Atomic Structure

:::info Board Coverage AQA Paper 1 | Edexcel Paper 1 | OCR A Gateway C1 | WJEC C1 :::

1. The Atom

1.1 Subatomic Particles

Atoms are the smallest particles of an element that can take part in chemical reactions. Each atom Consists of three subatomic particles:

Particle	Relative Mass	Relative Charge	Location
Proton	1	+1	Nucleus
Neutron	1	0	Nucleus
Electron	$\approx \frac{1}{1836}$	-1	Electron shells

The mass of an atom is concentrated almost entirely in the nucleus. The nucleus occupies a tiny Fraction of the atomic volume — roughly one part in $10^{12}$ to $10^{15}$ of the total volume. This means that atoms are overwhelmingly empty space, a fact with profound consequences for how we Understand scattering experiments and the behaviour of matter.

Key relationships:

$\mathrm{Atomic number (Z) = \mathrm{number of protons$ $\mathrm{Mass number (A) = \mathrm{number of protons + \mathrm{number of neutrons$ $\mathrm{In a neutral atom: number of protons = \mathrm{number of electrons$

These three equations are foundational. The atomic number uniquely identifies an element: no two Elements share the same value of $Z$ . The mass number, by contrast, can vary within a single element Because the number of neutrons is not fixed. This is the basis of isotopes.

Notation: An atom of element X with mass number $A$ and atomic number $Z$ is written as $\prescript{A}{Z}\mathrm{X$ .

Worked Example. An atom of sodium has 11 protons and 12 neutrons. State its atomic number, mass Number, and electron configuration.

$Z = 11, \quad A = 11 + 12 = 23$ $\mathrm{Electron configuration: 2, 8, 1$

Worked Example. An atom has 9 protons, 10 neutrons, and 9 electrons. Identify the element and Write its full notation.

The atomic number $Z = 9$ Which is fluorine. The mass number $A = 9 + 10 = 19$ . The notation is $\prescript{19}{9}\mathrm{F$ . Since the number of protons equals the number of electrons, this is a Neutral fluorine atom.

Worked Example. An ion has 8 protons, 8 neutrons, and 10 electrons. Identify the ion and give Its charge.

The element with 8 protons is oxygen ( $Z = 8$ ). There are 10 electrons but only 8 protons, so the Ion carries a charge of $10 - 8 = 2-$ . This is the oxide ion, $\mathrm{O^{2-}$ . The mass number is $A = 8 + 8 = 16$ .

1.2 History of the Atom

Scientist	Model	Key Discovery
Dalton (1803)	Solid sphere	All matter is made of atoms; atoms of the same element are identical
Thomson (1897)	Plum pudding	Discovered the electron; atoms contain smaller particles
Rutherford (1909)	Nuclear	Gold foil experiment showed a small, dense, positive nucleus
Bohr (1913)	Electron shells	Electrons orbit the nucleus in fixed energy levels
Chadwick (1932)	With neutrons	Discovered the neutron, completing the modern picture

Each model superseded its predecessor because new experimental evidence forced a revision. Dalton’s Model could not explain cathode rays (Thomson). Thomson’s plum pudding model could not explain the Gold foil results (Rutherford). Rutherford’s model was unstable by classical electrodynamics — a Charged electron orbiting a nucleus should radiate energy and spiral inward. Bohr resolved this by Postulating quantised orbits, although his model was itself superseded by the full quantum Mechanical treatment (beyond GCSE scope).

1.3 The Rutherford Scattering Experiment

Rutherford directed alpha particles at a thin gold foil. Most passed straight through, but some were Deflected at large angles, and a few bounced back.

Conclusions:

Most of the atom is empty space (most alpha particles passed through)
The nucleus is very small, dense, and positively charged (few alpha particles deflected back)
The mass is concentrated in the nucleus

The reasoning is worth stating carefully. Alpha particles carry a $+2$ charge and are relatively Heavy (helium nuclei). If the positive charge of the atom were spread diffusely (as in the plum Pudding model), the alpha particle would experience only a weak net repulsion and would not be Deflected through large angles. The observation that some alpha particles bounced straight back Implies a concentrated, massive, positively charged core — the nucleus.

