How to Count Lone Pairs in Lewis Structures for HSC Chemistry

Two O-H bonds might look as though they should point in opposite directions, making a water molecule linear. But gas-phase water has an H-O-H bond angle of approximately \(104.5^\circ\).

The missing piece is two lone pairs of electrons on oxygen.

The quickest way to count lone pairs is to count all the valence electrons, adjust the total for the molecule’s or ion’s charge, construct a Lewis structure, and identify the electron pairs that are not involved in bonding on the central atom.

Before reading on, make a prediction. If you counted only the two O-H bonds, what arrangement would you expect? Two electron regions spread as far apart as possible at \(180^\circ\). That would predict a linear molecule, which contradicts the observed structure of water.

The problem is that bonds aren’t the only electron regions around an atom. Lone pairs matter too.

01How experiments test the molecular shape model

How do we know water is bent rather than linear? We cannot simply look at a molecule through a microscope and count its electron pairs.

Instead, scientists use experimental techniques to determine the positions of atoms within molecules. One such technique is gas electron diffraction.

The experimental setup

Imagine a narrow beam of electrons travelling through a chamber containing water vapour.

The apparatus has four main parts:

• An electron source produces a beam.
• A collimator directs the electrons into a narrow path.
• A low-pressure gas chamber allows the beam to pass through water molecules.
• A detector records the electrons scattered at different angles.

The scattered electrons produce an interference pattern. Analysing that pattern allows scientists to determine molecular bond lengths and angles.

What should the experiment show?

Consider two competing models.

Model 1: Only the two O-H bonds matter.

If we ignore lone pairs, the two bonding regions should spread \(180^\circ\) apart. Water would be linear.

Model 2: Both bonds and lone pairs matter.

Oxygen has two bonding regions and two lone pairs. The electron regions arrange themselves differently, producing a bent molecule.

The experimental geometry supports the second model. The observed H-O-H angle is approximately \(104.5^\circ\).

We can estimate the physical difference between these arrangements.

Take the O-H bond length as approximately \(0.96\ \text{Å}\), where \(1\ \text{Å}=10^{-10}\ \text{m}\).

For two equal O-H bonds with an included angle \(\theta\), the separation between the hydrogen atoms is:

\[ d=2r\sin\left(\frac{\theta}{2}\right) \]

Using the approximate gas-phase bond length and angle:

\[ \begin{aligned} d&=2(0.96)\sin(52.25^\circ)\\ &\approx 1.52\ \text{Å} \end{aligned} \]

If the molecule were linear, the hydrogen atoms would instead be separated by:

\[ d=2(0.96)=1.92\ \text{Å} \]

That is a substantial difference in the positions of the nuclei, producing a different diffraction pattern.

These are representative approximate molecular dimensions used for a scale estimate, not measurements from a particular experiment.

What does this actually prove?

The experiment determines the positions of the nuclei. It does not directly photograph two lone pairs sitting beside oxygen.

The lone pairs belong to our electron-pair model. We use electron counting to construct that model, then compare its predictions with experimentally determined molecular structures.

The original bond-only prediction fails. Including non-bonding electron regions gives an explanation consistent with the observed geometry.

That distinction is important throughout this topic: Lewis structures are useful representations of electron distribution, not literal photographs of molecules.

02What exactly counts as a lone pair?

A lone pair is a pair of valence electrons associated with an atom that is not shared in a covalent bond.

In a Lewis structure, electrons are represented using dots, while covalent bonds are represented using lines.

There are two different things to recognise:

• A bonding pair contains two electrons shared between two atoms.
• A lone pair contains two non-bonding electrons assigned to one atom.

For example, a single bond represents one shared pair, a double bond represents two shared pairs, and a triple bond represents three shared pairs.

None of those shared pairs are lone pairs.

There is another distinction that catches students out.

When predicting molecular shape, we count the lone pairs on the central atom, not every lone pair in the molecule.

For example, oxygen atoms in carbon dioxide have lone pairs, but the central carbon has none. Those oxygen lone pairs are not counted as electron regions around the central carbon.

Keep asking yourself:

Which atom am I counting lone pairs on?

If you lose track of the central atom, your final electron-region count can be wrong even when your Lewis structure is correct.

03Count the total valence electrons first

Before drawing any dots, calculate how many valence electrons are available in the entire molecule or polyatomic ion.

For common main-group elements, use their periodic table groups to determine their valence electron counts.

Add the contributions from every atom.

Then adjust for the overall charge.

• A negative charge means the species has gained electrons. Add one electron for each negative charge.
• A positive charge means the species has lost electrons. Subtract one electron for each positive charge.

