How to Distinguish Shells, Subshells and Orbitals in Chemistry

A question gives you the electron configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^4\). How many shells does the atom have? How many subshells are occupied? And how many orbitals contain electrons?

These are three different questions. The key is to recognise that electrons are organised in three nested levels: shells contain subshells, and subshells contain orbitals. An orbital can hold a maximum of two electrons.

01Shell, subshell and orbital: what is the difference?

Think of an atom as having several layers of electron organisation. Each level describes something different.

1. A shell is the largest level

A shell is a principal energy level surrounding the nucleus. Shells are numbered using the principal quantum number, \(n\).

• Shell 1 has \(n=1\).
• Shell 2 has \(n=2\).
• Shell 3 has \(n=3\).
• Shell 4 has \(n=4\).

A larger shell number generally corresponds to electrons with greater energy and a greater average distance from the nucleus.

The important word is generally. Electrons do not travel around the nucleus in fixed circular tracks. The numbered shells are energy levels in the quantum model, not literal rings.

2. A subshell is a subdivision of a shell

Each shell contains one or more subshells. These are identified by the letters s, p, d and f.

The available subshells depend on the shell number:

Notice the pattern: shell 1 cannot contain a 1p subshell, and shell 2 cannot contain a 2d subshell.

Each type of subshell contains a fixed number of orbitals.

3. An orbital is the smallest level

An orbital is an individual quantum state that describes the spatial distribution of an electron around the nucleus.

It is not a fixed path that an electron follows. Instead, an orbital is commonly represented as a region where there is a high probability of detecting an electron.

Each orbital can contain a maximum of two electrons. If two electrons occupy the same orbital, they must have opposite spins. This is the Pauli exclusion principle.

The number of orbitals in each subshell is fixed:

For example, a 3p subshell contains three orbitals, not three electrons. Those three orbitals can hold up to six electrons altogether.

This is one of the most important distinctions to remember:

• \(3\) identifies the third shell.
• \(p\) identifies the subshell type.
• The three orbitals belong to that 3p subshell.
• A superscript of \(4\), as in \(3p^4\), means that four electrons occupy those three orbitals.

02How experiments revealed subshells and orbitals

There is a reason scientists use three levels of organisation rather than just describing electrons as belonging to shells.

The original shell model was useful, but it could not explain all the evidence from atomic experiments.

From electron shells to the quantum model

Early experiments with hydrogen produced a puzzle. When hydrogen gas was excited electrically, it emitted light at specific wavelengths rather than every possible wavelength.

The apparatus was straightforward:

1. A low-pressure hydrogen gas tube was connected to a high-voltage supply.
2. The electrical discharge excited electrons in the hydrogen atoms.
3. Light emitted by the gas passed through a narrow slit.
4. A diffraction grating separated the light into its component wavelengths.
5. A detector recorded distinct bright spectral lines.

What would you expect if electrons could have any energy whatsoever? You would expect a continuous range of emitted wavelengths.

Instead, only particular wavelengths appeared. This supported the idea that electrons occupy discrete energy levels.

The Bohr model used these observations to explain hydrogen’s emission spectrum. It described electrons as occupying permitted energy levels, labelled by the principal quantum number \(n\). These energy levels became the basis of the shell model.

However, the Bohr model had a limitation: it could not adequately explain the electronic structure of more complicated, multi-electron atoms.

Scientists needed a more detailed model.

The quantum mechanical model describes electrons using orbitals. These are mathematical descriptions of electron states, with associated probability distributions. Subshells group orbitals with related properties within a shell.

How photoelectron spectroscopy distinguishes subshells

One useful experimental technique is photoelectron spectroscopy (PES).

Instead of trying to photograph an electron’s position, PES measures how much energy is needed to remove electrons from an atom.

The apparatus works in a particular sequence:

1. A sample containing gaseous atoms is introduced into a high-vacuum chamber.
2. A source directs ultraviolet or X-ray photons towards the atoms.
3. The atoms absorb photons, ejecting electrons.
4. The ejected electrons travel into an energy analyser.
5. The analyser separates electrons according to their kinetic energies.
6. A detector counts the electrons and produces a spectrum.

The key observation is that electrons in different subshells have different binding energies.

