Why First Ionisation Energy Generally Increases Across a Period

Moving from lithium to neon, the first ionisation energy increases from approximately \(520\) to \(2081\ \mathrm{kJ\,mol^{-1}}\). Yet each step across Period 2 adds both a proton and an electron. If the extra electrons shielded the extra protons perfectly, the attraction on the outer electrons might stay roughly constant.

It doesn’t. First ionisation energy generally increases across a period because nuclear charge increases while shielding by inner-shell electrons remains relatively constant. The additional electrons enter the same principal energy level and do not shield one another as effectively as inner-shell electrons do. The effective nuclear attraction therefore increases, making the outer electron harder to remove.

There is one catch: the increase isn’t perfectly smooth. Two particular changes in electron arrangement produce small decreases. Understanding those exceptions helps explain why the overall trend occurs in the first place.

01Start with the experimental pattern

Before explaining the trend, consider a comparison.

Which atom should require more energy to remove its outermost electron: carbon or fluorine?

Both are in Period 2, and both lose an electron from the second principal energy level. However, fluorine has nine protons, while carbon has six.

You should predict that fluorine holds its outer electrons more strongly. The measured values confirm this.

Data: NIST Atomic Properties of the Elements. Values are rounded to the nearest \(\mathrm{kJ\,mol^{-1}}\).

The overall increase is obvious, but look more carefully at the numbers.

• From beryllium to boron, first ionisation energy decreases by approximately \(98\ \mathrm{kJ\,mol^{-1}}\).
• From nitrogen to oxygen, it decreases by approximately \(88\ \mathrm{kJ\,mol^{-1}}\).

These aren’t random errors. Both decreases have a physical explanation involving electron arrangement.

Before investigating them, we need to understand how ionisation energies are measured and what the numbers actually represent.

02How is first ionisation energy measured?

First ionisation energy describes the energy required to remove one electron from each atom in one mole of isolated gaseous atoms.

The process is represented by:

\[
\ce{X(g) -> X+(g) + e-}
\]

Here:

• \(\ce{X(g)}\) is a neutral gaseous atom.
• \(\ce{X+(g)}\) is the resulting positive ion.
• \(\ce{e-}\) is the removed electron.

The process requires energy, so first ionisation energy is positive. Its usual unit is \(\mathrm{kJ\,mol^{-1}}\).

The gas-state requirement matters. We are measuring the energy needed to remove an electron from an isolated atom, not the energy needed to break a metallic bond or separate an atom from a compound.

From threshold measurements to photoelectron spectroscopy

One experimental approach uses a beam of electrons to ionise gaseous atoms.

Picture an electron gun aimed at a chamber containing a dilute gas of neutral atoms. The electron gun produces a beam whose energy can be varied. As the beam passes through the chamber, some electrons collide with atoms and transfer enough energy to remove an electron.

The resulting positive ions travel towards a detector, which records how many ions are produced.

The experiment follows a straightforward prediction:

• Below the required energy, the beam cannot ionise the atoms.
• Once the beam reaches the ionisation threshold, positive ions begin to appear.
• Increasing the beam energy allows more ionisation events to occur.

By varying the beam energy and monitoring the ion signal, the minimum energy needed for ionisation can be determined.

There was an important measurement problem with early versions of this approach. Electron beams had a spread of energies, meaning not every electron had the same energy even when the apparatus was set to a particular value.

This made the exact ionisation threshold harder to identify.

Electron monochromators were developed to narrow the range of electron energies. Another approach used photons with precisely defined energies instead of an electron beam.

This led to an important experimental technique called photoelectron spectroscopy (PES).

How photoelectron spectroscopy works

In photoelectron spectroscopy, a beam of light with a known photon energy is directed at gaseous atoms.

When an atom absorbs a photon, it can eject an electron. The emitted electron travels into an energy analyser, which measures its kinetic energy. A detector then records the number of electrons detected at different energies.

The apparatus follows this sequence:

1. A sample of gaseous atoms enters a vacuum chamber.
2. A beam of monochromatic light (light with a defined photon energy) passes through the sample.
3. Some atoms absorb photons and eject electrons.
4. The electrons enter an energy analyser.
5. A detector records their kinetic energies.

