Notes · ChemistryUK · A-Levels

Atomic structure

Every idea in chemistry rests on the atom: a tiny, dense nucleus of protons and neutrons surrounded by electrons arranged in shells, sub-shells and orbitals. This chapter builds the model from the sub-atomic particles and isotopes, shows how the time-of-flight mass spectrometer weighs individual ions to give relative atomic mass, and develops electron configuration and ionisation energies as the direct experimental evidence for the shell structure.

5 sections~17 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

T·0111 / 18

Exam profile

AO1 · Describe the sub-atomic particles, isotopes, the stages of time-of-flight mass spectrometry and the rules governing electron configurationAO2 · Calculate relative atomic mass from isotopic abundance and write electron configurations of atoms and ionsAO3 · Interpret mass spectra and successive ionisation energies as evidence for isotopes, shells and sub-shells

Operators:statedescribeexplaincalculatededucepredict

basic level

AS-Level requires the fundamental particles, isotopes, the principle and stages of TOF mass spectrometry, relative-mass calculations, electron configuration in s, p and d sub-shells, and the trends in ionisation energy.

higher level

The full A-Level uses the same content more quantitatively and synoptically - flight-time calculations, successive ionisation energies as evidence for structure, and electron configuration underpinning periodicity and transition-metal chemistry later in the course.

Depth

Text

Contents · 5 sections

Atomic structure

Fundamental particles, mass number and isotopes#

FoundationAQA 7405 3.1.1.1DfE GCE Chemistry - atomic structure

Shell model of a sodium atom

Fig. 1The 11 electrons of sodium fill the shells 2, 8, 1; the nucleus holds 11 protons and 12 neutrons.

An atom has a central nucleus containing protons and neutrons, surrounded by electrons in shells. On the relative scale used throughout chemistry, a proton and a neutron each have a relative mass of

1

, while an electron has a relative mass of only about

\frac{1}{1836}

(taken as negligible). The proton carries a relative charge of

+1

, the electron

-1

, and the neutron is uncharged. Because almost all the mass sits in the nucleus but almost all the volume is empty space occupied by electrons, the atom is essentially a dense point of mass in a much larger cloud of charge.

Two numbers define any atom. The atomic (proton) number

Z

is the number of protons in the nucleus; it fixes the element's identity and, in a neutral atom, equals the number of electrons. The mass number

A

is the total number of protons and neutrons, so the number of neutrons is simply

A - Z

. These are written as a left superscript and subscript on the symbol, for example

^{23}_{11}\text{Na}

, which has

11

protons,

11

electrons and

23 - 11 = 12

neutrons.

Isotopes are atoms of the same element (same

Z

, same number of protons and electrons) that have different numbers of neutrons and therefore different mass numbers

A

. Chlorine, for instance, exists as

^{35}\text{Cl}

and

^{37}\text{Cl}

. Because chemical reactions involve only the electrons, isotopes of an element have identical chemical properties; their physical properties (such as density and rate of diffusion) differ slightly because they have different masses.

An ion forms when an atom gains or loses electrons, changing the electron count but never the proton or neutron count. Losing electrons gives a positive ion (a cation), gaining them a negative ion (an anion). It is a common trap to think that forming an ion changes the mass number - it does not, because electrons contribute negligible mass;

^{24}\text{Mg}

and

^{24}\text{Mg}^{2+}

have the same

12

protons and

12

neutrons and essentially the same mass.

\text{neutrons} = A - Z

Neutron count

The number of neutrons is the mass number minus the atomic number, for example $23 - 11 = 12$ in $^{23}\text{Na}$ .

Worked example

Particles in an ion

Give the number of protons, neutrons and electrons in one ion of $^{27}_{13}\text{Al}^{3+}$ .

Protons
The atomic number $Z = 13$ , so there are 13 protons; forming the ion does not change this.
Neutrons
Neutrons $= A - Z = 27 - 13 = 14$ .
$A - Z = 27 - 13 = 14$
Electrons
A neutral Al atom has 13 electrons; the $3+$ charge means 3 have been lost, leaving $13 - 3 = 10$ electrons.

Result: 13 protons, 14 neutrons and 10 electrons.

Typical mistakes

Saying an electron has zero mass rather than a negligible relative mass of about 1/1836.
Changing the number of neutrons or the mass number when an atom becomes an ion - only the electron count changes.

Active revision

An ion has the symbol $^{37}_{17}\text{Cl}^{-}$ . State the number of protons, neutrons and electrons it contains.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Time-of-flight mass spectrometry#

StandardAQA 7405 3.1.1.2DfE GCE Chemistry - mass spectrometry

Stages of a time-of-flight mass spectrometer

Fig. 2Ions are formed, accelerated to a common kinetic energy, drift through the flight tube (lighter ions arrive first) and are detected.

