How to Find an Element’s Block from Electron Configuration

An ion has the electron configuration \([\mathrm{Ar}]\,3d^{10}\). The last subshell shown is \(d\), so does that mean its element belongs to the \(d\)-block?

Not necessarily. A \(\mathrm{Ga}^{3+}\) ion has that configuration, but gallium belongs to the \(p\)-block. Its outer \(4p\) and \(4s\) electrons have been removed, leaving a configuration that no longer shows where the neutral atom sits in the periodic table.

To locate an element’s block, use the ground-state electron configuration of its neutral atom. Identify the element, then determine which subshell-filling region of the periodic table contains it:

• \(s\)-block: the two left-hand columns (Groups 1-2).
• \(p\)-block: the six right-hand columns (Groups 13-18), with helium as a special case.
• \(d\)-block: the ten central columns (Groups 3-12).
• \(f\)-block: the two series usually displayed below the main table.

For straightforward neutral configurations, the last subshell being filled usually tells you the block immediately. Ions, chromium, copper, and zinc are the cases that make this shortcut less reliable.

01Why the periodic table has blocks

Imagine filling the fourth period of the periodic table from left to right, starting with potassium.

Potassium begins with one electron in \(4s\). Calcium has two electrons in \(4s\). The next element, scandium, begins the series in which the \(3d\) subshell is populated.

After the central \(d\)-block series ends at zinc, gallium begins the \(4p\) series.

That’s why Period 4 has this arrangement:

\[ \underbrace{4s}_{2\text{ columns}} \quad \underbrace{3d}_{10\text{ columns}} \quad \underbrace{4p}_{6\text{ columns}} \]

The widths come from the maximum number of electrons in each subshell. An \(s\) subshell contains one orbital and holds two electrons. A \(p\) subshell contains three orbitals and holds six electrons. A \(d\) subshell contains five orbitals and holds ten electrons.

Each orbital holds a maximum of two electrons.

This is the first useful mental model: the blocks represent different regions of subshell filling.

But neighbouring elements are separate atoms, not one atom receiving electrons in a continuous experiment. When nuclear charge changes, the relative energies of subshells can also change. We’ll see why this qualification matters for copper and zinc.

A quick reference table

Here \(n\) is the principal shell number, such as 3 in \(3p\) or 4 in \(4s\). The superscript tells you how many electrons occupy the subshell.

The \(f\)-block completes the periodic-table picture, but the worked atom and ion configurations in this article stay within the first 36 elements, up to krypton.

Why \(4s\) fills before \(3d\)

A common prediction is that an atom should fill every part of shell 3 before starting shell 4.

That would give potassium a \(3d\) electron. But potassium’s neutral ground-state configuration is

\[ \ce{K}:\quad [\mathrm{Ar}]\,4s^1 \]

The usual filling sequence for neutral atoms through krypton is

\[ 1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p \]

The \(4s\) subshell begins filling before \(3d\) because of the relative energies of the available electron states.

However, this is a useful ground-state filling rule, not an unchanging energy ranking. Once \(3d\) electrons are present, electron interactions and the energies of the states change. That helps explain why transition-metal ions generally lose \(4s\) electrons before \(3d\) electrons.

If you want to revise the notation itself, spdf Notation: Electron Configurations and Valence Electrons explains the electron capacities and filling rules in more detail.

02How experiments revealed the structure behind electron configurations

The arrangement into blocks was not discovered simply by drawing columns on a periodic table.

Chemists first observed repeating chemical properties. Understanding why those patterns existed required evidence that electrons occupy distinct energy states.

The first clue: atoms emit particular wavelengths of light

Consider a gas-discharge experiment.

A transparent tube contains a low-pressure gas between two electrodes connected to a high-voltage supply. Electrical energy excites the gas, causing it to emit light.

That light passes through a narrow slit and then a diffraction grating, which separates the wavelengths. The separated light is observed as a spectrum.

If electrons could change energy by any amount, you might expect a continuous range of emitted wavelengths.

Instead, an excited atomic gas produces a line spectrum: distinct bright lines at particular wavelengths.

Each emitted photon carries energy equal to the difference between two atomic energy states:

\[ \Delta E=hf \]

Here \(\Delta E\) is the energy difference in joules (J), \(h\) is Planck’s constant in joule-seconds (J s), and \(f\) is the light frequency in hertz (Hz).

