How to Draw Lewis Structures for Polyatomic Ions

A carbonate ion, \(\ce{CO3^{2-}}\), contains one carbon atom and three oxygen atoms. You know the overall charge is \(2-\), but how do you show that charge in a Lewis structure without putting two extra electrons on the wrong oxygen?

There is another puzzle. Carbon normally forms four bonds, yet a diagram of carbonate can appear to contain one double bond and two single bonds. Does that mean the three C-O bonds are actually different?

To draw Lewis structures for polyatomic ions, you need to do more than draw bonds around a central atom. You must count the total valence electrons, adjust for the overall ionic charge, distribute the electrons, and check the formal charges. You also need to recognise when several valid Lewis structures represent the same ion.

## 1. Start with the total number of valence electrons

A polyatomic ion is a group of covalently bonded atoms that carries an overall electrical charge.

Examples include:

- \(\ce{NH4+}\), the ammonium ion.
- \(\ce{OH-}\), the hydroxide ion.
- \(\ce{NO3-}\), the nitrate ion.
- \(\ce{CO3^{2-}}\), the carbonate ion.
- \(\ce{SO4^{2-}}\), the sulfate ion.

The first step is always to count the total number of valence electrons available to the entire ion.

For common main-group elements, use the periodic table:

| Element | Valence electrons |
|---|---:|
| Hydrogen (H) | 1 |
| Carbon (C) | 4 |
| Nitrogen (N) | 5 |
| Oxygen (O) | 6 |
| Sulfur (S) | 6 |
| Phosphorus (P) | 5 |
| Fluorine (F) | 7 |
| Chlorine (Cl) | 7 |

Add the valence electrons from every atom. Then adjust the total for the ion's charge.

- For a negative charge, add one electron per unit of charge.
- For a positive charge, subtract one electron per unit of charge.

This is the rule students most often reverse. A negative ion has gained electrons, not lost them.

### Worked example: Count the electrons in ammonium

Calculate the total number of valence electrons in \(\ce{NH4+}\).

**Step 1**: Count the electrons contributed by the neutral atoms.

Nitrogen contributes five valence electrons. Each hydrogen contributes one.

\[
5+4(1)=9
\]

**Step 2**: Adjust for the positive charge.

The \(+1\) charge means the ion has one fewer electron than the corresponding neutral collection of atoms.

\[
9-1=8
\]

**Answer**: The ammonium ion has eight valence electrons available for its Lewis structure.

Notice that the four N-H bonds each contain two shared electrons. Therefore, all eight electrons are used in bonding, leaving no lone pairs on nitrogen.

[IMAGE: Annotated Lewis structure of the ammonium ion NH4+, with nitrogen in the centre, four single N-H bonds arranged around it, square brackets enclosing the complete structure, and a + charge outside the upper-right bracket. Show the eight bonding electrons represented by four lines and no lone pairs on nitrogen.]

### A useful general equation

You can write the electron-counting rule mathematically:

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

Here:

- \(N_{\mathrm{val}}\) 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 contributed by each atom of that element.
- \(q\) is the signed overall charge.

For example, a \(2-\) charge means \(q=-2\). Subtracting a negative number adds two electrons.

This equation is a compact way to avoid guessing where the extra electrons should go. At this stage, you are counting electrons for the whole ion, not assigning them to individual atoms.

## 2. Build the basic structure before placing the remaining electrons

Once you know the total electron count, you can start constructing the Lewis structure.

Use this sequence:

1. Choose the central atom.
2. Connect the surrounding atoms to it using single bonds.
3. Subtract two electrons for each single bond.
4. Complete the outer atoms' octets, except hydrogen, which needs only two electrons.
5. Place any remaining electrons on the central atom.
6. Check whether the central atom has an acceptable electron arrangement.
7. If necessary, convert a lone pair on an outer atom into an additional bonding pair.

The central atom is usually the least electronegative element, excluding hydrogen. Hydrogen forms only one bond in the common structures considered here, so it belongs on the outside.

A single bond represents two shared electrons. A double bond represents four shared electrons, and a triple bond represents six.

[IMAGE: A clear side-by-side electron-counting illustration showing a single covalent bond as two shared dots, a double bond as four shared dots, and a triple bond as six shared dots. Include the equivalent line-bond representations beneath each.]

### Worked example: Draw the hydroxide ion

Draw a Lewis structure for \(\ce{OH-}\).

**Step 1**: Count the valence electrons.

Oxygen contributes six, and hydrogen contributes one. The negative charge adds one more.

\[
6+1+1=8\text{ electrons}
\]

**Step 2**: Connect oxygen and hydrogen with a single bond.

The O-H bond uses two electrons.

\[
8-2=6\text{ electrons remaining}
\]

**Step 3**: Place the remaining electrons on oxygen.

The six remaining electrons form three lone pairs on oxygen.

**Step 4**: Check the electron count and charge.

Oxygen has three lone pairs and one bonding pair, giving it an octet. Hydrogen has its required duet.

The complete structure is:

\[
\boxed{\left[\ce{H-O}\right]^-}
\]

with three lone pairs on oxygen.

