Why does 4s fill before 3d if 3 is a lower energy level than 4?

Principal energy levels overlap once you pass level 2. As a neutral atom is built up, the 4s sublevel penetrates closer to the nucleus and sits at slightly lower energy than 3d, so it fills first. Once 3d orbitals contain electrons, though, they drop below 4s in energy — which is why cations of transition metals lose their 4s electrons first.

How do I count valence electrons when there are d electrons in the configuration?

Count only the electrons in the highest principal energy level n. For a main-group element like arsenic, [Ar]\,4s^2 3d^10 4p^3, the highest level is 4, so the valence count is 2 + 3 = 5. The 3d^10 electrons belong to level 3 and are core electrons. A quick check: for main-group elements, the valence count matches the group number pattern, so group 15 always gives 5.

What is the difference between an electron configuration and an orbital diagram?

An electron configuration is the compact notation such as 1s^2 2s^2 2p^3; it tells you how many electrons are in each sublevel. An orbital diagram draws a box or line for each individual orbital and shows arrows for spin direction. Only the orbital diagram reveals how many electrons are unpaired, which you need in order to predict paramagnetism.

Do I need to memorize the exceptions like chromium and copper?

For most chemistry courses, chromium ([Ar]\,4s^1 3d^5) and copper ([Ar]\,4s^1 3d^10) are the two worth knowing, along with the reason: a half-filled or completely filled d sublevel is extra stable, so one 4s electron shifts over. Silver, gold, and molybdenum follow the same pattern, but the underlying idea matters more than the list.

CHEM-2.4

Electron Configurations & Energy Levels

Learn to write ground-state electron configurations using the Aufbau order, Pauli exclusion principle, and Hund's rule, and use valence electrons to predict chemical behavior.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Electron Configurations & Energy Levels, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know that an atom's identity comes from its protons. But its chemistry — whether it explodes in water, refuses to react at all, or forms a

2-

ion — comes from where its electrons sit. Electrons are not scattered randomly around the nucleus. They occupy a strict, predictable set of energy levels, sublevels, and orbitals, and they fill those spaces according to three rules that never change.

In this lesson you will learn to write the ground-state electron configuration of any element in the first four rows of the periodic table, draw the matching orbital diagram, and shorten the notation using noble gases. Then you will do the part that actually matters to a chemist: read the outermost energy level and predict how that element behaves — what charge it forms, whether it is reactive, and why elements in the same column act alike.

Energy Levels, Sublevels, and Orbitals

Electrons live in energy levels numbered

n = 1, 2, 3, 4, \dots

, with higher

n

meaning higher energy and greater average distance from the nucleus. Each energy level contains one or more sublevels, labeled

s

p

d

, and

f

. Each sublevel is made of orbitals — regions where an electron is likely to be found — and every orbital holds a maximum of two electrons.

Sublevel	Number of orbitals	Maximum electrons	First level where it appears
$s$	1	2	$n = 1$
$p$	3	6	$n = 2$
$d$	5	10	$n = 3$
$f$	7	14	$n = 4$

Energy level

n

contains exactly

n

sublevels, so level 1 has only

1s

, level 2 has

2s

and

2p

, level 3 has

3s

3p

, and

3d

, and level 4 adds

4f

. The total capacity of a level is

2n^2

electrons: 2, 8, 18, 32.

A common misconception is that orbitals are orbits — little circular paths like planets. They are not. An orbital is a three-dimensional probability region. An

s

orbital is spherical; the three

p

orbitals are dumbbell-shaped and point along the

x

y

, and

z

axes. This is why the three

p

orbitals have equal energy: they are the same shape, just oriented differently. Orbitals in the same sublevel are called degenerate, and that idea becomes essential when you apply Hund's rule.

The Aufbau Order: Filling From the Bottom Up

Aufbau is German for "building up." The principle says electrons occupy the lowest-energy available orbital first. The order is not simply

1s, 2s, 2p, 3s, 3p, 3d

, because sublevel energies overlap once you reach the third level. The actual ground-state filling order is

1s\;2s\;2p\;3s\;3p\;4s\;3d\;4p\;5s\;4d\;5p\;6s\;4f\;5d\;6p\;7s\;5f\;6d\;7p

The surprise is

4s

filling before

3d

. The

4s

sublevel dips slightly below

3d

in energy for neutral atoms being built up, so potassium (19) is

1s^2 2s^2 2p^6 3s^2 3p^6 4s^1

, not

\dots 3d^1

.

