How to Distinguish Electronegativity from Ionisation Energy

Magnesium has a higher first ionisation energy than aluminium. Yet aluminium is more electronegative than magnesium. If both properties measured how strongly an atom attracts electrons, that would seem impossible.

The explanation is that they describe two different situations. First ionisation energy measures how much energy it takes to remove an electron from an isolated gaseous atom. Electronegativity measures how strongly an atom attracts a shared pair of electrons in a covalent bond.

Before going further, make a prediction: if you shine UV light at isolated gaseous atoms, which property are you investigating? What about comparing the electron distribution in an H-F bond?

The first experiment investigates ionisation energy. The second involves electronegativity. Understanding why requires looking at what happens to the electrons in each situation.

01The difference between electronegativity and ionisation energy

The easiest way to distinguish them is to ask what happens to the electron.

First ionisation energy: removing an electron

First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous neutral 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 electron that has been removed.

The energy must be supplied, so first ionisation energy is positive.

For example, sodium has a first ionisation energy of approximately \(496\ \mathrm{kJ\,mol^{-1}}\).

This means 496 kJ of energy is required to remove one electron from each atom in one mole of gaseous sodium atoms.

The word first matters. It describes removing an electron from a neutral atom, not removing another electron from an already positive ion.

Electronegativity: attracting shared electrons

Electronegativity describes how strongly an atom attracts the shared electrons in a covalent bond.

Consider hydrogen fluoride, \(\ce{HF}\).

Both atoms contribute to the covalent bond, but fluorine attracts the shared electron density more strongly than hydrogen does.

The electrons are therefore distributed unevenly:

\[
\ce{H^{\delta+}-F^{\delta-}}
\]

The symbols \(\delta+\) and \(\delta-\) represent partial charges.

Hydrogen becomes slightly positive because electron density shifts away from it. Fluorine becomes slightly negative because electron density shifts towards it.

Electronegativity is commonly represented using the Pauling scale. Fluorine has a value of approximately 3.98, making it the most electronegative element.

Unlike ionisation energy, electronegativity is not measured in units of energy.

The difference between two atoms’ electronegativities is calculated using:

\[
\Delta\chi=|\chi_A-\chi_B|
\]

Here, \(\chi_A\) and \(\chi_B\) are the electronegativities of the two bonded atoms, and \(\Delta\chi\) is their absolute difference.

A greater difference generally means more uneven sharing of the bonding electrons.

Why these are not the same measurement

Both properties involve electrostatic attraction between positively charged nuclei and negatively charged electrons.

However, the electron being considered is different.

For ionisation energy, the question is:

How much energy does it take to remove this electron completely from the atom?

For electronegativity, the question is:

When this atom shares electrons with another atom, how strongly does it attract the shared electron density?

An atom can therefore be relatively easy to ionise but still attract bonding electrons strongly.

This distinction becomes particularly important when comparing neighbouring elements in the periodic table.

02How ionisation energy is measured

To understand why ionisation energy has units of energy, consider how scientists can investigate the electrons in an isolated atom.

One experimental technique is photoelectron spectroscopy.

The aim is to determine how much energy is required to remove electrons from atoms.

Following the experiment

Imagine a low-pressure chamber containing gaseous atoms.

A beam of UV radiation is directed towards the atoms. The photons transfer energy to the electrons, and some electrons absorb enough energy to escape from their atoms.

The ejected electrons then travel into an energy analyser, which measures their kinetic energies. A detector records the electrons and produces a photoelectron spectrum.

What should happen if an electron is more strongly attracted to its nucleus?

It should require more energy to escape. If the incoming photons have the same energy, the escaping electron will therefore have less kinetic energy.

This follows from conservation of energy:

\[
E_{\mathrm{binding}}=h\nu-E_k
\]

Where:

• \(E_{\mathrm{binding}}\) is the energy required to remove the electron.
• \(h\) is Planck’s constant.
• \(\nu\) is the frequency of the incoming radiation.
• \(E_k\) is the measured kinetic energy of the ejected electron.

The experiment measures kinetic energy. The binding energy is then calculated.

For a neutral atom initially in its ground state, the binding energy of its least tightly held electron corresponds to its first ionisation energy.

The important limitation is that this experiment measures the energy needed to remove an electron. It does not directly measure how strongly an atom attracts a shared pair in a covalent bond.

That requires a different approach.

How electronegativity was developed

Linus Pauling developed his electronegativity scale using comparisons of chemical bond energies.

