How to Find Unpaired Electrons in Atoms and Ions

In 1922, scientists fired a narrow beam of silver atoms through a magnetic field. Instead of spreading into a continuous streak, the atoms formed two distinct traces. Something about the atoms was behaving in a way the classical model couldn’t explain.

Today, we can connect that behaviour to electron spin. Silver has one unpaired electron, but how would you recognise an unpaired electron from a configuration such as \(3p^4\) or \(3d^6\)?

To find unpaired electrons from an electron configuration, draw the orbitals in each subshell, place one electron in every equal-energy orbital before pairing electrons, and count the orbitals containing just one electron.

For example, a \(p\) subshell contains three orbitals. If it holds four electrons, would you expect two paired orbitals and one empty orbital?

Not in the ground state. Three electrons occupy separate orbitals first. The fourth pairs with one of them, leaving two unpaired electrons.

That single rule handles most unpaired-electron questions. The complications come when you form ions, encounter transition metals, or confuse a subshell’s electron count with its unpaired-electron count.

01What an unpaired electron actually means

An orbital is a quantum state associated with a region where an electron is likely to be found.

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

We represent electrons using arrows:

• ↑ means one electron with one spin orientation.
• ↓ means an electron with the opposite spin orientation.
• [↑↓] represents two paired electrons.
• [↑] represents one unpaired electron.
• [ ] represents an empty orbital.

The arrows describe spin states, not electrons physically travelling upwards or downwards.

How many orbitals are in each subshell?

The key distinction is that a subshell containing six electrons does not necessarily contain six unpaired electrons.

For instance, \(p^6\) contains six electrons, but all three orbitals are fully paired. It has zero unpaired electrons.

Hund’s rule: occupy separately before pairing

Imagine three empty seats on a train. If three people can each take a separate seat, there’s no need for anyone to share. The fourth person must share one of the occupied seats.

That is a useful picture for Hund’s rule: in the ground state, electrons occupy orbitals of equal energy singly, with parallel spins, before pairing.

The analogy has limits. Electrons aren’t people choosing seats, and orbitals aren’t physical compartments. Hund’s rule describes the energetically favoured arrangement of electrons in equivalent orbitals.

Consider the \(p\) subshell:

Notice the pattern: the number of unpaired electrons increases until the subshell is half-filled, then decreases as electrons begin pairing.

The same reasoning works for \(d\) subshells, except there are five orbitals rather than three.

A useful shortcut follows:

For a subshell with \(g\) orbitals and \(k\) electrons, the number of unpaired electrons, \(u\), is

\[
u=
\begin{cases}
k, & k\leq g\\
2g-k, & g<k\leq 2g
\end{cases}
\]

Here, \(g\) is the number of orbitals, \(k\) is the number of electrons in that subshell, and \(u\) is the number of unpaired electrons.

For \(3d^6\), there are five orbitals and six electrons:

\[
u=2(5)-6=4
\]

So \(3d^6\) contains four unpaired electrons.

This shortcut assumes the usual ground-state filling of equivalent orbitals. Drawing the boxes is still the safer method when you’re learning.

02How an experiment revealed that atomic magnetism was quantised

The connection between electron spin and magnetism helps explain why unpaired electrons matter.

Otto Stern and Walther Gerlach wanted to investigate whether atomic magnetic moments could take any orientation in a magnetic field, as classical physics suggested, or whether only particular orientations were allowed.

The apparatus: from silver source to detector

The 1922 experiment used a beam of neutral silver atoms.

Follow the apparatus from left to right:

1. A heated oven produced silver vapour.
2. Small apertures allowed atoms travelling in a narrow range of directions to form a beam.
3. The beam travelled through a vacuum and between specially shaped magnetic pole pieces.
4. The pole pieces produced a non-uniform magnetic field, meaning its strength varied across the beam.
5. A detector collected the silver atoms after they passed through the field.

The non-uniform field was essential. A magnetic moment in a uniform field can experience a turning effect, but it does not experience the net translational force needed to separate an ideal atomic beam.

What should the original model predict?