Quantitative intuition for the Rutherford experiment

The fraction of alpha particles scattered through an angle greater than $\theta$ is proportional to $\left(\frac{Z_1 Z_2 e^2}{4 E}\right)^2 \cot^2\!\left(\frac{\theta}{2}\right)$ Where $Z_1$ and $Z_2$ are the atomic numbers of the alpha particle and gold nucleus, $e$ is the elementary charge, And $E$ is the kinetic energy of the alpha particle. This shows that large-angle scattering is Extremely rare — consistent with a tiny nucleus.

2. Isotopes

2.1 Definition

Isotopes are atoms of the same element (same atomic number) with different numbers of neutrons (different mass numbers).

Example — Carbon:

Isotope	Protons	Neutrons	Electrons	Mass Number
Carbon-12	6	6	6	12
Carbon-13	6	7	6	13
Carbon-14	6	8	6	14

All three species have $Z = 6$ (carbon), but their mass numbers differ because the neutron count Differs. The chemical behaviour is essentially identical because chemical reactions involve Electrons, and all three isotopes have the same electron configuration ( $2, 4$ ).

2.2 Properties of Isotopes

Isotopes of the same element have:

The same chemical properties (same electron configuration)
Different physical properties (different mass, different density)

Radioactive isotopes have unstable nuclei and decay over time, emitting radiation. Radioactive Decay is a nuclear process, not a chemical one, so it is unaffected by temperature, pressure, or Chemical state.

2.3 Uses of Radioactive Isotopes

Isotope	Use	Reason
Carbon-14	Dating archaeological materials	Half-life of 5730 years matches human timescales
Cobalt-60	Radiotherapy for cancer	Emits gamma rays that kill cancer cells
Iodine-131	Treating thyroid conditions	Absorbed by the thyroid gland
Americium-241	Smoke detectors	Emits alpha particles; ionises air to detect smoke
Uranium-235	Nuclear fuel	Undergoes fission, releasing large amounts of energy

The choice of isotope for a given application depends critically on its half-life. A medical tracer Needs a short half-life (hours to days) so that it does not remain radioactive in the body. Carbon-14 dating works precisely because 5730 years is on a comparable timescale to human history.

2.4 Half-Life

The half-life ( $t_{1/2}$ ) is the time taken for half of the radioactive nuclei in a sample to Decay. It is a constant for a given isotope and is independent of the amount of sample.

$\mathrm{After n \mathrm{ half-lives: \frac{1}{2^n} \mathrm{ of the original remains$

Worked Example. A sample contains 80 g of iodine-131 (half-life = 8 days). How much remains After 32 days?

$n = \frac{32}{8} = 4 \mathrm{ half-lives$ $\mathrm{Remaining = 80 \times \frac{1}{2^4} = 80 \times \frac{1}{16} = 5 \mathrm{ g$

Worked Example. A sample of cobalt-60 has an initial activity of 800 counts per minute. Its Half-life is 5.3 years. What is the activity after 15.9 years?

$n = \frac{15.9}{5.3} = 3 \mathrm{ half-lives$ $\mathrm{Activity = 800 \times \frac{1}{2^3} = 800 \times \frac{1}{8} = 100 \mathrm{ counts/min$

Worked Example. A radioactive sample starts at 40 g. After 90 minutes, 5 g remain. Calculate the Half-life.

After $n$ half-lives: $40 \times \frac{1}{2^n} = 5$ So $\frac{1}{2^n} = \frac{5}{40} = \frac{1}{8}$ Which gives $2^n = 8$ So $n = 3$ . The half-life is $t_{1/2} = \frac{90}{3} = 30$ minutes.

