How to Find Valence Electrons from Successive Ionisation Energies

An atom has a set of successive ionisation energies:

• First ionisation energy: energy required to remove the first electron.
• Second ionisation energy: energy required to remove the second electron.
• Third ionisation energy: energy required to remove the third electron.

The key to finding the number of valence electrons is to look for a large jump between consecutive ionisation energies.

The number of electrons removed before this large jump is usually the number of valence electrons. The jump tells you that the next electron is being removed from an inner electron shell, where it experiences a stronger attraction to the nucleus.

For example, suppose an element has these illustrative successive ionisation energies:

There is a large jump between the third and fourth ionisation energies.

This means the first three electrons were relatively easy to remove compared with the fourth. The atom therefore has 3 valence electrons.

This is the central method. However, to use it confidently in an HSC Chemistry question, you need to understand why the jump occurs, how to recognise it in unfamiliar data, and when the simple rule needs qualification.

01Why does a large jump occur?

Imagine removing electrons from an atom one at a time.

The first electron comes from the outermost occupied electron shell. It is relatively far from the nucleus and is shielded by other electrons. The attraction holding it in place is therefore comparatively weak.

After removing that electron, the remaining electrons experience a greater effective attraction to the nucleus. Removing another electron generally requires more energy.

This continues as successive electrons are removed.

But something more significant happens when all the electrons in the outermost shell have been removed. The next electron must come from an inner shell.

Inner-shell electrons are generally:

• Closer to the nucleus.
• Less shielded from the full nuclear attraction.
• More strongly attracted to the positively charged nucleus.

As a result, removing an inner-shell electron requires substantially more energy.

The important distinction is that the large jump does not happen simply because the atom has fewer electrons. It happens because the electron being removed belongs to a different, more tightly bound shell.

02What are successive ionisation energies actually measuring?

Ionisation energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionisation energy describes this process:

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

The second ionisation energy describes removing another electron from the positively charged ion:

\[
\ce{X+(g) -> X^2+(g) + e-}
\]

The third ionisation energy removes another electron, and the process continues.

In general:

\[
\ce{X^{(n-1)+}(g) -> X^{n+}(g) + e-}
\]

Here, \(n\) identifies which electron is being removed.

All ionisation energies are positive because energy must be supplied to remove an electron. They are commonly reported in \(\mathrm{kJ\,mol^{-1}}\), meaning kilojoules per mole of ionisation events.

Why does each successive ionisation energy increase?

Consider a neutral magnesium atom. Its electron arrangement is \(2,8,2\).

After removing its first electron, the atom becomes a positively charged ion. There are now fewer electrons surrounding the same nucleus.

The nuclear charge has not changed. However, the remaining electrons experience a stronger effective attraction to the nucleus.

Three things contribute:

1. The number of protons remains constant, so the nuclear charge is unchanged.
2. There are fewer electrons, reducing electron-electron repulsion and shielding.
3. The remaining electrons are held more strongly by the positively charged ion.

Therefore:

\[
IE_1 < IE_2 < IE_3 < IE_4
\]

This pattern applies throughout the successive ionisation energy series of an element.

There is an important difference between the ordinary increases and the large jump. Removing another electron from the same outer shell generally requires more energy, but crossing into an inner shell produces a much greater increase.

03How are successive ionisation energies measured?

The numbers in an ionisation energy table aren’t simply calculated by counting electrons. They come from measurements of how much energy is needed to produce ions.

One experimental approach uses a threshold ionisation technique.

Picture the apparatus in this order:

1. A sample of the element is introduced as a gas into a low-pressure chamber.
2. A controlled-energy beam of electrons or photons passes through the chamber.
3. The beam transfers energy to the gaseous atoms or ions. If enough energy is supplied, an electron is removed.
4. The resulting positive ions travel towards a detector, often through a mass analyser that distinguishes different charge states.
5. The energy of the incoming beam is varied, and the appearance of the relevant ion is monitored.

What should the experiment show?

Below the required energy threshold, the relevant ionisation process cannot occur. As the energy supplied reaches the threshold, the corresponding ion begins to appear.

For the first ionisation energy, researchers monitor the formation of \(\mathrm{X}^{+}\). For the second, they start with \(\mathrm{X}^{+}\) and monitor the formation of \(\mathrm{X}^{2+}\).