In a classical picture, the atoms’ magnetic moments could point at a continuous range of angles.

Different orientations would experience different forces in the non-uniform field. A continuous distribution of orientations should therefore produce a continuous spread of deflections.

If the orientations were quantised, however, only certain deflections would occur.

The observed beam separated into two distinct components.

That supported the idea that atomic magnetic behaviour could have discrete outcomes rather than a continuous range.

Why the first apparatus needed improvement

The experiment was difficult because the atomic beam was extremely weak, and the separation was tiny.

An earlier arrangement used a small circular aperture. It restricted the beam, but too few atoms reached the detector, making the splitting difficult to resolve.

The researchers revised the apparatus, replacing the circular opening with a narrow rectangular slit. Its narrow dimension was approximately \(30\,\mu\mathrm{m}\), while its longer dimension allowed more atoms through.

This improved the beam intensity without sacrificing the required resolution across the magnetic-field gradient.

The historical review of the Stern-Gerlach experiment describes how this slit modification helped produce the decisive observation in February 1922.

A scale estimate: how small was the splitting?

Historical accounts report a separation of roughly \(60\)-\(100\,\mu\mathrm{m}\) between the two components.

Take \(80\,\mu\mathrm{m}\) as a representative value.

The narrow dimension of the revised slit was approximately \(30\,\mu\mathrm{m}\).

Comparing the two distances:

\[
\begin{aligned}
\text{Ratio}
&=\frac{80\,\mu\mathrm{m}}{30\,\mu\mathrm{m}}\\
&\approx 2.7
\end{aligned}
\]

The separation was roughly 2.7 times the slit’s narrow width.

A micrometre, written \(\mu\mathrm{m}\), is one-millionth of a metre. The ratio has no units because both measurements are lengths.

This is a scale comparison, not a calculation of the detector’s resolving power. The deposited traces could be wider than the original slit because atoms followed slightly different trajectories and speeds.

It does show why controlling the beam’s width was so important.

What we understand now that the original investigators didn’t

In 1922, electron spin had not yet been proposed. Stern and Gerlach interpreted their result using the atomic models available at the time.

Modern quantum mechanics provides a different explanation.

A neutral silver atom has the ground-state electron configuration

\[
[\mathrm{Kr}]\,4d^{10}\,5s^1
\]

The notation \([\mathrm{Kr}]\) represents the 36 inner electrons with krypton’s configuration.

The \(4d^{10}\) subshell is completely filled, with all electrons paired. The \(5s^1\) subshell contains one unpaired electron.

For the electronic magnetic moment of ground-state silver, that unpaired electron is crucial. Its spin has two possible projections along the measurement direction, which explains the two beam components in the modern interpretation.

Silver, atomic number 47, is a historical example rather than one of the first 30 elements used in the transition-metal calculations below.

There is also an important limitation: two beam components do not mean two unpaired electrons. The number of possible magnetic outcomes is not generally equal to the number of unpaired electrons.

The experiment provides evidence about quantised magnetic behaviour. Orbital diagrams provide a simpler chemical method for counting unpaired electrons in ground-state atoms and ions.

03Find the unpaired electrons in an atom

For an ordinary electron-configuration question, use the following sequence:

1. Determine the total number of electrons.
2. Write the ground-state electron configuration.
3. Identify the partially occupied subshells.
4. Draw one box for each orbital in those subshells.
5. Apply Hund’s rule and count singly occupied orbitals.

Filled inner subshells contribute zero unpaired electrons. You can usually ignore them once you’ve confirmed that they’re full.

If writing the electron configuration itself is the difficult part, spdf notation and valence electrons covers the filling sequence and notation in more detail.

04Finding unpaired electrons in ions

The method changes slightly for ions because you must first work out which electrons have been gained or lost.

For an ion with charge \(+q\), subtract \(q\) electrons from the neutral atom. For an ion with charge \(-q\), add \(q\) electrons, where \(q\) is the positive magnitude of the charge.

The charge changes the total electron count. You then determine the correct ground-state configuration before applying Hund’s rule.

05Transition-metal ions: why you remove 4s before 3d

Transition metals introduce a particularly common mistake.

