Why Oxygen Has a Lower First Ionisation Energy Than Nitrogen

Across Period 2, oxygen has one more proton than nitrogen. You would therefore expect oxygen to hold its outermost electron more strongly and have a higher first ionisation energy. But the measured values show the opposite: nitrogen requires approximately \(1402.3\ \mathrm{kJ\,mol^{-1}}\), while oxygen requires only \(1313.9\ \mathrm{kJ\,mol^{-1}}\).

Oxygen has a lower first ionisation energy because one of its \(2p\) orbitals contains a pair of electrons. The repulsion between these paired electrons makes one easier to remove. This effect outweighs the additional nuclear attraction from oxygen’s extra proton.

The interesting part is understanding why that small change in electron arrangement matters so much.

01The nitrogen-oxygen exception

Before explaining the electron arrangements, consider the measured values.

The values are rounded ground-state atomic data from the NIST Atomic Spectra Database.

Look at nitrogen and oxygen.

Oxygen has:

• One additional proton.
• One additional electron.
• The same number of occupied principal energy levels.
• The same \(1s^2\,2s^2\) inner electron configuration.

The extra proton should increase the attraction between the nucleus and the outer electrons.

So why doesn’t oxygen have the higher ionisation energy?

To answer that, we need to look beyond the number of electrons and consider where they are arranged.

First, though, it helps to understand how the ionisation-energy values are obtained.

02How do scientists measure first ionisation energy?

First ionisation energy is the energy required to remove one electron from each atom in one mole of neutral 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. We usually express it in \(\mathrm{kJ\,mol^{-1}}\).

An important detail: these values describe isolated nitrogen and oxygen atoms, not nitrogen gas (\(\ce{N2}\)) or oxygen gas (\(\ce{O2}\)). Removing an electron from a molecule involves a different electronic structure.

From an electron beam to a photon-based measurement

One way to investigate an ionisation threshold is to use an electron gun.

Imagine an electron gun on the left of an evacuated chamber containing gaseous atoms. The gun fires electrons into the chamber.

If an incoming electron has enough kinetic energy, it can transfer energy to an atom and remove one of its electrons. This produces a positive ion.

An electric field then directs the positive ions towards a detector, which records the ion signal.

The experiment follows a straightforward prediction:

• Below the ionisation threshold, the incoming electrons cannot supply enough energy to ionise the atoms.
• At and above the threshold, positive ions begin to appear.
• Increasing the electron energy allows ionisation to occur through more collisions.

By changing the electron beam energy and recording the ion signal, scientists can estimate the ionisation threshold.

However, there is a measurement problem. An electron gun does not produce electrons with perfectly identical kinetic energies. There is a spread of energies, which makes the exact point at which ionisation begins harder to identify.

A more precise approach uses photons with carefully controlled energies.

Photoionisation and ion detection

In a photon-based ionisation experiment, a beam of monochromatic ultraviolet light is directed towards a sample of gaseous atoms.

The apparatus works in the following order:

1. Neutral atoms enter an evacuated ionisation chamber.
2. A beam of photons with a known energy passes through the chamber.
3. Atoms absorb photons and may eject electrons, forming positive ions.
4. An electric field extracts the positive ions into a mass analyser.
5. A detector records the number of ions produced at each photon energy.

A mass analyser can distinguish nitrogen ions from oxygen ions because they have different mass-to-charge ratios.

The key idea is that the photon energy can be varied in small increments.

Below the threshold, the atom cannot be ionised by a single photon. Once the photon energy reaches the threshold, ionisation becomes possible.

The first appearance of the ion signal therefore identifies the first ionisation threshold, provided the atoms are initially in their ground states.

This produces an ion-yield curve: a graph showing the number of detected ions against photon energy.

Notice the prediction this experiment allows us to test.

At a photon energy of \(14.00\ \mathrm{eV}\):

• Oxygen can be ionised because \(14.00 > 13.6181\).
• Nitrogen cannot be ionised because \(14.00 < 14.5341\).

This is strong evidence that oxygen’s outermost electron requires less energy to remove.

