Electrons, Energy Levels and Atomic Orbitals

Where the electrons sit in an atom decides almost everything about how it behaves: its ionisation energies, the ions it forms, its bonding and its place in the Periodic Table. This note builds the modern picture of shells, sub-shells and orbitals, then shows how to write electronic configurations for any atom or ion from hydrogen to krypton, including the two famous exceptions (chromium and copper) and the rule for transition metal ions. Only ground-state atoms and ions from H to Kr are examined.

Shells, sub-shells and orbitals

Electrons are arranged in shells, also called energy levels. Each shell is labelled by its principal quantum number, $n = 1, 2, 3, 4, \ldots$ The higher $n$, the further the shell is (on average) from the nucleus and the higher the energy of its electrons.

Each shell is divided into sub-shells, labelled s, p, d and f. Each sub-shell is made of orbitals.

How many orbitals and electrons

sub-shell	number of orbitals	maximum electrons
s	1	2
p	3	6
d	5	10

The first shell has only an s sub-shell. Each shell after that adds one new type:

shell, $n$	sub-shells	maximum electrons
1	1s	2
2	2s, 2p	$2 + 6 = 8$
3	3s, 3p, 3d	$2 + 6 + 10 = 18$
4	4s, 4p, 4d, 4f	$2 + 6 + 10 + 14 = 32$

Shapes of s and p orbitals

An s orbital is spherical, centred on the nucleus. The 2s orbital is a larger sphere than the 1s orbital, and 3s is larger again.

A p orbital is dumbbell-shaped: two lobes on opposite sides of the nucleus, with zero probability of finding the electron at the nucleus itself. Each p sub-shell has three p orbitals at right angles to each other, along the $x$, $y$ and $z$ axes, called $p_x$, $p_y$ and $p_z$. The three have the same energy.

The order of filling

Electrons fill sub-shells in order of increasing energy in the ground state. For the first four shells, the order is:

The surprise is that 4s fills before 3d. For potassium and calcium, the 4s sub-shell is lower in energy than 3d, so the 19th and 20th electrons go into 4s. From scandium onwards electrons enter 3d.

Writing electronic configurations

The full electronic configuration lists every sub-shell with the number of electrons as a superscript. The shorthand form replaces the inner electrons with the symbol of the previous noble gas in square brackets.

element	electrons	full configuration	shorthand
N	7	$1s^2 2s^2 2p^3$	$[\text{He}]\,2s^2 2p^3$
Na	11	$1s^2 2s^2 2p^6 3s^1$	$[\text{Ne}]\,3s^1$
Cl	17	$1s^2 2s^2 2p^6 3s^2 3p^5$	$[\text{Ne}]\,3s^2 3p^5$
Ca	20	$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$	$[\text{Ar}]\,4s^2$
Fe	26	$1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$	$[\text{Ar}]\,3d^6 4s^2$
Br	35	$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^5$	$[\text{Ar}]\,3d^{10} 4s^2 4p^5$

Electrons in boxes

In electrons-in-boxes notation, each orbital is a box and each electron is a half-arrow; two electrons in one orbital point in opposite directions to show opposite spins. Orbitals of one sub-shell are drawn side by side.

atom	1s	2s	2p	2p	2p
C	↑↓	↑↓	↑	↑	(empty)
N	↑↓	↑↓	↑	↑	↑
O	↑↓	↑↓	↑↓	↑	↑

For iron, the syllabus shows $[\text{Ar}]$ followed by five 3d boxes holding ↑↓ ↑ ↑ ↑ ↑ and one 4s box holding ↑↓. So iron has four unpaired electrons.

Why the electrons go where they do

Two ideas explain every configuration: energy and repulsion between electrons.

1. Lowest energy first. In the ground state each electron enters the lowest-energy orbital available.
2. Singly before pairing. Within a sub-shell (for example the three 2p orbitals, which have the same energy), electrons occupy separate orbitals with parallel spins before any orbital gets a second electron. Two electrons in the same orbital are close together and repel more strongly (this is called spin-pair repulsion), so spreading out gives a lower energy. That is why nitrogen has three unpaired 2p electrons, and why oxygen's fourth 2p electron must pair up.
3. Opposite spins in a pair. Two electrons can share one orbital only if they have opposite spins.

The two exceptions: chromium and copper

The 3d and 4s sub-shells are very close in energy. For two elements this changes the ground state:

In chromium, moving one electron from 4s into the empty 3d orbital means all six outer electrons are unpaired in separate orbitals, so there is no spin-pair repulsion in 4s and the half-filled 3d sub-shell is a particularly stable, lower-energy arrangement. In copper, a completely filled 3d sub-shell with a single 4s electron is lower in energy than $3d^9 4s^2$.