For neutral atoms in Period 4, the \(4s\) subshell generally fills before \(3d\). However, when forming the common positive transition-metal ions, the \(4s\) electrons are removed before the \(3d\) electrons.

This is not a contradiction. The relative energies of these subshells change with electron occupancy and the charge on the atom.

The usual HSC approach is to begin with the correct neutral configuration, remove the outer \(4s\) electrons first, and then remove \(3d\) electrons if necessary.

06Chromium and copper: two configurations you must treat carefully

Most neutral Period 4 transition metals follow the familiar filling pattern, but chromium and copper are important exceptions.

Chromium has atomic number 24. Its actual ground-state configuration is

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

It is not \([\mathrm{Ar}]\,3d^4\,4s^2\).

Copper has atomic number 29:

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

It is not \([\mathrm{Ar}]\,3d^9\,4s^2\).

A half-filled \(d^5\) subshell and a filled \(d^{10}\) subshell are particularly important arrangements, but the complete explanation involves the small energy differences and electron interactions in these atoms. It is not a universal rule that an electron always moves to create a half-filled subshell.

These exceptions affect the unpaired-electron counts.

Consider the following examples, all using isolated ground-state species.

These configurations are consistent with the ground-state atomic data published by the National Institute of Standards and Technology.

Copper is especially instructive.

The neutral atom has one unpaired \(4s\) electron. Removing that electron produces \(\mathrm{Cu}^{+}\), which has no unpaired electrons.

Removing another electron gives \(\mathrm{Cu}^{2+}\), with a \(3d^9\) configuration and one unpaired electron.

The count changes from 1 to 0 to 1 as the charge changes from neutral to \(+1\) to \(+2\).

There is no general rule that unpaired-electron count must increase or decrease steadily with ionic charge.

When the orbital-box method has limits

These calculations use ground-state configurations for isolated atoms or ions.

That assumption matters in three situations:

• Excited states: An electron may occupy a higher-energy orbital, changing the unpaired-electron count. Use the specified excited-state configuration rather than automatically returning to the ground state.
• Transition-metal complexes: Surrounding ligands can split the energies of the \(d\) orbitals. Different electron-pairing arrangements may then become favourable. Predicting those arrangements requires information beyond the simple free-ion configuration.
• Bulk magnetic behaviour: Unpaired electrons help explain magnetic properties, but their number alone does not predict exactly how a solid, solution, or atomic beam will behave. Interactions between magnetic moments and the measurement conditions also matter.

For the following questions, treat all stated atoms and ions as isolated, ground-state species unless the question specifies otherwise. Any beam experiments in the questions are illustrative thought experiments.

07Questions and solutions

Question 1

A neutral chlorine atom has atomic number 17.

Determine its ground-state electron configuration and the number of unpaired electrons. Then determine how many unpaired electrons remain when chlorine gains one electron to form \(\mathrm{Cl}^{-}\).

Solution 1

Neutral chlorine has 1 unpaired electron, while chloride has 0 unpaired electrons.

A neutral chlorine atom contains 17 electrons:

\[
\ce{Cl}:\ [\mathrm{Ne}]\,3s^2\,3p^5
\]

The \(3s^2\) electrons are paired.

The \(3p^5\) orbitals have the arrangement

[↑↓] [↑↓] [↑]

One orbital is singly occupied, so chlorine has one unpaired electron.

When chlorine gains an electron, its electron count becomes

\[
17+1=18\text{ electrons}
\]

The chloride configuration is

\[
\mathrm{Cl}^{-}:\ [\mathrm{Ne}]\,3s^2\,3p^6
\]

Its three \(3p\) orbitals are all paired:

[↑↓] [↑↓] [↑↓]

Therefore chloride has zero unpaired electrons.

The tempting mistake is to treat the five electrons in \(3p^5\) as five unpaired electrons. Only one of its three orbitals is singly occupied.

Question 2

Chromium has atomic number 24 and manganese has atomic number 25.

A student claims that neutral chromium and neutral manganese must have the same number of unpaired electrons because both have five electrons in their \(3d\) subshells.