Experimental extension: The apparatus above illustrates how ionisation thresholds can be investigated. For an HSC response, the essential skill is interpreting the measured values and explaining them using electron configuration. You do not need to reproduce the engineering of a mass spectrometer.

03Why electron arrangement changes the result

Now we can return to the original puzzle.

Nitrogen and oxygen both have their outermost electrons in the second principal energy level.

Their electron configurations are:

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

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

The important difference is the number of electrons occupying the \(2p\) subshell.

Understanding the three \(2p\) orbitals

A \(p\) subshell contains three orbitals.

Each orbital can hold a maximum of two electrons, provided they have opposite spins.

The three \(2p\) orbitals have the same energy in an isolated atom. Electrons therefore occupy these orbitals according to Hund’s rule.

Hund’s rule states that electrons occupy orbitals of equal energy individually, with parallel spins, before pairing up.

Why?

Electrons have negative charges and repel one another. Keeping electrons in separate orbitals reduces the repulsion associated with occupying the same orbital.

Consider nitrogen first.

Nitrogen has three electrons in its \(2p\) subshell. Each electron occupies a separate orbital.

Its orbital arrangement is:

\[ 2p:\quad [\uparrow]\ [\uparrow]\ [\uparrow] \]

There are three unpaired electrons and no paired electrons in the \(2p\) subshell.

Now consider oxygen.

Oxygen has four electrons in its \(2p\) subshell. The first three occupy separate orbitals, but the fourth must pair with an electron already occupying one of them.

Its arrangement is:

\[ 2p:\quad [\uparrow\downarrow]\ [\uparrow]\ [\uparrow] \]

Oxygen therefore has one paired orbital and two singly occupied orbitals.

This is the crucial difference.

Why does pairing make oxygen easier to ionise?

Electrons repel one another because they carry the same negative charge.

When two electrons occupy the same orbital, their proximity produces additional electron-electron repulsion.

In oxygen, one of the \(2p\) electrons is therefore more easily removed than it would be if it occupied an otherwise equivalent singly occupied orbital.

Removing one of the paired electrons has two consequences:

1. The repulsion associated with that electron pair is relieved.
2. The resulting oxygen ion has a \(2p^3\) arrangement, with one electron in each of the three orbitals.

The resulting configuration is:

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

This is a half-filled \(2p\) subshell, which is a relatively favourable electron arrangement.

Nitrogen starts with this half-filled arrangement already:

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

Removing an electron from nitrogen produces:

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

This removes one of the three unpaired electrons and disrupts the half-filled arrangement.

Consequently, nitrogen’s outer electrons are harder to remove than oxygen’s, despite nitrogen having one fewer proton.

Don’t rely on the phrase “half-filled stability” alone

You might see the explanation that nitrogen has a particularly stable half-filled \(2p\) subshell.

That is useful, but it can become misleading if you stop there.

Ionisation energy is determined by the energy difference between the neutral atom and the resulting positive ion.

For oxygen, removing a paired electron relieves electron-electron repulsion and produces a relatively favourable half-filled arrangement.

For nitrogen, removing an electron disrupts an arrangement that is already relatively favourable.

Both effects help explain the difference.

The key is the overall energy change, not the idea that half-filled orbitals have some magical resistance to losing electrons.

04Nuclear charge, shielding and electron pairing: putting everything together

Let’s compare the competing effects directly.

The additional proton in oxygen increases nuclear attraction.

However, both elements have their outermost electrons in the same principal energy level, and their inner electron configurations are identical.

The major difference is the extra electron in oxygen’s \(2p\) subshell. Its pairing increases electron-electron repulsion, making one electron easier to remove.

This effect outweighs the increase in nuclear attraction.

Therefore:

\[ \boxed{IE_1(\mathrm{N})>IE_1(\mathrm{O})} \]

Oxygen has the lower first ionisation energy.

05A contrasting case: why fluorine does not follow the same decrease

There is an important trap here.

You might conclude that adding more paired electrons always decreases first ionisation energy.

That isn’t true.

Consider oxygen and fluorine.