Consider sodium, which has the electron configuration:

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

It has:

• 3 occupied shells.
• 4 occupied subshells.
• 6 occupied orbitals.

A simplified photoelectron spectrum would therefore show four peaks, corresponding to the four occupied subshells.

Importantly, the \(2p\) peak represents six electrons distributed across three orbitals. It does not represent one orbital containing six electrons.

There is an important limitation here. PES identifies subshells through their binding energies, but it does not directly photograph the three-dimensional shapes of orbitals. Those shapes come from the quantum mechanical model.

03How to read electron configurations

Now that the three levels are clear, you can use electron configuration notation to identify them in actual atoms.

Consider:

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

Every part of this notation has a specific meaning.

For example, \(3p^4\) means four electrons occupy the 3p subshell, which belongs to the third shell.

It does not mean four orbitals, and it does not mean four electrons in four separate shells.

The three counting rules

When an exam asks you to count shells, subshells or orbitals, use different methods for each.

Counting occupied shells

Count the distinct shell numbers that contain electrons.

For the example above, the occupied shells are 1, 2 and 3.

Therefore, there are 3 occupied shells.

Counting occupied subshells

Count the distinct shell-letter combinations that have electrons.

The example contains:

• 1s
• 2s
• 2p
• 3s
• 3p

Therefore, there are 5 occupied subshells.

Counting occupied orbitals

Count the orbitals containing electrons within each subshell.

Remember:

In a ground-state electron configuration, Hund’s rule means that electrons occupy separate orbitals within an equal-energy subshell before pairing.

This gives you a useful shortcut:

\[
\text{Occupied orbitals}
=
\min(\text{electrons},\text{available orbitals})
\]

Apply it separately to each occupied subshell, then add the results.

For the example:

The atom therefore has 9 occupied orbitals.

What does the 3p subshell actually look like when occupied?

The 3p subshell has three orbitals. Imagine three separate boxes, each representing one orbital.

For four electrons, the arrangement is:

\[
\boxed{\uparrow\downarrow}\quad
\boxed{\uparrow}\quad
\boxed{\uparrow}
\]

The first three electrons occupy separate orbitals. The fourth pairs with one of them.

This is Hund’s rule in action.

The two arrows in one box represent two electrons occupying the same orbital with opposite spins. The other two boxes each contain one electron.

This is why \(3p^4\) contains four electrons but only three occupied orbitals.

04How electrons fill shells, subshells and orbitals

Knowing the structure is only half the job. You also need to know how electrons are allocated.

For ground-state atoms, electrons occupy the lowest-energy available orbitals first.

Three principles govern this process.

1. The Aufbau principle

Electrons fill the available subshells in order of increasing energy.

For the first 36 elements, the filling sequence is:

\[
\begin{aligned}
&1s \rightarrow 2s \rightarrow 2p\\
&\rightarrow 3s \rightarrow 3p \rightarrow 4s\\
&\rightarrow 3d \rightarrow 4p
\end{aligned}
\]

Notice something unusual: 4s fills before 3d.

You might expect every subshell in the third shell to fill before electrons enter the fourth shell. That is not what happens.

The reason is that subshell energy depends on more than the shell number. In neutral atoms at the beginning of the fourth period, the 4s subshell is slightly lower in energy than the empty 3d subshell.

This is also why the maximum capacity of a shell does not tell you the order in which electrons fill.

For reference, the maximum electron capacity of a shell is given by:

\[
N_{\text{max}}=2n^2
\]

For example, calcium has 20 electrons and the configuration:

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

Its third shell contains only eight electrons, even though that shell can hold up to 18.

The 4s subshell fills before the 3d subshell.

2. The Pauli exclusion principle

An orbital can contain no more than two electrons.

If two electrons occupy the same orbital, they must have opposite spins.

In orbital box diagrams, we represent these electrons using arrows:

• One electron: \(\boxed{\uparrow}\)
• Two electrons: \(\boxed{\uparrow\downarrow}\)

You cannot place three electrons in one box.

3. Hund’s rule

When several orbitals have the same energy, electrons occupy them individually before pairing.

Consider a \(2p^3\) subshell.

The correct arrangement is:

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

Not:

\[
\boxed{\uparrow\downarrow}\quad
\boxed{\uparrow}\quad
\boxed{\phantom{\uparrow}}
\]

The second arrangement pairs electrons before every equivalent orbital has received one electron.