Evaluate the claim by determining:

• The ground-state configurations and unpaired-electron counts of neutral chromium and manganese.
• The configurations and unpaired-electron counts of \(\mathrm{Cr}^{+}\) and \(\mathrm{Mn}^{2+}\).
• Why the two neutral atoms give different results, even though their specified ions give the same result.

Solution 2

The student is incorrect about the neutral atoms. Chromium has 6 unpaired electrons and manganese has 5, but \(\mathrm{Cr}^{+}\) and \(\mathrm{Mn}^{2+}\) each have 5 unpaired electrons.

Chromium’s ground-state configuration is

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

The five \(3d\) orbitals each contain one electron, giving five unpaired \(d\) electrons.

The \(4s^1\) electron is also unpaired.

Therefore:

\[
u(\ce{Cr})=5+1=6
\]

Here, \(u\) denotes the number of unpaired electrons.

Manganese has the configuration

\[
\ce{Mn}:\ [\mathrm{Ar}]\,3d^5\,4s^2
\]

It also has five singly occupied \(3d\) orbitals, but its two \(4s\) electrons are paired.

Thus:

\[
u(\ce{Mn})=5+0=5
\]

Now consider the ions.

Chromium loses its single \(4s\) electron when forming \(\mathrm{Cr}^{+}\):

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

This leaves five unpaired electrons.

Manganese loses both \(4s\) electrons when forming \(\mathrm{Mn}^{2+}\):

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

It also has five unpaired electrons.

The ions each have 23 electrons:

\[
\begin{aligned}
\mathrm{Cr}^{+}:&\ 24-1=23\\
\mathrm{Mn}^{2+}:&\ 25-2=23
\end{aligned}
\]

They have the same ground-state configuration in this case.

The mistake was looking only at the \(3d\) electrons of the neutral atoms and ignoring the extra unpaired \(4s\) electron in chromium.

Question 3

A student examines a simplified Stern-Gerlach experiment using neutral silver atoms.

With the non-uniform magnetic field switched off, the detector shows one central deposit. When the field is switched on, two separated deposits appear.

The student makes three claims:

1. Two deposits prove that every silver atom contains two unpaired electrons.
2. Replacing the non-uniform field with an equally strong uniform field should produce the same separation.
3. If the original narrow-aperture apparatus failed to resolve two deposits, that would prove that the atoms did not have quantised magnetic behaviour.

Evaluate all three claims. Explain what the observation establishes and why the apparatus design matters.

Use the ground-state silver configuration \([\mathrm{Kr}]\,4d^{10}\,5s^1\).

Solution 3

All three claims are incorrect. The two deposits demonstrate discrete magnetic deflections under the experimental conditions, not two unpaired electrons. A uniform magnetic field would not separate an ideal beam by position, and an unresolved result from the first apparatus would not disprove quantisation.

First, silver’s configuration is

\[
[\mathrm{Kr}]\,4d^{10}\,5s^1
\]

All five \(4d\) orbitals contain paired electrons.

The \(5s^1\) orbital contains one unpaired electron.

Therefore silver has one unpaired electron, not two.

In the modern explanation, the two traces correspond to the two possible projections of the atom’s electron-spin magnetic moment along the measurement direction.

The first claim confuses the number of observed beam components with the number of unpaired electrons.

Second, a uniform magnetic field does not provide the field gradient needed for spatial separation.

A magnetic moment can experience a torque in a uniform field. However, without a magnetic-field gradient, it experiences no net translational force of the type required to separate the beam.

Assuming no other forces or apparatus effects, the beam would therefore not split into two spatially separated components.

The second claim confuses the strength of a magnetic field with the variation in field strength across space.

Third, the detector must be able to resolve the separation.

Suppose two beam components are separated by a very small distance, but the beam itself is too broad, or too few atoms reach the detector. Their deposits may overlap or remain too faint to distinguish.

That would limit the experiment’s ability to reveal splitting. It would not establish that the magnetic orientations were continuous.

The improved slit increased the useful beam intensity while maintaining a narrow spread across the field gradient.