**Answer**: The hydroxide ion has one O-H single bond and three lone pairs on oxygen. The entire structure is enclosed in brackets with a \(1-\) charge outside.

[IMAGE: Annotated hydroxide Lewis structure, showing one O-H single bond, three clearly separated lone pairs on oxygen, square brackets around the ion, and a minus sign outside the upper-right bracket.]

The important point is that the negative charge does not mean you should draw a minus sign next to one oxygen dot and leave the rest of the structure unchanged. The extra electron must be included in the total count and incorporated into the complete electron arrangement.

## 3. Check formal charges, not just octets

A Lewis structure can contain the correct number of electrons and still be an unhelpful representation.

You need another check: formal charge.

Formal charge is the charge assigned to an individual atom when the bonding electrons are divided equally between the bonded atoms. It helps you assess whether your proposed arrangement distributes electrons sensibly.

Use:

\[
\text{Formal charge}
=
V-L-\frac{B}{2}
\]

where:

- \(V\) is the number of valence electrons in the neutral atom.
- \(L\) is the number of non-bonding electrons (electrons in lone pairs).
- \(B\) is the number of bonding electrons around that atom.

For a single bond, \(B=2\). For a double bond, \(B=4\).

A useful check is that the sum of all formal charges must equal the overall charge of the ion.

### Worked example: Check the formal charges in hydroxide

For \(\ce{OH-}\), oxygen has six neutral valence electrons, six non-bonding electrons, and two bonding electrons.

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

Hydrogen has one neutral valence electron, no non-bonding electrons, and two bonding electrons.

\[
\begin{aligned}
\text{FC(H)}&=1-0-\frac{2}{2}\\
&=0
\end{aligned}
\]

The sum is:

\[
-1+0=-1
\]

This matches the overall charge of hydroxide.

The oxygen carries the formal charge of \(-1\), but the brackets still show that the entire O-H species is the ion.

[IMAGE: Hydroxide Lewis structure with the formal charge of -1 placed beside oxygen, the hydrogen marked 0, and the overall ionic charge shown outside the square brackets. Visually distinguish the atom's formal charge from the ion's bracketed overall charge.]

Do not confuse formal charge with oxidation state. They are different accounting systems. Formal charge assumes equal sharing of bonding electrons, whereas oxidation states assign bonding electrons to the more electronegative atom.

A carbonate ion, \(\ce{CO3^{2-}}\), contains one carbon atom and three oxygen atoms. You know the overall charge is \(2-\), but how do you show that charge in a Lewis structure without putting two extra electrons on the wrong oxygen?

There is another puzzle. Carbon normally forms four bonds, yet a diagram of carbonate can appear to contain one double bond and two single bonds. Does that mean the three C-O bonds are actually different?

To draw Lewis structures for polyatomic ions, you need to do more than draw bonds around a central atom. You must count the total valence electrons, adjust for the overall ionic charge, distribute the electrons, and check the formal charges. You also need to recognise when several valid Lewis structures represent the same ion.

## 1. Start with the total number of valence electrons

A polyatomic ion is a group of covalently bonded atoms that carries an overall electrical charge.

Examples include:

- \(\ce{NH4+}\), the ammonium ion.
- \(\ce{OH-}\), the hydroxide ion.
- \(\ce{NO3-}\), the nitrate ion.
- \(\ce{CO3^{2-}}\), the carbonate ion.
- \(\ce{SO4^{2-}}\), the sulfate ion.

The first step is always to count the total number of valence electrons available to the entire ion.

For common main-group elements, use the periodic table:

| Element | Valence electrons |
|---|---:|
| Hydrogen (H) | 1 |
| Carbon (C) | 4 |
| Nitrogen (N) | 5 |
| Oxygen (O) | 6 |
| Sulfur (S) | 6 |
| Phosphorus (P) | 5 |
| Fluorine (F) | 7 |
| Chlorine (Cl) | 7 |

Add the valence electrons from every atom. Then adjust the total for the ion's charge.

- For a negative charge, add one electron per unit of charge.
- For a positive charge, subtract one electron per unit of charge.

This is the rule students most often reverse. A negative ion has gained electrons, not lost them.

### Worked example: Count the electrons in ammonium

Calculate the total number of valence electrons in \(\ce{NH4+}\).

**Step 1**: Count the electrons contributed by the neutral atoms.

Nitrogen contributes five valence electrons. Each hydrogen contributes one.

\[
5+4(1)=9
\]

**Step 2**: Adjust for the positive charge.

The \(+1\) charge means the ion has one fewer electron than the corresponding neutral collection of atoms.

\[
9-1=8
\]

**Answer**: The ammonium ion has eight valence electrons available for its Lewis structure.

Notice that the four N-H bonds each contain two shared electrons. Therefore, all eight electrons are used in bonding, leaving no lone pairs on nitrogen.

[IMAGE: Annotated Lewis structure of the ammonium ion NH4+, with nitrogen in the centre, four single N-H bonds arranged around it, square brackets enclosing the complete structure, and a + charge outside the upper-right bracket. Show the eight bonding electrons represented by four lines and no lone pairs on nitrogen.]

