Electron-Pair Geometry vs Molecular Shape: HSC Chemistry

A Lewis structure for the nitrite ion, \(\mathrm{NO_{2}}^{-}\), can be drawn with nitrogen in the middle and two oxygen atoms attached. If you look only at the two bonds, you might predict a linear arrangement.

But the actual molecular shape is bent.

So which geometry should you give in an HSC Chemistry question: linear, bent, or trigonal planar?

The correct answer is trigonal planar electron-pair geometry and bent molecular shape. Both descriptions are correct because they describe different things:

• Electron-pair geometry includes every electron region around the central atom, including lone pairs.
• Molecular shape describes the arrangement of atoms only. Lone pairs are not included when naming the shape.

This distinction is one of the most important parts of using VSEPR theory correctly. Once you understand what each description includes, you can work out both without memorising two separate sets of shapes.

01The two descriptions answer different questions

Think about what you are actually trying to describe.

When you ask for electron-pair geometry, you are asking:

How are all the regions of electron density arranged around the central atom?

When you ask for molecular shape, you are asking:

Where are the surrounding atoms positioned relative to the central atom?

These questions produce different answers whenever lone pairs are present.

Electron-pair geometry

Electron-pair geometry describes the arrangement of all electron regions around a central atom.

An electron region can be:

• A single bond.
• A double bond.
• A triple bond.
• A lone pair.

The important detail is that a single, double, or triple bond each counts as one electron region.

Why? Because VSEPR theory considers the general direction in which the bonding electrons are concentrated. A double bond contains more electron density than a single bond, but it still occupies one general direction from the central atom.

Lone pairs also occupy regions around the central atom, even though they are not shared with another atom.

Molecular shape

Molecular shape describes the arrangement of the atoms around the central atom.

When naming molecular shape, you ignore the positions occupied by lone pairs.

This does not mean lone pairs have disappeared. They still influence the positions of the bonding regions. You simply leave them out when describing the arrangement of atoms.

A useful way to picture this is to imagine a seating arrangement.

Electron-pair geometry considers every occupied position, including positions taken by lone pairs. Molecular shape considers only the positions occupied by atoms.

The analogy is limited because electrons are not fixed objects sitting in individual seats. Their electron density is spread through space. However, it captures the distinction between the two descriptions.

The comparison table you need

For the common arrangements involving two, three, and four electron regions, use this table.

Notice the pattern.

When every electron region is a bonding region, electron-pair geometry and molecular shape are the same.

When one or more regions contain lone pairs, the two descriptions can differ.

02A reliable method for distinguishing the two

The biggest mistake students make is trying to name the molecular shape before they have worked out the electron arrangement.

Use this sequence instead.

Step 1: Count the total valence electrons

Start by counting the valence electrons contributed by every atom.

Remember to account for the overall charge:

• Add one electron for each negative charge.
• Subtract one electron for each positive charge.

For example, nitrogen contributes five valence electrons, and each oxygen contributes six.

The negative charge on \(\mathrm{NO_{2}}^{-}\) means you must add one electron:

\[
5+2(6)+1=18\text{ electrons}
\]

This gives you the total number of electrons available for the Lewis structure.

Step 2: Draw the Lewis structure

Use the available electrons to establish the bonds and lone pairs.

You need a reasonable Lewis structure before you can apply VSEPR theory.

Pay attention to:

• The number of bonds around the central atom.
• The number of lone pairs on the central atom.
• Any multiple bonds.
• The overall charge and formal charges.

If resonance is possible, consider the equivalent resonance contributors rather than treating one drawing as the complete physical structure.

Step 3: Count electron regions around the central atom

Count each bonding direction as one electron region.

For example:

• One single bond = one region.
• One double bond = one region.
• One triple bond = one region.
• One lone pair = one region.

Do not count the individual shared electron pairs in a double bond as separate regions.

This is particularly important when working with resonance structures.