You do not have to memorize that string. The periodic table is the filling order, read left to right, row by row. Groups 1 and 2 are the

s

block, groups 13 through 18 are the

p

block, the transition metals are the

d

block, and the lanthanides and actinides are the

f

block. The catch:

d

-block elements in row 4 fill the

3d

sublevel, one less than the row number, and

f

-block elements in row 6 fill

4f

, two less than the row number.

Check your work by adding the superscripts. For a neutral atom they must total the atomic number. If you write bromine (35) and your superscripts sum to 33, you dropped a sublevel somewhere. Also remember that when a written configuration is reordered by energy level for convenience — writing

3d

before

4s

— the atom is unchanged; only the bookkeeping order differs.

Pauli Exclusion and Hund's Rule in Orbital Diagrams

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of four quantum numbers. The practical consequence: an orbital holds at most two electrons, and those two must have opposite spins, drawn as one up arrow and one down arrow in the same box. Three arrows in one box is always wrong, and two arrows pointing the same direction in one box is always wrong.

Hund's rule governs how electrons spread out within a set of degenerate orbitals: put one electron in each orbital, all with parallel spin, before pairing any of them. Electrons repel each other, so occupying separate orbitals lowers the energy. For nitrogen,

2p^3

is drawn as three boxes each holding a single up arrow — not one paired box plus one single electron.

Situation	Correct	Incorrect
Two electrons, one orbital	one up, one down	two arrows same direction
$p^2$	two separate boxes, parallel spins	one box with a pair
$p^4$	one pair, then two singles	two pairs, one empty box
$d^5$	five singles, all parallel	two pairs plus one single

Orbital diagrams also tell you whether an atom is paramagnetic (has unpaired electrons and is attracted to a magnetic field) or diamagnetic (all electrons paired). Oxygen,

2p^4

, has two unpaired electrons and is paramagnetic — a fact you can predict on paper and observe with liquid oxygen sticking to a magnet.

Where students go wrong: filling a sublevel completely before applying Hund's rule, or forgetting that the rule only applies within one sublevel. The

2p

and

3s

orbitals are not degenerate, so nothing is shared between them.

Noble-Gas Notation, Valence Electrons, and Predicting Behavior

Writing all 35 electrons of bromine is tedious, so chemists abbreviate. Find the noble gas that comes immediately before the element, put its symbol in brackets, then continue. Bromine becomes

[\mathrm{Ar}]\,4s^2 3d^{10} 4p^5

. The bracketed part is the core; everything after it is what chemistry cares about.

Valence electrons are the electrons in the highest occupied energy level

n

. For main-group elements, count only the

s

and

p

electrons of that highest

n

. Bromine's highest level is 4, containing

4s^2 4p^5

, so it has 7 valence electrons — the

3d^{10}

electrons are inner electrons and do not count. This is the single most common error in this topic.

Valence count predicts behavior directly. Sodium,

[\mathrm{Ne}]\,3s^1

, has one loosely held electron and readily loses it to form

\mathrm{Na}^+

, reaching the stable configuration of neon. Chlorine,

[\mathrm{Ne}]\,3s^2 3p^5

, needs one electron and forms

\mathrm{Cl}^-

. Neon itself,

1s^2 2s^2 2p^6

, has a full outer level and essentially does not react.

This also explains the periodic table's shape. Every group 2 element ends in

ns^2

; every halogen ends in

ns^2 np^5

. Same outer configuration means same chemistry, which is exactly why magnesium and calcium both form

2+

ions.

Two notable exceptions: chromium is

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

and copper is

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

, because half-filled and completely filled

d

sublevels are unusually stable.

Building Ion Configurations and Checking Your Work

Ions follow the same rules with one adjustment. For an anion, add electrons in Aufbau order:

\mathrm{S}^{2-}

has 18 electrons, so

1s^2 2s^2 2p^6 3s^2 3p^6

, identical to argon. Species with matching configurations are called isoelectronic.