### A useful general equation

You can write the electron-counting rule mathematically:

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

Here:

- \(N_{\mathrm{val}}\) 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 contributed by each atom of that element.
- \(q\) is the signed overall charge.

For example, a \(2-\) charge means \(q=-2\). Subtracting a negative number adds two electrons.

This equation is a compact way to avoid guessing where the extra electrons should go. At this stage, you are counting electrons for the whole ion, not assigning them to individual atoms.

## 2. Build the basic structure before placing the remaining electrons

Once you know the total electron count, you can start constructing the Lewis structure.

Use this sequence:

1. Choose the central atom.
2. Connect the surrounding atoms to it using single bonds.
3. Subtract two electrons for each single bond.
4. Complete the outer atoms' octets, except hydrogen, which needs only two electrons.
5. Place any remaining electrons on the central atom.
6. Check whether the central atom has an acceptable electron arrangement.
7. If necessary, convert a lone pair on an outer atom into an additional bonding pair.

The central atom is usually the least electronegative element, excluding hydrogen. Hydrogen forms only one bond in the common structures considered here, so it belongs on the outside.

A single bond represents two shared electrons. A double bond represents four shared electrons, and a triple bond represents six.

[IMAGE: A clear side-by-side electron-counting illustration showing a single covalent bond as two shared dots, a double bond as four shared dots, and a triple bond as six shared dots. Include the equivalent line-bond representations beneath each.]

### Why you sometimes need a double bond

Consider a carbon atom connected to three oxygen atoms.

If you draw three single bonds, carbon has only six electrons around it. Each bond contributes two electrons, so:

\[
3(2)=6\text{ electrons}
\]

Carbon normally follows the octet rule in these structures. It needs eight electrons around it.

The solution is to convert one lone pair from a neighbouring oxygen into an additional bonding pair. This produces a C=O double bond.

You haven't created extra electrons. You've changed how an existing pair is represented.

[IMAGE: Three-stage electron-dot illustration showing carbon bonded to three oxygen atoms using single bonds, then an oxygen lone pair being converted into a shared pair, and finally the resulting carbon-oxygen double bond. Show the electron count remaining constant throughout.]

The octet rule is a useful starting point for carbon, nitrogen, and oxygen. Don't force it onto every element in the periodic table. Some elements have different stable electron arrangements, which is an important consideration for more advanced structures.

If you need to refresh the basic dot-and-line conventions before continuing, see [Lewis Dot Diagrams for Atoms, Ions and Covalent Molecules](https://tutorgum.com/learn/resources/hsc-chemistry-lewis-dot-diagrams/).

## 3. Check formal charges, not just octets

A Lewis structure can contain the correct number of electrons and still be an unhelpful representation.

You need another check: formal charge.

Formal charge is the charge assigned to an individual atom when the bonding electrons are divided equally between the bonded atoms. It helps you assess whether your proposed arrangement distributes electrons sensibly.

Use:

\[
\text{Formal charge}
=
V-L-\frac{B}{2}
\]

where:

- \(V\) is the number of valence electrons in the neutral atom.
- \(L\) is the number of non-bonding electrons (electrons in lone pairs).
- \(B\) is the number of bonding electrons around that atom.

For a single bond, \(B=2\). For a double bond, \(B=4\).

A useful check is that the sum of all formal charges must equal the overall charge of the ion.

### Worked example: Check the formal charges in hydroxide

For \(\ce{OH-}\), oxygen has six neutral valence electrons, six non-bonding electrons, and two bonding electrons.

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

Hydrogen has one neutral valence electron, no non-bonding electrons, and two bonding electrons.

\[
\begin{aligned}
\text{FC(H)}&=1-0-\frac{2}{2}\\
&=0
\end{aligned}
\]

The sum is:

\[
-1+0=-1
\]

This matches the overall charge of hydroxide.

The oxygen carries the formal charge of \(-1\), but the brackets still show that the entire O-H species is the ion.

[IMAGE: Hydroxide Lewis structure with the formal charge of -1 placed beside oxygen, the hydrogen marked 0, and the overall ionic charge shown outside the square brackets. Visually distinguish the atom's formal charge from the ion's bracketed overall charge.]

Do not confuse formal charge with oxidation state. They are different accounting systems. Formal charge assumes equal sharing of bonding electrons, whereas oxidation states assign bonding electrons to the more electronegative atom.

### Overall charge versus formal charge

These two ideas are related, but they are not interchangeable.

- Overall charge describes the net charge of the entire polyatomic ion.
- Formal charge describes the electron-accounting charge assigned to an individual atom in a particular Lewis structure.

For hydroxide, both happen to be \(-1\) on oxygen.

For carbonate, the overall charge is \(-2\), but no oxygen has a formal charge of \(-2\) in the usual Lewis contributors.

That difference becomes important when you draw ions with several equivalent oxygen atoms.

## 4. Worked example: Draw the carbonate ion

Construct a Lewis structure for \(\ce{CO3^{2-}}\). Show the total valence electron count, bonding arrangement, and formal charges.

