Electronic Structure & Ionisation Energy

Last updated: 15 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

At O-Level, an electronic structure such as 2,8,1 was enough to describe a sodium atom. It showed that sodium has three occupied electron shells and one electron in its outer shell. At A-Level, we look more closely inside those shells.

Electrons within the same principal shell do not all have exactly the same energy. They occupy different subshells and atomic orbitals. This extra detail explains several observations that the simple shell model cannot explain, including why 4s fills before 3d, why chromium and copper have unusual electronic configurations, and why first ionisation energy does not increase perfectly smoothly across a period.

The important idea is that electronic configuration and ionisation energy are not separate topics. The configuration tells us where the electrons are, while ionisation energy tells us how strongly those electrons are held. Both arise from the same extranuclear structure.

Diagnostic Quiz

What Do You Already Know About Electrons?

Try these five questions before reading the chapter. They check the ideas about shells, orbitals and electron removal that the A-Level model will build on.

5 quick questions
Spot common electronic-structure misconceptions
Every answer option is explained

Which statement best describes an atomic orbital?

A fixed circular path travelled by an electron around the nucleus

This is closer to the old idea of an electron orbit. In the modern atomic model, an electron is not assigned a definite circular path.

A region of space in which there is a high probability of finding an electron

Correct. An orbital describes the spatial probability distribution associated with an electron of a particular energy.

A complete principal electron shell

A principal shell contains one or more subshells, and each subshell contains orbitals. An orbital is therefore a more specific part of the electron structure.

The empty space between the nucleus and the outermost shell

An orbital is not defined as empty space. It describes where an electron has a significant probability of being found.

A p subshell contains three orbitals. What is the maximum number of electrons that can occupy a p subshell?

Two is the maximum number of electrons in one orbital, not in the entire p subshell.

Three is the number of orbitals in a p subshell. Each orbital can hold two electrons.

Five is the number of orbitals in a d subshell, not the electron capacity of a p subshell.

Correct. A p subshell has three orbitals, and each orbital can contain two electrons, giving a maximum of 3 × 2 = 6 electrons.

Which arrangement correctly represents the four electrons in a 2p⁴ subshell?

↑↓ ↑↓ empty

This pairs electrons in two orbitals while leaving the third orbital empty. Hund’s rule requires all three equal-energy p orbitals to be occupied singly before pairing occurs.

↑↓ ↑↓ ↑

This arrangement contains five electrons, not four.

↑↓ ↑ ↑

Correct. The first three electrons occupy separate p orbitals. The fourth electron then pairs with one of them.

↑↑ ↑ ↑

Two electrons in the same orbital cannot have the same spin. The Pauli exclusion principle requires paired electrons to have opposite spins.

Which equation represents the first ionisation of an element X?

X(g) → X⁺(g) + e⁻

Correct. First ionisation removes one electron from each gaseous atom, forming gaseous 1+ ions.

X⁺(g) → X²⁺(g) + e⁻

This represents the second ionisation because the electron is removed from an X⁺ ion.

X(s) → X⁺(g) + e⁻

First ionisation energy is defined using gaseous atoms. Starting with solid X would include other energy changes.

X(g) + e⁻ → X⁻(g)

This involves addition rather than removal of an electron and therefore is not an ionisation process.

An element shows a very large increase between its second and third ionisation energies. What does this most strongly suggest?

It has one electron in its outer principal shell

If there were one outer-shell electron, the major jump would usually occur between the first and second ionisation energies.

It has two electrons in its outer principal shell

Correct. Two electrons can be removed relatively easily from the outer shell. The third electron is then removed from a closer inner shell and requires much more energy.

Its second electron is in a different nucleus

All the electrons considered belong to the same atom and nucleus. The jump reflects a change in the principal shell from which an electron is removed.

Its nucleus contains only two protons

The position of the jump indicates the number of outer-shell electrons, not directly the total number of protons.

What is extranuclear structure?

The nucleus contains the protons and neutrons of the atom. The electrons outside the nucleus form its extranuclear structure.

At A-Level, we describe this structure using three levels:

principal shell → subshell → atomic orbital

A principal shell contains one or more subshells, and each subshell contains one or more orbitals.

For example, the third principal shell contains the 3s, 3p and 3d subshells. The 3p subshell then contains three separate 3p orbitals.

