How to Predict an Unfamiliar Element’s Properties from Its Position

In 1871, Dmitri Mendeleev predicted that an undiscovered element should exist directly below silicon in the periodic table. He called it eka-silicon and predicted that its density would be approximately \(5.5\ \mathrm{g\,cm^{-3}}\). Fifteen years later, germanium was discovered. Its measured density was approximately \(5.47\ \mathrm{g\,cm^{-3}}\).

How could Mendeleev predict a property of an element that nobody had identified yet?

The answer is that an element’s position reveals patterns in its atomic structure. By reading its period and group, then considering nuclear charge, electron shielding, and electron configuration, you can predict several important properties before knowing the element’s identity.

For HSC Chemistry, the most useful predictions involve atomic radius, first ionisation energy, electronegativity, metallic character, and similarities in chemical behaviour.

The key is to understand why these properties change, rather than memorising arrows on a periodic table.

01Why the position of an element tells you about its properties

Mendeleev’s original predictions came from observing repeating patterns in the properties of known elements. He did not know about protons, electron shells, or orbitals in their modern form.

The modern explanation comes from atomic structure.

An element’s position gives you two particularly useful pieces of information:

• Its period tells you the highest occupied principal energy level.
• Its group tells you the number and general arrangement of its valence electrons, provided you are dealing with a main-group element.

Valence electrons are the electrons in the outermost occupied energy level. They are particularly important because they are involved in chemical bonding and influence how an element behaves.

Reading the period

The period is the horizontal row of the periodic table.

For example:

An element in Period 3 has three occupied electron shells, with its outermost electrons in the third principal energy level.

An element in Period 4 has four occupied shells.

This distinction becomes important when comparing atomic sizes and the attraction between the nucleus and outer electrons.

Reading the group

For the main-group elements, the group tells you how many valence electrons an atom has.

Here, \(n\) represents the principal energy level, which is determined by the period.

There is one important exception: helium is in Group 18 but has only two electrons, with configuration \(1s^2\). Its first shell is full with two electrons.

For example, an element in Period 3, Group 16 has:

• Three occupied electron shells.
• Six valence electrons.
• An outer electron configuration of \(3s^2 3p^4\).

That already tells you quite a lot. It belongs to the non-metal region of the periodic table and has an outer electron arrangement similar to other Group 16 elements.

Important limitation: This group-number shortcut applies to the main-group elements. Transition metals in Groups 3-12 have more complicated electron configurations and variable oxidation states. Do not assume that their group number directly gives their number of valence electrons.

02The three factors that control periodic trends

Knowing an element’s electron arrangement is the first step. Now you need to understand why changing that arrangement changes atomic properties.

Three factors matter:

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

Nuclear charge

The nucleus contains positively charged protons and neutral neutrons.

The number of protons determines the nuclear charge. Each additional proton increases the positive charge of the nucleus.

For example:

• Magnesium has 12 protons.
• Aluminium has 13 protons.
• Silicon has 14 protons.

As you move across a period, nuclear charge increases because each successive element has one additional proton.

The positively charged nucleus attracts negatively charged electrons.

Therefore, increasing nuclear charge generally increases the attraction between the nucleus and its electrons.

Electron shielding

The outer electrons do not experience the full attractive force of the nucleus.

Why not?

Inner electrons repel outer electrons because they have the same negative charge. This partially reduces the net attraction experienced by the outer electrons.

This effect is called electron shielding.

Think of it as a partial screening effect. Inner electrons do not create a solid barrier around the nucleus, but their presence changes the net electrostatic attraction experienced by electrons further out.

Effective nuclear charge

The combination of nuclear charge and shielding determines the net attractive influence experienced by an electron.

This is described using effective nuclear charge, represented by \(Z_{\mathrm{eff}}\).

A simplified relationship is:

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

Where:

• \(Z_{\mathrm{eff}}\) is the effective nuclear charge.
• \(Z\) is the atomic number, equal to the number of protons.
• \(\sigma\) represents the shielding effect.

This is a simplified model, not an exact calculation obtained by simply counting inner electrons. The actual shielding depends on the electron distribution and orbital arrangement.

For HSC questions, the important skill is explaining how nuclear charge and shielding change when comparing elements.

What happens across a period?

Imagine moving from left to right across Period 3.