**Step 1**: Count the total valence electrons.

Carbon contributes four electrons. Each oxygen contributes six. The \(2-\) charge adds two more electrons.

\[
\begin{aligned}
N_{\mathrm{val}}
&=4+3(6)+2\\
&=24\text{ electrons}
\end{aligned}
\]

**Step 2**: Choose the central atom and draw the skeleton.

Carbon is less electronegative than oxygen, so place carbon in the centre.

Connect all three oxygen atoms to carbon using single bonds.

There are three single bonds, each containing two electrons.

\[
3(2)=6\text{ bonding electrons}
\]

The number of electrons remaining is:

\[
24-6=18
\]

**Step 3**: Complete the oxygen octets.

Each oxygen currently has one single bond to carbon. It needs three lone pairs to complete its octet.

Three oxygen atoms require:

\[
3(6)=18\text{ non\text{-}bonding electrons}
\]

All 18 remaining electrons are now used.

**Step 4**: Check carbon's octet.

Carbon has three single bonds, giving it only six electrons around it.

It needs another two electrons.

Convert one lone pair from one oxygen into an additional bonding pair. This creates one C=O double bond.

Carbon now has four shared electron pairs around it, giving it an octet.

**Step 5**: Calculate the formal charges.

There is one double-bonded oxygen and two single-bonded oxygens.

For carbon:

\[
\begin{aligned}
\text{FC(C)}&=4-0-\frac{8}{2}\\
&=0
\end{aligned}
\]

For the double-bonded oxygen:

\[
\begin{aligned}
\text{FC(O)}&=6-4-\frac{4}{2}\\
&=0
\end{aligned}
\]

For each single-bonded oxygen:

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

The total formal charge is:

\[
0+0-1-1=-2
\]

This matches the overall charge of carbonate.

[IMAGE: Full Lewis structure of carbonate showing central carbon, one C=O double bond, two C-O single bonds, two lone pairs on the double-bonded oxygen, three lone pairs on each singly bonded oxygen, and formal charges of -1 on both singly bonded oxygens. Enclose the complete ion in square brackets with 2- outside.]

**Answer**: A valid carbonate Lewis contributor has one C=O double bond, two C-O single bonds, and an overall charge of \(2-\). The carbon and double-bonded oxygen have formal charges of zero, while each singly bonded oxygen has a formal charge of \(-1\).

But there's something we haven't resolved.

Why did we choose one oxygen to form the double bond? Could we have chosen either of the other two?

Yes. And that leads to one of the most important ideas in drawing polyatomic ions.

## 5. Resonance: Why one Lewis structure isn't always enough

The three oxygen atoms in carbonate are chemically equivalent.

There is no reason for one oxygen to permanently have a double bond while the other two permanently have single bonds.

You can construct three equally valid Lewis structures by moving the double bond between the three oxygen atoms.

These are called resonance structures, or resonance contributors.

\[
\ce{
O=C(O^-)-O^-
<-> ^-O-C(=O)-O^-
<-> ^-O-C(O^-)=O
}
\]

Each contributor:

- Contains 24 valence electrons.
- Gives carbon an octet.
- Gives each oxygen an octet.
- Has a total formal charge of \(-2\).

[IMAGE: Three complete carbonate resonance contributors arranged horizontally. In the first, the left oxygen has a double bond; in the second, the upper oxygen has a double bond; in the third, the right oxygen has a double bond. Show all lone pairs, formal charges and the resonance arrows. Beneath them, show a simplified resonance hybrid with three equivalent C-O bonds and the overall 2- charge.]

### Does the double bond move between the oxygen atoms?

Not literally.

This is a common misconception. Students sometimes imagine that carbonate rapidly switches between three different structures.

The resonance drawings are alternative representations of the same ion. They are not three separate structures that the ion takes turns adopting.

The actual electron distribution is delocalised across the three C-O bonds. The three bonds are equivalent.

In a simple equal-contributor model, the average bond order is:

\[
\text{Average bond order}
=
\frac{2+1+1}{3}
=
\frac{4}{3}
\]

This means each C-O bond has an average bond order of approximately 1.33 in this model.

It does not mean that each bond is literally one-third double and two-thirds single at different moments. It is a way of representing the shared bonding across the ion.

### Why this matters when drawing Lewis structures

If you draw only one carbonate contributor, your electron count and formal charges can be correct.

However, you must not conclude that one C-O bond is physically different from the other two.

The Lewis structure is a model of electron distribution, not a photograph of the molecule.

This distinction becomes clearer when you compare the model with experimental evidence.

## 6. How X-ray diffraction helped resolve the carbonate structure

How can scientists tell whether the three C-O bonds are equivalent?

They cannot simply look at a carbonate ion and count its electron pairs. Instead, they can determine the positions of its atoms using experimental techniques such as X-ray diffraction.

The history of carbonate provides an excellent example of how physical evidence can challenge an existing chemical model.

### The original problem

A conventional Lewis representation of carbonate has:

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

If this were the complete physical description, you would expect one C-O bond to differ from the other two.

A double bond and a single bond generally have different bond lengths.

The question was whether the carbonate ion actually had this unequal arrangement.

### The apparatus

A simplified single-crystal X-ray diffraction arrangement contains four main components.