Configurations of ions

For main-group elements, add or remove electrons from the outermost sub-shell, following the filling order. $\ce{Cl-}$ is $1s^2 2s^2 2p^6 3s^2 3p^6$ and $\ce{Al^3+}$ is $1s^2 2s^2 2p^6$.

For transition metals there is a crucial rule:

Once the 3d orbitals are occupied, the 4s electrons are higher in energy and further from the nucleus than the 3d electrons, so they are lost first. "Last in, first out" does not apply.

ion	electrons	configuration
$\ce{Fe^2+}$	24	$[\text{Ar}]\,3d^6$
$\ce{Fe^3+}$	23	$[\text{Ar}]\,3d^5$
$\ce{Cu^2+}$	27	$[\text{Ar}]\,3d^9$
$\ce{Cr^3+}$	21	$[\text{Ar}]\,3d^3$
$\ce{Zn^2+}$	28	$[\text{Ar}]\,3d^{10}$

Blocks of the Periodic Table

The block of an element is named after the sub-shell its highest-energy electron enters: Groups 1 and 2 (and helium) are the s-block; Groups 13 to 18 the p-block; scandium to zinc the d-block. The outer-shell configuration tells you the group: an element ending $\ldots 3s^2 3p^4$ has six outer electrons and is in Group 16 (sulfur).

Free radicals

Chlorine atoms ($[\text{Ne}]\,3s^2 3p^5$, one unpaired electron), methyl radicals $\ce{CH3^.}$ and nitrogen monoxide $\ce{NO}$ (15 electrons, an odd number) are all free radicals. A radical is shown with a dot for the unpaired electron, $\ce{Cl^.}$. Radicals are very reactive because the unpaired electron readily pairs with an electron from another species. You will use this idea in free-radical substitution of alkanes.

Practice

Answers

1. Five orbitals; the third shell holds $2 \times 3^2 = 18$ electrons.
2. $\ce{K}$: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$. $\ce{K+}$: $1s^2 2s^2 2p^6 3s^2 3p^6$.
3. $\ce{Ni}$ (28): $[\text{Ar}]\,3d^8 4s^2$. $\ce{Ni^2+}$ (26): $[\text{Ar}]\,3d^8$. $\ce{Cu+}$ (28): copper is $[\text{Ar}]\,3d^{10} 4s^1$; remove the 4s electron to give $[\text{Ar}]\,3d^{10}$.
4. F, $1s^2 2s^2 2p^5$: 2s ↑↓; 2p ↑↓ ↑↓ ↑. $\ce{O+}$ has 7 electrons, $1s^2 2s^2 2p^3$: 2s ↑↓; 2p ↑ ↑ ↑ (three unpaired, parallel spins).
5. (a) Si, $[\text{Ne}]\,3s^2 3p^2$: two 3p electrons in separate orbitals, 2 unpaired. (b) V, $[\text{Ar}]\,3d^3 4s^2$: 3 unpaired. (c) Se, $[\text{Ar}]\,3d^{10} 4s^2 4p^4$: one 4p orbital paired, 2 unpaired.
6. The 3d and 4s sub-shells are very close in energy. Promoting one 4s electron into 3d gives six unpaired electrons, each in its own orbital, which removes the spin-pair repulsion in 4s and gives a half-filled 3d sub-shell. This arrangement is lower in energy, so it is the ground state.
7. $\ce{X^3+}$ has $18 + 5 = 23$ electrons, so X has $23 + 3 = 26$ protons: X is iron ($\ce{Fe^3+}$). Check: Fe is $[\text{Ar}]\,3d^6 4s^2$; remove $4s^2$ and one 3d electron to get $3d^5$.
8. A 2p orbital is dumbbell-shaped: two lobes on opposite sides of the nucleus, with the nucleus at the centre where the lobes meet. There are three 2p orbitals, $2p_x$, $2p_y$, $2p_z$, of equal energy, along three mutually perpendicular axes.
9. $\ce{CH3^.}$: $6 + 3 = 9$ electrons, odd, radical. $\ce{NO}$: $7 + 8 = 15$, odd, radical. $\ce{NH4+}$: $7 + 4 - 1 = 10$, all paired, not a radical. Br atom: 35 electrons, $4p^5$ has one unpaired, radical. $\ce{O2^2-}$: $16 + 2 = 18$, all paired, not a radical.
10. The student added electrons to the wrong places and failed to remove 4s first. Cobalt (27) is $[\text{Ar}]\,3d^7 4s^2$; $\ce{Co^2+}$ (25 electrons) loses both 4s electrons: $[\text{Ar}]\,3d^7$. Seven electrons in five d orbitals: five go in singly, two pair up, leaving 3 unpaired electrons.