The separate spectral lines support the conclusion that atoms have discrete energy states.

But emission spectra have a limitation: they reveal energy differences associated with transitions. They do not directly tell us how many electrons occupy every \(s\), \(p\), or \(d\) subshell in the ground state.

A later technique, photoelectron spectroscopy, can probe occupied electron states more directly.

Photoelectron spectroscopy: measuring how tightly electrons are held

In a simplified gas-phase photoelectron spectroscopy experiment:

1. A beam of photons with known energy enters an evacuated chamber containing gaseous atoms.
2. Some photons eject electrons from the atoms.
3. The emitted electrons travel into an energy analyser and detector.
4. Their measured kinetic energies allow scientists to infer the binding energies of the original electrons.

The binding energy tells us how much energy is required to remove an electron from a particular state.

For a simplified gas-phase measurement, neglecting atomic recoil and instrumental corrections, energy conservation gives

\[ E_{\mathrm{bind}} = E_{\mathrm{photon}}-E_{\mathrm{kinetic}} \]

All three energies are expressed in the same units.

Suppose an illustrative experiment uses photons with energies of \(100\ \mathrm{eV}\), and an emitted electron has a kinetic energy of \(78\ \mathrm{eV}\).

Then

\[ \begin{aligned} E_{\mathrm{bind}} &=100-78\\ &=22\ \mathrm{eV} \end{aligned} \]

The electron was bound by approximately \(22\ \mathrm{eV}\). An electronvolt (eV) is a unit of energy.

These are illustrative values, not a reported experimental measurement.

In simplified, sensitivity-corrected spectra, the relative areas of the peaks can indicate the populations of different subshells. Actual spectra require more careful interpretation because photoionisation probabilities, overlapping features, and detector response affect peak intensities.

The important conclusion is that the electron arrangement contains more structure than a simple count of electrons in each shell.

Spectroscopy supports models with distinct electron energies. Quantum mechanics provides the orbital descriptions, and the periodic arrangement of atomic numbers connects those configurations to the \(s\)-, \(p\)-, \(d\)-, and \(f\)-blocks.

It is not a photograph of electrons moving through fixed little compartments.

03The reliable method for identifying a neutral atom’s block

For a neutral atom, begin with its complete ground-state configuration.

Count the electrons if the element is unknown, identify its atomic number, and use its location in the periodic table to confirm the block.

For a simple \(s\)- or \(p\)-block atom, the final occupied subshell usually makes the classification obvious.

For a \(d\)-block atom, remember that the subshell being filled across the central series is generally \((n-1)d\), while \(ns\) electrons are also present.

The order in which someone writes \(4s\) and \(3d\) terms is not a reliable block test.

04Why ions can hide an element’s block

Ion formation changes the number of electrons, but not the number of protons.

That means an element does not move to a new periodic-table block when it forms an ion.

Consider iron:

\[ \ce{Fe}:\quad [\mathrm{Ar}]\,3d^6\,4s^2 \]

Iron is element 26 and belongs to the \(d\)-block.

When a typical iron atom forms an iron(II) ion, it loses its two \(4s\) electrons:

\[ \mathrm{Fe}^{2+}:\quad [\mathrm{Ar}]\,3d^6 \]

The neutral atom and ion have different occupied subshells, but both have 26 protons. The element remains iron.

A further electron can be removed from \(3d\):

\[ \mathrm{Fe}^{3+}:\quad [\mathrm{Ar}]\,3d^5 \]

Counting electrons in an ion

Use

\[ Z=N_e+q \]

where:

• \(Z\) is the atomic number, or number of protons.
• \(N_e\) is the number of electrons in the ion.
• \(q\) is the signed ionic charge expressed in units of the elementary charge: \(+2\) for a \(2+\) ion, or \(-1\) for a \(1-\) ion.

This is an equation involving particle counts. The numerical value of \(q\) is not a charge measured directly in coulombs.

For a cation, \(Z\) is greater than its electron count. For an anion, \(Z\) is smaller.

The crucial step comes next: once you identify \(Z\), reconstruct the neutral atom’s ground-state configuration. Do not simply add electrons to whichever subshell appears last in the ion.