The important feature is that we know how much energy the incoming photon carries and can measure how much kinetic energy the electron has after escaping.

For a simplified measurement in which the resulting ion is left in its ground electronic state, energy conservation gives:

\[
E_{\gamma}=IE_1+KE_e
\]

where:

• \(E_{\gamma}\) is the energy of the incoming photon, measured in eV.
• \(IE_1\) is the first ionisation energy per atom, measured in eV.
• \(KE_e\) is the kinetic energy of the emitted electron, measured in eV.

Rearranging:

\[
IE_1=E_{\gamma}-KE_e
\]

Some measurements also produce ions in excited electronic states. In those cases, some photon energy is stored in the excited ion, so the more complete energy balance must account for that additional energy.

For our introductory calculations, we assume the ion is produced in its ground state.

This method gives us a useful way to understand the measurements.

If two atoms are exposed to photons of identical energy, the atom with the more tightly bound outer electron produces a slower emitted electron. More of the photon’s energy has been used to overcome the attraction holding the electron in the atom.

03Why nuclear charge wins across a period

Now we can explain the main trend.

Moving from left to right across a period, the atomic number increases by one at each step. This means every successive element has one additional proton in its nucleus.

Protons are positively charged, and electrons are negatively charged. The electrostatic attraction between them is what holds electrons within an atom.

More protons generally mean a stronger attraction between the nucleus and its electrons.

However, there is another factor.

As we move across a period, electrons are also being added. These electrons repel one another and partially shield one another from the nucleus.

This is called electron shielding.

What actually happens to shielding?

Consider the electron configurations of lithium, carbon, and fluorine.

All three elements have the same two inner-shell electrons in the \(1s\) subshell.

As we move across Period 2:

• The number of protons increases.
• Additional electrons enter the second principal energy level.
• The number of inner-shell electrons remains two.
• Shielding increases somewhat because of the additional electrons, but it does not increase enough to cancel the extra nuclear charge.

The key distinction is that electrons in inner shells shield outer electrons much more effectively than electrons occupying the same principal energy level.

You can picture inner-shell electrons as a partial screen between the nucleus and the outer electron. Adding more electrons to the outer shell does create additional repulsion, but it does not provide an equally effective screen against the nucleus.

The analogy has a limit: electrons aren’t physical barriers. Shielding results from electrostatic interactions between the electrons and the nucleus.

Effective nuclear charge

Chemists use the term effective nuclear charge to describe the net positive attraction experienced by an electron after shielding is considered.

It is often represented by:

\[
Z_{\mathrm{eff}}
\]

where \(Z_{\mathrm{eff}}\) is the effective nuclear charge experienced by the electron.

A useful qualitative relationship is:

\[
Z_{\mathrm{eff}}\approx Z-S
\]

where:

• \(Z\) is the atomic number, or number of protons.
• \(S\) represents the shielding effect of the other electrons.

This is a simplified model, not an exact calculation obtained by subtracting the total number of electrons from the proton number. Different electrons shield one another by different amounts.

Across a period, the effective nuclear charge experienced by the outer electrons generally increases.

What does that actually do?

The stronger nuclear attraction pulls the electron cloud closer to the nucleus. The outer electrons become more tightly bound, so more energy is required to remove one.

This explains the general increase in first ionisation energy.

04Why the increase isn’t perfectly smooth

If nuclear charge increases across a period, why does the first ionisation energy sometimes decrease?

The answer lies in the arrangement of electrons within subshells.

Two changes are particularly important:

1. Moving from an \(s\) subshell to a \(p\) subshell.
2. Introducing electron pairing within a \(p\) subshell.

These changes explain the two dips in the Period 2 data.

Exception 1: beryllium to boron

Beryllium and boron have the following electron configurations:

\[
\mathrm{Be}:1s^2\,2s^2
\]

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

Boron has one more proton than beryllium. If nuclear charge were the only factor, we would expect boron to have the higher first ionisation energy.

But compare the electrons being removed.