For a cation, remove electrons from the highest

n

level first, not in reverse Aufbau order. This matters for transition metals. Iron is

[\mathrm{Ar}]\,4s^2 3d^6

, but

\mathrm{Fe}^{2+}

[\mathrm{Ar}]\,3d^6

— the

4s

electrons leave first even though

4s

filled first, because once

3d

is occupied it drops below

4s

in energy. Students who remove

3d

electrons from iron get a wrong configuration and a wrong magnetic prediction.

A reliable checking routine: confirm the superscripts sum to the electron count, confirm no sublevel exceeds its capacity (

s^2

p^6

d^{10}

f^{14}

), and confirm the last sublevel written matches the block the element sits in on the periodic table. If you write calcium ending in

3d^2

, the block check catches it immediately, since calcium is in the

s

block.

One more distinction worth keeping straight: a ground state configuration follows Aufbau exactly, while an excited state has an electron promoted to a higher orbital, leaving a gap below. A configuration like

1s^2 2s^2 2p^5 3s^1

for neon is not incorrect arithmetic — it is a real excited state, but it is not the ground state the question asks for.

Key terms

Aufbau principle.: Electrons occupy the lowest-energy available orbital first, producing the filling order in which $4s$ fills before $3d$ .
Pauli exclusion principle.: No two electrons in an atom share all four quantum numbers, so any orbital holds at most two electrons and they must have opposite spins.
Hund's rule.: Within a set of equal-energy orbitals, electrons occupy separate orbitals with parallel spins before any orbital gets a second electron.
Orbital.: A region of space where an electron is likely to be found; each orbital holds a maximum of two electrons.
Degenerate orbitals.: Orbitals within the same sublevel that have identical energy, such as the three $2p$ orbitals.
Valence electrons.: The electrons in the highest occupied principal energy level; for main-group atoms, the $s$ and $p$ electrons of that level.
Noble-gas notation.: A shorthand configuration that replaces the inner core electrons with the bracketed symbol of the preceding noble gas.
Paramagnetic.: Describes an atom or ion with one or more unpaired electrons, which is attracted to an external magnetic field.

Worked example

Write the full and noble-gas ground-state electron configurations for manganese (

Z = 25

). Draw the

3d

orbital diagram, state the number of unpaired electrons, identify the number of valence electrons, and predict the configuration of

\mathrm{Mn}^{2+}

Manganese has 25 protons, so a neutral atom has 25 electrons. Fill in Aufbau order:

1s^2

(2),

2s^2

(4),

2p^6

(10),

3s^2

(12),

3p^6

(18),

4s^2

(20), then

3d

takes the remaining 5.

Full configuration:

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5

. Check the sum:

2+2+6+2+6+2+5 = 25

. Correct.

Noble-gas notation: the noble gas before manganese is argon (

Z = 18

), so

[\mathrm{Ar}]\,4s^2 3d^5

.

Orbital diagram for

3d

: there are five degenerate

d

orbitals and five electrons. Hund's rule says spread them out one per orbital, all parallel, before pairing. So all five boxes hold a single up arrow and none are paired. That gives five unpaired electrons, the maximum possible for a

d

sublevel, which makes manganese strongly paramagnetic. Note that manganese is not an exception like chromium — the

4s^2 3d^5

arrangement already achieves a half-filled

d

sublevel without borrowing an

s

electron.

Valence electrons: the highest occupied principal level is

n = 4

, which holds only

4s^2

. So manganese has 2 valence electrons in the main-group sense, consistent with its common

2+

ion, though the

3d

electrons can also participate in bonding, which is why transition metals show multiple oxidation states.

For

\mathrm{Mn}^{2+}

, remove two electrons from the highest

n

level first — that is

4s

, not

3d

. The result is

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

, still five unpaired electrons and a particularly stable half-filled sublevel, which is one reason

\mathrm{Mn}^{2+}

is so common.