The experiment therefore illustrates two separate principles: the atomic magnetic response has quantised outcomes, and the apparatus must have sufficient resolution to distinguish them.

An experiment can fail to detect a predicted effect without that failure disproving the underlying model.

Question 4

Consider the following hypothetical one-electron-transfer process:

\[
\ce{Cr^2+ + Cu^2+ -> Cr^3+ + Cu+}
\]

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

Assume all four species have their isolated-ion ground-state configurations.

Determine:

• The number of electrons and unpaired electrons in each ion before and after the transfer.
• The total number of unpaired electrons before and after the process.
• Why the change in unpaired-electron count is not equal to the number of electrons transferred.
• Whether the same calculation would necessarily predict the measured magnetic behaviour of an aqueous solution containing these ions.

Solution 4

One electron is transferred, but the total number of unpaired electrons decreases from 5 to 3. This does not violate electron conservation because the number of unpaired electrons is not a conserved quantity.

Begin with chromium.

Neutral chromium has the configuration

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

Removing the \(4s\) electron and one \(3d\) electron gives

\[
\mathrm{Cr}^{2+}:\ [\mathrm{Ar}]\,3d^4
\]

Its electron count is

\[
24-2=22\text{ electrons}
\]

Four \(3d\) orbitals are singly occupied, giving four unpaired electrons.

Chromium loses another \(3d\) electron to form

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

Its electron count is

\[
24-3=21\text{ electrons}
\]

The \(3d^3\) configuration has three unpaired electrons.

Now consider copper.

Neutral copper has the configuration

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

The copper(II) ion has

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

It contains

\[
29-2=27\text{ electrons}
\]

A \(3d^9\) configuration has one unpaired electron.

After gaining an electron, the ion becomes

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

Its electron count is

\[
29-1=28\text{ electrons}
\]

All five \(3d\) orbitals are paired, so it has zero unpaired electrons.

Summarising:

Before the transfer:

\[
u_{\text{before}}=4+1=5
\]

After the transfer:

\[
u_{\text{after}}=3+0=3
\]

The change is

\[
\Delta u=3-5=-2
\]

The negative sign indicates a decrease of two unpaired electrons.

The total electron count is conserved:

\[
\begin{aligned}
N_{\text{before}}&=22+27=49\\
N_{\text{after}}&=21+28=49
\end{aligned}
\]

Chromium loses one electron, while copper gains one electron.

However, chromium loses an unpaired \(3d\) electron, decreasing its unpaired count by one. Copper gains an electron that completes an existing \(3d\) pair, decreasing its unpaired count by another one.

The combined effect is a reduction of two unpaired electrons from a transfer involving just one electron.

The tempting assumption is that each transferred electron must change the total number of unpaired electrons by exactly one. That would only follow if pairing arrangements were irrelevant, which they aren’t.

Finally, the calculation does not uniquely predict the magnetic response of an aqueous solution.

Ions in solution interact with surrounding water molecules and other ligands. These interactions can affect \(d\)-orbital energies and, in some systems, electron pairing. Magnetic measurements also depend on the conditions and interactions between magnetic moments.

The free-ion calculation establishes the predicted orbital occupancies under the stated assumptions. Describing a particular solution would require information about the chemical environment.

Question 5

Two separate atomic-beam systems contain isolated, ground-state transition-metal ions.

Beam A contains \(\mathrm{Mn}^{2+}\), with atomic number 25.

Beam B contains \(\mathrm{Fe}^{3+}\), with atomic number 26.

Both ions initially have 23 electrons and the configuration

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

In a hypothetical controlled process, each ion receives exactly one electron and relaxes into the ground state of its new charge state.

A student makes the following argument:

“The starting ions have identical electron configurations. Each gains one electron, so the resulting ions must also have identical electron configurations and identical numbers of unpaired electrons.”

Evaluate this argument.

Determine the configurations and unpaired-electron counts before and after electron capture. Explain why the added electrons do not necessarily occupy the same subshell, and assess whether the total number of unpaired electrons across both beams is conserved.