For a species with signed charge \(q\), the calculation can be written:

\[ N_{\text{val}}=\sum_i n_i v_i-q \]

Here:

• \(N_{\text{val}}\) is the total number of valence electrons.
• \(n_i\) is the number of atoms of element \(i\).
• \(v_i\) is the number of valence electrons per atom of that element.
• \(q\) is the signed overall charge, with positive values for cations and negative values for anions.

For example, a species with a \(2-\) charge has \(q=-2\), so we add two electrons.

Why the charge adjustment matters

Consider hydronium, \(\ce{H3O+}\).

Oxygen contributes six valence electrons, and the three hydrogen atoms contribute three more.

However, the ion has a \(+1\) charge, so we subtract one:

\[ N_{\text{val}}=6+3(1)-1=8 \]

Now compare this with nitrite, \(\mathrm{NO_{2}}^{-}\).

Nitrogen contributes five electrons, and the two oxygen atoms contribute twelve. The negative charge means we add one:

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

Notice that the extra electron belongs to the total electron count of the ion. You do not automatically place it on the central atom.

You must distribute all the electrons into a reasonable Lewis structure before deciding how many lone pairs the central atom has.

04A reliable method for counting central lone pairs

Use this sequence whenever a question asks you to determine the number of lone pairs before predicting molecular shape.

Step 1: Identify the central atom.

The central atom is usually the least electronegative atom in the structure, excluding hydrogen.

Hydrogen forms only one covalent bond in the molecules considered here, so it is a terminal atom.

If the question supplies a structural formula, use the given connectivity.

Step 2: Calculate the total valence electrons.

Add the valence electrons from every atom and adjust for the overall charge.

Keep this total visible while constructing your Lewis structure. You should account for every electron.

Step 3: Connect the atoms using single bonds.

Each single bond uses two electrons.

Subtract two electrons from your available total for every single bond drawn.

Step 4: Complete the outer atoms first.

Hydrogen needs two electrons in its valence shell.

Most other common main-group atoms in introductory Lewis structures are assigned eight electrons.

Add lone pairs to the terminal atoms until their valence shells are complete.

Step 5: Place the remaining electrons on the central atom.

After completing the outer atoms, any remaining electrons are generally placed on the central atom.

Count these non-bonding electrons in pairs.

Step 6: Check whether the central atom needs a multiple bond.

If a second-period central atom such as carbon, nitrogen, or oxygen has an incomplete octet, you may need to convert a lone pair on a neighbouring atom into a shared bonding pair.

This creates a double or triple bond without changing the total number of electrons.

Then check the formal charges to make sure your Lewis structure is reasonable.

There are exceptions to the octet rule. Beryllium and boron can form stable structures with incomplete octets, while some central atoms from period 3 and beyond can be represented with expanded valence shells.

Don’t force every atom into exactly eight electrons.

A useful formal charge check

Formal charge helps you check whether the electrons have been distributed sensibly.

The equation is:

\[ FC=V-L-\frac{B}{2} \]

Where:

• \(FC\) is the formal charge on an atom.
• \(V\) is the number of valence electrons in the neutral atom.
• \(L\) is the number of non-bonding electrons assigned to that atom.
• \(B\) is the total number of bonding electrons around that atom.

The formal charges across the structure must add up to the overall charge of the molecule or ion.

Formal charge is an electron-bookkeeping method. It does not mean the entire physical charge is necessarily concentrated at one particular atom.

Once your structure is checked, count the non-bonding electrons on the central atom and divide by two.

That gives you the number of central lone pairs.

07What physical evidence tells us about lone pairs

Counting lone pairs gives us a model of the electron regions around an atom. We can then compare its predictions with experimentally determined molecular geometries.

Consider three familiar molecules.

These are approximate gas-phase bond angles.

All three molecules have four electron regions around their central atom. However, the number of lone pairs increases from zero to two.

The bond angles become progressively smaller.

Why?

Lone pairs are generally more strongly repulsive towards neighbouring electron regions than bonding pairs are. Their electron density is concentrated around the central atom rather than being shared between two nuclei.

This pushes the bonding regions closer together, reducing the angles between the bonds.

The measurements support the usefulness of the electron-pair model, but they do not mean that lone pairs are rigid objects occupying fixed positions.

Real electron density is more complicated. Bond angles also depend on the surrounding atoms, bond order and molecular environment.

Use the Lewis structure to count lone pairs, then use molecular geometry and experimental evidence to assess the model.

08Questions and solutions

Question 1

A hydrogen sulfide molecule, \(\ce{H2S}\), contains one central sulfur atom bonded to two hydrogen atoms through single covalent bonds.