1. An X-ray source produces a beam of X-rays.
2. A collimator directs the beam towards a narrow target region.
3. A crystal containing carbonate ions, such as calcite (\(\ce{CaCO3}\)), is mounted in the path of the beam.
4. A detector records the X-rays scattered at different angles.

The crystal can be rotated so that different orientations produce useful diffraction patterns.

[IMAGE: Scientific apparatus illustration of single-crystal X-ray diffraction. Show an X-ray source on the left, a narrow collimated beam travelling towards a calcite crystal mounted centrally on a rotating stage, and diffracted beams travelling towards a detector on the right at different angles. Label the source, incident beam, crystal, rotating mount, diffracted beams and detector. Show the spatial path of the X-rays clearly.]

### What should the experiment show?

Consider two competing models.

**Model 1: One fixed double bond and two fixed single bonds.**

The three oxygen atoms would not be equivalent relative to carbon. One C-O bond should have a different length from the other two.

**Model 2: Delocalised bonding across three equivalent oxygen atoms.**

All three oxygen atoms should be equivalently related to carbon. The three C-O bonds should have the same length in the ideal carbonate ion.

### What was observed?

In 1914, W. H. and W. L. Bragg used X-ray diffraction to investigate calcite and related minerals.

Their structural work showed that the carbonate ion was planar and that its three oxygen atoms were equivalently related to the central carbon atom.

This evidence challenged the conventional picture of one permanently localised double bond and two single bonds. The development of resonance theory provided a way to represent the equivalent bonding using several Lewis contributors. The International Union of Crystallography records this development in its historical account of inorganic structures. citeturn912115view0

### A small calculation: Comparing bond lengths

Consider the following deliberately illustrative diffraction data. These are teaching values, not measurements reported in the 1914 Bragg study.

| Bond | Illustrative length | Uncertainty |
|---|---:|---:|
| C-O1 | 128 pm | ±2 pm |
| C-O2 | 129 pm | ±2 pm |
| C-O3 | 128 pm | ±2 pm |

Calculate the average bond length:

\[
\begin{aligned}
\bar d
&=\frac{128+129+128}{3}\\
&=128.3\text{ pm}
\end{aligned}
\]

The maximum difference between the three values is only:

\[
129-128=1\text{ pm}
\]

This difference is smaller than the stated uncertainty of each measurement.

The three bond lengths are therefore consistent with being equivalent within the precision of this illustrative dataset.

A clearly resolved difference between one bond and the other two would be needed to support the fixed one-double/two-single model.

### What does the experiment actually prove?

X-ray diffraction provides information about the arrangement of atoms and the distribution of electron density in a crystal.

It does not directly photograph individual bonding pairs or lone pairs.

The experimental evidence supports the equivalent-bond structure. The resonance model helps explain that observation.

This is an important distinction:

- Experimental observation: the three oxygen atoms are equivalently related to carbon.
- Lewis representation: three resonance contributors show how the bonding electrons can be distributed.
- Modern interpretation: the actual bonding is delocalised rather than a set of permanently localised single and double bonds.

## 7. A contrasting example: The hydrogen carbonate ion

Now consider the hydrogen carbonate ion, \(\ce{HCO3-}\).

It looks very similar to carbonate, but there is one important difference: one oxygen atom is bonded to hydrogen.

Does that mean all three C-O bonds are still equivalent?

No. The hydrogen atom makes one oxygen chemically different from the other two.

This is a useful test of whether you understand resonance rather than simply memorising that every polyatomic ion with three oxygen atoms has three equivalent bonds.

### Worked example: Draw the hydrogen carbonate ion

Construct a Lewis structure for \(\ce{HCO3-}\), showing the total valence electrons, formal charges, and possible resonance contributors.

**Step 1**: Count the valence electrons.

Hydrogen contributes one, carbon contributes four, and three oxygen atoms contribute eighteen.

The negative charge adds one electron.

\[
\begin{aligned}
N_{\mathrm{val}}
&=1+4+3(6)+1\\
&=24\text{ electrons}
\end{aligned}
\]

Notice that hydrogen carbonate and carbonate both have 24 valence electrons.

Their atoms are different, but their total electron counts are the same.

**Step 2**: Draw the basic connectivity.

Place carbon in the centre and connect it to all three oxygen atoms.

One of the oxygen atoms must also be connected to hydrogen.

There are four single bonds in this starting structure:

- Three C-O bonds.
- One O-H bond.

These bonds use:

\[
4(2)=8\text{ electrons}
\]

The remaining electrons are:

\[
24-8=16
\]

**Step 3**: Complete the oxygen octets.

The oxygen bonded to both carbon and hydrogen has two lone pairs.

Each of the other two oxygen atoms has three lone pairs.

The remaining 16 electrons are now accounted for.

**Step 4**: Check carbon's octet.

Carbon has three single bonds to oxygen, so it has only six electrons around it.

Convert a lone pair from one of the oxygen atoms not bonded to hydrogen into an additional bonding pair.

This creates one C=O double bond.

**Step 5**: Calculate the formal charges.

The carbon has a formal charge of zero.

The double-bonded oxygen has a formal charge of zero.

The oxygen bonded to hydrogen has a formal charge of zero.

