Electronic Configuration for HSC Chemistry: Atoms and Ions

Suppose you know an atom has 17 electrons. Where do those electrons actually go?

It’s tempting to imagine them stacking around the nucleus like people filling seats from the front row backwards. That idea is useful, but incomplete. Electrons occupy specific energy levels, subshells, and orbitals, and the order is not simply shell 1, then shell 2, then shell 3.

Before reading on, predict this: after filling \(3p\), does the next electron enter \(3d\) or \(4s\)?

It enters \(4s\). That slightly strange result is why electronic configuration needs more than just counting electrons.

01Start with shells: the broad energy levels

An electron in an atom can only have certain allowed energies. We organise these into shells, labelled by the principal quantum number \(n\):

• \(n = 1\): first shell
• \(n = 2\): second shell
• \(n = 3\): third shell
• \(n = 4\): fourth shell

As a first mental model, picture a multi-storey apartment building. The shell number tells you the floor.

That picture helps, but it soon breaks. Each floor is divided into different types of rooms, and some rooms can be lower in energy than rooms on the floor below. Electrons care about energy, not just the floor number.

The maximum number of electrons in a shell is:

\[
2n^2
\]

where \(n\) is the shell number.

For example, the second shell can contain:

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

The third shell can contain up to:

\[
2(3)^2 = 18 \text{ electrons}
\]

But this does not mean the first 18 electrons fill shells as \(2, 8, 8\), and then completely fill the third shell before starting the fourth. To understand the actual order, we need subshells.

02Subshells: splitting each shell into different energy regions

Each shell contains one or more subshells, labelled \(s\), \(p\), \(d\), and \(f\).

The number tells you the shell. The letter tells you the subshell.

So \(3p\) means the \(p\) subshell in the third shell.

Each type of subshell has a fixed maximum number of electrons:

This table is worth understanding rather than memorising as four unrelated numbers. Each orbital can hold at most two electrons, so the maximum number of electrons is twice the number of orbitals.

03Orbitals: where the two-electron limit comes from

An orbital is a region described by an electron’s quantum state. For HSC Chemistry, the useful point is that each orbital can contain a maximum of two electrons, and those electrons must have opposite spins.

We often draw an orbital as a box and the electrons as arrows:

\(1s:\ [\uparrow\downarrow]\)

The arrows do not mean electrons are literally little arrows spinning clockwise and anticlockwise. They represent two allowed spin states.

The rule that prevents two electrons in the same atom from having an identical set of quantum numbers is the Pauli exclusion principle. In an orbital, that gives us the two-electron maximum with opposite spins.

04The three rules that determine the ground-state configuration

A ground-state electronic configuration is the lowest-energy arrangement of an atom’s electrons.

Three main ideas control how we write it.

1. Fill lower-energy orbitals first

This is commonly called the Aufbau principle.

For the early parts of the periodic table, the filling order is:

\[
1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p,\ 5s,\ 4d,\ 5p,\ 6s,\ldots
\]

Notice the important part:

\[
3p \rightarrow 4s \rightarrow 3d
\]

So \(4s\) begins filling before \(3d\).

A student who only watches the shell numbers might predict \(3d\) comes before \(4s\). That feels reasonable because 3 is smaller than 4. The problem is that electrons fill according to orbital energy, not simply according to shell number.

2. Put no more than two electrons in one orbital

This follows from the Pauli exclusion principle.

An \(s\) subshell has one orbital, so it can contain at most two electrons:

\(s^2\)

A \(p\) subshell has three orbitals, so it can contain at most six:

\(p^6\)

A \(d\) subshell has five orbitals, so it can contain at most ten:

\(d^{10}\)

3. Spread electrons out before pairing them

This is Hund’s rule.

Suppose three electrons need to enter a \(p\) subshell. A \(p\) subshell contains three orbitals.

The ground-state arrangement is:

\[
[\uparrow]\ [\uparrow]\ [\uparrow]
\]

not:

\[
[\uparrow\downarrow]\ [\uparrow]\ [\ ]
\]

Electrons occupy equal-energy orbitals singly before any pairing occurs.

A slightly silly analogy is a nearly empty bus. If three people get on and there are three empty pairs of seats, they usually spread out before sitting shoulder-to-shoulder. Hund’s rule has a similar pattern.

The analogy breaks because electrons are not making social choices. The actual arrangement comes from quantum mechanics and electron interactions.