• Beryllium loses an electron from its \(2s\) subshell.
• Boron loses an electron from its \(2p\) subshell.

The \(2p\) electron is higher in energy than a \(2s\) electron. A \(2s\) orbital also penetrates closer to the nucleus than a \(2p\) orbital, meaning the \(2p\) electron experiences less effective nuclear attraction.

As a result, boron’s outermost electron is easier to remove.

This effect outweighs the increase in nuclear charge.

The measured values are:

\[
\begin{aligned}
IE_1(\mathrm{Be})&=899.5\\
IE_1(\mathrm{B})&=800.6
\end{aligned}
\]

Both values are in \(\mathrm{kJ\,mol^{-1}}\).

Boron therefore has a lower first ionisation energy than beryllium.

The important lesson is that moving across a period doesn’t always mean removing an electron from the same type of orbital. When the electron being removed changes from \(s\) to \(p\), the energy required can decrease.

Exception 2: nitrogen to oxygen

The second exception is slightly different.

Consider the electron configurations:

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

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

Both elements lose an electron from the \(2p\) subshell.

So why does oxygen have a lower first ionisation energy than nitrogen?

The answer involves how electrons occupy the three \(2p\) orbitals.

These orbitals have the same energy in an isolated atom. According to Hund’s rule, electrons occupy separate orbitals before pairing up.

Nitrogen has three \(2p\) electrons, so each occupies a separate orbital.

Oxygen has four \(2p\) electrons, meaning one orbital must contain a pair.

Electrons have the same negative charge, so they repel one another.

The two electrons occupying the same \(2p\) orbital in oxygen experience additional repulsion. This makes one of those paired electrons easier to remove.

Removing that electron also leaves an oxygen ion with the configuration:

\[
\mathrm{O^+}:1s^2\,2s^2\,2p^3
\]

This is a relatively stable half-filled \(2p\) arrangement.

The combined effect of electron pairing and the resulting electron arrangement makes oxygen’s first ionisation energy slightly lower than nitrogen’s.

The measured values confirm this:

\[
\begin{aligned}
IE_1(\mathrm{N})&=1402.3\\
IE_1(\mathrm{O})&=1313.9
\end{aligned}
\]

Both values are in \(\mathrm{kJ\,mol^{-1}}\).

A useful comparison of the two exceptions

The same two patterns appear in Period 3:

• Magnesium to aluminium: the electron being removed changes from \(3s\) to \(3p\).
• Phosphorus to sulfur: pairing begins in the \(3p\) subshell.

These repeated patterns show that the exceptions are systematic consequences of electron arrangement, not random irregularities in the measurements.

05A reliable way to explain the trend in an HSC response

When asked why first ionisation energy generally increases across a period, don’t just write that nuclear charge increases.

You need to explain what happens to the attraction experienced by the outer electron.

A strong explanation follows this sequence:

1. Nuclear charge increases because the number of protons increases.
2. Additional electrons enter the same principal energy level.
3. The number of inner-shell electrons remains constant, so inner-shell shielding changes very little.
4. Electrons in the same principal energy level shield one another less effectively than inner-shell electrons.
5. Effective nuclear charge therefore increases.
6. The outer electron experiences a stronger attraction to the nucleus, making it harder to remove.
7. Consequently, first ionisation energy generally increases.

For a comparison involving a local exception, check the electron configuration before applying this explanation.

In particular, identify whether the electron being removed:

• Changes from an \(s\) subshell to a \(p\) subshell.
• Becomes paired with another electron in the same orbital.

Those details can explain why a particular element breaks the general trend.

For a broader treatment of periodic trends, including changes down a group, see First Ionisation Energy Trends for HSC Chemistry.

06Questions and solutions

Question 1

Carbon and fluorine are both in Period 2.

Which element has the higher first ionisation energy? Explain your answer using nuclear charge, shielding, and electron arrangement.

Solution 1

Fluorine has the higher first ionisation energy.

Carbon has six protons, while fluorine has nine. Both elements have the same two inner-shell electrons, and their outermost electrons occupy the second principal energy level.