The remaining singly bonded oxygen has a formal charge of \(-1\).

Therefore:

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

This matches the overall charge of hydrogen carbonate.

**Step 6**: Consider resonance.

The double bond can be placed on either of the two oxygen atoms that are not bonded to hydrogen.

The two important equivalent contributors are:

\[
\ce{
HO-C(=O)-O^-
<-> HO-C(-O^-)=O
}
\]

[IMAGE: Hydrogen carbonate Lewis structure showing central carbon bonded to three oxygen atoms, with one oxygen also bonded to hydrogen. Show two resonance contributors side by side, with the double bond alternating between the two oxygen atoms not bonded to hydrogen. Include all lone pairs and the -1 formal charge. Clearly distinguish the O-H oxygen from the other two oxygen atoms.]

**Answer**: Hydrogen carbonate has 24 valence electrons. Its usual Lewis representation has one C=O double bond, one C-OH single bond, and one C-O single bond carrying a formal charge of \(-1\). The double bond can be represented on either of the two non-protonated oxygen atoms.

The two oxygen atoms not bonded to hydrogen are equivalent in these contributors. The oxygen bonded to hydrogen is different.

That is the important contrast:

| Feature | Carbonate, \(\ce{CO3^{2-}}\) | Hydrogen carbonate, \(\ce{HCO3-}\) |
|---|---|---|
| Number of oxygen atoms | 3 | 3 |
| Oxygen atoms bonded to H | 0 | 1 |
| Important equivalent contributors | 3 | 2 |
| Equivalent C-O bonds | All three | Two non-protonated C-O bonds |
| Overall charge | \(2-\) | \(1-\) |

The presence of hydrogen changes the symmetry of the electron distribution. You cannot assume that all oxygen atoms are equivalent just because they belong to the same element.

## 8. A reliable final checking routine

Before you finish any polyatomic ion Lewis structure, run through these checks.

| Check | What you should confirm |
|---|---|
| Valence electrons | You counted every atom's contribution. |
| Ionic charge | You added electrons for negative charge and subtracted for positive charge. |
| Bonding | Every single bond contains two electrons. |
| Outer atoms | Hydrogen has a duet; the usual oxygen and nitrogen structures have octets. |
| Central atom | Its electron arrangement is reasonable. |
| Formal charges | Their sum equals the overall ionic charge. |
| Brackets | The complete ion is enclosed in square brackets, with the overall charge outside. |
| Resonance | You have considered equivalent contributors where appropriate. |

A few mistakes are particularly easy to make.

**Mistake 1: Adding the charge after drawing the structure.**

The charge changes the number of available electrons. It must be included before you start distributing lone pairs and bonds.

**Mistake 2: Putting the overall charge on the central atom automatically.**

The overall charge does not have to sit on the central atom. Calculate the formal charges before deciding where to place them.

**Mistake 3: Counting a double bond as two electron regions when counting electrons.**

A double bond contains four electrons, not two. When counting valence electrons, include all four.

**Mistake 4: Assuming one resonance contributor is the complete physical structure.**

A correct Lewis contributor is a useful representation, but it may not show the full delocalised bonding.

**Mistake 5: Forgetting the brackets.**

The brackets enclose the entire polyatomic ion. The charge written outside them represents the overall charge, not necessarily the formal charge on one particular atom.

## 9. Questions and solutions

Try each question before reading its solution. The questions move from standard electron counting towards more challenging reasoning involving resonance and experimental evidence.

### Question 1

Draw the Lewis structure for the ammonium ion, \(\ce{NH4+}\).

Calculate its total number of valence electrons and determine the formal charge on nitrogen.

### Solution 1

**The ammonium ion has eight valence electrons, four N-H single bonds, and a formal charge of \(+1\) on nitrogen.**

Nitrogen contributes five valence electrons, and the four hydrogen atoms contribute four.

The positive charge means one electron must be subtracted.

\[
\begin{aligned}
N_{\mathrm{val}}
&=5+4(1)-1\\
&=8\text{ electrons}
\end{aligned}
\]

Place nitrogen in the centre and connect it to the four hydrogen atoms using single bonds.

The four bonds use:

\[
4(2)=8\text{ electrons}
\]

All available electrons have been used.

Nitrogen has an octet, and each hydrogen has a duet.

Calculate the formal charge on nitrogen:

\[
\begin{aligned}
\text{FC(N)}
&=5-0-\frac{8}{2}\\
&=+1
\end{aligned}
\]

Each hydrogen has a formal charge of zero.

The complete structure is enclosed in square brackets with a \(+1\) charge outside.

[IMAGE: Complete ammonium Lewis structure with nitrogen in the centre, four N-H single bonds, no lone pairs, and square brackets enclosing the ion with + outside. Show the four bonds clearly.]

### Question 2

Construct a Lewis structure for the nitrate ion, \(\ce{NO3-}\).

Your answer must include:

- The total number of valence electrons.
- The number and type of bonds around nitrogen.
- The formal charges on nitrogen and oxygen.
- An explanation of why three resonance contributors are possible.