Principal quantum number, n

Each principal electron shell is identified by its principal quantum number, n.

n = 1 is the principal shell closest to the nucleus.
n = 2 is the next principal shell.
n = 3 is farther from the nucleus.
n = 4 is farther still.

As n increases, electrons are generally found farther from the nucleus and have higher energies. However, this does not mean that every orbital in shell 4 must be higher in energy than every orbital in shell 3. Subshell energies can overlap, which becomes important when we compare 4s and 3d.

Subshells and atomic orbitals

Principal shells are divided into subshells. The subshells most important in this part of the syllabus are labelled s, p and d.

Subshell	Number of orbitals	Maximum electrons in each orbital	Maximum electrons in subshell
s	1	2	2
p	3	2	6
d	5	2	10

These maximum electron numbers are not three unrelated facts to memorise. They follow directly from the number of orbitals present because every orbital can contain a maximum of two electrons.

s: 1 orbital × 2 electrons = 2 electrons

p: 3 orbitals × 2 electrons = 6 electrons

d: 5 orbitals × 2 electrons = 10 electrons

What exactly is an atomic orbital?

An atomic orbital is a region of space around the nucleus in which there is a high probability of finding an electron with a particular energy.

The word orbital is important. An orbital is not a small physical box and it is not a track along which an electron travels.

Orbit: suggests a definite path, like a planet moving around the Sun.

Orbital: describes a three-dimensional probability distribution showing where an electron is likely to be found.

How to read notation such as 3p⁴

The notation used for a subshell contains three pieces of information.

Consider 3p⁴.
The number 3 identifies the third principal shell.
The letter p identifies the type of subshell.
The superscript 4 means that four electrons occupy the entire 3p subshell.

Since a p subshell contains three orbitals, the four electrons are distributed among those three orbitals rather than all being placed in a single orbital.

Shapes of s, p and d orbitals

Orbital shapes represent the regions in which an electron has a high probability of being found. The nucleus lies at the centre of the orbital diagrams.

s orbitals are spherical

An s orbital is spherical. The probability distribution is therefore the same in every direction around the nucleus.

The 1s, 2s and 3s orbitals are all spherical, although they differ in size, energy and their detailed internal probability distributions.

p orbitals are dumbbell-shaped

A p orbital has two lobes on opposite sides of the nucleus and is usually described as dumbbell-shaped.

Each p subshell contains three p orbitals:

p_x is oriented along the x-axis;
p_y is oriented along the y-axis; and
p_z is oriented along the z-axis.

Within an isolated atom, these three p orbitals have the same energy but differ in their orientation in space.

d orbitals have more complex shapes

A d subshell contains five orbitals. Four are commonly represented as four-lobed, cloverleaf-shaped orbitals.

The d_z² orbital looks different: it has two lobes along the z-axis together with a ring-shaped region around its centre.

The orbital drawing is not an electron path.

The boundary drawn around an orbital represents a region of high electron probability. The electron does not travel around that boundary as though it were moving on the surface of a balloon.

Relative energies and the order in which orbitals fill

Electrons tend to occupy the lowest-energy orbitals available. For constructing the ground-state electronic configurations needed here, use the following practical filling sequence:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p

One feature immediately stands out: 4s is occupied before 3d. The principal quantum number alone therefore does not determine the order in which subshells fill.

Treat this as a filling sequence, not a permanent energy ladder.

The relative energies of 4s and 3d depend on the electron configuration of the atom. This is why the statement “4s is always below 3d” causes problems when transition-metal ions are formed.

Three rules for placing electrons into orbitals

Knowing the order of the subshells is only part of the problem. We must also know how the electrons are arranged within individual orbitals.

1. Aufbau principle

The Aufbau principle states that electrons occupy the available orbitals of lowest energy first.

This is why 1s fills before 2s, 2s fills before 2p, and 4s begins to fill before 3d in potassium and calcium.

2. Pauli exclusion principle

The Pauli exclusion principle means that an atomic orbital can contain a maximum of two electrons, and those two electrons must have opposite spins.

permitted pair: ↑↓

Two electrons with the same spin cannot occupy the same orbital.

3. Hund’s rule

Hund’s rule applies when several orbitals of equal energy are available. Electrons occupy these orbitals singly, with parallel spins, before pairing occurs.