2.5 Derivation: The Half-Life Formula

The number of undecayed nuclei at time $t$ follows an exponential decay:

$N(t) = N_0 \cdot e^{-\lambda t}$

Where $\lambda$ is the decay constant. After one half-life, $N = \frac{N_0}{2}$ :

$\frac{N_0}{2} = N_0 \cdot e^{-\lambda t_{1/2}}$

$\frac{1}{2} = e^{-\lambda t_{1/2}}$

$\ln\!\left(\frac{1}{2}\right) = -\lambda t_{1/2}$

$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$

This derivation shows that half-life is inversely proportional to the decay constant. A large decay Constant (rapid decay) corresponds to a short half-life.

3. Electron Configuration

3.1 Energy Levels

Electrons occupy shells (energy levels) around the nucleus. The first shell can hold up to 2 Electrons, the second up to 8, and the third up to 8 (at GCSE level; in reality the third can hold 18).

Maximum electrons per shell: $2n^2$ where $n$ is the shell number.

This formula gives the maximum occupancy: shell 1 holds 2, shell 2 holds 8, shell 3 holds 18, and Shell 4 holds 32. At GCSE, the convention is to stop at 8 for the third shell because the 3d Subshell fills after the 4s subshell, and this detail is not required.

Proof of $2n^2$ : Each shell $n$ has $n$ subshells (labelled s, p, d, f, …), and each subshell Has $2(2\ell + 1)$ orbitals, each holding 2 electrons. Summing over all subshells from $\ell = 0$ to $\ell = n - 1$ :

$\sum_{\ell=0}^{n-1} 2(2\ell + 1) = 2\sum_{\ell=0}^{n-1}(2\ell + 1) = 2n^2$

The sum of the first $n$ odd numbers equals $n^2$ Which is a well-known result from arithmetic.

3.2 Rules for Filling Shells

Electrons fill the lowest energy shell first
Each shell must be full before the next one starts filling (except for transition metals at A-Level)
At GCSE, the first three shells hold a maximum of 2, 8, and 8 electrons

Worked Example. Write the electron configuration of calcium ( $Z = 20$ ).

$\mathrm{Ca: 2, 8, 8, 2$

Calcium has 20 electrons. The first three shells hold $2 + 8 + 8 = 18$ electrons, leaving 2 in the Fourth shell. This places calcium in group 2 (two outer electrons) and period 4 (four occupied Shells).

Worked Example. Write the electron configuration of phosphorus ( $Z = 15$ ).

$\mathrm{P: 2, 8, 5$

Worked Example. Write the electron configuration of sulfur ( $Z = 16$ ).

$\mathrm{S: 2, 8, 6$

Sulfur is two electrons short of a full outer shell, which explains why it readily gains two Electrons to form S $^{2-}$ Achieving the stable configuration of argon.

Worked Example. Write the electron configuration of argon ( $Z = 18$ ).

$\mathrm{Ar: 2, 8, 8$

Argon has a full outer shell, which explains why it is chemically inert.

Worked Example. Write the electron configuration of potassium ( $Z = 19$ ).

$\mathrm{K: 2, 8, 8, 1$

Potassium starts a new shell rather than filling the third shell to 18, because the fourth shell is Lower in energy once the third shell has 8 electrons.

3.3 Electron Configuration and the Periodic Table

The period number tells you the number of occupied shells.

The group number (for groups 1, 2, and 13—18) tells you the number of electrons in the outer Shell.

Example: Sodium is in period 3, group 1. Its electron configuration is 2, 8, 1 — three shells, One outer electron.

Example: Chlorine is in period 3, group 7. Its electron configuration is 2, 8, 7 — three Shells, seven outer electrons.

This connection between electron configuration and position in the periodic table is not a Coincidence. The periodic table was originally constructed by arranging elements in order of atomic Number; the pattern of properties repeats because the electron configuration repeats in a regular Way.

3.4 Electron Configuration of Ions

When atoms gain or lose electrons to form ions, the electron configuration changes.