You may use these neutral-atom configurations:

\[
\begin{aligned}
\ce{Mn}:&\ [\mathrm{Ar}]\,3d^5\,4s^2\\
\ce{Fe}:&\ [\mathrm{Ar}]\,3d^6\,4s^2
\end{aligned}
\]

Solution 5

The student’s argument is incorrect. Both ions begin with five unpaired electrons, but after gaining one electron, \(\mathrm{Mn}^{+}\) has six unpaired electrons while \(\mathrm{Fe}^{2+}\) has four.

Interestingly, the combined number remains ten. That equality is a consequence of these particular configurations, not a general conservation law.

Start with the initial species.

Manganese(II) has

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

Iron(III) has

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

Each contains 23 electrons:

\[
\begin{aligned}
\mathrm{Mn}^{2+}:&\ 25-2=23\\
\mathrm{Fe}^{3+}:&\ 26-3=23
\end{aligned}
\]

Both have five singly occupied \(d\) orbitals:

3d: [↑] [↑] [↑] [↑] [↑]

Thus, each starts with five unpaired electrons.

Now consider the electron added to manganese(II).

Neutral manganese has

\[
[\mathrm{Ar}]\,3d^5\,4s^2
\]

Removing one \(4s\) electron gives the ground-state manganese(I) ion:

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

We can reverse that relationship when adding an electron to \(\mathrm{Mn}^{2+}\).

The five \(3d\) electrons remain singly occupied, and the new \(4s\) electron is also unpaired.

Therefore:

\[
u(\mathrm{Mn}^{+})=5+1=6
\]

The resulting ion has 24 electrons:

\[
25-1=24
\]

Now consider the electron added to iron(III).

Neutral iron has

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

Removing its two \(4s\) electrons produces the ground-state iron(II) ion:

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

Therefore, adding an electron to \(\mathrm{Fe}^{3+}\) to form ground-state \(\mathrm{Fe}^{2+}\) produces a \(3d^6\) configuration.

The \(3d\) orbitals become

[↑↓] [↑] [↑] [↑] [↑]

The extra electron has completed a pair rather than occupying a previously empty \(4s\) orbital.

Thus:

\[
u(\mathrm{Fe}^{2+})=4
\]

Iron(II) also has 24 electrons:

\[
26-2=24
\]

The full comparison is:

The key insight is that having the same total number of electrons does not guarantee identical ground-state configurations for different transition-metal ions.

The nucleus of manganese contains 25 protons; iron contains 26. Their different nuclear charges, combined with electron-electron interactions, affect the relative energies of the \(3d\) and \(4s\) arrangements.

The added electron therefore occupies different subshells in the two resulting ground-state ions.

This is where the student’s reasoning fails. They treat electron count as though it uniquely determines the lowest-energy orbital arrangement regardless of nuclear charge.

Finally, compare the total number of unpaired electrons.

Initially:

\[
u_{\text{initial}}=5+5=10
\]

After electron capture:

\[
u_{\text{final}}=6+4=10
\]

The total is unchanged.

But that does not establish conservation of unpaired-electron number.

In Beam A, gaining an electron increases the number of unpaired electrons by one. In Beam B, gaining an electron decreases it by one.

Those changes happen to cancel.

Question 4 provides a counterexample to any proposed conservation law: its one-electron transfer reduced the total number of unpaired electrons from five to three.

The deeper lesson is that three quantities must be kept separate:

• Total number of electrons, determined by atomic number and charge.
• Ground-state electron configuration, determined by the available quantum states and their energies.
• Number of unpaired electrons, determined by how those electrons occupy the orbitals.

Knowing one does not always determine the other two.

That is why the orbital-box method is more reliable than guessing from electron count, charge, or a familiar-looking configuration.

Once you can determine the unpaired-electron count, the next useful question is what that configuration predicts about the species’ chemical behaviour. For atomic and ionic size, the important factors include occupied shells, nuclear attraction, and electron shielding. These are explored further in atomic radius trends and electron configuration. For magnetic behaviour, the next step is to examine how the electron arrangement changes when the atom or ion interacts with its surroundings.