Step 4: Name the electron-pair geometry

Use the total number of electron regions to identify their arrangement.

The main arrangements are:

These are ideal arrangements. Actual bond angles can differ because lone pairs and different bonding regions do not repel equally.

Step 5: Name the molecular shape

Now look only at the atoms bonded to the central atom.

Ignore the lone-pair positions when naming the shape.

For example, if there are three electron regions consisting of two bonding regions and one lone pair:

• Electron-pair geometry: trigonal planar.
• Molecular shape: bent.

The lone pair occupies the third position in the electron arrangement, but it is not an atom.

That is the entire distinction.

04What physical evidence tells us about molecular shape

There is another important distinction to understand.

A Lewis structure is a representation of bonding and electron bookkeeping. It is not a photograph of a molecule.

Chemists use experimental techniques to investigate actual molecular structures. One example is X-ray diffraction.

How an X-ray diffraction experiment works

Imagine a small crystal containing sodium nitrite.

The experiment follows this sequence:

1. The crystal is mounted in an X-ray beam.
2. The X-rays interact with the electron density in the crystal.
3. The crystal produces a diffraction pattern.
4. A detector records the intensities of the diffracted X-rays.
5. The pattern is analysed to reconstruct electron density and refine the positions of the atoms.

The resulting structural model provides information about bond lengths and bond angles.

It does not provide a photograph of a lone pair as two little dots.

What should the experiment show?

Consider two competing predictions.

If the oxygen atoms in nitrite were arranged linearly, the O-N-O bond angle would be:

\[
180^\circ
\]

If they were arranged in a bent shape, the angle would be substantially smaller.

For practice, consider the following illustrative structural data. These values are not presented as a particular published crystal refinement.

The bond angle is clearly not \(180^\circ\).

We can calculate the difference:

\[
180.0^\circ-115.5^\circ=64.5^\circ
\]

A difference of \(64.5^\circ\) is far greater than the stated measurement uncertainty of \(0.5^\circ\).

The measured arrangement is therefore bent, not linear.

This supports the molecular-shape prediction from VSEPR theory.

The approximately equal N-O bond lengths also provide evidence for resonance. The two Lewis contributors do not represent two different structures that the ion rapidly switches between. They are alternative representations of one delocalised structure.

An O-N-O angle around \(115^\circ\) is also consistent with reported nitrite structural data; the University of Bristol discusses this result in its chemistry explanation of nitrogen dioxide species.

What does the experiment not tell us?

The measured bond angle tells us where the oxygen atoms are positioned.

It does not directly establish the number of lone pairs on nitrogen.

We infer the lone pair by combining:

1. The total valence electron count.
2. A suitable Lewis structure.
3. Formal charge calculations.
4. The VSEPR model.

There is also an experimental limitation worth recognising.

A crystal contains neighbouring ions, and their interactions can influence the exact bond angles and distances. A gas-phase measurement, where feasible, can reduce the influence of the crystal lattice.

This is why you should not treat a particular measured angle as an exact universal value.

The important distinction remains: experimental measurements establish the arrangement of atoms, while electron-pair geometry is a model of the electron regions that helps explain that arrangement.

06Structural representations: What are you actually looking at?

A lot of confusion comes from treating different chemical representations as though they show exactly the same information.

They don’t.

For example, a Lewis structure might be written with the atoms in a straight horizontal line.

That does not prove that the molecule is linear.

The drawing is primarily showing which atoms are connected. The three-dimensional arrangement must be established using the electron regions and an appropriate structural model.

Similarly, a resonance contributor showing one double bond does not mean that one particular bond is permanently double.

Always ask what information the representation is designed to communicate.

07Common mistakes to avoid

Mistake 1: Counting a double bond as two electron regions.

A double bond contains two shared electron pairs, but both bonding pairs are concentrated in the same general direction. Count one region.

Mistake 2: Ignoring lone pairs when finding electron-pair geometry.