Hund’s rule explains why the arrangement of electrons across a subshell matters, not just the total number of electrons.

What about chromium and copper?

The Aufbau sequence is an excellent starting point, but it does not predict every ground-state configuration perfectly.

Chromium and copper are two important exceptions in the first transition series.

The simple filling sequence would suggest \(4s^2\,3d^4\) for chromium and \(4s^2\,3d^9\) for copper.

However, the actual ground-state arrangements have a half-filled \(3d^5\) subshell for chromium and a completely filled \(3d^{10}\) subshell for copper.

The 3d and 4s subshells are close in energy, and the balance of electron interactions makes these alternative arrangements slightly lower in energy.

Treat these as actual configurations to learn, rather than trying to force every atom into the basic filling pattern.

05Applying electron configurations to ions

This is where a lot of students make a mistake.

An ion does not necessarily form by simply reversing the order in which an atom’s electrons were added.

You need to distinguish between adding electrons and removing electrons.

When an atom forms a negative ion

A negative ion forms when an atom gains electrons.

Add the electrons to the next available orbitals, following the ground-state filling rules.

For example, oxygen has 8 electrons:

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

An oxide ion, \(\mathrm{O}^{2-}\), has gained two electrons:

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

It now has 10 electrons, the same electron configuration as neon.

When an atom forms a positive ion

A positive ion forms when an atom loses electrons.

For main-group atoms, remove electrons from the outermost occupied shell first.

For example, aluminium has 13 electrons:

\[
\mathrm{Al}: [\mathrm{Ne}]\,3s^2\,3p^1
\]

To form \(\mathrm{Al}^{3+}\), remove three electrons:

• Remove one from 3p.
• Remove two from 3s.

This gives:

\[
\mathrm{Al}^{3+}: [\mathrm{Ne}]
\]

The ion has 10 electrons.

The important transition-metal exception: 4s electrons are removed first

Return to the earlier observation that 4s fills before 3d.

You might predict that if 4s fills first, it should also be the last subshell to lose electrons.

That prediction is incorrect.

For transition-metal cations, the 4s electrons are removed before the 3d electrons. Once the 3d subshell becomes occupied, the relative energies change, and the 4s electrons are generally less tightly held.

Consider iron.

A neutral iron atom has 26 electrons:

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

When iron forms \(\mathrm{Fe}^{2+}\), it loses the two 4s electrons:

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

When it forms \(\mathrm{Fe}^{3+}\), it loses one additional 3d electron:

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

The important distinction is:

• Filling neutral atoms: 4s before 3d.
• Forming transition-metal cations: remove 4s before 3d.

Do not use the filling sequence as a simple reverse removal sequence.

Useful configurations to recognise

The following examples cover important parts of the first 36 elements and the first transition series.

Here, \([\mathrm{Ar}]\) represents the electron configuration of argon:

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

It is a shorthand for the first 18 electrons.

Notice that iron(III) has no electrons in shell 4. Its highest occupied shell is shell 3, even though neutral iron originally had 4s electrons.

06Successive ionisation energies: using evidence to identify electron arrangements

There is another way to investigate the arrangement of electrons: measure how much energy is required to remove them one at a time.

The first ionisation energy is the energy needed to remove one electron from a gaseous atom:

\[
X(g)\rightarrow X^+(g)+e^-
\]

The second ionisation energy removes an electron from the resulting positive ion:

\[
X^+(g)\rightarrow X^{2+}(g)+e^-
\]

Each successive ionisation energy is generally larger than the previous one. As electrons are removed, the remaining electrons experience a stronger effective attraction to the nucleus.

The most useful clue is a sudden, unusually large increase.

Example: Identifying an element from successive ionisation energies

Consider the following illustrative data for an unknown element in Period 3.

The major increase occurs between the third and fourth ionisation energies.

Why?

The first three electrons are removed from the outermost shell. The fourth electron must come from an inner shell, where it is much more strongly attracted to the nucleus.

Therefore, the atom has three outer-shell electrons.

Since it is in Period 3, its configuration is:

\[
[\mathrm{Ne}]\,3s^2\,3p^1
\]

This is aluminium.