Their configurations are:

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

\[ \mathrm{F}:1s^2\,2s^2\,2p^5 \]

Fluorine has one more proton and one more electron than oxygen.

Its \(2p\) arrangement contains two paired orbitals and one singly occupied orbital.

Fluorine therefore has more electron pairing than oxygen, yet its first ionisation energy is much higher:

\[ \begin{aligned} IE_1(\mathrm{O})&=1313.9\\ IE_1(\mathrm{F})&=1681.0 \end{aligned} \]

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

Why?

The additional proton in fluorine increases the nuclear attraction experienced by its outer electrons. The additional electron does create more electron-electron repulsion, but this does not outweigh the increase in nuclear attraction.

This gives us a more accurate rule:

Electron pairing can produce a local decrease in first ionisation energy, but its effect must be considered alongside nuclear charge and shielding.

It explains why nitrogen to oxygen is an exception without suggesting that the general periodic trend stops working.

For the broader explanation of periodic trends, see First Ionisation Energy Trends for HSC Chemistry.

06Worked examples

07Questions and solutions

Question 1

Nitrogen and oxygen are consecutive elements in Period 2.

(a) Write the electron configurations of both atoms using subshell notation.

(b) Describe how their \(2p\) electrons are arranged in their respective orbitals.

(c) Explain why nitrogen has a higher first ionisation energy than oxygen.

Solution 1

Nitrogen has the higher first ionisation energy because oxygen has additional electron-electron repulsion within a paired \(2p\) orbital.

(a) The electron configurations are:

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

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

(b) According to Hund’s rule, nitrogen’s three \(2p\) electrons occupy three separate orbitals:

\[ \mathrm{N}:\quad [\uparrow]\ [\uparrow]\ [\uparrow] \]

Oxygen has four \(2p\) electrons, so one orbital contains a pair:

\[ \mathrm{O}:\quad [\uparrow\downarrow]\ [\uparrow]\ [\uparrow] \]

(c) Oxygen has one additional proton, which increases its nuclear charge and would normally increase its first ionisation energy.

However, oxygen also has a paired \(2p\) orbital. Repulsion between the paired electrons makes one of them easier to remove.

Removing this electron relieves the additional repulsion and leaves an oxygen ion with a relatively favourable half-filled \(2p^3\) arrangement.

This effect outweighs the increased nuclear attraction, resulting in oxygen having the lower first ionisation energy.

Question 2

A photoionisation experiment uses photons with an energy of \(14.20\ \mathrm{eV}\).

The first ionisation energies of nitrogen and oxygen are:

• Nitrogen: \(14.5341\ \mathrm{eV}\)
• Oxygen: \(13.6181\ \mathrm{eV}\)

Assume both atoms begin in their ground states and the resulting ions remain in their ground states.

(a) Determine which atom can be ionised by absorbing one photon.

(b) Calculate the kinetic energy of the ejected electron.

(c) Express this kinetic energy per mole of ionisation events in \(\mathrm{kJ\,mol^{-1}}\).

Solution 2

Only oxygen can be ionised, and its ejected electron has a kinetic energy of \(0.5819\ \mathrm{eV}\), equivalent to approximately \(56.1\ \mathrm{kJ\,mol^{-1}}\).

(a) Compare the photon energy with the ionisation thresholds.

For nitrogen:

\[ 14.20<14.5341 \]

There is insufficient energy to ionise nitrogen.

For oxygen:

\[ 14.20>13.6181 \]

Oxygen can be ionised.

(b) Calculate the kinetic energy of the ejected oxygen electron.

The energy supplied by the photon is divided between ionisation and the kinetic energy of the electron:

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

Rearranging:

\[ \begin{aligned} KE_e&=E_{\gamma}-IE_1\\ &=14.20-13.6181\\ &=0.5819\ \mathrm{eV} \end{aligned} \]

(c) Convert to molar energy.

\[ \begin{aligned} KE_e&=0.5819\times96.485\\ &=56.14\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{KE_e=56.1\ \mathrm{kJ\,mol^{-1}}} \]

This is the kinetic energy available per mole of emitted electrons, assuming the oxygen ions are formed in their ground states.