05How to write an electronic configuration

Start with the number of electrons, then fill the subshells in energy order without exceeding each subshell’s capacity.

For a neutral atom:

\[
\text{number of electrons} = \text{atomic number}
\]

For an ion, adjust the number of electrons for the charge.

A \(2+\) ion has lost two electrons.

A \(1-\) ion has gained one electron.

Electronic configuration depends on the number of electrons and nuclear charge, not on the number of neutrons. Two isotopes of the same neutral element therefore have the same ground-state electronic configuration. If you need to refresh the difference between atomic number, mass number, and isotopes, see Stable and Unstable Isotopes Explained for HSC Chemistry.

06Noble gas shorthand

Long configurations become awkward, so we often replace the completed inner shells with the symbol of the previous noble gas.

Sulfur has:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^4
\]

The first ten electrons are the same as neon, so we can write:

\[
\boxed{\ce{[Ne]}\,3s^2\,3p^4}
\]

The square brackets mean “the complete electron configuration of this noble gas”.

This shorthand does not change the configuration. It simply avoids rewriting filled inner subshells.

07Writing configurations for ions

For main-group ions, the safest method is:

1. determine the number of electrons in the ion,
2. fill those electrons into the lowest available energy levels.

For example, chlorine has atomic number 17.

Neutral chlorine has 17 electrons:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^5
\]

The chloride ion, \(\mathrm{Cl}^{-}\), has gained one electron, giving 18:

\[
\boxed{1s^2\,2s^2\,2p^6\,3s^2\,3p^6}
\]

or:

\[
\boxed{\ce{[Ar]}}
\]

Calcium has atomic number 20. Neutral calcium is:

\[
\ce{[Ar]}\,4s^2
\]

The calcium ion \(\mathrm{Ca}^{2+}\) has lost those two outer electrons:

\[
\boxed{\ce{[Ar]}}
\]

So far, “remove electrons from the end of the written configuration” seems to work.

Then transition metals arrive and ruin that shortcut.

08The transition-metal trap: remove \(4s\) before \(3d\)

Consider iron, atomic number 26.

Its neutral ground-state configuration is:

\[
\boxed{\ce{[Ar]}\,4s^2\,3d^6}
\]

You may also see this written as:

\[
\ce{[Ar]}\,3d^6\,4s^2
\]

Both notations describe the same occupied subshells.

Now predict the configuration of \(\mathrm{Fe}^{2+}\). Would you remove two electrons from \(3d\), because \(3d\) appears after \(4s\) in the filling sequence?

No. The \(4s\) electrons are removed first.

So:

\[
\boxed{\mathrm{Fe}^{2+}: [Ar]\,3d^6}
\]

For \(\mathrm{Fe}^{3+}\), one additional electron is removed from \(3d\):

\[
\boxed{\mathrm{Fe}^{3+}: [Ar]\,3d^5}
\]

This looks contradictory only if the filling order is treated as a rigid staircase that never changes.

The simplified rule “\(4s\) fills before \(3d\)” is useful for building many neutral atoms. Once the \(3d\) subshell contains electrons, however, the relative orbital energies are affected by the electron arrangement. For common transition-metal cations, the \(4s\) electrons are lost before the \(3d\) electrons.

A reliable HSC rule is therefore:

Fill \(4s\) before \(3d\) when constructing the usual neutral configuration, but remove \(4s\) electrons before \(3d\) when forming transition-metal cations.

09Chromium and copper: where the simple filling pattern needs revision

The usual filling sequence is a model that works very well, but it has important exceptions.

A basic Aufbau prediction for chromium, atomic number 24, would be:

\[
\ce{[Ar]\,4s^2\,3d^4}
\]

Its observed ground-state configuration is instead:

\[
\boxed{\ce{Cr}: [Ar]\,4s^1\,3d^5}
\]

For copper, atomic number 29, the simple prediction would be:

\[
\ce{[Ar]\,4s^2\,3d^9}
\]

Its ground-state configuration is:

\[
\boxed{\ce{Cu}: [Ar]\,4s^1\,3d^{10}}
\]

You will sometimes hear this explained by saying “half-filled and fully filled \(d\) subshells are more stable”. That is a useful memory aid, but it is not a complete physical explanation. The actual energies depend on several contributions, including electron-electron interactions, and the energy differences between these arrangements are small.