### Solution 2

**Nitrate has 24 valence electrons and three equivalent resonance contributors, each containing one N=O double bond and two N-O single bonds.**

Nitrogen contributes five electrons, and the three oxygen atoms contribute eighteen. The negative charge adds one electron.

\[
\begin{aligned}
N_{\mathrm{val}}
&=5+3(6)+1\\
&=24\text{ electrons}
\end{aligned}
\]

Place nitrogen in the centre and connect it to the three oxygen atoms using single bonds.

These bonds use six electrons, leaving eighteen.

Give each oxygen three lone pairs to complete its octet. Nitrogen now has only six electrons around it.

Convert a lone pair from one oxygen into an additional bonding pair. Nitrogen now has an octet.

Calculate the formal charges.

For nitrogen:

\[
\begin{aligned}
\text{FC(N)}
&=5-0-\frac{8}{2}\\
&=+1
\end{aligned}
\]

For the double-bonded oxygen:

\[
\begin{aligned}
\text{FC(O)}
&=6-4-\frac{4}{2}\\
&=0
\end{aligned}
\]

For each single-bonded oxygen:

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

The total is:

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

This matches the overall charge.

The double bond can be placed between nitrogen and any of the three oxygen atoms. This produces three equivalent resonance contributors.

The important trap is to assume that the oxygen with the double bond in your drawing is permanently different from the other two. The actual nitrate ion has three equivalent N-O bonds in its ideal structure.

### Question 3

A student is asked to draw the hydrogen carbonate ion, \(\ce{HCO3-}\).

They correctly calculate 24 valence electrons but draw three equivalent C-O bonds, arguing that all three oxygen atoms are identical elements.

Explain why this conclusion is incorrect. Include the resonance contributors and formal charges in your answer.

[IMAGE: Unfinished hydrogen carbonate connectivity diagram for a question. Show a central C atom connected to three O atoms, with one O also connected to H. Include only atom labels and skeletal single bonds. Do not show lone pairs, multiple bonds or formal charges.]

### Solution 3

**The three C-O bonds are not all equivalent because one oxygen is bonded to hydrogen. Only the two non-protonated oxygen atoms are equivalent in the usual important resonance contributors.**

First, confirm the electron count:

\[
1+4+3(6)+1=24
\]

Connect carbon to all three oxygen atoms and connect one oxygen to hydrogen.

After completing the oxygen octets, carbon has only six electrons around it. Convert a lone pair from one of the non-protonated oxygen atoms into a double bond.

This produces two important contributors:

\[
\ce{
HO-C(=O)-O^-
<-> HO-C(-O^-)=O
}
\]

In the first contributor, the left non-protonated oxygen has a double bond. In the second, the other non-protonated oxygen has the double bond.

The formal charges in either contributor are:

| Atom | Formal charge |
|---|---:|
| Carbon | 0 |
| Oxygen bonded to hydrogen | 0 |
| Double-bonded oxygen | 0 |
| Single-bonded non-protonated oxygen | \(-1\) |
| Hydrogen | 0 |

The sum is \(-1\).

The student's mistake is assuming that identical elements must occupy identical chemical environments. They don't.

One oxygen is part of an O-H group. That makes it different from the other two oxygen atoms, which can share the double-bond character through resonance.

### Question 4

A crystallography investigation examines the C-O bond lengths in a hydrogen carbonate ion.

The following values are illustrative teaching data, not results from a published experiment.

| Bond | Measured length |
|---|---:|
| C-O1, not bonded to H | \(129\pm1\text{ pm}\) |
| C-O2, not bonded to H | \(128\pm1\text{ pm}\) |
| C-O3, bonded to H | \(136\pm1\text{ pm}\) |

**(a)** Calculate the average length of the two C-O bonds not bonded to hydrogen.

**(b)** Explain what the data suggest about resonance in hydrogen carbonate.

**(c)** Explain why the data do not support the claim that all three C-O bonds are equivalent.

### Solution 4

**The two non-protonated C-O bonds are consistent with being equivalent, while the C-OH bond is clearly longer in this illustrative dataset.**

**(a)** Calculate the average length of the two non-protonated bonds.

\[
\begin{aligned}
\bar d
&=\frac{129+128}{2}\\
&=128.5\text{ pm}
\end{aligned}
\]

The average is \(128.5\text{ pm}\).

**(b)** Compare the two non-protonated bonds.

Their measured lengths differ by:

\[
129-128=1\text{ pm}
\]

The uncertainty ranges overlap, so the small difference is not enough to establish that the bonds are genuinely different.

This is consistent with resonance between the two non-protonated oxygen atoms.

**(c)** Compare the C-OH bond with the other two bonds.

The C-OH bond is:

\[
136-128.5=7.5\text{ pm}
\]

longer than their average.

The uncertainty ranges do not overlap, so the difference is clearly resolved in this illustrative dataset.

This supports the distinction between the two non-protonated oxygen atoms and the oxygen bonded to hydrogen.

The tempting conclusion is that all three oxygen atoms must behave identically because they are all oxygen atoms. The measurement shows why that is incorrect: the O-H bond changes the chemical environment of one oxygen.

The data are illustrative, and actual crystal measurements can also be influenced by the surrounding ions and crystal environment. Bond lengths provide structural evidence, not a direct measurement of formal charges.