Moving from carbon to fluorine increases nuclear charge without introducing an additional inner shell.

The extra outer-shell electrons provide some additional shielding, but this does not cancel the increase in nuclear charge.

Fluorine therefore experiences a greater effective nuclear attraction. Its outer electrons are held more tightly, so more energy is required to remove one.

The key point is that fluorine has a greater effective nuclear charge, not simply a greater number of protons.

Question 2

Consider the following first ionisation energy measurements:

(a) Calculate the percentage decrease in first ionisation energy from beryllium to boron.

(b) Calculate the percentage decrease from nitrogen to oxygen.

(c) Explain why both decreases occur, despite the increase in nuclear charge across each comparison.

Solution 2

The first ionisation energy decreases by approximately \(11.0\%\) from Be to B and \(6.3\%\) from N to O. Both decreases result from changes in electron arrangement that outweigh the additional nuclear attraction.

(a) Beryllium to boron

The percentage decrease is calculated using:

\[
\%\text{ decrease}=
\frac{\text{initial}-\text{final}}
{\text{initial}}\times100
\]

Substituting:

\[
\begin{aligned}
\%\text{ decrease}
&=\frac{899.5-800.6}{899.5}\times100\\
&=\frac{98.9}{899.5}\times100\\
&=11.0\%
\end{aligned}
\]

(b) Nitrogen to oxygen

\[
\begin{aligned}
\%\text{ decrease}
&=\frac{1402.3-1313.9}{1402.3}\times100\\
&=\frac{88.4}{1402.3}\times100\\
&=6.30\%
\end{aligned}
\]

(c) Explain the two decreases.

For beryllium and boron, the electron removed changes from the \(2s\) subshell to the \(2p\) subshell.

The \(2p\) electron in boron is higher in energy and less strongly attracted to the nucleus than the \(2s\) electron in beryllium. This makes boron’s outer electron easier to remove, despite its greater nuclear charge.

For nitrogen and oxygen, both electrons are removed from the \(2p\) subshell.

However, nitrogen has three unpaired \(2p\) electrons, while oxygen has one pair of electrons occupying the same orbital. The additional electron-electron repulsion in oxygen makes one of its paired electrons easier to remove.

Both decreases are therefore caused by electron arrangement, rather than a decrease in nuclear charge.

Question 3

Two gaseous elements, A and B, are investigated using photoelectron spectroscopy.

The following measurements are obtained:

Assume the resulting ions are in their ground electronic states.

(a) Calculate the first ionisation energy of each element in eV.

(b) Convert both values into \(\mathrm{kJ\,mol^{-1}}\).

(c) A student claims that element B must have the lower first ionisation energy because its emitted electrons have greater kinetic energy. Explain why this conclusion is incorrect.

Solution 3

Element B has the higher first ionisation energy, approximately \(1679\ \mathrm{kJ\,mol^{-1}}\), compared with approximately \(801\ \mathrm{kJ\,mol^{-1}}\) for element A. The student has incorrectly compared electron kinetic energies without accounting for the different photon energies.

(a) Calculate the first ionisation energies.

Using:

\[
IE_1=E_{\gamma}-KE_e
\]

For element A:

\[
\begin{aligned}
IE_1(A)&=18.0-9.7\\
&=8.3\ \mathrm{eV}
\end{aligned}
\]

For element B:

\[
\begin{aligned}
IE_1(B)&=30.0-12.6\\
&=17.4\ \mathrm{eV}
\end{aligned}
\]

(b) Convert into \(\mathrm{kJ\,mol^{-1}}\).

For element A:

\[
\begin{aligned}
IE_1(A)&=8.3\times96.485\\
&=800.8\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

For element B:

\[
\begin{aligned}
IE_1(B)&=17.4\times96.485\\
&=1678.8\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

(c) Explain the student’s mistake.

The student’s mistake is comparing the kinetic energies without considering the incoming photon energies.

Element B’s photons carry considerably more energy than element A’s photons.

Although B’s emitted electrons have greater kinetic energy, a larger amount of the incoming photon energy was required to remove them from their atoms.

Therefore, B has the higher first ionisation energy.