05Exceptions that can defeat the last-subshell shortcut

Chromium and copper: the simple filling prediction needs correcting

The usual filling sequence predicts a convenient pattern for the early \(3d\) elements. But the actual ground-state configurations of chromium and copper differ from the simplest Aufbau predictions.

For chromium, the simple prediction would be \(3d^4\,4s^2\). Instead, its ground state has \(3d^5\,4s^1\).

For copper, the simple prediction would be \(3d^9\,4s^2\). Instead, copper has \(3d^{10}\,4s^1\).

The actual arrangements result from the total energies of the competing electron configurations, including electron-electron interactions. Saying that half-filled and filled \(d\) subshells are relatively favourable is a useful first explanation, but it is not a universal rule that overrides all other energy effects.

These are established ground-state configurations, not alternative ways of writing the same arrangement. They are included in the NIST atomic electron-configuration reference.

Zinc: the more subtle exception

Zinc belongs to the \(d\)-block, even though its neutral configuration has a completely filled \(3d^{10}\) subshell and a filled \(4s^2\) subshell.

It is the Group 12 element at the end of the first \(d\)-block series.

Compare copper and zinc:

\[ \begin{aligned} \ce{Cu}:&\quad [\mathrm{Ar}]\,3d^{10}\,4s^1\\ \ce{Zn}:&\quad [\mathrm{Ar}]\,3d^{10}\,4s^2 \end{aligned} \]

Zinc has one more \(4s\) electron than copper. Nevertheless, zinc remains in the \(d\)-block.

This shows why the statement ‘an element’s block is always the subshell that receives the extra electron compared with the previous neutral atom’ is too simplistic.

Block classification refers to the element’s place in the periodic table and the associated subshell-filling series, not a literal reconstruction of how each electron entered its atom.

It also helps to distinguish a \(d\)-block element from a transition metal. Under the usual chemical definition, zinc is not classified as a transition metal because its atom and common \(\mathrm{Zn}^{2+}\) ion both have filled \(3d\) subshells.

Helium: an \(s\)-type configuration in Group 18

Helium has

\[ \ce{He}:\quad 1s^2 \]

Its electrons occupy an \(s\) subshell. By electron-configuration classification, helium is an \(s\)-block element.

It is positioned in Group 18 because its shell is complete and its chemical behaviour resembles that of the other noble gases.

So the position of helium above the right-hand side of many periodic tables is not evidence that it contains \(p\) electrons.

06Questions and solutions

Assume ground-state configurations throughout, unless a question explicitly describes a competing hypothetical model. Use the conventional periodic-table blocks and the standard school-level conventions for forming common ions.

Question 1

A neutral atom has the electron configuration

\[ [\mathrm{Ne}]\,3s^2\,3p^3 \]

Identify the element, its atomic number, and its block.

Solution 1

The element is phosphorus, atomic number 15, and it belongs to the \(p\)-block.

The neon core contains 10 electrons. Adding the remaining electrons gives

\[ \begin{aligned} N_e&=10+2+3\\ &=15\text{ electrons} \end{aligned} \]

A neutral atom has equal numbers of electrons and protons, so \(Z=15\). This identifies phosphorus.

Its outer subshells are \(3s^2\,3p^3\). The \(3p\) subshell is being filled in Period 3, placing phosphorus in Group 15 of the \(p\)-block.

The complete shell contains five outer electrons. Looking at only the \(3p^3\) term would miss the two \(3s\) electrons, although it still identifies the correct block.

Question 2

Two unidentified ions, \(\mathrm{X}^{2+}\) and \(\mathrm{Y}^{-}\), both have the electron configuration

\[ [\mathrm{Ar}] \]

A student concludes that both original elements must be in the \(p\)-block because their ions have a filled \(3p^6\) subshell.

Another student concludes that both original elements must belong to Period 3 because their ions have no occupied fourth shell.

Identify X and Y, give their neutral ground-state configurations, and determine their blocks and periods. Explain why both students’ arguments fail.

Solution 2

X is calcium, an \(s\)-block element in Period 4. Y is chlorine, a \(p\)-block element in Period 3. Their ions have identical electron configurations, but their neutral atoms do not.

For X:

The argon configuration contains 18 electrons.