The important distinction is that the photon must first supply enough energy to overcome the ionisation threshold. Only the remaining energy becomes kinetic energy.

Question 3

An experiment is designed to measure the ionisation thresholds of nitrogen and oxygen using an electron gun.

The nominal thresholds are:

The electron gun produces a beam with an energy spread of \(\pm0.50\ \mathrm{eV}\).

(a) Calculate the possible threshold measurement intervals for nitrogen and oxygen, treating the spread as a maximum absolute uncertainty.

(b) Explain whether this apparatus can reliably distinguish the two thresholds.

(c) Propose an improved experimental method and explain how it would address the limitation.

Solution 3

The electron gun cannot reliably distinguish the two thresholds using this uncertainty because the possible measurement intervals overlap. A narrow-band tunable photon source provides a better method.

(a) Calculate the uncertainty intervals.

For nitrogen:

\[ \begin{aligned} E_{\mathrm{min}}&=14.53-0.50=14.03\\ E_{\mathrm{max}}&=14.53+0.50=15.03 \end{aligned} \]

Therefore, the nitrogen interval is:

\[ 14.03\text{-}15.03\ \mathrm{eV} \]

For oxygen:

\[ \begin{aligned} E_{\mathrm{min}}&=13.62-0.50=13.12\\ E_{\mathrm{max}}&=13.62+0.50=14.12 \end{aligned} \]

Therefore, the oxygen interval is:

\[ 13.12\text{-}14.12\ \mathrm{eV} \]

(b) The intervals overlap between \(14.03\) and \(14.12\ \mathrm{eV}\).

This means that the uncertainty in the measurements is large enough to make the two thresholds difficult to distinguish confidently.

Although the actual thresholds differ by approximately \(0.91\ \mathrm{eV}\), the electron gun’s energy spread blurs the onset of ionisation.

(c) A better approach is to use a tunable, narrow-band ultraviolet photon source with a photoionisation mass spectrometer.

The photon energy can be adjusted in small increments, and the positive-ion signal can be recorded at each energy.

The narrower energy distribution makes the ionisation onset easier to identify. A mass analyser can also distinguish nitrogen ions from oxygen ions.

The revised method therefore reduces uncertainty in the incoming energy and provides a clearer comparison of the two ionisation thresholds.

Question 4

The following first ionisation energies are recorded:

A student argues:

“If electron pairing makes oxygen easier to ionise, fluorine should be even easier to ionise because it has more paired electrons.”

(a) Calculate the increase in first ionisation energy from oxygen to fluorine.

(b) Explain why the student’s prediction is incorrect.

(c) Explain what this comparison tells us about the relative importance of nuclear charge and electron-electron repulsion.

Solution 4

Fluorine has a higher first ionisation energy than oxygen by \(367.1\ \mathrm{kJ\,mol^{-1}}\), because the increase in nuclear attraction outweighs the additional electron-electron repulsion.

(a) Calculate the difference.

\[ \begin{aligned} \Delta IE_1 &=1681.0-1313.9\\ &=367.1\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

(b) Fluorine has nine protons, compared with oxygen’s eight.

Both elements have their outermost electrons in the second principal energy level, and their inner electron configurations are identical.

Fluorine’s greater nuclear charge therefore increases the attraction between its nucleus and its outer electrons.

Although fluorine has more electron pairing, the increased nuclear attraction outweighs the additional electron-electron repulsion.

(c) This comparison demonstrates that electron pairing is only one factor affecting first ionisation energy.

In the nitrogen-to-oxygen comparison, the additional repulsion from pairing outweighs the effect of increased nuclear charge.

In the oxygen-to-fluorine comparison, increased nuclear attraction dominates.

The result depends on the combined effects of nuclear charge, shielding, and electron arrangement rather than one factor acting independently.

Question 5

A chemistry student investigates the first and second ionisation energies of nitrogen and oxygen.

The following data are obtained from atomic ionisation-energy measurements:

The student notices something unexpected.