Determine the number of lone pairs on sulfur.

Solution 1

Sulfur has two lone pairs.

Sulfur contributes six valence electrons, and the two hydrogen atoms contribute one each.

\[ N_{\text{val}}=6+2(1)=8 \]

The two S-H single bonds use four electrons:

\[ 2\times2=4 \]

The remaining electrons are:

\[ 8-4=4 \]

Four non-bonding electrons form two lone pairs on sulfur.

There are therefore two bonding pairs and two lone pairs around the central sulfur atom.

The common mistake is to assume that two bonds mean only two electron regions. The two lone pairs must also be included when considering the electron arrangement.

Question 2

A nitrate ion, \(\mathrm{NO_{3}}^{-}\), contains a central nitrogen atom bonded to three oxygen atoms.

Determine the number of lone pairs on nitrogen.

Your answer must:

• Show the total valence electron calculation.
• Explain how the nitrogen atom achieves an octet.
• Verify that the formal charges add up to the overall ionic charge.

Solution 2

The central nitrogen has zero lone pairs.

Start by calculating the total number of valence electrons:

\[ N_{\text{val}}=5+3(6)+1=24 \]

The extra electron comes from the overall negative charge.

The three initial N-O single bonds use six electrons:

\[ 3\times2=6 \]

Each oxygen needs six additional non-bonding electrons to complete its octet.

For three oxygen atoms:

\[ 3\times6=18 \]

All 24 electrons have now been accounted for:

\[ 24-6-18=0 \]

There are no remaining electrons to place as a lone pair on nitrogen.

However, nitrogen has only six electrons around it from the three single bonds. We need to improve this Lewis structure.

One lone pair from an oxygen atom can be converted into an additional shared bonding pair, producing one N=O double bond.

Nitrogen now has eight shared electrons around it and still has zero lone pairs.

In this resonance contributor, the formal charges are:

The total is:

\[ (+1)+0+(-1)+(-1)=-1 \]

This matches the overall charge of the nitrate ion.

There are three equivalent resonance contributors because any of the three oxygen atoms can form the double bond in an individual Lewis representation.

The important trap is assuming that the negative charge automatically gives nitrogen a lone pair. The total electron count and the completed Lewis structure show otherwise.

Question 3

Hydrogen cyanide, \(\ce{HC\bond{#}N}\), contains a hydrogen atom single-bonded to carbon and a carbon atom triple-bonded to nitrogen.

Determine:

1. The number of lone pairs on the central carbon atom.
2. The number of lone pairs on the terminal nitrogen atom.
3. The number of electron regions around carbon.

Explain why the triple bond does not count as three lone pairs.

Solution 3

Carbon has zero lone pairs, while nitrogen has one lone pair. There are two electron regions around carbon.

First, calculate the total valence electrons:

\[ N_{\text{val}}=1+4+5=10 \]

The H-C single bond uses two electrons, while the C≡N triple bond uses six.

\[ 2+6=8 \]

There are two electrons remaining.

Place these two electrons on nitrogen to give it one lone pair.

The final Lewis structure is:

\[ \ce{H-C\bond{#}N:} \]

The central carbon has four shared electron pairs:

• One pair in the H-C bond.
• Three pairs in the C≡N bond.

Therefore, carbon has no lone pairs.

Nitrogen has one non-bonding pair.

Now consider the electron regions around carbon.

The H-C bond points in one direction, and the C≡N bond points in another. Although the triple bond contains three shared pairs, it counts as one electron region in VSEPR.

Therefore:

\[ 2\text{ bonding regions}+0\text{ lone pairs} =2\text{ electron regions} \]

The misconception is counting individual shared electron pairs as separate electron regions or treating them as lone pairs. A multiple bond is still one bonding region around the central atom.

Question 4

Chlorine trifluoride, \(\ce{ClF3}\), has a central chlorine atom bonded to three terminal fluorine atoms.

Assume a conventional Lewis representation in which chlorine, a period 3 element, can have an expanded valence shell.

Determine:

1. The total number of valence electrons.
2. The number of lone pairs on chlorine.
3. The total number of electron regions around chlorine.

Explain why applying the octet rule rigidly to chlorine gives an incorrect result.

Solution 4

Chlorine has two lone pairs, giving five electron regions around the central atom.

First, calculate the total valence electrons:

\[ N_{\text{val}}=7+3(7)=28 \]

The three Cl-F single bonds use six electrons:

\[ 3\times2=6 \]

Each fluorine atom requires six non-bonding electrons to complete its octet.