Instead of studying an isolated atom, the approach examines bonds between atoms.

Imagine three types of bonds:

• A-A, between two identical atoms.
• B-B, between two identical atoms of another element.
• A-B, between the two different elements.

The bond dissociation energies of these bonds can be determined using thermochemical data.

Pauling compared the measured strength of the A-B bond with the strength expected from the A-A and B-B bonds.

If the A-B bond was stronger than expected, the additional stabilisation was associated with unequal electron sharing and the polar character of the bond.

This comparison helped establish relative electronegativity values.

Notice the difference between the two approaches:

• Photoelectron spectroscopy investigates how difficult it is to remove an electron from an isolated atom.
• Pauling’s approach uses bond-energy comparisons to establish how strongly different atoms attract shared electrons.

Both involve electron attraction, but they answer different questions.

03Why both properties follow similar periodic trends

Although electronegativity and ionisation energy are different, they generally follow similar trends across the periodic table.

Both are influenced by three important factors:

1. Nuclear charge.
2. Electron shielding.
3. Distance between the nucleus and the electrons.

Understanding these factors makes the periodic trends much easier to explain.

Nuclear charge

Nuclear charge refers to the positive charge of the nucleus.

Each proton contributes a positive charge. Increasing the number of protons generally increases the attraction between the nucleus and negatively charged electrons.

For example, magnesium has 12 protons, while aluminium has 13.

If their outer electrons experienced otherwise similar conditions, aluminium’s greater nuclear charge would produce a stronger attraction.

This tends to increase both first ionisation energy and electronegativity.

Electron shielding

Outer electrons are not attracted to an entirely unshielded nucleus.

Inner-shell electrons repel outer electrons, reducing the net positive attraction experienced by the outer electrons.

This effect is called electron shielding.

An atom with more occupied inner shells generally experiences greater shielding of its outer electrons.

Distance from the nucleus

Electrostatic attraction becomes weaker as the distance between charged particles increases.

Electrons in higher principal energy levels are generally further from the nucleus.

This means that an outer electron in the fourth shell is generally less strongly attracted to the nucleus than an equivalent electron in the third shell, assuming other factors are similar.

Distance and shielding therefore work together to reduce the effective nuclear attraction experienced by electrons further from the nucleus.

Comparing the periodic trends

These three factors explain the general patterns:

Across a period:

• Nuclear charge increases because proton number increases.
• Added electrons enter the same principal energy level.
• Shielding increases relatively little compared with the increase in nuclear charge.
• The attraction between the nucleus and outer electrons generally becomes stronger.

Consequently, removing an electron requires more energy, and the atom generally attracts shared bonding electrons more strongly.

Down a group:

• Additional electron shells are occupied.
• The outer electrons are further from the nucleus.
• Shielding increases.
• The effective attraction experienced by the outer electrons generally decreases.

Consequently, first ionisation energy and electronegativity generally decrease.

These are general trends, not rules that must hold between every pair of neighbouring elements.

The electron arrangement can create exceptions, particularly for first ionisation energy.

04Important exceptions: when the trends disagree

This is where students often make mistakes.

If electronegativity and ionisation energy both generally increase across a period, it is tempting to assume that an element with a higher electronegativity must always have a higher first ionisation energy.

That is incorrect.

Two neighbouring comparisons demonstrate why.

Exception 1: Magnesium and aluminium

Consider the following approximate values:

Notice the reversal:

• Magnesium has the higher first ionisation energy.
• Aluminium has the higher electronegativity.

Why?

Start by examining their electron configurations.

Magnesium:

\[
\ce{Mg}: 1s^2 2s^2 2p^6 3s^2
\]

Aluminium:

\[
\ce{Al}: 1s^2 2s^2 2p^6 3s^2 3p^1
\]

Both elements have their outer electrons in the third principal energy level.

However, their first electrons are removed from different subshells.

Magnesium loses a \(3s\) electron, while aluminium loses a \(3p\) electron.

The \(3p\) electron in aluminium is higher in energy and less penetrating than a \(3s\) electron. It is also more shielded by the inner electron density, including the \(3s\) electrons.

As a result, aluminium’s \(3p\) electron is easier to remove.

This effect outweighs the increase in nuclear charge.

But electronegativity asks a different question.

Aluminium has a greater nuclear charge than magnesium, and its outer electrons experience stronger effective nuclear attraction across the period.

Therefore, aluminium has the higher Pauling electronegativity.