The same process can be repeated for subsequent ionisations.

One difficulty is that an energy source with a broad range of energies can make the threshold less precise. Using a more precisely defined photon energy, for example, can improve the measurement.

The actual instruments used vary, but the important idea is that ionisation energy is determined from the energy threshold needed to remove an electron. The describes several experimental approaches.

A useful energy conversion

Suppose a first ionisation energy is reported as \(580\ \mathrm{kJ\,mol^{-1}}\).

How much energy does this represent for one atom?

First, convert kilojoules into joules:

\[
580\ \mathrm{kJ\,mol^{-1}}
=580\,000\ \mathrm{J\,mol^{-1}}
\]

Divide by Avogadro’s constant:

\[
E=\frac{580\,000}{6.022\times10^{23}}
\]

\[
E=9.63\times10^{-19}\ \mathrm{J}
\]

Convert to electronvolts using \(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}\):

\[
E=\frac{9.63\times10^{-19}}{1.602\times10^{-19}}
\]

\[
\boxed{E=6.01\ \mathrm{eV\ per\ atom}}
\]

This is the same ionisation energy expressed using a different unit. It doesn’t change the chemistry or the electron arrangement.

04A reliable method for finding the valence-electron count

Use the following method when given a table of successive ionisation energies.

Step 1: Confirm what species is being ionised.

Check whether the series starts with a neutral gaseous atom or an already charged ion.

The standard method for finding the valence-electron count of a neutral element assumes the series starts with its first ionisation energy, \(IE_1\).

Step 2: Compare consecutive ionisation energies.

Don’t just read the numbers individually. Look at how much the energy increases between each removal.

For example, calculate:

\[
IE_2-IE_1
\]

\[
IE_3-IE_2
\]

\[
IE_4-IE_3
\]

You are looking for a particularly large increase compared with the surrounding changes.

Step 3: Find the first major jump.

This is the crucial step.

For a neutral main-group atom, the first major jump occurs when the outer-shell electrons have all been removed and the next electron must come from an inner shell.

Step 4: Count the electrons removed before the jump.

That number is the valence-electron count.

Step 5: Check your answer against the electron arrangement.

For a main-group element, you can use the periodic table as a cross-check.

Helium is an exception. It has two electrons in its only occupied shell.

One warning: there is no universal numerical ratio that defines a large jump. Compare the increase with the surrounding values and consider the electron arrangement.

05How nuclear charge and shielding affect ionisation energy

The shell-jump method becomes easier to understand when you connect it to periodic trends.

Two factors are especially important:

• Nuclear charge: the positive charge of the nucleus, determined by its number of protons.
• Shielding: the reduction in the attraction experienced by an outer electron because of repulsion from other electrons, particularly those in inner shells.

The combined effect is often described using effective nuclear charge. This is the net positive attraction experienced by an electron after shielding is considered.

Across a period: ionisation energy generally increases

Consider elements moving from left to right across a period.

The number of protons increases, so nuclear charge increases.

At the same time, electrons are added to the same principal electron shell. Shielding increases, but it does not completely cancel the increase in nuclear charge.

The effective nuclear charge generally increases.

As a result:

• Electrons experience a stronger attraction to the nucleus.
• Atomic radius generally decreases.
• More energy is required to remove an outer electron.

Therefore, first ionisation energy generally increases across a period.

Down a group: ionisation energy generally decreases

Now compare elements moving down a group.

Each step down adds another occupied electron shell.

The outermost electrons are generally:

• Further from the nucleus.
• More shielded by inner electrons.
• Less strongly attracted to the nucleus.

Even though nuclear charge increases down the group, the additional distance and shielding usually have a greater effect.

Consequently, less energy is generally required to remove an outer electron.

Two important exceptions to the across-period trend

The general trend is useful, but it isn’t perfectly smooth.

Two common exceptions involve electron subshells and electron pairing.

Exception 1: Beryllium and boron

Beryllium has the electron configuration:

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

Boron has:

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

You might predict that boron has a higher first ionisation energy because it has one more proton.

However, the first electron removed from boron occupies the \(2p\) subshell, whereas the first electron removed from beryllium occupies the \(2s\) subshell.

The \(2p\) electron is higher in energy and less penetrating than a \(2s\) electron. It is therefore easier to remove.

As a result, boron’s first ionisation energy is lower than beryllium’s.