For HSC work, the important point is practical: do not force chromium and copper into the simplest Aufbau pattern when you are asked for their ground-state configurations.

Their ions still follow the transition-metal removal rule.

For example:

\[
\ce{Cu}: [Ar]\,4s^1\,3d^{10}
\]

Remove the \(4s\) electron to form \(\mathrm{Cu}^{+}\):

\[
\boxed{\mathrm{Cu}^{+}: [Ar]\,3d^{10}}
\]

Then remove one \(3d\) electron to form \(\mathrm{Cu}^{2+}\):

\[
\boxed{\mathrm{Cu}^{2+}: [Ar]\,3d^9}
\]

10Shell notation and subshell notation are not the same thing

You may sometimes describe electrons only by shells.

For chlorine:

\[
2,\ 8,\ 7
\]

This tells us there are two electrons in the first shell, eight in the second, and seven in the third.

Its subshell configuration is much more precise:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^5
\]

Both descriptions contain useful information, but they answer different questions.

Shell notation tells you how many electrons occupy each principal energy level.

Subshell notation tells you which \(s\), \(p\), \(d\), or \(f\) subshells the electrons occupy.

If a question asks for an electronic configuration “using subshell notation”, writing only \(2,8,7\) is not enough.

11A compact method for exam questions

When you need a ground-state configuration, work in this order:

1. Find the electron count. Start from the atomic number, then adjust for charge.
2. Fill the correct subshells. Use the energy order, not just shell number.
3. Respect the capacities. \(s^2\), \(p^6\), \(d^{10}\), \(f^{14}\).
4. Use Hund’s rule if orbitals are shown individually.
5. For transition-metal cations, remove \(4s\) before \(3d\).
6. Remember genuine ground-state exceptions such as Cr and Cu.
7. Add the superscripts at the end. They should equal the total number of electrons.

That final check catches a surprising number of mistakes.

12Questions and solutions

Question 1

A neutral phosphorus atom has atomic number 15.

Write:

a) its full ground-state electronic configuration, and
b) its shell configuration.

Solution 1

The configurations are \(1s^2\,2s^2\,2p^6\,3s^2\,3p^3\) and \(2,8,5\).

Phosphorus is neutral, so its 15 protons are matched by 15 electrons. Filling the subshells in energy order gives:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^3
\]

The total is:

\[
2 + 2 + 6 + 2 + 3 = 15
\]

To convert this to shell notation, group electrons with the same principal shell number:

• \(n=1\): \(1s^2\), giving 2 electrons
• \(n=2\): \(2s^2 + 2p^6\), giving 8 electrons
• \(n=3\): \(3s^2 + 3p^3\), giving 5 electrons

Therefore:

\[
\boxed{2,8,5}
\]

A common mistake is to confuse the \(3p^3\) superscript with the total number of electrons in shell 3. The \(3s^2\) electrons must also be included.

Question 2

An oxide ion is written as \(\mathrm{O}^{2-}\). Oxygen has atomic number 8.

Write the ion’s full electronic configuration and explain why it contains more electrons than a neutral oxygen atom.

Solution 2

The oxide ion has the configuration \(1s^2\,2s^2\,2p^6\) because it contains 10 electrons.

Neutral oxygen has 8 electrons:

\[
\ce{O}: 1s^2\,2s^2\,2p^4
\]

A \(2-\) charge means the atom has gained two electrons:

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

Filling 10 electrons gives:

\[
\boxed{\mathrm{O}^{2-}: 1s^2\,2s^2\,2p^6}
\]

The negative charge does not mean electrons are negative in number. It means there are two more negatively charged electrons than there are positively charged protons.

Question 3

A student gives the ground-state orbital arrangement for the three \(2p\) electrons in nitrogen as:

\[
[\uparrow\downarrow]\ [\uparrow]\ [\ ]
\]

Explain what is wrong with this arrangement and draw the correct \(2p\) arrangement.

Solution 3

The arrangement is incorrect because Hund’s rule requires the three equal-energy \(2p\) orbitals to be occupied singly before any electron pairing occurs.

Nitrogen has the configuration:

\[
1s^2\,2s^2\,2p^3
\]

The three \(2p\) electrons therefore occupy separate \(p\) orbitals:

\[
\boxed{[\uparrow]\ [\uparrow]\ [\uparrow]}
\]

The student’s arrangement does not violate the two-electron capacity of an orbital, so it can look plausible. The problem is that it is not the lowest-energy arrangement. Pairing begins only after each orbital in that subshell contains one electron.