A time-of-flight (TOF) mass spectrometer weighs individual particles by turning them into ions, giving them all the same kinetic energy, and timing how long each takes to drift down a tube to a detector. It measures the mass-to-charge ratio

m/z

of each ion; because most ions carry a single positive charge (

z = +1

), the reading is effectively the mass of the particle. The instrument works under a high vacuum so that ions are not knocked off course by collisions with air molecules.

There are four stages. Ionisation converts the sample into positive ions by one of two methods: in electrospray ionisation the sample is dissolved, pushed through a fine nozzle held at high voltage and each molecule gains a proton to form

\text{[M+H]}^+

(so the recorded mass is

M_r + 1

); in electron impact ionisation the sample is bombarded with high-energy electrons that knock one electron off each particle to form

\text{M}^+

. Electrospray is a 'soft' method used for large, fragile molecules; electron impact is used for elements and smaller molecules and often causes fragmentation.

In acceleration, an electric field accelerates all the positive ions through the same potential difference, giving every ion the same kinetic energy

E_k = \tfrac{1}{2}mv^2

. Because the energy is the same for all, lighter ions reach a higher velocity than heavier ions. In the flight tube (the ion-drift region) the ions travel a fixed distance

d

at constant speed, so the time of flight

t = d/v

is shorter for the lighter, faster ions. Finally the ions strike a detector, where each gains an electron; this generates a tiny current whose size is proportional to the abundance of that ion.

Rearranging the kinetic-energy equation gives the working relationship for calculations. Since

E_k = \tfrac{1}{2}mv^2

, the speed is

v = \sqrt{2E_k/m}

and the flight time is

t = d\sqrt{m/(2E_k)}

. This shows explicitly that time of flight increases with the square root of the ion's mass: a heavier ion takes proportionally longer. Given a flight time you can find a mass, and given two ions accelerated to the same energy the ratio of their flight times equals the square root of the ratio of their masses.

E_k = \tfrac{1}{2}mv^2

Kinetic energy of an accelerated ion

Every ion is given the same kinetic energy $E_k$ during acceleration, so lighter ions ( $m$ small) reach a larger speed $v$ .

t = d\,\sqrt{\dfrac{m}{2E_k}}

Time of flight

The flight time over a fixed distance $d$ is proportional to $\sqrt{m}$ , so heavier ions arrive later.

Worked example

Comparing flight times

In a TOF spectrometer, a $^{24}\text{Mg}^{+}$ ion and a $^{96}\text{X}^{+}$ ion are accelerated to the same kinetic energy. How many times longer does the heavier ion take to reach the detector?

Use the time-of-flight relation
For a fixed distance and equal kinetic energy, $t \propto \sqrt{m}$ , so the ratio of times equals the square root of the ratio of masses.
Substitute the masses
The mass ratio is $96/24 = 4$ , so the time ratio is $\sqrt{4} = 2$ .
$\dfrac{t_X}{t_{Mg}} = \sqrt{\dfrac{96}{24}} = \sqrt{4} = 2$

Result: The mass-96 ion takes exactly twice as long as the mass-24 ion to reach the detector.

Typical mistakes

Forgetting that electrospray adds a proton, so the recorded mass is one greater than the true Mr.
Saying heavier ions have more kinetic energy - all ions are given the same kinetic energy; the heavier ones are simply slower.

Active revision

Two ions are accelerated to the same kinetic energy in a TOF spectrometer. Explain, in terms of speed and time of flight, why the ion of mass 20 reaches the detector before the ion of mass 80.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Relative atomic mass from mass spectra#

StandardAQA 7405 3.1.1.2DfE GCE Chemistry - relative atomic mass

Mass spectrum of magnesium

Fig. 3The three isotopes of magnesium and their percentage abundances; the weighted mean gives Ar = 24.3.

The relative atomic mass

A_r

of an element is the weighted mean mass of its atoms measured on a scale where an atom of carbon-12 has a mass of exactly

12

. 'Weighted' means each isotope contributes in proportion to how common it is: a rare isotope shifts the average very little, a common one dominates it. The relative molecular mass

M_r

of a molecule (or relative formula mass of a giant/ionic substance) is the sum of the relative atomic masses of all the atoms in the formula.

A mass spectrum of an element shows a peak for each isotope, positioned at its

m/z

(its isotopic mass) with a height proportional to its abundance. To find

A_r

you multiply each isotopic mass by its percentage abundance, add these products, and divide by the total abundance (usually

100

). The answer must lie between the lightest and heaviest isotope, and closest to whichever isotope is most abundant - a quick sense-check that catches arithmetic slips.