Exception 2: Nitrogen and oxygen

Nitrogen has:

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

Oxygen has:

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

Nitrogen has three electrons distributed across its three \(2p\) orbitals, with one electron in each orbital.

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

The two electrons in that orbital experience additional electron-electron repulsion. This makes one of them easier to remove.

Consequently, oxygen has a slightly lower first ionisation energy than nitrogen, despite having a greater nuclear charge.

These exceptions are useful because they remind you that electron arrangement matters just as much as nuclear charge.

When the simple valence-electron method needs care

The first-major-jump rule is most straightforward for neutral main-group elements.

There are two situations where you should slow down.

1. The data begin with an ion rather than a neutral atom.

If the first ionisation energy supplied is for \(\mathrm{X}^{+}\), the jump tells you about the electron shells of \(\mathrm{X}^{+}\).

It does not automatically give you the number of valence electrons in the original neutral element.

You must account for the electrons already lost.

2. The element is a transition metal.

Transition metals have \(ns\) and \((n-1)d\) subshells that are relatively close in energy.

The outer \(ns\) electrons are generally removed before the \(d\) electrons, even though the \(d\) electrons can also participate in chemical bonding.

Therefore, the first large jump in a transition-metal ionisation series should not automatically be treated as the total number of chemically available valence electrons.

For more detail about these electron configurations, see spdf Notation: Electron Configurations and Valence Electrons.

06Questions and solutions

Try each question before reading its solution. The questions progress from straightforward data interpretation to more demanding reasoning about electron arrangement.

Question 1

An unknown neutral gaseous element has the following successive ionisation energies.

Determine the number of valence electrons in the element. Explain your reasoning.

Solution 1

The element has two valence electrons.

The increase between the first and second ionisation energies is:

\[
1200-600=600\ \mathrm{kJ\,mol^{-1}}
\]

The increase between the second and third is:

\[
7500-1200=6300\ \mathrm{kJ\,mol^{-1}}
\]

This is the first major jump.

The first two electrons were removed from the outermost shell. The third electron must come from an inner shell, where it experiences a stronger attraction to the nucleus.

Therefore:

\[
\boxed{2\text{ valence electrons}}
\]

The element is consistent with Group 2.

Question 2

Beryllium and boron are adjacent elements in Period 2.

Boron has a greater nuclear charge than beryllium. However, its first ionisation energy is lower.

Their electron configurations are:

• Beryllium: \(1s^2\,2s^2\)
• Boron: \(1s^2\,2s^2\,2p^1\)

Explain why boron has a lower first ionisation energy despite having more protons.

Solution 2

Boron has a lower first ionisation energy because its outermost electron occupies a higher-energy \(2p\) subshell.

Although boron has one more proton, the electron being removed is less tightly bound than beryllium’s outermost \(2s\) electron.

The \(2p\) electron is less penetrating and experiences greater shielding than a \(2s\) electron in the same principal shell.

This makes it easier to remove.

The important point is that increasing nuclear charge does not guarantee that every successive element will have a higher first ionisation energy. The subshell occupied by the electron also matters.

Question 3

Magnesium and calcium are both Group 2 elements.

Their approximate first ionisation energies are:

Calcium has more protons than magnesium.

Explain why calcium nevertheless has a lower first ionisation energy. Your answer should refer to nuclear charge, shielding, and electron-shell arrangement.

Solution 3

Calcium has a lower first ionisation energy because its outermost electron is further from the nucleus and experiences greater shielding.

Magnesium has the electron arrangement:

\[
2,8,2
\]

Calcium has:

\[
2,8,8,2
\]

Both elements have two valence electrons, but calcium has an additional occupied shell.

Its outermost electrons are further from the nucleus and are shielded by more inner electrons.

Although calcium has a greater nuclear charge, the increased distance and shielding reduce the effective attraction experienced by its outermost electrons.

Therefore, less energy is needed to remove an electron from calcium.

This explains why first ionisation energy generally decreases down a group.

Question 4

A neutral gaseous atom contains 18 electrons.

Its successive ionisation energy data show the following pattern. Intermediate values are omitted, and all values are illustrative.

The ionisation energies increase throughout the series.

(a) Determine the number of valence electrons.

(b) Explain why the jump between the sixteenth and seventeenth ionisation energies is much larger than the jump between the eighth and ninth.