The number of protons increases, but the added electrons enter the same principal energy level.

Shielding increases slightly, but it does not increase enough to cancel the additional nuclear charge.

The effective nuclear attraction therefore generally increases.

The outer electrons are pulled closer to the nucleus.

What happens down a group?

Now imagine moving from lithium to sodium to potassium.

Each successive element has an additional occupied electron shell.

Its outer electrons are further from the nucleus and experience greater shielding from inner electrons.

Although nuclear charge increases, the increased distance and shielding reduce the net attraction experienced by the outer electrons.

This gives us the foundation for predicting periodic trends.

03Use the trends to predict an unfamiliar element’s properties

The following table is one of the most useful references for making periodic predictions.

Let’s connect each trend to the underlying atomic structure.

Atomic radius

Atomic radius describes the approximate size of an atom.

An atom does not have a perfectly defined outer edge because its electrons occupy probability distributions. Atomic radius is therefore determined using an agreed measurement convention, such as covalent radius.

Across a period: atomic radius generally decreases.

As you move from left to right:

• Nuclear charge increases.
• Electrons enter the same principal energy level.
• Shielding increases relatively little.
• Effective nuclear attraction increases.

The stronger attraction pulls the electron cloud closer to the nucleus.

Therefore, atomic radius decreases.

Down a group: atomic radius generally increases.

As you move down:

• Additional electron shells are occupied.
• Outer electrons are further from the nucleus.
• Shielding increases.
• The outer electron cloud extends further out.

Therefore, atomic radius increases.

A useful way to remember this is that the largest main-group atoms are generally found towards the bottom-left of the periodic table.

First ionisation energy

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:

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

It is measured in \(\mathrm{kJ\,mol^{-1}}\).

Think about what happens when an electron is removed.

If the nucleus attracts the outer electron strongly, more energy is required to remove it.

If the electron is further from the nucleus and more heavily shielded, less energy is generally required.

Across a period: first ionisation energy generally increases because the increasing effective nuclear attraction makes the outer electrons harder to remove.

Down a group: first ionisation energy generally decreases because the outer electrons are further away and more shielded.

There are important exceptions to the across-period trend, which we will examine shortly.

Electronegativity

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

It is not the same as first ionisation energy.

Ionisation energy concerns removing an electron completely from a gaseous atom. Electronegativity concerns attracting electrons that are being shared in a chemical bond.

Across a period: electronegativity generally increases because increasing effective nuclear attraction makes the atom better able to attract shared bonding electrons.

Down a group: electronegativity generally decreases because the outer electron region is further from the nucleus and experiences greater shielding.

Fluorine is the most electronegative element.

One limitation is that conventional Pauling electronegativity values are generally not assigned to noble gases such as neon and argon because they rarely form ordinary covalent bonds.

Metallic character

Metallic character describes an element’s tendency to exhibit metallic behaviour, particularly its tendency to lose electrons.

Across a period: metallic character generally decreases.

The increasing effective nuclear attraction makes it harder for atoms to lose electrons.

Down a group: metallic character generally increases.

The outer electrons become easier to remove because they are further from the nucleus and more heavily shielded.

This is why the lower elements in a metallic group generally show stronger metallic character than those above them.

Do not confuse metallic character with every physical property of a metal. Melting point, density, and conductivity depend on additional factors, including bonding and structure.

04Important exceptions: when the simple trends fail

If you memorise that first ionisation energy increases across a period, you might confidently predict that aluminium has a higher first ionisation energy than magnesium.

That prediction is wrong.

Magnesium actually has a higher first ionisation energy than aluminium.

The reason is that the arrangement of electrons within subshells matters, not just the number of protons.

There are two particularly important exceptions.

Exception 1: Magnesium and aluminium

Consider their electron configurations.

Magnesium:

\[ \ce{Mg}: [Ne]\,3s^2 \]

Aluminium:

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

Magnesium’s first electron is removed from the \(3s\) subshell.

Aluminium’s first electron is removed from the \(3p\) subshell.

The \(3p\) electron in aluminium is higher in energy and less penetrating than a \(3s\) electron. It is therefore easier to remove.

This effect outweighs the additional nuclear charge in aluminium.

Consequently:

\[ IE_1(\mathrm{Mg})>IE_1(\mathrm{Al}) \]

Exception 2: Phosphorus and sulfur

Now consider phosphorus and sulfur.