Example: Sodium ion (Na $^+$ )

$\mathrm{Na: 2, 8, 1 \implies \mathrm{Na^+: 2, 8$

The sodium ion has the same electron configuration as neon. This is why sodium forms a $1+$ ion: by Losing one electron, it achieves a full outer shell.

Example: Oxide ion (O $^{2-}$ )

$\mathrm{O: 2, 6 \implies \mathrm{O^{2-}: 2, 8$

The oxide ion has the same electron configuration as neon. By gaining two electrons, oxygen achieves A full outer shell.

Example: Calcium ion (Ca $^{2+}$ )

$\mathrm{Ca: 2, 8, 8, 2 \implies \mathrm{Ca^{2+}: 2, 8, 8$

Calcium loses its two outer electrons to achieve the configuration of argon.

3.5 Why Atoms Form Ions: An Energy Argument

Atoms gain or lose electrons to achieve a lower energy state. A full outer shell is the lowest Energy electron configuration for the atom. The energy released when the ion is formed (ionisation Energy for cations, electron affinity for anions) and the lattice energy when the ions come together Make the ionic compound more stable than the separated neutral atoms.

This is why group 1 metals readily lose one electron (low first ionisation energy, high reactivity) And group 7 non-metals readily gain one electron (high electron affinity, high reactivity).

4. Relative Atomic Mass

4.1 Definition

The relative atomic mass ( $A_r$ ) of an element is the weighted mean mass of an atom of the Element relative to 1/12 the mass of a carbon-12 atom.

$A_r = \frac{\sum (\mathrm{isotope mass \times \mathrm{abundance)}{\sum \mathrm{abundance}$

The word “weighted” is crucial. Relative atomic mass is not a simple average; it accounts for the Fact that some isotopes are much more abundant than others. A rare but heavy isotope contributes Less to the average than a common but lighter one.

4.2 Calculating Relative Atomic Mass

Worked Example. Chlorine has two isotopes: Cl-35 (75% abundance) and Cl-37 (25% abundance). Calculate the relative atomic mass of chlorine.

$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = \frac{3550}{100} = 35.5$

Worked Example. Magnesium has three isotopes: Mg-24 (79%), Mg-25 (10%), and Mg-26 (11%). Calculate $A_r$ .

$A_r = \frac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100} = \frac{1896 + 250 + 286}{100} = \frac{2432}{100} = 24.32$

Worked Example. Boron has two isotopes: B-10 (20%) and B-11 (80%). Calculate $A_r$ .

$A_r = \frac{(10 \times 20) + (11 \times 80)}{100} = \frac{200 + 880}{100} = \frac{1080}{100} = 10.8$

Worked Example. Neon has three isotopes: Ne-20 (90.5%), Ne-21 (0.3%), and Ne-22 (9.2%). Calculate $A_r$ .

$A_r = \frac{(20 \times 90.5) + (21 \times 0.3) + (22 \times 9.2)}{100} = \frac{1810 + 0.63 + 202.4}{100} = \frac{2013.03}{100} = 20.13$

Worked Example. A sample of boron has $A_r = 10.81$ . Given that boron has two isotopes, B-10 and B-11, calculate the percentage abundance of each.

Let $x\%$ be the abundance of B-10. Then $(100 - x)\%$ is the abundance of B-11.

$10.81 = \frac{10x + 11(100 - x)}{100} = \frac{10x + 1100 - 11x}{100} = \frac{1100 - x}{100}$

$1081 = 1100 - x$

$x = 19\%$

So B-10 is 19% and B-11 is 81%.