(c) Determine the electron arrangement of the neutral atom.

Solution 4

(a) The atom has eight valence electrons.

The first major jump occurs between the eighth and ninth ionisation energies.

\[
40800-13800=27000\ \mathrm{kJ\,mol^{-1}}
\]

This indicates that the first eight electrons were removed from the outermost occupied shell.

Therefore:

\[
\boxed{8\text{ valence electrons}}
\]

(b) The jump between the sixteenth and seventeenth ionisation energies occurs when removal moves from the second shell into the first shell.

The first eight electrons come from the third shell.

The next eight come from the second shell.

After 16 electrons have been removed, only two electrons remain. Both occupy the first shell, very close to the nucleus.

The seventeenth electron is therefore much more strongly attracted to the nucleus than the electrons removed earlier.

This produces the enormous increase in ionisation energy.

(c) The electron arrangement is \(2,8,8\).

The atom originally contained 18 electrons:

\[
2+8+8=18
\]

It has eight electrons in its outermost shell and is consistent with argon, a Group 18 element.

The important lesson is that the largest jump in the complete series occurs much later than the first shell boundary. The first major jump is the one used to determine the neutral atom’s valence-electron count.

Question 5

Two gaseous positive ions, \(A^+\) and \(B^{2+}\), are investigated.

Both contain ten electrons, but they have different nuclear charges.

The following illustrative data are obtained:

For both ions, the first major jump occurs between the eighth and ninth ionisation energies.

A student concludes:

“Both elements must have eight valence electrons because eight electrons are removed before the large jump.”

Evaluate this conclusion.

In your answer:

(a) Identify the neutral elements from which \(A^+\) and \(B^{2+}\) were formed.

(b) Determine the number of valence electrons in each neutral element.

(c) Explain why both ions have their first major jump after the eighth electron, but \(B^{2+}\) has higher successive ionisation energies.

Solution 5

The student’s conclusion is incorrect. The neutral elements have different valence-electron counts: \(A\) has one, while \(B\) has two.

The mistake is applying the neutral-atom valence-electron rule directly to data that begin with positive ions.

(a) Identify the neutral elements.

For \(A^+\):

• Number of protons = 11.
• Number of electrons in the ion = 10.
• The neutral atom must have 11 electrons.

An element with 11 protons is sodium, Na.

For \(B^{2+}\):

• Number of protons = 12.
• Number of electrons in the ion = 10.
• The neutral atom must have 12 electrons.

An element with 12 protons is magnesium, Mg.

Therefore:

\[
\boxed{A=\mathrm{Na},\qquad B=\mathrm{Mg}}
\]

(b) Determine the valence-electron counts of the neutral elements.

Neutral sodium has the electron arrangement:

\[
2,8,1
\]

It therefore has one valence electron.

Neutral magnesium has:

\[
2,8,2
\]

It therefore has two valence electrons.

Thus:

\[
\boxed{\mathrm{Na}:1,\qquad \mathrm{Mg}:2}
\]

(c) Explain the matching jump positions and different ionisation energies.

Both ions have ten electrons.

Their electron configurations are identical:

\[
1s^2\,2s^2\,2p^6
\]

Their shell arrangement is:

\[
2,8
\]

This means both ions have eight electrons in their outermost occupied shell.

The first eight electrons removed therefore come from the second shell. The ninth electron must come from the first shell, producing a large increase in ionisation energy.

This explains why the jump occurs at the same position for both ions.

However, their nuclear charges are different.

• \(A^+\) has 11 protons.
• \(B^{2+}\) has 12 protons.

Both ions have the same number of electrons, but \(B^{2+}\) has one additional proton.

Its electrons therefore experience a stronger effective attraction to the nucleus. More energy is needed to remove them.

This explains why the successive ionisation energies of \(B^{2+}\) are higher.

The deeper lesson is that a large jump identifies a boundary in the electron arrangement of the species actually being ionised.

It does not automatically tell you the valence-electron count of a different species from which that ion was formed.

07What to remember when interpreting ionisation energy data

When answering an HSC Chemistry question, keep these distinctions clear:

Once you can interpret the first major jump, the next useful step is connecting the result to the full spdf electron configuration. That lets you move beyond simply counting valence electrons and explain why particular elements form particular ions and show different chemical behaviour.