Question 4

Iron has atomic number 26. A student writes:

\[
\ce{Fe}: [Ar]\,4s^2\,3d^6
\]

They then argue that \(\mathrm{Fe}^{3+}\) must be:

\[
\ce{[Ar]\,4s^2\,3d^3}
\]

because “the \(3d\) electrons were added after the \(4s\) electrons, so they should be removed first”.

Give the correct configuration of \(\mathrm{Fe}^{3+}\), and explain the flaw in the student’s reasoning.

Solution 4

The correct configuration is \(\ce{[Ar]\,3d^5}\).

Neutral iron is:

\[
\ce{Fe}: [Ar]\,4s^2\,3d^6
\]

To form \(\mathrm{Fe}^{3+}\), three electrons must be removed. For a transition-metal cation, the \(4s\) electrons are removed before the \(3d\) electrons.

First remove two \(4s\) electrons:

\[
\ce{[Ar]\,3d^6}
\]

Then remove one \(3d\) electron:

\[
\boxed{\mathrm{Fe}^{3+}: [Ar]\,3d^5}
\]

The student has treated the neutral-atom filling sequence as though it also determines the removal sequence. It does not. The relative energies of the occupied orbitals are not fixed independently of the electron arrangement, and in these transition-metal cations the \(4s\) electrons are removed first.

The electron-count check confirms the result. Iron has 26 electrons when neutral, so \(\mathrm{Fe}^{3+}\) has:

\[
26 – 3 = 23
\]

The configuration \(\ce{[Ar]\,3d^5}\) contains:

\[
18 + 5 = 23
\]

electrons.

Question 5

Copper has atomic number 29 and the ground-state configuration:

\[
\ce{Cu}: [Ar]\,4s^1\,3d^{10}
\]

A hypothetical student makes two claims:

1. \(\mathrm{Cu}^{+}\) should be \(\ce{[Ar]\,4s^1\,3d^9}\) because the \(d\) subshell is written last.
2. \(\mathrm{Cu}^{2+}\) should have a completely filled \(3d\) subshell because copper “prefers” \(3d^{10}\).

Evaluate both claims and give the correct configurations of \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}^{2+}\).

Solution 5

Both claims are incorrect. The correct configurations are \(\mathrm{Cu}^{+}: [Ar]\,3d^{10}\) and \(\mathrm{Cu}^{2+}: [Ar]\,3d^9\).

Start from the observed ground-state configuration:

\[
\ce{Cu}: [Ar]\,4s^1\,3d^{10}
\]

For \(\mathrm{Cu}^{+}\), one electron is removed. The \(4s\) electron is removed before a \(3d\) electron:

\[
\boxed{\mathrm{Cu}^{+}: [Ar]\,3d^{10}}
\]

Copper has 29 electrons when neutral, so \(\mathrm{Cu}^{+}\) must have 28 electrons. The configuration above contains:

\[
18 + 10 = 28
\]

For \(\mathrm{Cu}^{2+}\), one more electron must be removed. With the \(4s\) subshell now empty, that electron comes from \(3d\):

\[
\boxed{\mathrm{Cu}^{2+}: [Ar]\,3d^9}
\]

This contains:

\[
18 + 9 = 27
\]

electrons, as required for a \(2+\) ion.

The first claim fails because the order in which subshells are written is not a rule for electron removal.

The second claim turns the useful observation about copper’s neutral ground state into an absolute law. A filled \(3d\) subshell does not prevent copper from losing another electron. Once \(\mathrm{Cu}^{2+}\) has formed, its electron count requires \(3d^9\).

This is the more useful way to think about exceptions: use the experimentally established ground-state configuration, apply the ionisation rule correctly, and check the final electron count rather than trying to force every species into a preferred-looking pattern.

13What electronic configuration explains next

Once you can move confidently between electron count, shells, subshells, orbitals, and ions, several later ideas stop looking like separate facts.

The position of an element in the periodic table reflects the structure of its ground-state configuration. Valence electrons help explain common ion charges and chemical bonding. Unpaired electrons help explain magnetic behaviour, and changes between electron energy levels connect electronic structure to atomic emission spectra.

So when a later chemistry question asks why an element reacts in a particular way, electronic configuration is often the first place to look. It tells you which electrons are present, which are easiest to remove or share, and which arrangements are available next.