Consider a p subshell containing three equal-energy orbitals:

Configuration	Distribution among the three p orbitals	Unpaired electrons
p¹	↑ empty empty	1
p²	↑ ↑ empty	2
p³	↑ ↑ ↑	3
p⁴	↑↓ ↑ ↑	2
p⁵	↑↓ ↑↓ ↑	1
p⁶	↑↓ ↑↓ ↑↓	0

Example: nitrogen and oxygen
Nitrogen has the outer configuration 2p³:
N: 2p³ = ↑ ↑ ↑
Each 2p orbital is singly occupied.
Oxygen has the outer configuration 2p⁴:
O: 2p⁴ = ↑↓ ↑ ↑
The fourth 2p electron has to pair with one of the first three electrons. The resulting electron–electron repulsion will later help us explain why oxygen has a lower first ionisation energy than nitrogen.

Writing electronic configurations

An electronic configuration shows how an atom’s electrons are distributed among its occupied subshells.

A reliable method is:

Determine the total number of electrons.
Fill the subshells in order of increasing energy.
Do not exceed the capacity of any subshell.
Apply Hund’s rule when filling orbitals of equal energy.
Add the superscripts at the end to check that every electron has been included.

Examples of electronic configurations

Atom	Number of electrons	Electronic configuration
Na	11	1s² 2s² 2p⁶ 3s¹
Cl	17	1s² 2s² 2p⁶ 3s² 3p⁵
K	19	1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Ca	20	1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Fe	26	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Br	35	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵

Noble-gas notation

Long configurations can be shortened by replacing the completed inner-shell configuration with the symbol of the preceding noble gas in square brackets.

Atom	Full electronic configuration	Shortened configuration
Na	1s² 2s² 2p⁶ 3s¹	[Ne] 3s¹
Ca	1s² 2s² 2p⁶ 3s² 3p⁶ 4s²	[Ar] 4s²
Fe	1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²	[Ar] 3d⁶ 4s²
Br	1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵	[Ar] 3d¹⁰ 4s² 4p⁵

You may notice that iron is often written as [Ar] 3d⁶ 4s² even though 4s filled before 3d. The written order of occupied subshells does not change the electron arrangement. What matters is knowing which subshells are occupied and how many electrons each contains.

Why chromium and copper are exceptions to the simple filling pattern

If we followed the simple filling sequence mechanically, we might initially predict:

predicted Cr: [Ar] 3d⁴ 4s²

predicted Cu: [Ar] 3d⁹ 4s²

The observed ground-state configurations are instead:

Cr: [Ar] 3d⁵ 4s¹

Cu: [Ar] 3d¹⁰ 4s¹

The 3d and 4s subshells are very close in energy. In chromium and copper, transferring one electron from 4s into 3d gives an overall electron arrangement that is slightly lower in energy.

Chromium obtains a half-filled 3d⁵ subshell.
Copper obtains a completely filled 3d¹⁰ subshell.

Do not turn “half-filled is stable” into a new filling rule.

Chromium and copper are important exceptions because the 3d and 4s energies are unusually close. The experimentally observed configurations are the lower-energy arrangements for these atoms.

Electronic configurations of ions

When an atom forms an ion, the number of protons does not change. The number of electrons changes.

A positive ion has lost electrons.
A negative ion has gained electrons.

Main-group ions

Example: Na⁺ and O²⁻
A sodium atom has 11 electrons. Na⁺ has lost one electron, so it has 10:
Na⁺: 1s² 2s² 2p⁶
An oxygen atom has 8 electrons. O²⁻ has gained two electrons, so it also has 10:
O²⁻: 1s² 2s² 2p⁶
Na⁺ and O²⁻ are isoelectronic: they have the same number and arrangement of electrons.

Transition-metal ions: remove 4s before 3d

The formation of transition-metal ions causes an apparent contradiction: 4s fills before 3d, but 4s empties before 3d.

The key is that the relative energies of 4s and 3d change as the 3d subshell becomes occupied. Once 3d electrons are present, the occupied 4s level lies higher in energy and the 4s electrons are removed first when a cation forms.

Species	Electron change	Electronic configuration
Fe	Neutral atom	[Ar] 3d⁶ 4s²
Fe²⁺	Remove two 4s electrons	[Ar] 3d⁶
Fe³⁺	Remove two 4s electrons, then one 3d electron	[Ar] 3d⁵
Cu	Neutral atom	[Ar] 3d¹⁰ 4s¹
Cu⁺	Remove the 4s electron	[Ar] 3d¹⁰
Cu²⁺	Remove the 4s electron, then one 3d electron	[Ar] 3d⁹

When building a neutral atom: 4s is occupied before 3d.