Phosphorus:

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

Sulfur:

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

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

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

The two electrons in that orbital repel each other because they have the same charge.

This additional repulsion makes one of the paired electrons easier to remove.

Therefore:

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

Despite sulfur having an additional proton.

These exceptions are important because they show that the periodic trend is a general pattern, not an absolute rule.

They also demonstrate why you should examine electron configuration when a question asks you to explain an unexpected first ionisation energy.

The same types of exceptions occur between beryllium and boron, and between nitrogen and oxygen.

Importantly, these exceptions concern the energy required to remove a particular electron. They do not mean that the general trends in atomic radius and electronegativity automatically reverse.

For a deeper comparison of these two measurements, see How to Distinguish Electronegativity from Ionisation Energy.

05A practical method for predicting unfamiliar elements

When an exam gives you an unfamiliar element’s position, use this sequence.

Step 1: Identify its period and group.

Determine its number of occupied shells and valence electrons. First check whether it is a main-group element.

Step 2: Write its outer electron configuration.

Use the period to identify the principal energy level and the group to determine the outer electron arrangement.

Step 3: Identify the property being compared.

Are you predicting atomic radius, first ionisation energy, electronegativity, or metallic character?

Each property responds to nuclear attraction in a slightly different way.

Step 4: Decide whether you are moving across a period or down a group.

Consider how nuclear charge, shielding, and distance change.

Step 5: Check for exceptions.

For first ionisation energy, examine the subshell and whether electrons are paired.

For electronegativity, remember that it concerns bonding electrons, not the removal of an electron.

Step 6: State a cause-and-effect explanation.

Do not simply write that a property increases or decreases.

Explain what happens to the attraction between the nucleus and the relevant electrons, then connect that change to the property.

For example:

Element A has a smaller atomic radius than element B because both occupy the same period, but A is further to the right. Its greater nuclear charge, combined with relatively little additional shielding, produces a stronger effective nuclear attraction. This pulls its electron cloud closer to the nucleus.

That is the difference between identifying a trend and explaining it.

06Worked examples

07Questions and solutions

Question 1

An unfamiliar main-group element X is located in Period 3, Group 17.

Determine:

a) The number of occupied electron shells.

b) The number of valence electrons.

c) Its outer electron configuration.

d) Whether it is most likely a metal or non-metal.

Solution 1

X has three occupied shells, seven valence electrons, an outer configuration of \(3s^2 3p^5\), and is a non-metal.

The period determines the number of occupied shells. Since X is in Period 3, its outer electrons occupy the third principal energy level.

Group 17 elements have seven valence electrons.

Their general outer configuration is \(ns^2np^5\).

Substituting \(n=3\):

\[ 3s^2 3p^5 \]

Group 17 is located on the right-hand side of the periodic table, where the non-metals are found.

This is a useful starting prediction because the seven valence electrons indicate that X is one electron short of a filled outer shell.

Question 2

Two unfamiliar main-group elements are located in the same period.

• Element A is in Period 3, Group 14.
• Element B is in Period 3, Group 17.

Predict which element has:

a) The larger atomic radius.

b) The higher first ionisation energy.

c) The higher electronegativity.

Explain all three predictions using nuclear charge and shielding.

Solution 2

Element A has the larger atomic radius, while element B has the higher first ionisation energy and electronegativity.

Both elements are in Period 3, so their outer electrons occupy the same principal energy level.

However, B is further to the right.

Moving from A to B increases the number of protons in the nucleus. Although additional electrons are added, they enter the same principal energy level, so shielding does not increase enough to cancel the greater nuclear charge.

Part a: Atomic radius

B experiences greater effective nuclear attraction.

Its electron cloud is pulled closer to the nucleus.

Therefore:

\[ r_A>r_B \]

where \(r_A\) and \(r_B\) represent the atomic radii of A and B.

Part b: First ionisation energy

The stronger nuclear attraction makes B’s outer electrons harder to remove.

Therefore:

\[ IE_1(B)>IE_1(A) \]

There is no relevant Group 2-to-13 or Group 15-to-16 exception in this comparison that reverses the overall ranking.

Part c: Electronegativity

B is further to the right and has a greater effective nuclear attraction.

It therefore generally attracts shared bonding electrons more strongly.

