spdf Notation: Electron Configurations and Valence Electrons

A chlorine atom has 17 electrons. You might already write its shell arrangement as \(2,8,7\). That tells you something important: chlorine has seven electrons in its outer shell. But it hides a detail that becomes essential in HSC Chemistry. Those electrons aren’t simply dumped into three featureless shells.

Inside each shell are subshells, labelled \(s\), \(p\), \(d\), and \(f\). The more precise electron configuration of chlorine is

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^5
\]

Before going further, make a prediction. In \(3p^5\), what do you think the 3, the \(p\), and the 5 each mean?

They describe three different things: which shell, which type of subshell, and how many electrons are in that subshell. Once you can read those three pieces separately, spdf notation becomes much less intimidating.

01What spdf notation is actually showing

Picture an apartment building.

The shell number tells you the floor. The subshell letter tells you the type of room on that floor. The superscript tells you how many electrons are occupying that type of room.

So in \(3p^5\):

• \(3\) means the third electron shell
• \(p\) means the \(p\) subshell
• \(5\) means five electrons occupy that \(3p\) subshell

The apartment analogy is useful because it separates the three jobs performed by the notation. It isn’t exact, though. Electrons aren’t tiny balls sitting in fixed little rooms. Orbitals describe regions associated with particular electron energies and probability distributions.

For HSC Chemistry, the important first step is being able to decode and construct the notation correctly.

02Shells, subshells, and orbitals

A shell is identified by its principal quantum number, \(n\). For the first shell, \(n=1\). For the second shell, \(n=2\), and so on.

Each shell can contain one or more subshells.

The letters \(s\), \(p\), \(d\), and \(f\) are labels for different types of subshell.

Each subshell contains a certain number of orbitals, and each orbital can contain a maximum of two electrons.

This gives the capacities worth knowing:

\[
s^2,\qquad p^6,\qquad d^{10},\qquad f^{14}
\]

The superscripts are electron counts, not powers in the normal mathematical sense. \(3p^4\) doesn’t mean “3p to the power of 4”. It means four electrons in the \(3p\) subshell.

03Reading an electron configuration

Consider sulfur:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^4
\]

You can check the configuration by adding the superscripts:

\[
2+2+6+2+4=16
\]

A neutral sulfur atom has atomic number 16, so it should contain 16 electrons. The count matches.

This is one of the quickest error checks you can use. For a neutral atom:

\[
\text{number of electrons}=\text{atomic number}
\]

For example, if you wrote an electron configuration for phosphorus, atomic number 15, and your superscripts added to 16, something has gone wrong.

If you need a refresher on how electron number changes when atoms become ions, see Electronic Configuration for HSC Chemistry: Atoms and Ions.

04Electrons do not simply fill one shell completely before starting the next

Here’s the part that catches people.

You might predict that electrons fill every available subshell in shell 3 before entering shell 4. That would seem logical if the shell number alone determined energy.

It doesn’t.

Electrons occupy available subshells in order of increasing energy. For the elements you’ll commonly meet, the start of the filling order is

\[
1s,\ 2s,\ 2p,\ 3s,\ 3p,\ 4s,\ 3d,\ 4p,\ 5s,\ldots
\]

Notice the strange-looking part:

\[
4s \text{ fills before } 3d
\]

So potassium, which has 19 electrons, ends with \(4s^1\):

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1
\]

It does not continue directly into \(3d\).

A useful way to think about this is that shell number tells you something about the electron state, but it does not by itself give the complete energy ranking. The \(4s\) and \(3d\) subshells overlap in energy in a way that makes a simple “finish floor 3 before floor 4” model fail.

05How to identify valence electrons from spdf notation

For the main-group elements you usually meet when first learning electron configurations, valence electrons are the electrons in the atom’s outermost occupied shell.

The decision rule is simple:

1. Write or inspect the electron configuration.
2. Find the largest shell number, \(n\).
3. Count the electrons in subshells carrying that shell number.

