VSEPR Practice Questions: Predict the Shape and Bond Angle

Predicting molecular shapes with VSEPR (valence shell electron pair repulsion) theory is a skill that appears in almost every chemistry exam at this level. The method is always the same: count electron pairs around the central atom, find the arrangement that keeps them furthest apart, then name the shape from the positions of the atoms only. These 15 questions start simple and work up to the tricky five- and six-pair cases. Try each one before checking the answer key.

Quick reminder of the method

1. Find the central atom and its number of valence electrons (group number).
2. Add one electron for each atom bonded to it by a single bond (each bond contributes one electron from the other atom). Adjust for charge: add one electron per negative charge, subtract one per positive charge.
3. Divide by two to get the number of electron pairs (or count bonded atoms + lone pairs directly from a Lewis structure). Treat a double or triple bond as one region of electron density.
4. Pick the electron-pair geometry for that number of regions.
5. Name the molecular shape from the atom positions, and adjust angles: lone pairs repel more than bonding pairs, squeezing bond angles by about 2–2.5° per lone pair.

Reference table:

Regions	Bonding	Lone	Shape	Angle
2	2	0	Linear	180°
3	3	0	Trigonal planar	120°
3	2	1	Bent	about 118°
4	4	0	Tetrahedral	109.5°
4	3	1	Trigonal pyramidal	about 107°
4	2	2	Bent (V-shaped)	about 104.5°
5	5	0	Trigonal bipyramidal	90°, 120°
5	4	1	Seesaw	about 90°, below 120°
5	3	2	T-shaped	about 90°
5	2	3	Linear	180°
6	6	0	Octahedral	90°
6	5	1	Square pyramidal	about 90°
6	4	2	Square planar	90°

For the full theory, see VSEPR and molecular geometry.

Questions

For each species, give the number of bonding pairs and lone pairs on the central atom, the shape, and the bond angle.

1. BeCl₂ (gas phase)
2. BCl₃
3. SiH₄
4. PH₃
5. H₂S
6. CO₂
7. SO₂
8. NH₄⁺
9. NO₃⁻
10. PF₅
11. SF₄
12. ClF₃
13. XeF₂
14. SF₆
15. ICl₄⁻

Answer key

1. BeCl₂ Be has 2 valence electrons + 2 from Cl = 4 → 2 pairs, both bonding. Linear, 180°. (Be has an incomplete octet.)

2. BCl₃ B: 3 + 3 = 6 → 3 bonding pairs, 0 lone. Trigonal planar, 120°.

3. SiH₄ Si: 4 + 4 = 8 → 4 bonding pairs. Tetrahedral, 109.5°.

4. PH₃ P: 5 + 3 = 8 → 4 pairs: 3 bonding, 1 lone. Trigonal pyramidal. VSEPR predicts slightly less than 109.5°; the measured angle is about 93.5° — much smaller, because phosphorus’s bonding pairs are further from the nucleus and larger atoms hybridise less. Exams usually accept “less than 109.5°, about 107°” for a simple VSEPR answer, but know that heavier atoms give smaller angles.

5. H₂S S: 6 + 2 = 8 → 2 bonding, 2 lone. Bent. VSEPR predicts about 104.5° by analogy with water; the measured angle is about 92°, for the same reason as PH₃.

6. CO₂ C has two double bonds and no lone pairs → 2 regions. Linear, 180°. (Each double bond counts as one region.)

7. SO₂ S forms two bonds (drawn as S=O, or one double and one single with resonance) and has one lone pair → 3 regions: 2 bonding, 1 lone. Bent, about 119°.

8. NH₄⁺ N: 5 + 4 − 1 (positive charge) = 8 → 4 bonding pairs, 0 lone. Tetrahedral, 109.5°. (Compare NH₃: trigonal pyramidal, 107°.)

9. NO₃⁻ N forms bonds to three O atoms, no lone pairs → 3 regions. Trigonal planar, 120°. All three N–O bonds are identical because of resonance (see resonance structures).

10. PF₅ P: 5 + 5 = 10 → 5 bonding pairs. Trigonal bipyramidal: three equatorial F at 120°, two axial F at 90° to the equator (180° to each other).