When interpreting photoelectron measurements, always calculate the energy required for ionisation rather than ranking elements using the emitted electrons’ kinetic energies alone.

Question 4

A student proposes the following explanation for the increase in first ionisation energy across Period 2:

Each new element has one additional proton and one additional electron. The new electron shields the extra proton perfectly, so the effective nuclear charge experienced by the outer electrons should remain constant.

(a) Identify the incorrect assumption in this model.

(b) Explain why electrons in the same principal energy level do not shield one another as effectively as inner-shell electrons.

(c) Use the measured first ionisation energies of lithium and fluorine to assess the student’s prediction.

(d) Explain what happens to the outer electron distribution as effective nuclear charge increases across the period.

Solution 4

The student’s model is incorrect because electrons in the same principal energy level do not provide perfect shielding. The effective nuclear charge experienced by outer electrons generally increases across the period.

(a) Identify the incorrect assumption.

The incorrect assumption is that every additional electron perfectly cancels the attraction of one additional proton.

Shielding is not a simple one-electron-for-one-proton cancellation.

Electrons repel one another, but their ability to reduce the nucleus’s attraction depends on their distribution and the orbitals they occupy.

(b) Explain the difference between inner-shell and same-shell shielding.

Inner-shell electrons are located closer to the nucleus than the outer electrons being considered.

They therefore provide substantial shielding by reducing the net positive attraction experienced by the outer electrons.

Electrons in the same principal energy level are distributed in a similar region of the atom. They provide some shielding, but this effect is less effective than shielding from inner-shell electrons.

As a result, adding electrons to the same principal energy level does not fully cancel the effect of adding protons.

(c) Assess the prediction using measured data.

The first ionisation energies are approximately:

\[
IE_1(\mathrm{Li})=520\ \mathrm{kJ\,mol^{-1}}
\]

\[
IE_1(\mathrm{F})=1681\ \mathrm{kJ\,mol^{-1}}
\]

The difference is:

\[
\begin{aligned}
\Delta IE_1&=1681-520\\
&=1161\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

This substantial increase contradicts the prediction that effective nuclear charge remains constant.

The measurements support the conclusion that increasing nuclear charge outweighs the additional shielding from same-shell electrons.

(d) Explain the change in electron distribution.

As effective nuclear charge increases, the nucleus attracts the electron cloud more strongly.

The outer electron distribution becomes more contracted, meaning the electrons are generally closer to the nucleus.

Consequently, more energy is needed to remove an outer electron.

Question 5

A chemistry research group investigates the first ionisation energies of nitrogen and oxygen using photoelectron spectroscopy.

Both gaseous samples are exposed to monochromatic photons with an energy of \(21.218\ \mathrm{eV}\).

The measured kinetic energies of the outermost electrons are:

Assume the resulting ions are in their ground electronic states and that electron recoil is negligible.

The research group also doubles the intensity of the photon beam while keeping its photon energy constant. The detector records approximately twice as many emitted electrons, but the positions of the kinetic-energy peaks remain unchanged.

Answer the following.

(a) Calculate the first ionisation energy of nitrogen and oxygen in both eV and \(\mathrm{kJ\,mol^{-1}}\).

(b) Calculate the difference between their first ionisation energies. Which atom holds its outermost electron more strongly?

(c) Explain why doubling the photon intensity increases the number of detected electrons without changing the measured first ionisation energy.

(d) Nitrogen has seven protons, while oxygen has eight. Explain why oxygen nevertheless has a lower first ionisation energy.

(e) A student argues that oxygen’s lower first ionisation energy is explained entirely by the fact that removing an electron produces an oxygen ion with a half-filled \(2p\) subshell. Evaluate this explanation. Your answer should consider both the initial neutral atoms and the resulting positive ions.

Solution 5

Nitrogen has the higher first ionisation energy, approximately \(1402.3\ \mathrm{kJ\,mol^{-1}}\), compared with oxygen’s \(1313.9\ \mathrm{kJ\,mol^{-1}}\). The result arises from the different \(2p\) electron arrangements, despite oxygen having the greater nuclear charge.

