VSEPR Shape Problem Set

Introduction

VSEPR questions can appear deceptively simple: draw a Lewis structure and name a shape. The challenge is keeping two geometries separate. Electron geometry includes lone pairs, while molecular shape names only atomic positions. Multiple bonds count as one domain around the central atom in the basic model.

Core explanation

For two electron domains, a central atom has linear electron geometry. CO₂ has two C=O double bonds; each double bond counts as one region, so carbon has two domains and the molecule is linear, with an ideal 180° O–C–O angle. Counting four individual bond lines would falsely give tetrahedral geometry. The double-bond electron density is substantial, but VSEPR still groups it as one directional region.

For three domains, the electron geometry is trigonal planar. BF₃ has three B–F bonding regions and no central lone pair, so its molecular shape is trigonal planar. SO₂ has two S–O bonding regions and one central lone pair in the common model, so its molecular shape is bent despite a trigonal-planar electron-domain arrangement. Resonance and the precise bonding model should not be mistaken for a third bonded atom.

For four domains, the electron geometry is tetrahedral. CH₄ has four bonding regions and no lone pair, so its shape is tetrahedral. NH₃ has three bonds plus one lone pair: electron geometry tetrahedral, shape trigonal pyramidal. H₂O has two O–H bonds plus two lone pairs: electron geometry tetrahedral, shape bent. The ideal tetrahedral angle is about 109.5°, but repulsion and bonding details shift the measured H–N–H and H–O–H angles. VSEPR is qualitative; do not derive exact angles merely by counting pairs.

Five domains make trigonal-bipyramidal electron geometry, with three equatorial and two axial positions. SF₄ has four bonds and one lone pair; placing the lone pair equatorially reduces its 90° contacts, producing a seesaw molecular shape. ClF₃ has three bonds and two lone pairs, commonly described as T-shaped. XeF₂ has two bonds and three equatorial lone pairs, leaving axial F atoms in a linear molecular shape. These examples reveal why “linear” does not always imply two domains: XeF₂ has five around the xenon centre in the traditional VSEPR account.

Six domains give an octahedral electron arrangement. SF₆ has six bonds and no central lone pairs, so molecular shape is octahedral. BrF₅ has five bonds and one lone pair, so it is square pyramidal. XeF₄ has four bonds and two lone pairs opposite one another; the atoms lie in a square plane. Electron-domain count remains six even though the observed atom count varies. For electron-deficient and hypervalent species, VSEPR labels are geometric descriptions; they do not settle the detailed orbital explanation of bonding.

Always build the Lewis electron budget first. A central atom's valence electrons, net charge and bond pattern determine whether lone pairs remain. An ion can have a different shape from its neutral counterpart: NH₄⁺ is tetrahedral with four bonds and no lone pair, whereas NH₃ is pyramidal with one lone pair. A formula alone does not licence guessing the shape without electron accounting.

Step-by-step reasoning

1. Draw a valid Lewis representation and identify the chosen central atom. 2. Count one domain per sigma-bond direction, regardless of multiple bond lines, plus each lone pair. 3. Name electron geometry from the total number of domains. 4. Omit lone-pair positions when naming molecular shape. 5. Check whether lone pairs or multiple bonds qualify ideal-angle predictions.

Visual explanation

Draw a table with rows CO₂ 2+0, BF₃ 3+0, SO₂ 2+1, NH₃ 3+1 and H₂O 2+2, where the first count is bonding regions and the second is lone pairs. The last two rows share tetrahedral electron geometry but have different atom-only shapes.

Real-world analogy

People and empty reserved seats can determine spacing around a round table, but a photograph showing only people gives a different outline. Lone pairs influence electron-region arrangement while remaining invisible in the atom-only molecular shape.

Real-world example

Water's bent shape makes its O–H bond dipoles reinforce rather than cancel, contributing to its polarity. Geometry therefore connects a Lewis electron count to measurable electrical and intermolecular properties.

Why?

Why is NH₄⁺ tetrahedral instead of pyramidal? Protonation uses ammonia's lone pair to form a fourth N–H bond; the ion has four bonding domains and no central lone pair in its Lewis description.

Common misconception

“A double bond counts as two VSEPR domains.” It is two shared electron pairs but one directional bonding region around the centre. CO₂ is the decisive counterexample.

Worked example

Determine BrF₅'s shape. Bromine has seven valence electrons; five Br–F single bonds leave one central lone pair in the conventional Lewis representation. This gives six domains total, an octahedral electron geometry. With one position occupied by a lone pair, five F atoms form a square-pyramidal molecular shape. Ideal 90° and 180° reference angles may be distorted.

Quick check

1. What is H₂O's electron geometry and molecular shape? Answer: Tetrahedral electron geometry and bent molecular shape.

Exam focus

Report both geometries when asked. Make the bond-region plus lone-pair count visible, especially for ions and multiple bonds. Treat VSEPR angle values as approximate unless supplied or measured.

Advanced insight

The simple claim that lone pairs always repel more strongly than bond pairs is a heuristic, not a complete predictive law. Ligand electronegativity, multiple bonding and electronic structure can change angles, especially for heavy or hypervalent centres.

Summary

VSEPR names an arrangement of all electron domains, then an atom-only molecular shape. CO₂ is linear with two domains; water is bent with four; five- and six-domain examples can produce several atom shapes depending on lone-pair positions.

Practice questions

1. How many central domains does CO₂ have? Answer: Two, because each C=O counts as one region. 2. What is NH₃'s molecular shape? Answer: Trigonal pyramidal. 3. Which positions do XeF₂'s three lone pairs occupy in simple VSEPR? Answer: The three equatorial positions, leaving a linear F–Xe–F arrangement. 4. Why is XeF₄ square planar? Answer: Six domains are arranged octahedrally with two opposed lone pairs and four F atoms in one plane.