Lone pairs must be included when determining electron-pair geometry. They are excluded only when naming molecular shape.

Mistake 3: Assuming the molecular shape must match the electron-pair geometry.

This is true when there are no lone pairs on the central atom, but not when lone pairs occupy electron-region positions.

Mistake 4: Assuming the overall charge tells you the number of lone pairs directly.

The charge affects the total valence electron count. You must still construct a Lewis structure and determine where those electrons are located.

Mistake 5: Treating resonance structures as separate physical molecules.

Equivalent resonance contributors represent one delocalised structure. They are not different molecules rapidly switching between drawings.

Mistake 6: Treating a flat Lewis structure as a 3D diagram.

A Lewis structure shows connectivity and electron bookkeeping. It does not necessarily show actual bond angles or molecular shape.

For a broader explanation of the electron-region arrangements and how to predict molecular shapes, see Molecular Shapes: Predicting Geometry with VSEPR for HSC Chemistry.

08Questions and solutions

Question 1

Formaldehyde, \(\ce{CH2O}\), has a central carbon atom bonded to two hydrogen atoms and double-bonded to one oxygen atom.

Determine the electron-pair geometry and molecular shape around carbon. Explain why the C=O double bond counts as one electron region.

Solution 1

Both the electron-pair geometry and molecular shape are trigonal planar.

Carbon has three electron regions around it:

• Two C-H single bonds.
• One C=O double bond.

The double bond counts as one electron region because it occupies one general direction from the carbon atom.

There are no lone pairs on the central carbon.

Therefore:

\[
3\text{ bonding regions}+0\text{ lone pairs}
=3\text{ regions}
\]

Three electron regions produce a trigonal planar electron-pair geometry.

Because all three regions contain bonded atoms, the molecular shape is also trigonal planar.

The important point is that the number of shared electron pairs is not the same as the number of electron regions.

Question 2

The hydronium ion, \(\ce{H3O+}\), contains one oxygen atom bonded to three hydrogen atoms.

Determine:

a) The total number of valence electrons.

b) The number of bonding regions and lone pairs around oxygen.

c) The electron-pair geometry and molecular shape.

d) The formal charge on the central oxygen atom.

Solution 2

The electron-pair geometry is tetrahedral, the molecular shape is trigonal pyramidal, and the central oxygen has a formal charge of \(+1\).

a) Total valence electrons

Oxygen contributes six electrons, the three hydrogen atoms contribute three, and the positive charge removes one electron.

\[
6+3(1)-1=8
\]

The total valence electron count is 8.

Correction: the positive charge removes one electron from the neutral atom total, so the correct count is:

\[
6+3-1=8
\]

b) Bonding regions and lone pairs

The oxygen forms three O-H single bonds.

These use six electrons, leaving two electrons as one lone pair.

Therefore, oxygen has:

• Three bonding regions.
• One lone pair.

c) Electron-pair geometry and molecular shape

The total number of electron regions is:

\[
3+1=4
\]

Four electron regions arrange tetrahedrally.

Therefore, the electron-pair geometry is tetrahedral.

Only the three hydrogen atoms are included when naming the molecular shape. Their arrangement is trigonal pyramidal.

d) Formal charge

For oxygen:

\[
\begin{aligned}
\mathrm{FC}(O)
&=6-2-\frac{6}{2}\\
&=+1
\end{aligned}
\]

The central oxygen has a formal charge of \(+1\), matching the overall charge of the ion.

This example shows why a positive charge does not mean that the central atom has no lone pairs.

Question 3

Compare the nitronium ion, \(\ce{NO2+}\), and the nitrite ion, \(\mathrm{NO_{2}}^{-}\).

For each species:

a) Calculate the total number of valence electrons.

b) Identify the bonding regions and lone pairs around nitrogen.

c) State the electron-pair geometry and molecular shape.

d) Explain why the two ions have different geometries despite containing the same types and numbers of atoms.