Because \(\mathrm{X}^{2+}\) has a charge of \(+2\),

\[ \begin{aligned} Z_X&=18+2\\ &=20 \end{aligned} \]

Element 20 is calcium.

Its neutral ground-state configuration is

\[ \ce{Ca}:\quad [\mathrm{Ar}]\,4s^2 \]

Calcium is in Group 2, Period 4, and the \(s\)-block.

When calcium loses its two \(4s\) electrons, it becomes \(\mathrm{Ca}^{2+}\), with the argon configuration.

For Y:

The ion \(\mathrm{Y}^{-}\) also contains 18 electrons, but its charge is \(-1\).

\[ \begin{aligned} Z_Y&=18+(-1)\\ &=17 \end{aligned} \]

Element 17 is chlorine.

Its neutral ground-state configuration is

\[ \ce{Cl}:\quad [\mathrm{Ne}]\,3s^2\,3p^5 \]

Chlorine is in Group 17, Period 3, and the \(p\)-block.

It gains one electron to complete \(3p^6\), forming \(\mathrm{Cl}^{-}\).

The first student’s argument fails because the configuration of an ion is not necessarily the configuration that determines its parent’s block.

The second student’s argument fails because ionisation can remove an entire occupied outer shell. Calcium begins in Period 4, even though \(\mathrm{Ca}^{2+}\) has no fourth-shell electrons.

The two ions are isoelectronic: they have the same electron count and configuration. Their nuclei still contain different numbers of protons, so they remain different elements.

Question 3

A hypothetical photoelectron spectroscopy experiment investigates a neutral atom X in Period 4.

The inner electrons match an argon core. Analysis of the outer-region spectrum gives the following simplified, sensitivity-corrected subshell populations.

The \(3d\) signal accounts for the largest number of electrons.

An analyst therefore identifies X as a \(d\)-block element.

Determine the actual element and block. Then predict the configurations of \(\mathrm{X}^{2+}\) and \(\mathrm{X}^{4+}\), assuming electrons are removed from \(4p\) before \(4s\), with the filled \(3d\) subshell unchanged.

Explain what the experiment reveals and why the analyst’s interpretation is incorrect.

Solution 3

X is germanium, atomic number 32, and it belongs to the \(p\)-block. Its \(2+\) ion has \([\mathrm{Ar}]\,3d^{10}\,4s^2\), while its \(4+\) ion has \([\mathrm{Ar}]\,3d^{10}\).

First, count all the electrons using the corrected subshell populations:

\[ \begin{aligned} N_e&=18+10+2+2\\ &=32\text{ electrons} \end{aligned} \]

Because X is neutral, \(Z=32\).

Element 32 is germanium, with the configuration

\[ \ce{Ge}:\quad [\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^2 \]

Germanium lies in Group 14, Period 4, so it is a \(p\)-block element.

For the \(2+\) ion, remove the two \(4p\) electrons:

\[ \mathrm{Ge}^{2+}:\quad [\mathrm{Ar}]\,3d^{10}\,4s^2 \]

The ion contains

\[ 18+10+2=30\text{ electrons} \]

which agrees with \(32-2=30\).

For the \(4+\) ion, remove the two remaining \(4s\) electrons:

\[ \mathrm{Ge}^{4+}:\quad [\mathrm{Ar}]\,3d^{10} \]

This leaves

\[ 18+10=28\text{ electrons} \]

which agrees with \(32-4=28\).

The experiment provides evidence about the number of electrons occupying different subshells, provided the spectral signals are correctly assigned and calibrated.

The analyst’s mistake is assuming that the most populated subshell determines the block.

The \(3d\) subshell happens to contain ten electrons because it is full. But germanium lies in the \(4p\) series of the periodic table.

Ion formation makes the mistake even more tempting: the \(\mathrm{Ge}^{4+}\) ion has no \(4p\) electrons left, yet its parent element remains in the \(p\)-block.

Question 4

Chromium has atomic number 24.

A student predicts the neutral configuration

\[ \text{Model A: } [\mathrm{Ar}]\,3d^4\,4s^2 \]

However, the experimentally established neutral ground-state configuration is

\[ \text{Model B: } [\mathrm{Ar}]\,3d^5\,4s^1 \]

Both models account for 24 electrons.