The difference comes down to the type of electron being considered:

• Ionisation energy depends on the particular electron being removed.
• Electronegativity concerns the attraction of shared bonding electrons.

Exception 2: Phosphorus and sulfur

A second important exception involves electron pairing.

Consider phosphorus and sulfur:

Phosphorus has a slightly higher first ionisation energy, even though sulfur has a greater nuclear charge.

To understand this, examine their outer electron configurations.

Phosphorus:

\[
\ce{P}: 3s^2 3p^3
\]

Sulfur:

\[
\ce{S}: 3s^2 3p^4
\]

The three \(3p\) electrons in phosphorus occupy three separate \(p\) orbitals.

In sulfur, the fourth \(3p\) electron must pair with another electron in one of these orbitals.

Electrons have negative charges, so they repel one another.

The paired electrons in sulfur experience additional electron-electron repulsion. This makes one of the paired electrons slightly easier to remove.

Consequently, sulfur has a lower first ionisation energy than phosphorus.

However, sulfur is still more electronegative.

Why doesn’t the pairing effect reverse electronegativity as well?

Because the additional repulsion is particularly relevant to removing a specific electron from the isolated sulfur atom. Electronegativity concerns the attraction of shared bonding electrons.

Sulfur’s greater nuclear charge across the period means it generally attracts shared electrons more strongly than phosphorus.

This is another example of why the two properties cannot be treated as interchangeable.

For a more detailed explanation of the periodic pattern and its subshell exceptions, see First Ionisation Energy Trends for HSC Chemistry.

05A useful boundary case: noble gases

Noble gases provide another way to recognise the difference.

Consider neon and fluorine.

Neon has a higher first ionisation energy than fluorine.

Its electrons are held very strongly, partly because neon has a small atomic radius and a stable, filled outer electron shell.

However, neon is often left without a value on conventional Pauling electronegativity tables used in school chemistry.

Why?

Electronegativity describes an atom’s attraction for shared electrons in a covalent bond. Neon rarely forms ordinary covalent bonds, so a conventional Pauling value is generally not assigned.

This does not mean neon has zero electronegativity, or that its nucleus cannot attract electrons.

It means that the usual Pauling comparison is not available in the same way.

Some alternative electronegativity scales assign values to selected noble gases, so always use the scale specified in a question.

The key distinction is that an isolated gaseous atom can have a measurable first ionisation energy even when a conventional electronegativity value is not available.

06A quick decision method for HSC questions

When a question mentions electronegativity or ionisation energy, identify what happens to the electron before attempting to explain the trend.

Once you have identified the property, use the appropriate reasoning.

For first ionisation energy:

1. Identify the electron being removed.
2. Compare nuclear charge.
3. Consider the number of occupied electron shells and shielding.
4. Check whether the electron is in a different subshell.
5. Check for additional electron-electron repulsion caused by pairing.

For electronegativity:

1. Identify the two bonded atoms.
2. Compare their positions in the periodic table.
3. Consider nuclear charge, shielding, and atomic radius.
4. Determine which atom attracts the shared electrons more strongly.
5. If numerical values are supplied, calculate the absolute difference.

Remember that electronegativity difference predicts bond polarity, not the energy required to remove an electron.

For the next step, see Bond Polarity and Electronegativity: HSC Chemistry Guide.

07Questions and solutions

The following questions progress from identifying the correct property to combining electron configurations, periodic trends, and molecular behaviour.

Question 1

A chemistry student makes the following observations:

Observation A: Removing one electron from each atom in one mole of gaseous sodium atoms requires \(496\ \mathrm{kJ}\) of energy.

Observation B: In a hydrogen fluoride molecule, fluorine attracts the shared bonding electrons more strongly than hydrogen.

For each observation:

(a) Identify whether it describes ionisation energy or electronegativity.

(b) State the appropriate unit, if applicable.

(c) Explain what happens to the electrons in each situation.

Solution 1

Observation A describes first ionisation energy, measured in \(\mathrm{kJ\,mol^{-1}}\). Observation B describes electronegativity, which is dimensionless on the Pauling scale.

For Observation A, an electron is completely removed from each neutral gaseous sodium atom.

The process is:

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

The energy supplied is the first ionisation energy.

For Observation B, the electron pair remains shared between hydrogen and fluorine, but electron density shifts towards fluorine.

Fluorine becomes partially negative, while hydrogen becomes partially positive:

\[
\ce{H^{\delta+}-F^{\delta-}}
\]

The key difference is complete electron removal versus unequal sharing.