Take phosphorus:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^3
\]

The largest shell number is 3.

The electrons in shell 3 are

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

so phosphorus has

\[
2+3=5
\]

valence electrons.

Don’t just look at the final superscript. If you looked only at the \(3\) in \(3p^3\), you might incorrectly say phosphorus has three valence electrons. You must include the \(3s^2\) electrons as well.

06A useful shortcut from the periodic table

The periodic table and spdf notation are closely connected.

The broad regions of the table correspond to the subshell being filled:

For example, chlorine sits in the \(p\) block. Its configuration ends in

\[
3p^5
\]

Calcium is in the \(s\) block. Its configuration ends in

\[
4s^2
\]

This isn’t just a memorisation trick. The shape of the periodic table reflects the capacities of the subshells. The \(s\) subshell holds 2 electrons, so the \(s\) block is two columns wide. The \(p\) subshell holds 6 electrons, so the \(p\) block is six columns wide.

07The main misconception: “valence electrons are just the last electrons written”

That works often enough to become dangerous.

Consider chlorine:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^5
\]

The last term is \(3p^5\), but chlorine does not have five valence electrons.

Its outermost shell is shell 3, containing both

\[
3s^2 \text{ and } 3p^5
\]

Therefore chlorine has seven valence electrons.

The problem becomes even clearer with calcium:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2
\]

Here, looking only at the last term happens to give the correct valence-electron count of two. The shortcut worked, but only by coincidence. The reliable method is always to identify the outermost occupied shell first.

08Be careful with transition metals

For main-group elements, defining valence electrons as the electrons in the highest occupied shell works neatly.

Transition metals need more care.

Consider scandium:

\[
[\ce{Ar}]\,4s^2\,3d^1
\]

The notation \([\ce{Ar}]\) means “the same inner-electron configuration as argon”. It is a shortened way of avoiding repeatedly writing

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^6
\]

If you use the simple outermost-shell definition, scandium has two electrons with the highest shell number, \(n=4\).

However, the \(3d\) electron can also participate in the chemistry of transition metals. For that reason, more advanced discussions of transition-metal valence electrons can include both the outer \(s\) electrons and nearby \(d\) electrons.

So don’t silently apply a main-group shortcut to every element on the periodic table. In an HSC question, use the definition and chemical context being tested. For main-group electron configurations, “highest occupied shell” is the clean rule.

09A fast method for checking spdf notation

When you’re given or have written an electron configuration, check four things.

1. Does the electron count match?

For a neutral atom, the superscripts should add to the atomic number.

2. Has any subshell exceeded its capacity?

Remember:

\[
s^2,\quad p^6,\quad d^{10},\quad f^{14}
\]

So \(2p^7\) is impossible.

3. Does the filling order make sense?

For example, \(4s\) begins filling before \(3d\) in the usual ground-state configurations of neutral atoms.

4. If you’re finding valence electrons, have you counted the whole outer shell?

Don’t automatically use the final superscript.

Those four checks catch a surprisingly large fraction of mistakes.

10Questions and solutions

Question 1

A neutral magnesium atom has atomic number 12. Write its complete spdf electron configuration and state the number of valence electrons.

Solution 1

The electron configuration is \(\boxed{1s^2\,2s^2\,2p^6\,3s^2}\), and magnesium has \(\boxed{2}\) valence electrons.

A neutral magnesium atom has 12 electrons. Filling the available subshells in order gives

\[
1s^2\,2s^2\,2p^6\,3s^2
\]

Checking the electron count:

\[
2+2+6+2=12
\]

The highest occupied shell is \(n=3\). That shell contains \(3s^2\), so it contains two electrons.

Therefore magnesium has two valence electrons.

Question 2

A neutral atom has the electron configuration

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^3
\]

Identify the element and determine its number of valence electrons.

Solution 2

The element is \(\boxed{\text{phosphorus}}\), and it has \(\boxed{5}\) valence electrons.