For this question, apply the usual rule of removing \(4s\) electrons before \(3d\) electrons, without additional rearrangement after removal.

(a) Predict the configurations of \(\mathrm{Cr}^{+}\) and \(\mathrm{Cr}^{2+}\) from Model B.

(b) Show what Models A and B would each predict for \(\mathrm{Cr}^{+}\) and \(\mathrm{Cr}^{2+}\). Explain why observing only the \(2+\) ion would not distinguish the models under these assumptions.

(c) Identify chromium’s block and explain why the rule ‘4s is always lower in energy than 3d’ is not a sufficient explanation of its electronic structure.

Solution 4

Chromium belongs to the \(d\)-block. Its actual neutral ground state is Model B, which gives \(\mathrm{Cr}^{+}\) as \([\mathrm{Ar}]\,3d^5\) and \(\mathrm{Cr}^{2+}\) as \([\mathrm{Ar}]\,3d^4\) under the specified removal convention.

Part (a): Calculate the ions from Model B.

Neutral chromium is

\[ [\mathrm{Ar}]\,3d^5\,4s^1 \]

The first electron is removed from \(4s\):

\[ \mathrm{Cr}^{+}:\quad [\mathrm{Ar}]\,3d^5 \]

That ion contains \(18+5=23\) electrons, as expected for \(24-1=23\).

A second removal now involves a \(3d\) electron:

\[ \mathrm{Cr}^{2+}:\quad [\mathrm{Ar}]\,3d^4 \]

This ion contains \(18+4=22\) electrons, matching \(24-2=22\).

Part (b): Test the competing models.

The two proposed neutral arrangements give:

Both models predict the same \(2+\) configuration using the specified removal procedure.

That means the observation

\[ \mathrm{Cr}^{2+}:\quad[\mathrm{Ar}]\,3d^4 \]

would not, by itself, prove which proposed neutral arrangement was correct.

The \(1+\) predictions differ under this simplified model, so additional information about the neutral atom and its first ion can help distinguish the arrangements.

For real atoms, ionisation can also involve electronic relaxation and rearrangement. This is why configurations of neutral ground states are established using experimental evidence and energy calculations rather than inferred by reversing an oversimplified electron-removal rule.

Part (c): Determine the block and interpret the energy ordering.

Chromium is element 24, in Group 6 of the central \(3d\) series. It is a \(d\)-block element.

The usual filling sequence places \(4s\) before \(3d\), but the actual total energy of an electron arrangement depends on interactions between electrons as well as their attraction to the nucleus.

The relative energies of \(4s\) and \(3d\) are not fixed independently of the atom’s configuration and charge.

The tempting mistake is to treat the Aufbau sequence as a permanent ranking of subshell energies. It is a useful starting model for constructing neutral configurations, not a guarantee that the same order applies throughout ionisation.

Question 5

A student proposes a more sophisticated rule for identifying periodic-table blocks:

‘Compare each neutral atom with the preceding neutral atom. The block of the new element is whichever subshell has gained an electron.’

The student argues that this avoids the problem of reading the final written term of an electron configuration.

Consider the following experimentally established ground-state configurations of four consecutive neutral elements.

(a) Identify the four elements. For each step from one atomic number to the next, determine the net changes in the \(3d\), \(4s\), and \(4p\) electron populations.

(b) Identify the blocks of elements 29, 30, and 31. Determine exactly where the student’s rule fails.

(c) The elements with atomic numbers 29, 30, and 31 form ions with charges \(+1\), \(+2\), and \(+3\), respectively. All three ions have the configuration \([\mathrm{Ar}]\,3d^{10}\). Explain how this is possible.

(d) Propose a reliable classification method that works for all these species and explain why it is more defensible than either using the final occupied subshell of an ion or comparing neighbouring neutral configurations.

Assume the supplied configurations are ground-state configurations of isolated neutral atoms. A comparison of occupancies between different atoms does not represent electrons physically travelling between those atoms.

Solution 5

The student’s rule fails at zinc, atomic number 30. Zinc belongs to the \(d\)-block even though its \(4s\) population increases from one electron in copper to two electrons in zinc. The three resulting ions can also have identical configurations despite coming from elements in different blocks.

Part (a): Identify the atoms and compare their electron populations.