Oxygen has a lower first ionisation energy than nitrogen, but a much higher second ionisation energy.

The student concludes:

“If oxygen holds its first electron less strongly, it should also hold its second electron less strongly. The measurements must be inconsistent.”

Evaluate this conclusion.

Your explanation must:

• Identify the electron configurations of neutral nitrogen and oxygen.
• Show the configurations of both ions after their first ionisation.
• Explain why the first ionisation-energy ordering differs from the second ionisation-energy ordering.
• Consider both electron-electron repulsion and nuclear attraction.

Solution 5

The measurements are consistent. Oxygen has a lower first ionisation energy but a higher second ionisation energy because removing the first electron changes its electron arrangement. The second electron is being removed from a different species with a different configuration.

This is the crucial distinction: first and second ionisation energies do not describe the removal of electrons from the same neutral atom.

Step 1: Examine the neutral electron configurations.

Nitrogen:

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

Oxygen:

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

Nitrogen has three unpaired \(2p\) electrons.

Oxygen has one paired \(2p\) orbital and two singly occupied orbitals.

Step 2: Explain the first ionisation-energy ordering.

For the first ionisation of oxygen, one of the paired electrons is removed.

This relieves the additional electron-electron repulsion and leaves:

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

The resulting ion has a half-filled \(2p\) subshell.

For nitrogen, removing one electron produces:

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

This disrupts nitrogen’s original half-filled arrangement.

Consequently, oxygen is easier to ionise initially:

\[ IE_1(\mathrm{N})>IE_1(\mathrm{O}) \]

Step 3: Consider what happens during the second ionisation.

The second ionisation removes an electron from an already positively charged gaseous ion.

For nitrogen:

\[ \ce{N+ -> N^2+ + e-} \]

The electron configuration changes from \(2p^2\) to \(2p^1\).

For oxygen:

\[ \ce{O+ -> O^2+ + e-} \]

The electron configuration changes from \(2p^3\) to \(2p^2\).

Notice the difference.

Oxygen’s first ionisation produced an especially favourable half-filled \(2p^3\) arrangement.

The second ionisation now removes an electron from this arrangement, disrupting it.

There is also a stronger nuclear attraction in oxygen because it has eight protons, compared with nitrogen’s seven.

Both effects make the second electron harder to remove from \(\mathrm{O}^{+}\).

Step 4: Compare the measured second ionisation energies.

The data show:

\[ IE_2(\mathrm{N})=29.6013\ \mathrm{eV} \]

\[ IE_2(\mathrm{O})=35.1211\ \mathrm{eV} \]

Therefore:

\[ \boxed{IE_2(\mathrm{O})>IE_2(\mathrm{N})} \]

Oxygen requires substantially more energy to lose its second electron.

The reason is that the second electron is removed from \(\mathrm{O}^{+}\), which has a greater nuclear charge and a half-filled \(2p\) subshell.

The original pairing effect that made neutral oxygen easier to ionise has disappeared after the first electron is removed.

Step 5: Evaluate the student’s conclusion.

The student’s reasoning incorrectly assumes that an element has one fixed tendency to lose electrons.

In reality, every ionisation changes the number of electrons and the overall charge of the species.

The first and second ionisation energies therefore depend on different electron configurations and different electron-nucleus attractions.

This is why the ordering can reverse:

\[ \boxed{IE_1(\mathrm{N})>IE_1(\mathrm{O})} \]

but

\[ \boxed{IE_2(\mathrm{N})<IE_2(\mathrm{O})} \]

The deeper lesson is that ionisation energy depends on the electronic arrangement of the species being ionised, not simply the identity of the element.

08Where to go next

The nitrogen-oxygen exception shows why electron configuration matters when explaining periodic trends. It also introduces an important idea for successive ionisation energies: removing an electron changes the atom’s electronic structure, so the next electron may require a very different amount of energy.

This becomes particularly useful when identifying the number of valence electrons from a sequence of ionisation energies.

Continue with How to Find Valence Electrons from Successive Ionisation Energies.