Practice questions

Which ground-state electron configuration is written correctly for a neutral sulfur atom (

Z = 16

$1s^2 2s^2 2p^6 3s^2 3p^4$
$1s^2 2s^2 2p^6 3s^2 3d^4$
$1s^2 2s^2 2p^8 3s^2 3p^2$
$1s^2 2s^2 2p^6 3s^1 3p^5$

Answer: $1s^2 2s^2 2p^6 3s^2 3p^4$

Sulfur has 16 electrons. Filling in Aufbau order gives

1s^2 2s^2 2p^6 3s^2 3p^4

, and the superscripts sum to 16. The second option skips

3p

entirely and jumps to

3d

, which violates the Aufbau order. The third option puts 8 electrons in

2p

, but a

p

sublevel has only three orbitals and holds at most 6 electrons — that breaks the Pauli exclusion principle. The fourth option sums to 16 but leaves

3s

half-empty while filling

3p

, so it is an excited state, not the ground state.

Selenium has atomic number 34. Write its noble-gas electron configuration, determine its number of valence electrons, and predict the charge of the ion it most commonly forms. Explain your reasoning.

Answer: $[\mathrm{Ar}]\,4s^2 3d^{10} 4p^4$ ; 6 valence electrons; it forms $\mathrm{Se}^{2-}$ .

Argon accounts for 18 electrons, leaving 16 to place:

4s^2

(2),

3d^{10}

(12), and

4p^4

(16). The highest occupied principal level is

n = 4

, which contains

4s^2 4p^4

, giving 6 valence electrons. The

3d^{10}

electrons are in level 3 and are core electrons, so they are not counted — this is the step students most often get wrong. With 6 valence electrons, selenium needs 2 more to reach the eight-electron arrangement of krypton, so it gains two electrons and forms a

2-

anion. That matches its position directly below sulfur in group 16.

Nitrogen's

2p

sublevel contains three electrons. Explain, using Hund's rule and the Pauli exclusion principle, why the orbital diagram shows three unpaired electrons rather than one pair and one single electron.

Answer: Hund's rule requires one electron in each of the three degenerate $2p$ orbitals with parallel spins before any pairing occurs, so nitrogen has three unpaired electrons.

The three

2p

orbitals are degenerate — identical in energy, differing only in spatial orientation. Electrons carry negative charge and repel one another, so placing two in the same orbital raises the energy of the atom. Hund's rule captures this: singly occupy every orbital in a sublevel, with spins aligned, before doubling up. The Pauli principle then adds the constraint that when pairing finally does happen at

2p^4

and beyond, the two electrons sharing an orbital must have opposite spins. The result for nitrogen is three unpaired electrons, which makes the atom paramagnetic and helps explain why nitrogen forms three covalent bonds in ammonia.

FAQ

Why does $4s$ fill before $3d$ if 3 is a lower energy level than 4?: Principal energy levels overlap once you pass level 2. As a neutral atom is built up, the $4s$ sublevel penetrates closer to the nucleus and sits at slightly lower energy than $3d$ , so it fills first. Once $3d$ orbitals contain electrons, though, they drop below $4s$ in energy — which is why cations of transition metals lose their $4s$ electrons first.
How do I count valence electrons when there are $d$ electrons in the configuration?: Count only the electrons in the highest principal energy level $n$ . For a main-group element like arsenic, $[\mathrm{Ar}]\,4s^2 3d^{10} 4p^3$ , the highest level is 4, so the valence count is $2 + 3 = 5$ . The $3d^{10}$ electrons belong to level 3 and are core electrons. A quick check: for main-group elements, the valence count matches the group number pattern, so group 15 always gives 5.
What is the difference between an electron configuration and an orbital diagram?: An electron configuration is the compact notation such as $1s^2 2s^2 2p^3$ ; it tells you how many electrons are in each sublevel. An orbital diagram draws a box or line for each individual orbital and shows arrows for spin direction. Only the orbital diagram reveals how many electrons are unpaired, which you need in order to predict paramagnetism.
Do I need to memorize the exceptions like chromium and copper?: For most chemistry courses, chromium ( $[\mathrm{Ar}]\,4s^1 3d^5$ ) and copper ( $[\mathrm{Ar}]\,4s^1 3d^{10}$ ) are the two worth knowing, along with the reason: a half-filled or completely filled $d$ sublevel is extra stable, so one $4s$ electron shifts over. Silver, gold, and molybdenum follow the same pattern, but the underlying idea matters more than the list.

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