Solution 3

The nitronium ion has linear electron-pair geometry and linear molecular shape. The nitrite ion has trigonal planar electron-pair geometry and bent molecular shape.

The difference comes from their total valence electron counts and the resulting Lewis structures.

a) Total valence electrons

For nitronium:

\[
5+2(6)-1=16
\]

For nitrite:

\[
5+2(6)+1=18
\]

The nitrite ion has two more valence electrons than the nitronium ion.

b) Bonding regions and lone pairs

For \(\ce{NO2+}\), a suitable Lewis structure contains two N=O double bonds.

Nitrogen has:

• Two bonding regions.
• No lone pairs.

For \(\mathrm{NO_{2}}^{-}\), a suitable Lewis contributor contains one N=O double bond, one N-O single bond, and one lone pair on nitrogen.

Nitrogen has:

• Two bonding regions.
• One lone pair.

c) Geometries and shapes

For nitronium:

\[
2\text{ regions}\rightarrow\text{linear}
\]

Both electron-pair geometry and molecular shape are linear.

For nitrite:

\[
3\text{ regions}\rightarrow\text{trigonal planar}
\]

The third region is a lone pair, leaving two bonded oxygen atoms.

Therefore, its molecular shape is bent.

d) Why are the structures different?

The overall charge changes the number of available valence electrons.

The nitronium ion has 16 electrons and forms a structure with two bonding regions around nitrogen.

The nitrite ion has 18 electrons, allowing a lone pair on nitrogen in its Lewis structure.

That additional electron pair changes the electron-region arrangement and therefore changes the molecular shape.

The tempting assumption is that two species with the same atom arrangement in a formula must have the same geometry. Their electron counts show why that assumption fails.

Question 4

The carbonate ion, \(\mathrm{CO_{3}}^{2-}\), and nitrate ion, \(\mathrm{NO_{3}}^{-}\), both contain three oxygen atoms bonded to a central atom.

Interestingly, both species have 24 total valence electrons.

For each ion:

a) Verify the total valence electron count.

b) Describe one valid resonance contributor and calculate the formal charge on the central atom.

c) Determine the electron-pair geometry and molecular shape.

d) Explain why having the same total number of valence electrons does not mean the two ions have identical formal charge distributions.

Solution 4

Both ions have trigonal planar electron-pair geometry and trigonal planar molecular shape. However, their formal charge distributions are different.

a) Total valence electrons

For carbonate:

\[
4+3(6)+2=24
\]

For nitrate:

\[
5+3(6)+1=24
\]

Both ions contain 24 valence electrons.

b) Lewis contributors and formal charges

One carbonate resonance contributor contains:

• One C=O double bond.
• Two C-O single bonds.
• No lone pair on carbon.

The two singly bonded oxygen atoms each have a formal charge of \(-1\).

For carbon:

\[
4-0-\frac{8}{2}=0
\]

The central carbon has a formal charge of zero.

One nitrate resonance contributor contains:

• One N=O double bond.
• Two N-O single bonds.
• No lone pair on nitrogen.

The two singly bonded oxygen atoms each have a formal charge of \(-1\).

For nitrogen:

\[
5-0-\frac{8}{2}=+1
\]

The central nitrogen has a formal charge of \(+1\).

c) Electron-pair geometry and molecular shape

For both ions, the central atom has three bonding regions and no lone pairs.

Therefore:

\[
3\text{ bonding regions}+0\text{ lone pairs}
=3\text{ regions}
\]

Both have trigonal planar electron-pair geometry.

Because all three regions contain oxygen atoms, both also have trigonal planar molecular shapes.

d) Why are the formal charge distributions different?

The central atoms have different valence electron contributions.

Carbon contributes four valence electrons, while nitrogen contributes five.

Even though the total electron counts are equal, the formal charge calculations produce different results.

Both ions have three equivalent resonance contributors, so their actual bonding is delocalised across the three oxygen atoms.