Consequently:

\[ \chi_B>\chi_A \]

where \(\chi\) represents electronegativity.

The important point is that all three predictions follow from the same change in effective nuclear attraction, but the properties describe different aspects of the atom.

Question 3

Two neighbouring elements in Period 3 are labelled X and Y.

• X is in Group 2.
• Y is in Group 13.

A student predicts that Y must have a higher first ionisation energy because it has one additional proton.

a) Is the prediction correct?

b) Which element has the smaller atomic radius?

c) Which element has the higher electronegativity?

Explain why the first ionisation energy comparison produces an unexpected result.

Solution 3

The student’s prediction is incorrect. X has the higher first ionisation energy, but Y has the smaller atomic radius and higher electronegativity.

The relevant electron configurations are:

\[ X: [Ne]\,3s^2 \]

\[ Y: [Ne]\,3s^2 3p^1 \]

Part a: First ionisation energy

Although Y has one additional proton, the first electron removed from Y occupies a \(3p\) subshell.

The \(3p\) electron is higher in energy and less penetrating than the \(3s\) electron removed from X.

It is therefore easier to remove.

The resulting relationship is:

\[ IE_1(X)>IE_1(Y) \]

The important point is that ionisation energy depends on the particular electron being removed, not just the total nuclear charge.

Part b: Atomic radius

Y is further to the right in the same period.

The greater effective nuclear attraction pulls its electron cloud closer to the nucleus.

Therefore, Y has the smaller atomic radius.

Part c: Electronegativity

Y also has the greater effective nuclear attraction across the period.

It therefore generally attracts shared bonding electrons more strongly.

Hence:

\[ \chi_Y>\chi_X \]

There is no contradiction here.

First ionisation energy and electronegativity measure different things. The higher-energy \(3p\) electron makes Y easier to ionise, but Y still has a stronger general attraction for shared bonding electrons.

This is a good example of why you should not use one periodic trend as a substitute for another.

Question 4

Two unfamiliar main-group elements, P and Q, are located in Group 16.

• P is in Period 3.
• Q is in Period 4.

Compare their:

a) Number of valence electrons.

b) Atomic radii.

c) First ionisation energies.

d) Electronegativities.

e) Likely chemical similarities.

Explain each comparison using electron shells and shielding.

Solution 4

P and Q both have six valence electrons, but Q has a larger atomic radius, a generally lower first ionisation energy, and a generally lower electronegativity.

Part a: Number of valence electrons

Both elements belong to Group 16.

Therefore, both have six valence electrons and the general outer configuration:

\[ ns^2np^4 \]

Their chemical behaviour is expected to show some similarities because their outer electron arrangements are similar.

Part b: Atomic radius

Q is one period below P.

It has an additional occupied electron shell, increasing the distance between its nucleus and its outer electrons.

Therefore:

\[ r_Q>r_P \]

Part c: First ionisation energy

The outer electrons of Q are further from the nucleus and experience greater shielding.

They are therefore generally easier to remove.

Consequently:

\[ IE_1(P)>IE_1(Q) \]

Part d: Electronegativity

P is higher in the group.

Its smaller atomic radius and lower shielding generally allow it to attract shared bonding electrons more strongly.

Therefore:

\[ \chi_P>\chi_Q \]

Part e: Chemical similarities

Both elements have six valence electrons.

This means they have similar outer electron arrangements and may show similar patterns in the compounds they form.

However, their properties will not be identical. Q has a larger atomic radius and experiences different shielding and nuclear attraction.

The key distinction is that group position helps predict similarities in electron arrangement, while period position helps explain differences in the strength of electron attraction.

Question 5

Two neutral, ground-state main-group elements are labelled A and B.

Their positions are:

• A: Period 2, Group 14.
• B: Period 3, Group 16.

The following approximate reference data are supplied.

The radii are covalent radii, measured using the same convention. The ionisation energies refer to gaseous neutral atoms, and electronegativity values use the Pauling scale.

These are rounded reference values for the elements occupying the specified positions. See the Royal Society of Chemistry’s carbon data and sulfur data.

A student makes the following argument:

B is further to the right, so it must have a smaller atomic radius and a higher first ionisation energy. A is higher in the periodic table, so it must also have a higher electronegativity.