When forming a transition-metal cation: remove 4s electrons before removing 3d electrons.

What is ionisation energy?

Ionisation energy measures how much energy is required to remove an electron from an atom or ion in the gaseous state. It therefore gives us evidence about how strongly an electron is attracted to the nucleus.

First ionisation energy

The first ionisation energy of an element is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions.

X(g) → X⁺(g) + e⁻

Ionisation energies are normally measured in kJ mol⁻¹.

Every part of the definition matters.

The particles must be gaseous atoms, one electron is removed from each atom, and the products are gaseous 1+ ions.

Energy must be supplied because the negatively charged electron is attracted electrostatically to the positively charged nucleus. The electron must be pulled away against this attraction.

Second and later ionisation energies

Once one electron has been removed, another electron can be removed from the resulting ion.

Second ionisation: X⁺(g) → X²⁺(g) + e⁻

Third ionisation: X²⁺(g) → X³⁺(g) + e⁻

Notice that the second ionisation energy removes an electron from X⁺, not from another neutral X atom.

What determines the magnitude of ionisation energy?

The central question is: how strongly is the electron being removed attracted to the nucleus?

Several factors affect this attraction.

1. Nuclear charge

A nucleus containing more protons has a greater positive charge. If the other factors are similar, a greater nuclear charge produces a stronger attraction for an electron and therefore a higher ionisation energy.

2. Distance from the nucleus

Electrostatic attraction becomes weaker as the distance between opposite charges increases.

An electron in a higher principal shell is generally farther from the nucleus, so it is usually easier to remove.

3. Shielding

Electrons between the nucleus and the electron being removed repel that electron and reduce the full attractive effect of the nucleus. This is called shielding.

Inner-shell electrons are especially important in shielding outer electrons from the nucleus.

4. The subshell occupied

Electrons in different subshells of the same principal shell do not have exactly the same energy or experience the nucleus in exactly the same way.

For example, a 3p electron is higher in energy and less penetrating than a 3s electron. It is therefore generally easier to remove a 3p electron than a comparable 3s electron.

5. Repulsion between paired electrons

If two electrons occupy the same orbital, they repel one another. This additional electron–electron repulsion can make it easier to remove one of the paired electrons.

The stronger the nucleus holds the electron, the higher the ionisation energy.

When explaining an ionisation-energy trend, consider the nuclear charge, distance from the nucleus, shielding and the particular subshell or orbital occupied by the electron.

First ionisation energy across the Periodic Table

Across a period: the general trend is upwards

First ionisation energy generally increases across a period.

Across Period 3, for example:

proton number increases;
nuclear charge increases;
the additional electrons enter the same principal shell;
the number of occupied inner shells stays the same; and
shielding by the inner shells therefore remains approximately constant.

The increasing nuclear charge produces a greater effective attraction for the outer electrons. The atomic radius generally decreases, and more energy is needed to remove the outer electron.

Exam-style reasoning
Across Period 3, nuclear charge increases while electrons are added to the same principal shell. Shielding by inner-shell electrons remains approximately constant, so the effective nuclear attraction increases. The outer electron is held more strongly and the first ionisation energy generally increases.

Why the trend is not perfectly smooth

The general increase contains small decreases. These are useful rather than troublesome because they provide evidence for subshell energies and electron pairing.

Why first ionisation energy decreases from Mg to Al

Mg: [Ne] 3s²

Al: [Ne] 3s² 3p¹

The electron removed from magnesium is a 3s electron. The electron removed from aluminium is a 3p electron.

The 3p electron is higher in energy and less penetrating than the 3s electron. It is therefore more shielded from the nucleus and easier to remove.

This subshell effect outweighs the increase in nuclear charge from Mg to Al, so aluminium has the lower first ionisation energy.

Weak explanation: Aluminium has more electrons, so there is more shielding.

Better explanation: Aluminium loses a higher-energy 3p electron, whereas magnesium loses a lower-energy, more penetrating 3s electron.