Therefore:

\[ 3\times6=18 \]

The electrons remaining are:

\[ 28-6-18=4 \]

Four electrons form two lone pairs on chlorine.

The central chlorine therefore has:

\[ 3\text{ bonding pairs}+2\text{ lone pairs} \]

This gives five electron regions.

The tempting mistake is to force chlorine to have exactly eight electrons because of the octet rule.

However, chlorine is a period 3 element, and the conventional Lewis representation of \(\ce{ClF3}\) allows an expanded valence shell.

The central atom has ten electrons around it: six from three bonding pairs and four from two lone pairs.

The formal charges are zero on chlorine and all three fluorine atoms, so the structure is consistent with the neutral molecule.

Do not apply the octet rule without considering the element involved.

Question 5

A nitrite ion, \(\mathrm{NO_{2}}^{-}\), contains one central nitrogen atom bonded to two oxygen atoms.

Suppose structural measurements show that the two N-O bond lengths are equal within experimental uncertainty.

A student proposes the following Lewis structure:

\[ \ce{O=N=O} \]

They place one lone pair on nitrogen and argue that two equivalent double bonds explain the equal bond lengths.

The student also claims that resonance is unnecessary.

Evaluate this proposal.

Your answer must:

• Calculate the total valence electrons.
• Determine the correct number of lone pairs on nitrogen.
• Assess whether the proposed structure satisfies the octet rule.
• Explain how resonance accounts for the equal bond lengths.

Solution 5

The proposed structure is not a valid Lewis structure for nitrite because it gives second-period nitrogen ten electrons around it, exceeding its octet.

The correct structures have one lone pair on the central nitrogen, and resonance explains why the two N-O bonds are equivalent.

Start with the total valence electron calculation:

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

There are 18 valence electrons because the ion has an overall charge of \(-1\).

Begin with the O-N-O skeleton containing two single bonds.

These bonds use:

\[ 2\times2=4\text{ electrons} \]

Each terminal oxygen needs six additional non-bonding electrons to complete its octet.

\[ 2\times6=12\text{ electrons} \]

The remaining electrons are:

\[ 18-4-12=2 \]

These two electrons form one lone pair on nitrogen.

At this stage, nitrogen has only six electrons around it: four from the two bonding pairs and two from its lone pair.

To complete nitrogen’s octet, one lone pair from either oxygen can form an additional shared bonding pair.

This produces two equivalent resonance contributors:

\[ \ce{O=N-O^-} \]

and

\[ \ce{^-O-N=O} \]

In both contributors, nitrogen retains one lone pair.

The formal charges in either contributor are:

The overall charge is therefore \(-1\).

Now consider the student’s proposed structure:

\[ \ce{O=N=O} \]

with one lone pair on nitrogen.

This structure accounts for 18 electrons, but the electron arrangement around nitrogen is incorrect.

Each N=O double bond contributes four shared electrons around nitrogen. Two double bonds therefore contribute eight electrons.

The additional lone pair contributes another two:

\[ 8+2=10\text{ electrons} \]

Nitrogen is a period 2 element and cannot expand its octet in this Lewis model.

Removing the lone pair would not solve the problem either. The resulting structure would contain only 16 electrons, rather than the required 18, and would not represent the correct ion.

So why are the two N-O bonds experimentally equivalent?

Because the two resonance contributors are alternative Lewis representations of the same ion. The actual electronic structure is a resonance hybrid, with bonding electron density and negative charge delocalised across the two oxygen atoms.

Neither oxygen permanently has a single bond while the other permanently has a double bond.

The equal measured bond lengths are consistent with this delocalisation.

This is the deeper lesson: a Lewis structure is an electron-counting model, not a photograph of a molecule frozen in one arrangement. You must preserve the total electron count, respect the relevant valence-shell limitations, and use resonance when one structure cannot represent the observed bonding adequately.

09What to do after counting lone pairs

Once you know the number of lone pairs on the central atom, the next step is to count all the electron regions around it.

Use:

\[ \text{Electron regions} = \text{Bonding regions} + \text{Lone pairs} \]

Remember that a single, double or triple bond each counts as one bonding region for VSEPR, even though they contain different numbers of shared electron pairs.

Then determine the electron-domain geometry before naming the molecular shape. Lone pairs influence the arrangement but are not included when naming the positions of the atoms.

For the complete process, including electron-domain geometry, molecular shapes, bond angles and worked VSEPR problems, continue to Molecular Shapes: Predicting Geometry with VSEPR for HSC Chemistry.

Getting the lone-pair count right is the first decision. Everything that follows depends on it.