The student has applied three familiar periodic trends. However, the predictions do not all agree with the supplied data.

a) Determine the outer electron configuration of A and B.

b) Use the data to rank their atomic radii, first ionisation energies, and electronegativities.

c) Explain why moving from A to B creates competing effects involving nuclear charge, shielding, and electron-shell distance. Identify which effect dominates each property.

d) Explain why neither element’s density can be confidently predicted from its position alone.

Solution 5

A has the smaller atomic radius and higher first ionisation energy. B has a slightly higher electronegativity. The apparent disagreement occurs because moving diagonally across the periodic table changes both nuclear charge and electron-shell distance.

This question is more difficult because neither element is simply further right in the same period or further down in the same group.

You must consider both directions simultaneously.

Part a: Electron configurations

A is in Period 2, Group 14.

It has four valence electrons, giving the outer configuration:

\[ 2s^2 2p^2 \]

B is in Period 3, Group 16.

It has six valence electrons, giving the outer configuration:

\[ 3s^2 3p^4 \]

Notice that B has an additional occupied electron shell.

It also has a greater nuclear charge.

These two changes have competing effects on the properties we are comparing.

Part b: Ranking the properties

The supplied data give the following results.

Atomic radius:

\[ r_A<r_B \]

First ionisation energy:

\[ IE_1(A)>IE_1(B) \]

Electronegativity:

\[ \chi_B>\chi_A \]

The electronegativity difference is small:

\[ 2.58-2.55=0.03 \]

Therefore, the two elements have very similar electronegativities, even though B is slightly higher on the supplied scale.

Part c: Why the predictions differ

Consider each property separately.

Atomic radius

Moving from A to B introduces an additional occupied electron shell.

B’s outer electrons are further from the nucleus and experience more shielding.

Although B has a greater nuclear charge, the additional shell has the dominant effect on atomic size.

Therefore, B has the larger atomic radius.

The student’s prediction that B must be smaller simply because it is further right is incorrect.

First ionisation energy

B’s outer electrons occupy the third principal energy level, rather than the second.

They are further from the nucleus and more heavily shielded.

This makes them easier to remove, despite B having more protons.

Consequently, A has the higher first ionisation energy.

The student’s prediction ignores the effect of the additional shell.

Electronegativity

This comparison is particularly interesting.

Moving to the right generally increases electronegativity because nuclear charge increases.

Moving down generally decreases electronegativity because the outer electrons are further from the nucleus and more shielded.

These effects oppose one another in this diagonal comparison.

The supplied data show that B is slightly more electronegative, but the difference is only 0.03 on the Pauling scale.

You should not confidently predict this small difference from the position alone.

This is the important lesson: a trend that works clearly across a period or down a group may not provide a reliable ranking when two elements are separated diagonally.

Part d: Predicting density

Density cannot be confidently predicted from the periodic-table position alone.

Density is defined as:

\[ \rho=\frac{m}{V} \]

where:

• \(\rho\) is density.
• \(m\) is mass.
• \(V\) is volume.

The periodic table gives information about atomic structure and approximate atomic mass, but density also depends on how atoms are arranged and packed in a material.

Carbon provides a useful example. It occurs as both graphite and diamond, which have substantially different densities despite being forms of the same element.

Therefore, knowing an element’s position is not enough to calculate its density without additional structural information.

The tempting mistake in this question is to treat the periodic table as though every property follows one universal arrow.

It doesn’t.

You need to identify the property, determine which structural factors influence it, and decide which effect is dominant. When the effects compete closely, additional data may be necessary.

08Taking your predictions further

The periodic table is useful because it lets you make predictions before you know every detail about an element.

The most reliable approach is to move through three levels of reasoning:

1. Use the period and group to predict electron arrangement.
2. Use nuclear charge, shielding, and distance to predict electron attraction.
3. Apply those predictions to the particular property being investigated, checking for exceptions.

Remember that position provides a starting model, not a complete description of an element.

Once you can predict how strongly an atom holds its outer electrons, the next step is to predict what it does with them in a chemical reaction: lose, gain, or share electrons.

That is where periodic trends connect directly to Chemical Properties of Elements for HSC Chemistry.

Syllabus note: The NSW Chemistry 11-12 Syllabus (2025) is scheduled for implementation from 2028, with its first HSC examination in 2029. Check the NESA syllabus implementation information for the requirements applying to your cohort.