(a) Calculate both first ionisation energies.

Use the energy relationship:

\[
IE_1=E_{\gamma}-KE_e
\]

For nitrogen:

\[
\begin{aligned}
IE_1(\mathrm{N})
&=21.218-6.684\\
&=14.534\ \mathrm{eV}
\end{aligned}
\]

For oxygen:

\[
\begin{aligned}
IE_1(\mathrm{O})
&=21.218-7.600\\
&=13.618\ \mathrm{eV}
\end{aligned}
\]

Convert both values into \(\mathrm{kJ\,mol^{-1}}\).

For nitrogen:

\[
\begin{aligned}
IE_1(\mathrm{N})
&=14.534\times96.485\\
&=1402.3\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

For oxygen:

\[
\begin{aligned}
IE_1(\mathrm{O})
&=13.618\times96.485\\
&=1313.9\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

(b) Calculate the difference and compare the elements.

\[
\begin{aligned}
\Delta IE_1
&=1402.3-1313.9\\
&=88.4\ \mathrm{kJ\,mol^{-1}}
\end{aligned}
\]

Nitrogen has a first ionisation energy approximately \(88.4\ \mathrm{kJ\,mol^{-1}}\) greater than oxygen.

Therefore, nitrogen holds its outermost electron more strongly.

Notice that oxygen’s emitted electrons have greater kinetic energy, even though both samples received photons of identical energy. This confirms that less energy was needed to remove oxygen’s outermost electron.

(c) Explain the effect of increasing photon intensity.

Doubling the photon intensity increases the number of photons arriving at the sample per second.

Under ordinary single-photon conditions, this produces more ionisation events per second, so the detector records more emitted electrons.

However, the energy carried by each individual photon remains unchanged.

The kinetic energy of each emitted electron is determined by the photon energy and the energy required for ionisation:

\[
KE_e=E_{\gamma}-IE_1
\]

Increasing the number of photons does not increase the energy of each photon.

Therefore, the detector records more electrons, but the kinetic-energy peak positions remain unchanged. The measured first ionisation energies do not change.

This is the same distinction between light intensity and individual photon energy explored in Why Brighter Light Doesn’t Increase Photoelectron Energy.

(d) Explain why oxygen has a lower first ionisation energy.

The electron configurations are:

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

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

Nitrogen has three unpaired electrons occupying the three \(2p\) orbitals.

Oxygen has four \(2p\) electrons, meaning one orbital contains a pair.

The two electrons in oxygen’s paired orbital experience additional electron-electron repulsion.

This makes one of the paired electrons easier to remove, reducing oxygen’s first ionisation energy.

Although oxygen has an additional proton, the effect of electron pairing outweighs the increase in nuclear attraction for this particular comparison.

(e) Evaluate the student’s explanation.

The student’s explanation identifies an important part of the effect, but it is incomplete.

When nitrogen loses an electron, its configuration changes from:

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

to:

\[
\mathrm{N^+}:1s^2\,2s^2\,2p^2
\]

When oxygen loses one electron, its configuration changes from:

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

to:

\[
\mathrm{O^+}:1s^2\,2s^2\,2p^3
\]

Oxygen therefore forms a positive ion with a half-filled \(2p\) subshell, which is a relatively stable electron arrangement.

However, first ionisation energy is determined by the energy difference between the initial neutral atom and the resulting positive ion.

It cannot be explained by considering only the stability of the product.

The paired-electron repulsion in the original neutral oxygen atom is also important. It makes the starting electron arrangement easier to ionise, while the resulting half-filled configuration contributes to the favourable energy change.

The extra proton in oxygen still increases nuclear attraction. It simply isn’t enough to overcome the effects of electron pairing and the resulting electron arrangement.

This is why the nitrogen-to-oxygen decrease is a genuine exception to the general trend rather than a contradiction of the underlying model.

The key idea is that first ionisation energy depends on the electron being removed and how strongly it is held, not just the number of protons in the nucleus. Once you can explain that distinction, the next useful step is to examine successive ionisation energies, where the size of the energy jumps can reveal how many electrons an atom has in its outer shell.