### Question 5

Hydrogen carbonate can lose a proton to form carbonate:

\[
\ce{HCO3^- -> CO3^{2-} + H+}
\]

A student draws one resonance contributor of hydrogen carbonate:

\[
\ce{HO-C(=O)-O^-}
\]

They argue:

"The oxygen that was bonded to hydrogen must remain permanently different after the hydrogen leaves. Therefore, carbonate should retain one C-O single bond from that oxygen, one C-O single bond from the other oxygen, and one C=O double bond."

Evaluate this argument.

Your answer must:

- Explain what happens to the electrons in the O-H bond.
- Calculate the valence electron totals before and after the reaction.
- Explain how the formal charges change.
- Determine how many equivalent resonance contributors are possible for carbonate.
- Explain why the student's conclusion is incorrect.

Assume an isolated carbonate ion, so that the three oxygen atoms are not distinguished by an external crystal environment.

[IMAGE: Starting-state schematic for the conceptual question. Show hydrogen carbonate with H bonded to one oxygen, carbon connected to three oxygen atoms, and an arrow indicating removal of H+ from the O-H group. Do not show the resulting carbonate resonance structures or electron redistribution.]

### Solution 5

**The student is incorrect. After the proton leaves, all three oxygen atoms can participate in equivalent carbonate resonance contributors. The electron pair from the broken O-H bond remains with oxygen, so no electron is removed with the proton.**

This is the crucial detail.

**Step 1**: Compare the valence electron totals.

For hydrogen carbonate:

\[
\begin{aligned}
N_{\mathrm{val}}
&=1+4+3(6)+1\\
&=24\text{ electrons}
\end{aligned}
\]

For carbonate:

\[
\begin{aligned}
N_{\mathrm{val}}
&=4+3(6)+2\\
&=24\text{ electrons}
\end{aligned}
\]

Both ions have 24 valence electrons.

This seems surprising because carbonate contains one fewer atom. However, the species loses \(\ce{H+}\), not a neutral hydrogen atom.

A proton contains no electrons. Therefore, the electron pair in the O-H bond remains with the oxygen when the proton leaves.

**Step 2**: Track the electrons from the O-H bond.

Before deprotonation, the oxygen bonded to hydrogen has:

- One O-H single bond.
- One C-O single bond.
- Two lone pairs.

When \(\ce{H+}\) leaves, the bonding pair from the O-H bond stays with oxygen.

The oxygen is now bonded only to carbon and has three lone pairs.

It carries a formal charge of \(-1\).

The former hydroxyl oxygen has therefore become a negatively charged, singly bonded oxygen in the carbonate ion.

**Step 3**: Check the formal charges after deprotonation.

One carbonate contributor has:

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

The double-bonded oxygen has formal charge zero.

Each singly bonded oxygen has formal charge \(-1\).

Carbon has formal charge zero.

The total is:

\[
0+0-1-1=-2
\]

This matches the charge of carbonate.

**Step 4**: Consider the possible resonance contributors.

In hydrogen carbonate, the oxygen bonded to hydrogen is chemically different. The two other oxygen atoms can share the double-bond character.

After deprotonation, there is no longer an O-H bond distinguishing that oxygen.

All three oxygen atoms are now equivalent in the ideal isolated carbonate ion.

The double bond can therefore be represented on any of the three oxygen atoms, giving three equivalent resonance contributors.

**Step 5**: Identify the flaw in the student's argument.

The student has correctly noticed that one oxygen was originally bonded to hydrogen.

However, they have incorrectly assumed that this difference must remain after the proton leaves.

The chemical environment has changed. The oxygen that previously formed the O-H bond now carries a lone pair and a formal negative charge, and it can participate in the same resonance system as the other oxygen atoms.

The final carbonate ion has three equivalent C-O bonds in its resonance hybrid.

This question brings together three important ideas:

1. Ionic charge changes the total electron count.
2. When a proton leaves, its bonding electrons remain with the atom that was bonded to it.
3. Resonance depends on the structure and electron distribution of the final ion, not on which atom happened to carry hydrogen beforehand.

That is why you should always reconstruct and check the Lewis structure after a chemical change rather than simply carrying every bond from the original structure into the product.

## 10. Where to go next

Once you can construct a correct Lewis structure, you can use it to work out the arrangement of electron regions around the central atom.

This is the starting point for predicting molecular shape using VSEPR theory. Remember that lone pairs count as electron regions even though they are not included when naming the molecular shape.

Continue with [Electron-Pair Geometry vs Molecular Shape: HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-distinguish-electron-pair-geometry-from-molecular-shape/).

For another useful skill, practise identifying the number of lone pairs on a central atom before predicting its geometry.

**Syllabus note:** As of 2026, the Chemistry Stage 6 Syllabus (2017) remains in use for Year 12 until the new Chemistry 11-12 Syllabus (2025) begins implementation in 2028. The first HSC examination under the new syllabus is scheduled for 2029. Formal-charge and resonance explanations in this article provide additional structural reasoning; this article does not claim that every detail is a separately prescribed NESA examination requirement. citeturn912115view1