The elements are:

• \(Z=28\): nickel, \(\ce{Ni}\).
• \(Z=29\): copper, \(\ce{Cu}\).
• \(Z=30\): zinc, \(\ce{Zn}\).
• \(Z=31\): gallium, \(\ce{Ga}\).

Now compare the occupied subshells.

From nickel to copper:

\[ \begin{aligned} 3d:&\quad 8\rightarrow10\\ 4s:&\quad 2\rightarrow1 \end{aligned} \]

The net change in the \(3d\) population is \(+2\) electrons, while the net change in the \(4s\) population is \(-1\) electron.

Total change:

\[ +2-1=+1\text{ electron} \]

This agrees with the one-electron difference between neutral atoms of atomic numbers 28 and 29.

However, it is not simply a case of placing one additional electron into an unchanged configuration. The electron populations have reorganised.

From copper to zinc:

\[ \begin{aligned} 3d:&\quad 10\rightarrow10\\ 4s:&\quad 1\rightarrow2 \end{aligned} \]

The only net occupation increase is in \(4s\).

From zinc to gallium:

\[ \begin{aligned} 3d:&\quad10\rightarrow10\\ 4s:&\quad2\rightarrow2\\ 4p:&\quad0\rightarrow1 \end{aligned} \]

The net increase is in \(4p\).

Part (b): Test the proposed rule.

The correct blocks are:

For copper, the large increase in \(3d\) occupation is consistent with its membership of the \(d\)-block, although the redistribution between \(3d\) and \(4s\) already makes a simple one-electron addition model misleading.

For zinc, the proposed rule predicts the \(s\)-block because the net extra electron appears in \(4s\).

But zinc occupies Group 12, the final column of the \(3d\)-block series. Its actual block is \(d\), not \(s\).

For gallium, the rule again agrees with the correct classification: the new electron occupies \(4p\), and gallium begins the \(p\)-block region in Period 4.

The zinc case proves the rule is not universally valid. A classification method must handle every relevant case, not merely the majority.

Part (c): Explain the identical ionic configurations.

Copper begins with

\[ [\mathrm{Ar}]\,3d^{10}\,4s^1 \]

Removing one \(4s\) electron produces

\[ \mathrm{Cu}^{+}:\quad[\mathrm{Ar}]\,3d^{10} \]

Zinc begins with

\[ [\mathrm{Ar}]\,3d^{10}\,4s^2 \]

Removing its two \(4s\) electrons produces

\[ \mathrm{Zn}^{2+}:\quad[\mathrm{Ar}]\,3d^{10} \]

Gallium begins with

\[ [\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^1 \]

Removing the \(4p\) electron and the two \(4s\) electrons produces

\[ \mathrm{Ga}^{3+}:\quad[\mathrm{Ar}]\,3d^{10} \]

Each ion now contains

\[ 18+10=28\text{ electrons} \]

Their electron counts can also be checked directly from their atomic numbers and charges:

\[ \begin{aligned} \mathrm{Cu}^{+}:&\quad29-1=28\\ \mathrm{Zn}^{2+}:&\quad30-2=28\\ \mathrm{Ga}^{3+}:&\quad31-3=28 \end{aligned} \]

All three ions are isoelectronic, but their nuclei contain 29, 30, and 31 protons, respectively.

Their identical electron arrangements do not make them the same element, nor do they place their parents in the same block.

Part (d): State the reliable method.

The most defensible method is to identify the element by its atomic number and classify it using the neutral ground-state electronic structure together with its location in the periodic table.

If only an ion configuration is provided, first use the ionic charge and electron count to recover the number of protons. Then identify the neutral element and its correct ground-state configuration, accounting for known exceptions.

This method works because periodic-table blocks describe patterns across elements arranged by atomic number.

They are not labels that change whenever an atom gains or loses electrons.

The conceptual hinge is zinc: even a rule based on changes between neutral ground-state configurations can fail when the pattern of orbital occupations does not match a simple one-electron filling story.

This is also why electron configuration and periodic position need to be used together. The configuration explains the subshell occupations; the periodic table identifies which block the element belongs to.

Once that distinction is secure, you can use the same neutral-atom and ion reasoning to explain changes in electron attraction, ionisation energy, and atomic size. Those relationships are developed further in Atomic Radius Trends and Electron Configuration for HSC Chemistry.