Why first ionisation energy decreases from P to S

P: [Ne] 3s² 3p³

S: [Ne] 3s² 3p⁴

The three 3p electrons in phosphorus occupy separate p orbitals:

P: 3p³ = ↑ ↑ ↑

Sulfur has one additional 3p electron. This electron must pair with one of the existing electrons:

S: 3p⁴ = ↑↓ ↑ ↑

Repulsion between the paired electrons makes one of the electrons in the doubly occupied orbital easier to remove. This effect outweighs the increase in nuclear charge, so sulfur has a slightly lower first ionisation energy than phosphorus.

Decrease	Main explanation
Mg → Al	The electron removed changes from a lower-energy 3s subshell to a higher-energy 3p subshell.
P → S	Electron pairing in a 3p orbital introduces additional electron–electron repulsion.

The corresponding decreases from Be to B and from N to O in Period 2 arise from the same underlying ideas.

Down a group: the general trend is downwards

First ionisation energy generally decreases down a group.

Down a group:

the outer electron occupies a higher principal shell;
its average distance from the nucleus increases;
there are more occupied inner shells; and
shielding increases.

Nuclear charge also increases, but the effects of increased distance and shielding outweigh this. The outer electron is held less strongly and is easier to remove.

Exam-style reasoning
Down a group, the outer electron is farther from the nucleus and experiences greater shielding by inner-shell electrons. These effects outweigh the increase in nuclear charge, so the attraction between the nucleus and the outer electron decreases. Less energy is required to remove the electron.

Successive ionisation energies reveal electron structure

Successive ionisation energies give us an experimental way to work backwards from measured energies to the electron arrangement of an atom.

Why successive ionisation energies increase

After each electron is removed, the species becomes more positively charged. The remaining electrons are therefore increasingly strongly attracted to the same nucleus.

As a result, successive ionisation energies generally become progressively larger.

What causes a very large jump?

A particularly large jump occurs when all the electrons in one outer principal shell have already been removed and the next electron must be taken from an inner principal shell.

Compared with the outer-shell electron, an inner-shell electron:

is closer to the nucleus;
experiences less shielding; and
is therefore much more strongly attracted to the nucleus.

A much larger amount of energy is needed to remove it.

Example: a large jump after the third ionisation energy
Suppose the first three ionisation energies increase gradually, but the fourth value is dramatically larger.
The first three electrons were removed from the outer principal shell.
The fourth electron must come from a closer inner principal shell.
The atom therefore has three electrons in its outer principal shell.
For a main-group element, this is characteristic of Group 13.

Large shell jumps and smaller subshell changes

Not every increase in successive ionisation energy means that a new principal shell has been reached.

A smaller but noticeable increase can occur when electron removal changes from a higher-energy p subshell to a lower-energy s subshell within the same principal shell.

A change to a new inner principal shell produces a much larger increase because the electron is both significantly closer to the nucleus and less shielded.

Pattern	Likely interpretation
Gradual increase	Electrons are being removed from similar outer energy levels.
Noticeable but moderate increase	Electron removal may have changed from one subshell to another within the same principal shell.
Very large increase	Electron removal has begun from a new inner principal shell.

How to use successive ionisation-energy data

When interpreting a sequence of ionisation energies:

Look first for the very large jumps.
Count the electrons removed before the first major jump.
This gives the number of electrons in the outer principal shell.
Look for later major jumps to identify the sizes of inner shells.
Combine this information with the period or total number of electrons when given.

Example: identifying aluminium
An element in Period 3 shows a very large jump between its third and fourth ionisation energies.
Three electrons are therefore removed from the third principal shell before an electron must be removed from the second principal shell.
outer configuration = 3s² 3p¹
The Period 3 element with this configuration is aluminium.

Be careful with transition elements.

The simple rule linking the first major jump directly to a main-group group number works most straightforwardly for s- and p-block elements. Transition elements require more care because the 4s and 3d subshells are close in energy.