Question 2

Four elements have the following approximate properties:

(a) Arrange the elements in increasing order of first ionisation energy.

(b) Arrange the elements in increasing order of electronegativity.

(c) Explain why potassium has a lower first ionisation energy than sodium, despite having a greater nuclear charge.

(d) Explain why magnesium has a higher first ionisation energy than sodium.

(e) Does the matching order of these two properties mean that they measure the same thing? Justify your answer.

Solution 2

Both properties increase in the order K, Na, Mg, Cl, but they measure different physical situations.

(a) Increasing first ionisation energy:

\[
\mathrm{K<Na<Mg<Cl}
\]

This follows directly from the supplied values:

\[
419<496<738<1251
\]

(b) Increasing electronegativity:

\[
\mathrm{K<Na<Mg<Cl}
\]

The Pauling values give:

\[
0.82<0.93<1.31<3.16
\]

(c) Potassium has a lower first ionisation energy than sodium because its outer electron occupies a higher principal energy level.

Sodium has the electron configuration:

\[
2,8,1
\]

Its outer electron occupies the third shell.

Potassium has:

\[
2,8,8,1
\]

Its outer electron occupies the fourth shell.

Although potassium has more protons, its outer electron is further from the nucleus and experiences greater shielding from the additional inner shell.

These effects outweigh the increase in nuclear charge, making the outer electron easier to remove.

(d) Magnesium has a higher first ionisation energy than sodium because both lose an electron from the third shell, but magnesium has a greater nuclear charge.

Sodium has 11 protons, while magnesium has 12.

Both have the same inner-shell configuration, \(2,8\), so their outer electrons experience broadly similar core-electron shielding.

The additional proton in magnesium increases the effective nuclear attraction.

Therefore, more energy is required to remove magnesium’s first electron.

(e) No. The matching order does not mean the properties are identical.

The similar rankings arise because both properties are generally influenced by nuclear charge, shielding, and atomic radius.

However, first ionisation energy measures the energy required to remove an electron from an isolated atom, while electronegativity measures attraction for shared bonding electrons.

Their units, definitions, and applications remain different.

Question 3

Phosphorus and sulfur have the following approximate properties:

(a) Identify which element has the higher first ionisation energy.

(b) Identify which element has the higher electronegativity.

(c) Explain why the first ionisation energy decreases from phosphorus to sulfur, despite the increase in nuclear charge.

(d) Explain why the electronegativity trend does not show the same decrease.

Solution 3

Phosphorus has the higher first ionisation energy, while sulfur has the higher electronegativity.

(a) Phosphorus has the higher first ionisation energy.

\[
1012>1000
\]

The difference is:

\[
1012-1000=12\ \mathrm{kJ\,mol^{-1}}
\]

(b) Sulfur has the higher electronegativity.

\[
2.58>2.19
\]

Sulfur therefore attracts shared bonding electrons more strongly.

(c) The decrease in first ionisation energy is caused by electron pairing in sulfur’s \(3p\) subshell.

Phosphorus has three electrons occupying three separate \(3p\) orbitals.

Sulfur has four \(3p\) electrons, meaning one orbital contains a pair of electrons.

The paired electrons repel one another. This additional repulsion makes one of sulfur’s paired electrons easier to remove.

This effect outweighs the increase in nuclear charge, producing a slightly lower first ionisation energy for sulfur.

(d) Sulfur’s electronegativity is higher because its greater nuclear charge increases its attraction for shared bonding electrons.

Although electron pairing affects the energy required to remove a particular electron from an isolated sulfur atom, electronegativity concerns the attraction of shared electrons in a bond.

Sulfur’s stronger effective nuclear attraction across the period therefore produces a higher electronegativity.

The exception in ionisation energy does not require an equivalent exception in electronegativity.

Question 4

A student is comparing fluorine and neon using the following data:

The student concludes:

“Neon has a higher first ionisation energy, so it must also have a higher electronegativity than fluorine.”

Evaluate this conclusion.

In your answer, explain why neon’s first ionisation energy is high and why a conventional Pauling electronegativity value is generally not assigned.

Solution 4

The student’s conclusion is incorrect. Neon has a higher first ionisation energy, but this does not establish that it has a higher electronegativity than fluorine.

First, consider ionisation energy.

Neon has a filled outer electron shell:

\[
\ce{Ne}: 1s^2 2s^2 2p^6
\]

Its small atomic radius and stable electron arrangement mean that removing an electron requires considerable energy.

