's additive content rather than a conceptual leap.
What extra shapes does HL Chemistry add for VSEPR?
HL adds shapes built from five electron domains (trigonal bipyramidal, see-saw, T-shaped, linear) and six electron domains (octahedral, square pyramidal, square planar), used for expanded-octet central atoms such as phosphorus, sulfur and chlorine — think PCl5, SF6 and ClF3 rather than the simple molecules seen at SL.
Worked example — SF4 (HL):
- Central atom S has 6 valence electrons; 4 form bonds to F, 1 pair remains as a lone pair → 5 electron domains.
- Electron domain geometry = trigonal bipyramidal.
- The lone pair occupies an equatorial position (less repulsion there) → molecular shape = see-saw.
- Bond angles are slightly less than 90° and 120° due to lone pair repulsion pushing the axial and equatorial F atoms inward.
How to Work Out Shapes
How do I work out the shape of a molecule using VSEPR?
Draw the Lewis structure first, then count the electron domains — bonding regions plus lone pairs — around the central atom. Apply repulsion rules to arrange those domains as far apart as possible, name that arrangement as the electron domain geometry, then remove the lone pairs from view to state the actual molecular shape.
Worked example — NH3:
- Lewis structure: N bonded to 3 H atoms, with one lone pair on N.
- Count domains: 3 bonding + 1 lone pair = 4 total.
- Electron domain geometry: tetrahedral.
- Remove the lone pair visually → molecular shape: trigonal pyramidal.
- Bond angle: compressed from the ideal 109.5° to roughly 107°, because the lone pair repels the three N-H bonding pairs more strongly than they repel each other.
What are the most common mistakes students make with VSEPR shapes?
The single most common error I see in mock scripts is naming the electron domain geometry when the question asks for the molecular shape — writing "tetrahedral" for water instead of "bent". The next most frequent slip is forgetting that lone pairs count as domains at all, which throws off the whole geometry.
3 things to check before your next mock:
- Did you draw the full Lewis structure first, including lone pairs on the central atom?
- Did you name the molecular shape (not the domain geometry) if that's what's asked?
- Did you justify the bond angle using repulsion strength, not just state a number?
How do I predict exact bond angles for molecules with lone pairs?
Start from the parent electron domain geometry's ideal angle, then reduce it for each lone pair squeezing the bonding pairs closer together. CH4 with no lone pairs stays at 109.5°, NH3 with one lone pair drops to about 107°, and H2O with two lone pairs compresses further to about 104.5°.
| Molecule | Lone pairs | Bond angle |
|---|---|---|
| CH4 | 0 | 109.5° |
| NH3 | 1 | ~107° |
| H2O | 2 | ~104.5° |
You won't be asked to calculate these angles from scratch — IB expects you to know the trend and quote the standard values, then explain the direction of the shift using repulsion strength.
Exam, Syllabus & Grades
What molecular shapes do I need to know for IB Chemistry exams?
For SL you need shapes from two to four electron domains: linear (180°), trigonal planar (120°), bent (around 117° or 104.5°), tetrahedral (109.5°) and trigonal pyramidal (107°). HL adds five- and six-domain shapes with angles around 90° and 120° for expanded-octet molecules such as PCl5 and SF6.
| Domains | Shape | Angle | Level |
|---|---|---|---|
| 2 | Linear | 180° | SL |
| 3, no lone pair | Trigonal planar | 120° | SL |
| 3, 1 lone pair | Bent | ~117° | SL |
| 4, no lone pair | Tetrahedral | 109.5° | SL |
| 4, 1 lone pair | Trigonal pyramidal | 107° | SL |
| 4, 2 lone pairs | Bent | 104.5° | SL |
| 5, no lone pair | Trigonal bipyramidal | 90°/120° | HL |
| 5, 1 lone pair | See-saw | <90°/<120° | HL |
| 5, 2 lone pairs | T-shaped | <90° | HL |
| 5, 3 lone pairs | Linear | 180° | HL |
| 6, no lone pair | Octahedral | 90° | HL |
| 6, 1 lone pair | Square pyramidal | <90° | HL |
| 6, 2 lone pairs | Square planar | 90° | HL |
Is VSEPR linked to polarity questions in IB Chemistry exams?
Yes — examiners very often follow a shape question with "hence deduce whether the molecule is polar", so you need the 3D shape and the vector sum of bond dipoles, not just the shape name. A molecule can have polar bonds but be non-polar overall if its shape is symmetric, like CO2 or CCl4.
Worked example — CO2 vs H2O: Both have polar C=O or O-H bonds. CO2 is linear, so the two bond dipoles point in exactly opposite directions and cancel — non-polar overall. H2O is bent, so the two O-H dipoles don't cancel — the molecule is polar. Same idea, opposite outcome, purely because of shape.
Is VSEPR & molecular shapes a hard topic in IB Chemistry?
VSEPR itself isn't conceptually difficult, but it's a frequent source of dropped marks because it demands precise vocabulary — the shape name, the bond angle and the reasoning all correctly matched in one answer. "State and explain" shape questions consistently show weaker performance than pure recall questions, mostly from vague or missing reasoning.
In my own marking of internal mocks, students who lose marks here almost never get the shape itself wrong — they lose the explanation mark by writing "lone pairs repel more" without naming which pairs are repelling which, or by what strength. That single missing clause is usually the difference between full and partial credit.
Revision & Resources
How can my child revise VSEPR and molecular shapes effectively for IB Chemistry?
The most effective approach is repeated low-stakes practice: have your child draw the Lewis structure, name the shape, state the angle and justify it out loud for fifteen to twenty different molecules until the pattern becomes automatic. Isolated memorisation of shape names without this repeated reasoning practice tends to fall apart under exam pressure.
On RevisionPrep, the Topical Worksheets and question bank for DP Chemistry give graded VSEPR practice with full mark-scheme-style answers, so mistakes get corrected as they happen rather than being discovered for the first time in a mock exam. Revision Notes cover the SL and HL shape lists side by side, which helps when a student is deciding how far into the HL content they need to go.
SL vs HL: VSEPR Shapes Coverage
| Electron domains | Shape name | Bond angle | Level |
| 2 | Linear | 180° | SL |
| 3 (no lone pair) | Trigonal planar | 120° | SL |
| 3 (1 lone pair) | Bent | ~117° | SL |
| 4 (no lone pair) | Tetrahedral | 109.5° | SL |
| 4 (1 lone pair) | Trigonal pyramidal | 107° | SL |
| 4 (2 lone pairs) | Bent | 104.5° | SL |
| 5 (no lone pair) | Trigonal bipyramidal | 90°/120° | HL |
| 5 (1 lone pair) | See-saw | <90°/<120° | HL |
| 5 (2 lone pairs) | T-shaped | <90° | HL |
| 5 (3 lone pairs) | Linear | 180° | HL |
| 6 (no lone pair) | Octahedral | 90° | HL |
| 6 (1 lone pair) | Square pyramidal | <90° | HL |
| 6 (2 lone pairs) | Square planar | 90° | HL |
For graded VSEPR practice with full mark-scheme answers, work through the DP Chemistry Topical Worksheets and question bank on revisionprep.com.