4.3 Summary Table: Relative Atomic Mass Calculations

Element	Isotopes	$A_r$	Method
Chlorine	Cl-35 (75%), Cl-37 (25%)	35.5	Two-isotope weighted mean
Magnesium	Mg-24 (79%), Mg-25 (10%), Mg-26 (11%)	24.32	Three-isotope weighted mean
Boron	B-10 (19%), B-11 (81%)	10.81	Reverse calculation from $A_r$

5. The Periodic Table

5.1 Development of the Periodic Table

Scientist	Contribution
Newlands	Law of octaves (every 8th element had similar properties) — limited to known elements
Mendeleev	Arranged elements by atomic mass, leaving gaps for undiscovered elements; predicted their properties
Modern table	Arranged by atomic number (proton number), not mass

Mendeleev’s genius was leaving gaps, which allowed him to predict the properties of undiscovered Elements (e.g. Gallium, germanium). When these elements were later discovered, their properties Matched Mendeleev’s predictions.

The modern table orders elements by atomic number rather than mass. This resolves problems that Mendeleev encountered: for instance, tellurium (Te, $A_r \approx 127.6$ ) and iodine (I, $A_r \approx 126.9$ ) would be swapped if ordered by mass, but ordering by atomic number places Te (Z = 52) before I (Z = 53), which is consistent with their chemical properties.

5.2 Structure of the Periodic Table

Periods (rows): Elements in the same period have the same number of electron shells
Groups (columns): Elements in the same group have the same number of outer electrons and similar chemical properties

Groups 1 and 2 are the s-block. Groups 13—18 are the p-block. The transition metals (the central Block) are the d-block. At GCSE, you need not use this terminology, but recognising the pattern Helps to explain why transition metals have variable oxidation states.

5.3 Group 1: The Alkali Metals

Property	Trend
Reactivity	Increases down the group
Melting/boiling point	Decreases down the group
Density	Increases down the group
Softness	Softer down the group

Why does reactivity increase down Group 1?

As you descend the group, the outer electron is in a shell further from the nucleus. Even though the Nuclear charge increases, the extra inner shells provide significant shielding. The net effect is That the outer electron is less strongly held and is more lost. Since Group 1 metals react by Losing their outer electron, lower ionisation energy means greater reactivity.

Reactions with water:

$2\mathrm{Na + 2\mathrm{H_2\mathrm{O \to 2\mathrm{NaOH + \mathrm{H_2$

Lithium fizzes steadily; sodium melts and moves vigorously; potassium ignites with a lilac flame.

Reactions with chlorine:

$2\mathrm{Na + \mathrm{Cl_2 \to 2\mathrm{NaCl$

These form ionic compounds called metal halides.

Worked Example. Predict the products and write a balanced equation for the reaction of potassium With water.

$2\mathrm{K + 2\mathrm{H_2\mathrm{O \to 2\mathrm{KOH + \mathrm{H_2$

Potassium hydroxide and hydrogen gas are produced. The reaction is more vigorous than that of sodium Because potassium is more reactive.

Worked Example. Write the balanced equation for lithium reacting with chlorine.

$2\mathrm{Li + \mathrm{Cl_2 \to 2\mathrm{LiCl$

5.4 Group 7: The Halogens

Property	Trend
Reactivity	Decreases down the group
Melting/boiling point	Increases down the group
Physical state at room temperature	Gas to liquid to solid
Colour	Gets darker down the group

Why does reactivity decrease down Group 7?

Reactivity for halogens involves gaining an electron. As you descend the group, the outer shell is Further from the nucleus. The incoming electron is less strongly attracted, so electron affinity Decreases. This makes the halogen less eager to accept an electron, hence less reactive.

Displacement reactions: A more reactive halogen displaces a less reactive halogen from its Compound.

$\mathrm{Cl_2 + 2\mathrm{KBr \to 2\mathrm{KCl + \mathrm{Br_2$

Chlorine displaces bromine from potassium bromide solution (chlorine is more reactive).

Bromine cannot displace chlorine: $\mathrm{Br_2 + 2\mathrm{KCl \to$ no reaction.

Worked Example. Predict whether a reaction occurs when bromine is added to potassium iodide Solution.