This is an important distinction: electron-pair geometry depends on the number and arrangement of electron regions, not simply on the total number of valence electrons or the overall charge.

Question 5

Water, \(\ce{H2O}\), and the nitrite ion, \(\mathrm{NO_{2}}^{-}\), both have bent molecular shapes.

Consider the following approximate structural angles:

The two N-O bonds in nitrite are also equivalent in length, consistent with resonance.

A student makes two claims:

1. Because both species are bent, they must have the same electron-pair geometry.
2. Because the two N-O bonds in nitrite have equal lengths, nitrogen cannot have a lone pair.

Evaluate both claims.

Your answer must include:

a) The total valence electron count for each species.

b) The bonding regions and lone pairs around each central atom.

c) The electron-pair geometry and molecular shape of both species.

d) An explanation of why the measured bond angles and equal N-O bond lengths do not support the student’s conclusions.

Solution 5

Both claims are incorrect.

Water and nitrite have the same molecular shape, but different electron-pair geometries. The equal N-O bond lengths in nitrite are explained by resonance, not by the absence of a lone pair.

a) Total valence electrons

For water:

\[
6+2(1)=8
\]

For nitrite:

\[
5+2(6)+1=18
\]

Therefore:

• Water has 8 valence electrons.
• Nitrite has 18 valence electrons.

b) Bonding regions and lone pairs

For water, oxygen forms two O-H single bonds and has two lone pairs.

Therefore, it has:

• Two bonding regions.
• Two lone pairs.

For nitrite, nitrogen has two N-O bonding regions and one lone pair.

Therefore, it has:

• Two bonding regions.
• One lone pair.

Remember that the N=O double bond in a resonance contributor still counts as one bonding region.

c) Electron-pair geometry and molecular shape

For water:

\[
2+2=4\text{ electron regions}
\]

Four electron regions give tetrahedral electron-pair geometry.

Only two hydrogen atoms are bonded to oxygen, so the molecular shape is bent.

For nitrite:

\[
2+1=3\text{ electron regions}
\]

Three electron regions give trigonal planar electron-pair geometry.

Only two oxygen atoms are bonded to nitrogen, so the molecular shape is bent.

The comparison is:

d) Interpreting the physical evidence

The student’s first claim confuses molecular shape with electron-pair geometry.

Both molecules are bent because each has two bonded atoms around its central atom.

However, the number of electron regions differs.

Water has four electron regions, while nitrite has three. Therefore, their electron-pair geometries must differ.

The bond angles are consistent with this distinction.

For water, the ideal tetrahedral angle is approximately \(109.5^\circ\), but two lone pairs compress the H-O-H angle to approximately \(104.5^\circ\).

For nitrite, the ideal trigonal planar angle is \(120^\circ\). One lone pair produces greater repulsion than the bonding regions, reducing the O-N-O angle to approximately \(115.5^\circ\).

The second claim confuses bond equivalence with lone-pair presence.

In nitrite, there are two equivalent resonance contributors. One shows a double bond to the first oxygen, while the other shows a double bond to the second oxygen.

The actual structure is a resonance hybrid, with equivalent N-O bonds.

Equal bond lengths therefore do not mean nitrogen has no lone pair.

The important limitation of the experimental evidence is that bond angles and bond lengths describe the arrangement and bonding of atoms. They do not directly count the lone pairs on the central atom.

You need both the structural evidence and the Lewis/VSEPR model to explain the complete picture.

This is the deeper lesson: two species can have the same molecular shape while having different electron-pair geometries. Molecular shape alone does not tell you everything about the electron arrangement.

09Where to go next

Once you can distinguish electron-pair geometry from molecular shape, the next useful step is predicting how lone pairs affect bond angles and molecular polarity.

These ideas connect directly to why molecules with similar structures can have different dipole moments, and why a molecule’s shape matters when explaining its physical and chemical properties.