First, add the electrons:

\[
2+2+6+2+3=15
\]

A neutral atom with 15 electrons has atomic number 15, which is phosphorus.

The highest occupied shell is \(n=3\). It contains

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

so the number of valence electrons is

\[
2+3=5
\]

A tempting mistake is to use only the final superscript and answer three. The \(3s\) electrons are in the same outer shell and must also be counted.

Question 3

A student proposes this ground-state electron configuration for a neutral atom:

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^7
\]

They argue that the superscripts add to 19, so the configuration must represent potassium. Explain whether the configuration is valid, and write the correct configuration if necessary.

Solution 3

The configuration is \(\boxed{\text{invalid}}\). Potassium’s correct ground-state configuration is

\[
\boxed{1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1}
\]

The student’s electron count is correct:

\[
2+2+6+2+7=19
\]

but counting electrons isn’t the only check.

A \(p\) subshell contains three orbitals, with at most two electrons in each orbital. It can therefore contain no more than

\[
3\times2=6
\]

electrons.

The term \(3p^7\) is impossible.

After \(3p\) reaches its capacity of six electrons, the next electron occupies \(4s\). Therefore potassium is

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^1
\]

The trap is assuming that a correct total electron count guarantees a correct electron configuration.

Question 4

Two students are discussing the configuration

\[
1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^3
\]

Student A says the atom has three valence electrons because the final term is \(4p^3\).

Student B says it has five valence electrons.

Determine which electron count is correct, identify the element, and explain the reasoning.

Solution 4

The atom is \(\boxed{\text{arsenic}}\), and for this main-group atom the correct valence-electron count is \(\boxed{5}\).

First, count all the electrons:

\[
2+2+6+2+6+2+10+3=33
\]

A neutral atom with 33 electrons has atomic number 33, so the atom is arsenic.

The highest occupied shell is \(n=4\). The subshells belonging to that shell are

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

Therefore the number of valence electrons is

\[
2+3=5
\]

Student A’s mistake is treating “the last subshell written” as though it means “the entire valence shell”. The \(4s^2\) electrons are also in shell 4.

The \(3d^{10}\) electrons aren’t counted as outer-shell electrons because their shell number is \(n=3\), despite appearing later than \(4s^2\) in the written filling sequence.

Question 5

A student makes the following rule:

“To find an atom’s valence electrons, locate the largest number in its spdf configuration and add every superscript beside that number.”

They test the rule on selenium:

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

and obtain six valence electrons.

They then claim that the same rule gives a complete description of the chemically available valence electrons for every element, including transition metals.

Evaluate the rule.

Solution 5

The rule is \(\boxed{\text{reliable for identifying the outer-shell electrons of main-group atoms, but it is not a complete general rule for transition-metal valence electrons}}\).

For selenium, the rule works perfectly. The largest shell number is \(n=4\), and the occupied \(n=4\) subshells are

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

so selenium has

\[
2+4=6
\]

outer-shell valence electrons.

The hidden assumption is that only electrons in the shell with the highest value of \(n\) can participate significantly in chemical behaviour.

That is a useful model for main-group elements, but transition metals complicate it. Their \((n-1)d\) electrons can be sufficiently close in energy to the outer \(ns\) electrons that both can participate in bonding and ion formation.

For example, scandium has the configuration

\[
[\ce{Ar}]\,4s^2\,3d^1
\]

The highest-shell rule identifies the two \(4s\) electrons as its outer-shell electrons. However, the \(3d\) electron is also relevant to scandium’s chemistry.

So the rule should be stated more carefully:

For main-group atoms, find the highest occupied shell and count its electrons to determine the usual valence-electron count. For transition metals, the meaning of “valence electrons” depends on the chemical context and can include \(d\) electrons as well.

That distinction becomes useful when you move from electron configurations into periodic trends, ion formation, and chemical bonding. spdf notation isn’t just a longer way to write \(2,8,7\). It shows where the electrons are arranged within each shell, which is the detail needed to explain why different parts of the periodic table behave differently.