Successive ionisation energies can reveal subshells too

Consider magnesium and aluminium.

Their ground-state configurations are:

\[
\mathrm{Mg}: [\mathrm{Ne}]\,3s^2
\]

\[
\mathrm{Al}: [\mathrm{Ne}]\,3s^2\,3p^1
\]

Magnesium loses a 3s electron first. Aluminium loses a 3p electron first.

The 3p electron in aluminium is at a higher energy and is less penetrating than a 3s electron. It is therefore easier to remove.

This explains an interesting observation: aluminium has a lower first ionisation energy than magnesium, despite having one more proton.

The general trend of increasing ionisation energy across a period is not a perfectly smooth increase. Changes in subshell occupancy can create exceptions.

The key is to consider both nuclear attraction and the subshell occupied by the electron being removed.

Don’t confuse two types of ionisation evidence

Both methods provide evidence about electron organisation, but they measure different things.

07Worked examples

08Questions and solutions

The following questions move from basic electron counting to interpreting experimental evidence, applying ionisation rules and evaluating more subtle differences between ions.

Question 1

A neutral chlorine atom has atomic number 17.

(a) Write its ground-state electron configuration.

(b) Determine the number of occupied shells and occupied subshells.

(c) Calculate the number of occupied orbitals.

Solution 1

Chlorine has 3 occupied shells, 5 occupied subshells and 9 occupied orbitals.

(a) A neutral chlorine atom has 17 electrons, so its ground-state configuration is:

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

The electron total is:

\[
2+2+6+2+5=17
\]

(b) The occupied shell numbers are 1, 2 and 3.

Therefore, there are 3 occupied shells.

The occupied subshells are 1s, 2s, 2p, 3s and 3p.

Therefore, there are 5 occupied subshells.

(c) Count the orbitals occupied in each subshell:

The five 3p electrons occupy all three p orbitals. Two orbitals contain pairs, and the remaining orbital contains one electron.

This follows Hund’s rule.

Question 2

An unknown neutral atom produces an idealised photoelectron spectrum with five peaks.

Reading from highest to lowest binding energy, the relative electron counts represented by the peaks are:

\[
2,\quad 2,\quad 6,\quad 2,\quad 3
\]

Assume the relative counts are proportional to the number of electrons in each occupied subshell.

(a) Determine the electron configuration of the atom.

(b) Identify the element.

(c) Determine the number of occupied shells, occupied subshells and occupied orbitals.

(d) Explain why the final peak represents three electrons in one subshell rather than three separate subshells.

Solution 2

The unknown atom is phosphorus, with the configuration \(1s^2\,2s^2\,2p^6\,3s^2\,3p^3\). It has 3 occupied shells, 5 occupied subshells and 9 occupied orbitals.

(a) Each peak corresponds to an occupied subshell. Starting with the most tightly bound electrons, assign the peaks:

Therefore:

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

(b) Add the electrons:

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

A neutral atom with 15 electrons has atomic number 15.

The element is phosphorus (P).

(c) The occupied shell numbers are 1, 2 and 3, giving 3 shells.

There are five occupied subshells:

1s, 2s, 2p, 3s and 3p.

For the occupied orbitals:

\[
1+1+3+1+3=9
\]

Therefore, there are 9 occupied orbitals.

(d) The final peak represents the 3p subshell.

A p subshell contains three orbitals. Its three electrons occupy these orbitals individually, with parallel spins, according to Hund’s rule.

The spectrum shows one peak for the subshell, not three separate peaks for its individual orbitals.

This question highlights an important experimental distinction: the number of PES peaks tells you the number of occupied subshells in the simplified model, not the number of occupied orbitals.

Question 3

Calcium has atomic number 20.

A student argues that calcium should fill the 3d subshell before the 4s subshell because the third shell can hold 18 electrons and currently contains only eight.

(a) Explain why this reasoning is incorrect.

(b) Write the electron configurations of neutral calcium, \(\mathrm{Ca}^{+}\) and \(\mathrm{Ca}^{2+}\).

(c) Determine the number of occupied shells, subshells and orbitals in \(\mathrm{Ca}^{2+}\).

Solution 3

The student’s reasoning is incorrect because electron filling depends on subshell energy, not on completing the maximum capacity of one shell before entering another.