Fluorine, meanwhile, has seven outer-shell electrons and a first ionisation energy of approximately \(1681\ \mathrm{kJ\,mol^{-1}}\).

Neon’s higher value therefore reflects how difficult it is to remove an electron from its isolated atom.

Electronegativity describes something different.

The Pauling scale is based on comparisons involving covalent bonds. Neon rarely forms ordinary covalent bonds, so a conventional Pauling value is generally not assigned.

This does not mean neon has no attraction for electrons. It means the usual Pauling comparison is not available.

The student’s mistake is assuming that a higher ionisation energy automatically produces a higher electronegativity.

The two properties cannot be ranked against each other using their numerical values.

Question 5

Phosphorus and sulfur have the following approximate properties:

Consider a gaseous molecule containing a covalent P-S bond.

A student makes the following argument:

“Phosphorus has a higher first ionisation energy, so it holds electrons more strongly. Therefore, the shared bonding electrons must be attracted towards phosphorus. Also, because sulfur has a lower first ionisation energy, an electron must be removed from sulfur when the molecule is ionised.”

Assess both claims.

In your answer:

(a) Calculate the electronegativity difference between phosphorus and sulfur.

(b) Identify the direction of electron-density displacement in the P-S bond and the resulting partial charges.

(c) Explain why phosphorus has a higher first ionisation energy than sulfur.

(d) Determine whether the supplied atomic data are sufficient to identify which atom contributes the electron removed when the molecule is ionised. Justify your conclusion.

Solution 5

The P-S bond is polar towards sulfur, but the supplied atomic data cannot identify which atom contributes the electron removed when the molecule is ionised. The student’s two claims are therefore incorrect.

(a) Calculate the electronegativity difference.

The formula is:

\[
\Delta\chi=|\chi_{\mathrm{P}}-\chi_{\mathrm{S}}|
\]

Substituting the values:

\[
\Delta\chi=|2.19-2.58|
\]

\[
\boxed{\Delta\chi=0.39}
\]

The electronegativity difference is 0.39 on the Pauling scale.

(b) Electron density shifts towards sulfur.

Sulfur has a higher electronegativity than phosphorus:

\[
2.58>2.19
\]

Therefore, sulfur attracts the shared bonding electrons more strongly.

The bond can be represented as:

\[
\ce{P^{\delta+}-S^{\delta-}}
\]

Phosphorus carries a partial positive charge, while sulfur carries a partial negative charge.

The student’s first claim is incorrect because first ionisation energy does not determine which atom attracts a shared bonding pair more strongly.

(c) Phosphorus has a higher first ionisation energy because of its more stable half-filled \(3p\) subshell.

Phosphorus has:

\[
3p^3
\]

Its three \(3p\) electrons occupy separate orbitals.

Sulfur has:

\[
3p^4
\]

One of its \(3p\) orbitals contains a pair of electrons.

The additional electron-electron repulsion in this paired orbital makes one of sulfur’s electrons easier to remove.

Consequently:

\[
IE_1(\mathrm{P})>IE_1(\mathrm{S})
\]

This explains why phosphorus has a slightly higher first ionisation energy, despite sulfur having a greater nuclear charge.

(d) No. The atomic data are insufficient to identify which atom contributes the electron removed from the molecule.

This is the most important distinction in the question.

The supplied first ionisation energies describe isolated, neutral gaseous phosphorus and sulfur atoms.

However, when the molecule is ionised, the electron is removed from the bonded molecular system.

In a molecule, electrons occupy molecular orbitals that can extend across more than one atom. They are not necessarily assigned exclusively to either phosphorus or sulfur.

The fact that sulfur attracts the shared bonding electron density more strongly does not prove that the electron removed during molecular ionisation must come from sulfur.

Similarly, phosphorus having a higher isolated-atom first ionisation energy does not establish which molecular orbital is easiest to ionise.

Additional information about the molecule’s electronic structure, such as molecular photoelectron data or molecular orbital information, would be needed.

The two measurements therefore tell us different things:

• Electronegativity identifies which atom attracts shared bonding electron density more strongly.
• First ionisation energy identifies how much energy is needed to remove an electron from an isolated gaseous atom.

There is no contradiction between phosphorus being harder to ionise as an isolated atom and sulfur attracting the shared electrons more strongly in a bond.

That is precisely why electronegativity and ionisation energy must be treated as separate chemical properties.