The same weighted-mean method runs in reverse. If you are told

A_r

and the masses of two isotopes, you can set the unknown abundance to

x

and the other to

(100 - x)

, write the weighted-mean equation and solve for

x

. This is a favourite examination style for elements such as chlorine (

A_r = 35.5

) and bromine (

A_r = 79.9

), where the near-equal abundances of two isotopes produce the familiar half-integer relative atomic masses.

For a molecular element the spectrum can also show a molecular-ion peak. Chlorine gas,

\text{Cl}_2

, gives peaks at

m/z = 70,\,72,\,74

for the combinations

^{35}\text{Cl}{-}^{35}\text{Cl}

^{35}\text{Cl}{-}^{37}\text{Cl}

and

^{37}\text{Cl}{-}^{37}\text{Cl}

, in the ratio

9:6:1

that follows from the

3:1

abundance of the two atomic isotopes. Recognising these patterns lets you read isotopic composition straight from a spectrum.

A_r = \dfrac{\sum (\text{isotopic mass} \times \text{abundance})}{\sum \text{abundance}}

Relative atomic mass

The weighted mean of the isotopic masses; dividing by the total abundance (usually 100) gives the value on the carbon-12 scale.

Worked example

Relative atomic mass of magnesium

Magnesium has isotopes of mass 24, 25 and 26 with abundances 78.99%, 10.00% and 11.01%. Calculate its relative atomic mass to three significant figures.

Multiply mass by abundance
$24 \times 78.99 = 1895.76$ ; $25 \times 10.00 = 250.0$ ; $26 \times 11.01 = 286.26$ .
Add and divide by 100
Total $= 1895.76 + 250.0 + 286.26 = 2432.02$ ; divide by 100.
$A_r = \dfrac{2432.02}{100} = 24.3202$
Round
To three significant figures this is 24.3, which sensibly lies just above 24 (the most abundant isotope).

Result: Ar(Mg) = 24.3 (3 s.f.).

Typical mistakes

Averaging the isotopic masses without weighting by abundance.
Dividing by the number of isotopes instead of by the total abundance.

Active revision

Boron consists of $^{10}\text{B}$ and $^{11}\text{B}$ only, and has $A_r = 10.8$ . Calculate the percentage abundance of each isotope.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Electron configuration: shells, sub-shells and orbitals#

StandardAQA 7405 3.1.1.3DfE GCE Chemistry - electron configuration

Relative energies and filling order of the sub-shells

Fig. 4Sub-shells fill in order of increasing energy; note that 4s lies just below 3d.

Electrons occupy quantum shells (principal energy levels) numbered

n = 1, 2, 3, \dots

, and each shell holds a maximum of

2n^2

electrons (

2, 8, 18, 32

). Within a shell the electrons are divided into sub-shells labelled s, p, d and f, which hold

2, 6, 10

and

14

electrons respectively. Each sub-shell is made of orbitals - regions where there is a high probability of finding an electron - with one s-orbital, three p-orbitals, five d-orbitals, and so on; every orbital holds a maximum of two electrons of opposite spin.

The order in which sub-shells fill follows increasing energy, and the key subtlety at A-Level is that the

4s

sub-shell is slightly lower in energy than

3d

, so it fills first. The filling order is therefore

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p

. Configurations are written by listing each sub-shell with the number of electrons as a superscript; sodium is

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

and iron is

1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2

. Note that once occupied,

3d

is written before

4s

in the final configuration by convention, even though

4s

filled first.

Two rules complete the picture. The Aufbau (build-up) principle fills the lowest-energy sub-shells first, and Hund's rule states that within a sub-shell electrons occupy separate orbitals singly, with parallel spins, before pairing up - because pairing two negatively charged electrons in one orbital costs energy. So the three

2p

electrons of nitrogen sit one in each

2p

orbital. Drawing the electrons-in-boxes diagram (arrows in boxes) makes both rules visible and is often the clearest way to answer a configuration question.

When atoms form ions the electrons are removed from, or added to, the highest occupied shell. For the transition metals there is an important quirk: because

4s

is at higher energy than

3d

once the

3d

orbitals are occupied, the

4s

electrons are lost first. Thus

\text{Fe}

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

but

\text{Fe}^{2+}

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

(not

3d^4\,4s^2

) and

\text{Fe}^{3+}

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

. Two elements break the tidy filling pattern: chromium is

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

and copper is

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

, because a half-filled or fully-filled

3d

sub-shell is extra stable.

\text{shell capacity} = 2n^2

Maximum electrons in shell n

Shell 1 holds 2, shell 2 holds 8, shell 3 holds 18, shell 4 holds 32 electrons.

Worked example

Configuration of an iron(III) ion

Write the electron configuration of Fe (Z = 26) and hence of the Fe3+ ion.