Common mistakes in electronic structure and ionisation energy

Common mistake	Why it is wrong	Better approach
Describing an orbital as a fixed path travelled by an electron	An orbital is a probability distribution, not an electron orbit	Describe it as a region in which there is a high probability of finding the electron
Saying a p subshell contains six orbitals	A p subshell contains three orbitals; six is its maximum number of electrons	Remember: 3 p orbitals × 2 electrons each = 6 electrons
Pairing p electrons before all three orbitals are singly occupied	This violates Hund’s rule	Occupy equal-energy orbitals singly before pairing
Writing Cr as [Ar] 3d⁴ 4s² and Cu as [Ar] 3d⁹ 4s²	These are simple predictions, not the observed ground-state configurations	Use Cr: [Ar] 3d⁵ 4s¹ and Cu: [Ar] 3d¹⁰ 4s¹
Removing 3d electrons before 4s when forming transition-metal cations	Once 3d is occupied, the 4s electrons are removed first	For Fe²⁺, remove the two 4s electrons from [Ar] 3d⁶ 4s²
Leaving out the gaseous state in an ionisation-energy definition or equation	Ionisation energy is defined for gaseous atoms or ions	Write X(g) → X⁺(g) + e⁻ for first ionisation
Explaining Mg → Al using only “more shielding”	The important difference is the higher-energy 3p electron in aluminium	Compare the subshells from which the electrons are removed
Explaining P → S using a change of subshell	Both electrons being compared are removed from 3p	Explain the additional repulsion caused by electron pairing in sulfur
Saying shielding increases greatly across a period because electron number increases	The additional electrons enter the same principal shell and the number of occupied inner shells stays the same	State that inner-shell shielding remains approximately constant while nuclear charge increases
Treating every increase in successive ionisation energy as a new shell	Some smaller changes arise from differences between subshells	Look for a very large jump when identifying a change of principal shell

Electronic structure and ionisation-energy checklist

Understand the orbitals

Can I distinguish a principal shell, subshell and orbital?
Do I know that s, p and d contain 1, 3 and 5 orbitals?
Can I recognise the general shapes of s, p and d orbitals?
Can I explain why an orbital is not a fixed electron path?

Write configurations correctly

Can I use the correct filling sequence up to 4p?
Can I apply the Aufbau principle, Pauli principle and Hund’s rule?
Do I remember the configurations of Cr and Cu?
Can I form ions correctly, including removing 4s before 3d?

Explain ionisation energies

Can I state the first ionisation-energy definition precisely?
Can I write equations for first and successive ionisations?
Can I use nuclear charge, distance and shielding in explanations?
Can I explain the Mg → Al and P → S decreases separately?

Interpret the evidence

Can I explain the trend across a period?
Can I explain the trend down a group?
Can I identify a major shell jump in successive ionisation energies?
Can I use the position of that jump to deduce outer-shell electron number?

Mastery Quiz

Electronic Structure and Ionisation Energy Mastery Quiz

Now apply the chapter to unfamiliar configurations and ionisation-energy patterns. These questions focus on reasoning rather than simple definition recall.

10 application-focused questions
Electronic configurations and ionisation-energy reasoning
Every answer option is explained

Magnesium and aluminium are adjacent in Period 3, yet aluminium has a lower first ionisation energy. Which comparison best explains this?

Aluminium has a smaller nuclear charge than magnesium

Aluminium has one more proton than magnesium and therefore has the larger nuclear charge.

Magnesium loses a 3p electron while aluminium loses a 3s electron

This is reversed. Mg is [Ne] 3s², while Al is [Ne] 3s² 3p¹.

Aluminium loses a higher-energy 3p electron, whereas magnesium loses a 3s electron

Correct. The 3p electron in aluminium is higher in energy and less penetrating, so it is easier to remove despite aluminium’s greater nuclear charge.

Aluminium’s outer electron is in a lower principal shell than magnesium’s

Both electrons being compared belong to the third principal shell. The relevant difference is 3s versus 3p.

Why is the first ionisation energy of sulfur slightly lower than that of phosphorus?

Sulfur has fewer protons than phosphorus

Sulfur has one more proton than phosphorus, so its nuclear charge is greater.

Sulfur’s electron is removed from 4s rather than 3p

Both phosphorus and sulfur lose an electron from the 3p subshell.

Sulfur has an additional occupied principal shell

Both phosphorus and sulfur are in Period 3 and have the same number of occupied principal shells.

One 3p orbital in sulfur contains a pair of electrons whose mutual repulsion makes removal easier

Correct. Phosphorus has three singly occupied 3p orbitals, whereas sulfur has one doubly occupied 3p orbital. Repulsion within the pair makes one electron easier to remove.

Iron has the ground-state configuration [Ar] 3d⁶ 4s². What is the correct electronic configuration of Fe³⁺?

[Ar] 3d³ 4s²

This incorrectly removes three 3d electrons while leaving both 4s electrons. The 4s electrons are removed first.