11. SF₄ S: 6 + 4 = 10 → 5 pairs: 4 bonding, 1 lone. The lone pair sits in an equatorial position (where it has more room: only two neighbours at 90°, rather than three). Seesaw. Axial F–S–F about 173° (instead of 180°); equatorial F–S–F about 102° (instead of 120°).

12. ClF₃ Cl: 7 + 3 = 10 → 5 pairs: 3 bonding, 2 lone. Both lone pairs go equatorial. T-shaped, F–Cl–F angles about 87.5° (slightly less than 90° because of lone-pair repulsion).

13. XeF₂ Xe: 8 + 2 = 10 → 5 pairs: 2 bonding, 3 lone. All three lone pairs occupy the equatorial positions, leaving the two F atoms axial. Linear, 180°. (The electron-pair geometry is trigonal bipyramidal, but the atoms are in a line.)

14. SF₆ S: 6 + 6 = 12 → 6 bonding pairs. Octahedral, 90°. Non-polar because of its symmetry.

15. ICl₄⁻ I: 7 + 4 + 1 (negative charge) = 12 → 6 pairs: 4 bonding, 2 lone. The two lone pairs sit opposite each other (180° apart) to minimise repulsion. Square planar, 90°.

Why lone pairs go where they do

In the trigonal bipyramid (5 regions), there are two kinds of position:

• Axial: three neighbours at 90°.
• Equatorial: two neighbours at 90° (and two at 120°).

Lone pairs take equatorial positions because repulsions at 90° are the strongest, and equatorial positions have fewer of them. That’s why SF₄ is a seesaw, ClF₃ is T-shaped and XeF₂ is linear.

In the octahedron (6 regions), all positions are equivalent at first. A second lone pair goes opposite the first (180°), giving square planar shapes like ICl₄⁻ and XeF₄.

Worked walkthrough: ClF₃ in full

The five-region questions cause the most trouble, so here is question 12 written out as a model answer.

Chlorine is in group 17, so it has 7 valence electrons. Each of the three fluorine atoms forms a single bond and contributes one electron, giving 7 + 3 = 10 electrons, or five pairs. Three pairs are bonding pairs (one to each F) and the remaining two are lone pairs. Five regions of electron density arrange themselves as a trigonal bipyramid.

Next, place the lone pairs. An axial position has three neighbours at 90°, while an equatorial position has only two. Lone pairs repel most strongly, so both go equatorial, where they suffer fewer 90° repulsions. That leaves one fluorine in the equatorial plane and two in the axial positions. Looking only at the atoms, the four atoms form a T shape. Finally, the two lone pairs push the axial fluorines slightly towards the equatorial one, so the F–Cl–F angles are a little under 90°, measured at about 87.5°.

Every five- and six-region question can be answered with the same four moves: count electrons, count regions, place lone pairs where they suffer the fewest 90° repulsions, then name the shape from the atoms.

Scoring

• 14–15: excellent — try bond angles explained for the deeper reasons.
• 10–13: solid; revisit the five- and six-region cases.
• Below 10: review lone pairs and molecular shape and redo questions 1–9.

Common mistakes

• Counting double bonds as two regions (CO₂ would wrongly become tetrahedral).
• Forgetting the charge on ions (NH₄⁺, ICl₄⁻).
• Naming the shape from the electron-pair geometry instead of the atoms: XeF₂ is linear, not trigonal bipyramidal (see electron geometry vs molecular geometry).
• Putting lone pairs in axial positions in five-region molecules.
• Assuming all tetrahedral-based molecules have 109.5° — lone pairs squeeze angles.

Key takeaways

• Count regions of electron density (bonded atoms + lone pairs) around the central atom; multiple bonds count once.
• The electron-pair geometry comes from the number of regions; the shape comes from the atom positions.
• Lone pairs repel more, compressing bond angles; in five-region molecules they go equatorial, in six-region molecules opposite each other.
• Heavier central atoms (P, S) often have angles near 90°, smaller than simple VSEPR predicts.