(a) The third shell can hold a maximum of 18 electrons, but this does not mean all 18 must be present before shell 4 begins filling.

For neutral atoms at this stage, the 4s subshell is lower in energy than the empty 3d subshell.

Therefore, 4s fills first.

(b) Calcium has 20 electrons.

Its ground-state configuration is:

\[
\mathrm{Ca}: [\mathrm{Ar}]\,4s^2
\]

When it loses one electron, it loses an outer 4s electron:

\[
\mathrm{Ca}^{+}: [\mathrm{Ar}]\,4s^1
\]

When it loses a second electron:

\[
\mathrm{Ca}^{2+}: [\mathrm{Ar}]
\]

(c) The \(\mathrm{Ca}^{2+}\) ion has the same electron configuration as argon.

Its occupied shells are 1, 2 and 3, giving 3 occupied shells.

Its occupied subshells are 1s, 2s, 2p, 3s and 3p, giving 5 occupied subshells.

Its occupied orbitals are:

\[
1+1+3+1+3=9
\]

Therefore, \(\mathrm{Ca}^{2+}\) has 3 occupied shells, 5 occupied subshells and 9 occupied orbitals.

The tempting mistake is to treat shell capacity as a filling instruction. It is only a maximum capacity. The actual filling sequence depends on the relative energies of available subshells.

Question 4

Chromium has atomic number 24, while copper has atomic number 29.

(a) Write the actual ground-state electron configurations of both neutral atoms.

(b) Determine the electron configurations of \(\mathrm{Cr}^{+}\) and \(\mathrm{Cu}^{2+}\).

(c) For each ion, calculate the number of occupied shells, occupied subshells and occupied orbitals.

(d) Explain why the 4s electrons are removed before the 3d electrons when these atoms form positive ions.

Solution 4

Chromium has the configuration \([\mathrm{Ar}]\,4s^1\,3d^5\), while copper has \([\mathrm{Ar}]\,4s^1\,3d^{10}\). Their ions are \(\mathrm{Cr}^{+}=[\mathrm{Ar}]\,3d^5\) and \(\mathrm{Cu}^{2+}=[\mathrm{Ar}]\,3d^9\).

(a) The actual ground-state configurations are:

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

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

These are exceptions to the simple Aufbau prediction because the 3d and 4s subshells are close in energy.

(b) For chromium, remove the single 4s electron:

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

For copper, remove the 4s electron first, followed by one 3d electron:

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

(c) Both ions have occupied shells 1, 2 and 3.

Therefore, each has 3 occupied shells.

Both have six occupied subshells:

1s, 2s, 2p, 3s, 3p and 3d.

The argon core contains 9 occupied orbitals.

The 3d subshell has five orbitals. Both \(3d^5\) and \(3d^9\) occupy all five.

Therefore:

\[
9+5=14
\]

Both ions have 14 occupied orbitals.

Their electron arrangements within the 3d subshell differ:

• \(\mathrm{Cr}^{+}\): five singly occupied orbitals.
• \(\mathrm{Cu}^{2+}\): four paired orbitals and one singly occupied orbital.

(d) The filling and removal orders are not exact reverses of each other.

The 4s subshell fills before 3d in the relevant neutral atoms. Once the 3d subshell is occupied, the relative energy ordering changes. The 4s electrons are generally less tightly held and are removed first.

The important exam skill is to start with the actual neutral configuration, then apply the ionisation rule. Do not try to construct the ion by simply reversing the Aufbau sequence.

Question 5

Two isolated gaseous ions are being compared:

• Ion A is \(\mathrm{Fe}^{3+}\), containing 26 protons.
• Ion B is \(\mathrm{Mn}^{2+}\), containing 25 protons.

Both ions contain 23 electrons.

Each ion then loses one additional electron, producing \(\mathrm{Fe}^{4+}\) and \(\mathrm{Mn}^{3+}\), respectively.

A student makes the following claim:

“Because the original ions have the same number of electrons, they must have identical electron configurations, identical orbital energies and identical ionisation energies. The same must remain true after both ions lose another electron.”

Evaluate this claim.

In your response:

(a) Determine the electron configurations of the original ions.

(b) Calculate the number of occupied shells, subshells and orbitals in each original ion.