Fill in energy order
Filling 1s to 4s then 3d gives $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$ ; written in shell order this is $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2$ .
Remove three electrons for Fe3+
The 4s electrons go first (two of them), then one 3d electron, leaving 3d5.
$\text{Fe}^{3+}:\ 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5$

Result: Fe is [Ar]3d6 4s2 and Fe3+ is [Ar]3d5, a stable half-filled d sub-shell.

Typical mistakes

Removing 3d electrons before 4s when forming a transition-metal ion.
Forgetting the chromium ([Ar]3d5 4s1) and copper ([Ar]3d10 4s1) exceptions.

Active revision

Write the full electron configuration of a $\text{Cu}^{2+}$ ion and of a chromium atom, and state the rule that the neutral copper and chromium atoms appear to break.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Ionisation energies as evidence for structure#

AdvancedAQA 7405 3.1.1.3DfE GCE Chemistry - ionisation energy

First ionisation energy across Period 3

Fig. 5The general rise across Period 3 is broken by dips at Al (3p starts) and S (3p electrons begin to pair).

The first ionisation energy is the energy needed to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous

1+

ions:

\text{X}(g) \rightarrow \text{X}^+(g) + e^-

. The state symbols and the 'per mole' are essential in the definition. Successive ionisation energies remove the second, third, ... electrons; each is larger than the last, because after the first electron leaves, every remaining electron is held by the same nuclear charge but there is one fewer electron to share the repulsion, so the effective pull on each is greater.

Three factors control the size of an ionisation energy: the nuclear charge (more protons pull the electrons in more strongly, raising the energy), the distance of the electron from the nucleus (electrons in shells further out are held less tightly, lowering the energy), and shielding (inner shells of electrons repel the outer electron and screen it from the full nuclear charge, lowering the energy). A larger ionisation energy always means the electron was more strongly held.

Successive ionisation energies are powerful evidence for shells. Plotting them (best on a logarithmic scale, because the values span orders of magnitude) shows small rises within a shell and a large jump each time an electron must be taken from a shell closer to the nucleus. Sodium's successive ionisation energies jump sharply after the first electron and again after the ninth, the pattern

1\,|\,8\,|\,2

that confirms the

2, 8, 1

arrangement. Counting electrons before the first big jump tells you the number of electrons in the outer shell - and hence the group.

Across a period the first ionisation energy generally rises, because the nuclear charge increases while the electrons are added to the same shell with similar shielding. Two dips reveal the sub-shell structure. From magnesium (

3s^2

) to aluminium (

3s^2\,3p^1

) the value falls, because aluminium's outer electron is in a

3p

orbital that is higher in energy and slightly shielded by the

3s

pair. From phosphorus (

3p^3

, three singly-occupied p orbitals) to sulfur (

3p^4

) it falls again, because sulfur's fourth

3p

electron must pair up in an orbital, and the repulsion between the paired electrons makes it easier to remove. These anomalies are direct experimental support for the s and p sub-shells.

\text{X}(g) \rightarrow \text{X}^{+}(g) + e^{-}

First ionisation energy

One mole of gaseous atoms each lose one electron to form one mole of gaseous 1+ ions; the state symbols are part of the definition.

Successive ionisation energies of sodium

Fig. 6The 1 | 8 | 2 grouping of the jumps is direct evidence for the 2, 8, 1 shell structure of sodium.

Worked example

Deducing a group from ionisation-energy jumps

An element has successive ionisation energies of 590, 1145, 4912, 6491, 8153 kJ mol^-1. Which group is it in?

Find the big jump
There is a large increase between the 2nd (1145) and 3rd (4912) ionisation energies - a factor of over four.
Count the outer electrons
Two electrons are removed relatively easily before the jump, so there are 2 electrons in the outer shell.
Assign the group
Two outer electrons means the element is in Group 2. (The values are in fact those of calcium.)

Result: The element is in Group 2, because the jump after the 2nd ionisation energy shows two outer-shell electrons.

Typical mistakes

Omitting state symbols or the 'one mole' from the definition of ionisation energy.
Explaining the Mg-to-Al dip by nuclear charge instead of the 3p electron being higher in energy and shielded by 3s.

Active revision

The first six ionisation energies of an element (kJ mol^-1) are 738, 1451, 7733, 10540, 13630, 17995. Deduce the group of the element and explain your reasoning.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Atomic structure

5 sections~17 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

Atomic structure

Key points

Particles in an ion

Key points

Comparing flight times

Key points

Relative atomic mass of magnesium

Key points

Configuration of an iron(III) ion

Key points

Deducing a group from ionisation-energy jumps

Sources

Atomic structure

Key points

Particles in an ion

Key points

Comparing flight times

Key points

Relative atomic mass of magnesium

Key points

Configuration of an iron(III) ion

Key points

Deducing a group from ionisation-energy jumps

Sources