[Ar] 3d⁵

Correct. Remove the two 4s electrons first to form Fe²⁺, then remove one 3d electron to form Fe³⁺.

[Ar] 3d⁶ 4s¹

This represents loss of only one electron from neutral iron and would not correspond to Fe³⁺.

[Ar] 3d⁴ 4s¹

Although three electrons have been removed in total, this incorrectly leaves a 4s electron while removing electrons from 3d.

Which is the ground-state electronic configuration of a chromium atom?

[Ar] 3d⁵ 4s¹

Correct. Chromium is an exception to the simple filling prediction and has the observed configuration [Ar] 3d⁵ 4s¹.

[Ar] 3d⁴ 4s²

This is the simple configuration predicted before accounting for the chromium anomaly, but it is not the observed ground-state configuration.

[Ar] 3d⁶

This contains the correct number of electrons beyond [Ar], but it places all six into 3d and leaves 4s empty. That is not chromium’s ground state.

[Ar] 3d³ 4s³

An s subshell can hold a maximum of only two electrons, so 4s³ is impossible.

A Period 3 element shows a very large jump between its third and fourth successive ionisation energies. What is its most likely outer electronic configuration?

3s¹

A 3s¹ element has one outer-shell electron, so its major jump would occur after the first ionisation.

3s²

A 3s² element has two outer-shell electrons, so the major jump would occur after the second ionisation.

3s² 3p¹

Correct. Three electrons are removed before the large jump, so the element has three electrons in its outer third shell. In Period 3 this corresponds to 3s² 3p¹.

3s² 3p³

This configuration has five outer-shell electrons, so its first principal-shell jump would occur after five electrons had been removed.

Magnesium has a higher first ionisation energy than sodium. Which explanation best accounts for this?

Magnesium has fewer protons, so its outer electron is attracted more strongly

Magnesium has more protons than sodium, not fewer.

Magnesium has a greater nuclear charge while the electron removed is in the same principal shell and experiences similar inner-shell shielding

Correct. Both outer electrons are in the third principal shell, but magnesium has the greater nuclear charge. Its outer electron is therefore held more strongly.

Magnesium’s outer electron is in the second principal shell

Magnesium has the outer configuration 3s², so its outer electrons are in the third principal shell.

Sodium has no inner-shell electrons to provide shielding

Sodium has the [Ne] core and therefore has ten inner electrons that shield its 3s electron.

First ionisation energy generally decreases from lithium to sodium to potassium. Which change is mainly responsible?

Nuclear charge decreases down the group

Nuclear charge actually increases down the group because proton number increases.

The outer electron moves into a lower principal shell

The opposite occurs. The outer electron occupies progressively higher principal shells down the group.

Shielding decreases because there are fewer inner electrons

More inner shells are present down the group, so shielding increases.

Increased distance and shielding outweigh the increase in nuclear charge

Correct. The outer electron is farther from the nucleus and more strongly shielded, so it is held less strongly despite the larger nuclear charge.

Which statement correctly describes the relationship between the 4s and 3d subshells in the first-row transition elements?

4s is occupied before 3d when constructing the neutral atoms, but 4s electrons are removed before 3d electrons when cations form

Correct. The relative energies change once the 3d subshell is occupied, so the occupied 4s electrons are lost first during ion formation.

3d always fills before 4s and 3d electrons are always removed first

Potassium and calcium begin filling 4s before 3d, and transition-metal cations lose 4s electrons first.

4s is permanently lower in energy than 3d in every atom and ion

The relative energies are not fixed in this simple way. Their ordering changes as the electronic environment changes.

4s and 3d are the same subshell because their energies are similar

They are different subshells with different principal and angular quantum characteristics, even though their energies are close.

Which pair of species is isoelectronic?

Na and Na⁺

Neutral Na has 11 electrons, while Na⁺ has 10. They do not have the same electron count.

K⁺ and Ar

Correct. K has 19 electrons, so K⁺ has 18. Ar also has 18 electrons. Both therefore have the same electronic configuration.

Mg²⁺ and Ne⁻

Mg²⁺ has 10 electrons, while a hypothetical Ne⁻ species would have 11. They are not isoelectronic.

Cl and Cl⁻

Neutral Cl has 17 electrons and Cl⁻ has 18, so their electronic configurations are different.

A student says, “First ionisation energy increases across Period 3 because shielding decreases from sodium to argon.” What is the best correction?