(c) Determine the configurations and occupied orbital counts after each ion loses another electron.

(d) Explain whether the two ions should have identical electron binding energies and sizes, both before and after the additional ionisation.

Assume ground-state configurations and ignore external fields and chemical bonding effects.

Solution 5

The student’s claim is only partly correct. The two ions have identical electron configurations and orbital occupancies at each matching stage, but their orbital energies, electron binding energies and sizes are not identical because their nuclei contain different numbers of protons.

(a) Both original ions contain 23 electrons.

Iron has atomic number 26:

\[
26-3=23
\]

Manganese has atomic number 25:

\[
25-2=23
\]

Their configurations are therefore:

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

\[
\mathrm{Mn}^{2+}: [\mathrm{Ar}]\,3d^5
\]

They are isoelectronic, meaning they have the same number of electrons and the same electron configuration.

(b) Both ions have the same occupied shells: 1, 2 and 3.

Therefore, each has 3 occupied shells.

Both have the same six occupied subshells:

1s, 2s, 2p, 3s, 3p and 3d.

The argon core occupies 9 orbitals. The five electrons in 3d occupy five separate orbitals according to Hund’s rule.

Therefore:

\[
9+5=14
\]

Each original ion has 14 occupied orbitals.

(c) Each ion loses one electron from its 3d subshell.

The resulting configurations are:

\[
\mathrm{Fe}^{4+}: [\mathrm{Ar}]\,3d^4
\]

\[
\mathrm{Mn}^{3+}: [\mathrm{Ar}]\,3d^4
\]

Both now contain 22 electrons.

The four 3d electrons occupy four separate orbitals, leaving the fifth 3d orbital empty.

The number of occupied orbitals becomes:

\[
9+4=13
\]

Both ions still have:

• 3 occupied shells.
• 6 occupied subshells.
• 13 occupied orbitals.

Notice that losing one electron has reduced the number of occupied orbitals from 14 to 13. It has not removed an entire subshell.

(d) This is where the student’s reasoning breaks down.

The ions have the same electron arrangements, but they do not have the same nuclear charge.

Compare the original ions:

Iron(III) has one more proton than manganese(II), while both have the same number of electrons.

The greater nuclear charge produces a stronger attraction between the nucleus and the electrons. Consequently, \(\mathrm{Fe}^{3+}\) is smaller and generally holds its electrons more strongly than \(\mathrm{Mn}^{2+}\).

The same reasoning applies after the additional ionisation:

• \(\mathrm{Fe}^{4+}\) has 26 protons and 22 electrons.
• \(\mathrm{Mn}^{3+}\) has 25 protons and 22 electrons.

Both have the configuration \([\mathrm{Ar}]\,3d^4\), but the iron ion has the greater nuclear attraction.

Therefore, under the same gas-phase conditions, the next electron is more strongly bound in the iron ion, and removing it requires more energy.

The tempting mistake is to assume that identical electron configurations mean identical atoms or ions. They do not.

An electron configuration tells you how electrons are arranged. It does not tell you the complete nuclear charge or all the energies associated with that arrangement.

This is the distinction that makes isoelectronic comparisons more interesting than simply counting electrons.

09A note on the NSW HSC syllabus

This article follows the Chemistry Stage 6 Syllabus (2017), which remains applicable to students studying the current course in 2026 and 2027.

The 2017 syllabus explicitly covers:

• The Pauli exclusion principle and the maximum of two electrons per orbital.
• Subshells and their filling order.
• Hund’s rule and ground-state electron configurations.
• The relationship between ionisation energies and electron orbitals.
• Electron configurations of the first transition series.

The Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. This article does not assume that every detail covered here is prescribed in that future course.

For the official requirements, consult the NESA Chemistry Stage 6 Syllabus (2017) and the 2025 syllabus implementation information.

One final terminology distinction: an atomic orbital is a quantum description of an electron, not the physical path of an object travelling around another body. The latter is a different use of the word, explored in tutorgum’s HSC Physics guide to orbital velocity.

Once you can distinguish shells, subshells and orbitals, you have the foundation for understanding periodic trends, successive ionisation energies and the variable oxidation states of transition metals. Those topics all rely on knowing exactly where electrons are arranged and how strongly they are held.