The trend is caused only by atomic mass increasing

Atomic mass is not the main factor determining the electrostatic attraction between the nucleus and the electron being removed.

Shielding increases dramatically because each added electron creates a new shell

The additional electrons across Period 3 enter the same third principal shell. A new principal shell is not added at each step.

Inner-shell shielding remains approximately constant while nuclear charge increases, so effective nuclear attraction generally increases

Correct. This is the main reason for the general increase across the period. Small deviations arise from subshell energy and electron pairing.

Nuclear charge stays constant while electron number decreases

Proton number and therefore nuclear charge increase across the period, while the neutral atoms also contain increasing numbers of electrons.

Frequently asked questions

Questions students often ask about electronic structure

The hardest parts of this topic usually come from ideas that appear contradictory at first. These questions bring those ideas together.

Are electron shells, subshells and orbitals three different models?

They are different levels of detail within the same description. A principal shell contains subshells, and a subshell contains orbitals. Saying that an electron is in the third shell is therefore less specific than saying it is in a 3p orbital.

Does an orbital show where the electron definitely is?

No. An orbital describes a probability distribution. It identifies a region in which the electron has a high probability of being found rather than specifying one definite position or path.

Why are there three p orbitals but only one p subshell?

A subshell is a collection of orbitals of the same type. One p subshell contains three differently oriented orbitals, p_x, p_y and p_z. Together they can hold a maximum of six electrons.

Why can only two electrons occupy one orbital?

The Pauli exclusion principle limits an orbital to two electrons. When two electrons occupy the same orbital, they must have opposite spins.

Why does 4s fill before 3d if 4s electrons are removed first?

The relative energies of 4s and 3d are very close and depend on the electron configuration. Before 3d becomes occupied, 4s is the lower-energy available subshell and fills first. Once 3d electrons are present, the occupied 4s level lies higher in energy, so 4s electrons are removed first when transition-metal cations form.

Should iron be written as [Ar] 4s² 3d⁶ or [Ar] 3d⁶ 4s²?

Both orders may be encountered because they describe the same occupied subshells. Writing [Ar] 3d⁶ 4s² is common when discussing transition metals. The important point is that 4s filled before 3d during construction of the neutral atom and is removed before 3d when the ion forms.

Why do chromium and copper not follow the simple filling prediction?

The 3d and 4s subshells are very close in energy. For chromium and copper, moving one 4s electron into 3d gives a slightly lower-energy overall arrangement. The resulting configurations are [Ar] 3d⁵ 4s¹ for Cr and [Ar] 3d¹⁰ 4s¹ for Cu.

Is shielding the same as effective nuclear charge?

No. Shielding describes the reduction in nuclear attraction caused by other electrons, especially inner-shell electrons. Effective nuclear attraction is the resulting attraction experienced by the electron after shielding and electron–electron interactions are taken into account.

Why does a larger nuclear charge not always mean a larger ionisation energy?

Nuclear charge is only one factor. The electron may be farther from the nucleus, more strongly shielded or in a higher-energy subshell. For example, aluminium has a greater nuclear charge than magnesium, yet its outer 3p electron is easier to remove than magnesium’s 3s electron.

Why are successive ionisation energies useful for finding electron arrangements?

Electrons in the same outer principal shell are removed at broadly comparable energy scales. Once those electrons have been removed, taking an electron from a closer inner shell requires dramatically more energy. The positions of these large jumps therefore reveal how electrons are distributed among principal shells.

Does every jump in successive ionisation energy represent a new principal shell?

No. Successive values rise anyway because electrons are being removed from increasingly positive ions. Smaller changes can also arise when removal moves between subshells. A change to a new inner principal shell is identified by a much larger jump.

What is the most important idea to remember for ionisation-energy explanations?

Always compare how strongly the electron being removed is attracted to the nucleus. Then identify why that attraction differs: nuclear charge, distance, shielding, subshell energy or electron pairing. This is much stronger than trying to memorise each trend as an isolated fact.

The extranuclear structure of an atom gives one connected explanation for many apparently separate observations. Principal shells divide into subshells and orbitals; orbital energies determine how electrons are arranged; and nuclear attraction determines how difficult those electrons are to remove. Once you can move confidently between orbital structure → electronic configuration → nuclear attraction → ionisation energy, the trends and so-called exceptions become consequences of the same atomic model rather than facts that have to be memorised separately.