Bromine is more reactive than iodine, so it will displace iodine:

$\mathrm{Br_2 + 2\mathrm{KI \to 2\mathrm{KBr + \mathrm{I_2$

The solution would turn brown as iodine is formed.

Worked Example. Predict whether a reaction occurs when chlorine is added to potassium fluoride Solution.

Chlorine is more reactive than fluorine is false — fluorine is the most reactive halogen. So Chlorine cannot displace fluorine: $\mathrm{Cl_2 + 2\mathrm{KF \to$ no reaction.

5.5 Group 0: The Noble Gases

Colourless gases at room temperature
Very unreactive (full outer shell of electrons)
Monatomic (exist as single atoms)
Boiling point increases down the group

Uses: helium in balloons, neon in advertising signs, argon in light bulbs (provides an inert Atmosphere).

The increasing boiling point down Group 0 is explained by increasing London dispersion forces as the Number of electrons increases. Helium ( $A_r = 4$ ) has very weak intermolecular forces and boils at $-269^{\circ}\mathrm{C$ Whereas xenon ( $A_r = 131$ ) boils at $-108^{\circ}\mathrm{C$ .

5.6 Transition Metals

Transition metals are found in the central block of the periodic table. They have the following Properties:

Good conductors of heat and electricity
High melting points and high densities
Form coloured compounds
Often have more than one oxidation state (e.g. Iron can be Fe $^{2+}$ or Fe $^{3+}$ )
Can act as catalysts (e.g. Iron in the Haber process, nickel in hydrogenation)

The variable oxidation states arise because transition metals have electrons in both the 4s and 3d Subshells, and electrons from both can be lost. Iron, for example, has the configuration $[\mathrm{Ar]\,4s^2 3d^6$ . It can lose the two 4s electrons to form Fe $^{2+}$ Or both 4s electrons And one 3d electron to form Fe $^{3+}$ .

5.7 Comparing Group 1 Metals with Transition Metals

Property	Group 1 metals	Transition metals
Melting point	Relatively low	High
Density	Low	High
Reactivity with water	Very reactive	Much less reactive (or unreactive)
Oxidation states	Always $1+$	Variable (e.g. $2+$ , $3+$ for iron)
Coloured compounds	No	Yes
Catalytic activity	No	Common

5.8 Predicting Properties from Position in the Periodic Table

Worked Example. Element X is in period 3, group 2. Predict its electron configuration and the Charge of its ion.

Period 3 means 3 shells. Group 2 means 2 outer electrons. Electron configuration: 2, 8, 2. It will Lose 2 electrons to form a $2+$ ion, achieving the configuration of argon (2, 8, 8).

Worked Example. Element Y is in period 4, group 7. Predict its electron configuration and the Charge of its ion.

Period 4 means 4 shells. Group 7 means 7 outer electrons. Electron configuration: 2, 8, 8, 7. It Will gain 1 electron to form a $1-$ ion, achieving the configuration of krypton (2, 8, 18, 8).

6. Atomic Structure and Chemical Properties

6.1 Why Elements React

Elements react in order to achieve a full outer shell of electrons (a stable electron configuration Like the noble gases). This drive towards stability is the fundamental explanation for all chemical Bonding and reactivity.

Metals lose electrons to form positive ions (cations)
Non-metals gain electrons to form negative ions (anions)

6.2 Ion Formation

Group 1 elements lose 1 electron to form $1+$ ions:

$\mathrm{Na \to \mathrm{Na^+ + e^-$

Group 2 elements lose 2 electrons to form $2+$ ions:

$\mathrm{Mg \to \mathrm{Mg^{2+} + 2e^-$

Group 6 elements gain 2 electrons to form $2-$ ions:

$\mathrm{O + 2e^- \to \mathrm{O^{2-}$

Group 7 elements gain 1 electron to form $1-$ ions:

$\mathrm{Cl + e^- \to \mathrm{Cl^-$

6.3 Ionic Charges

The charge of an ion is determined by its group number:

Group	Ions formed	Example
1	$1+$	Na $^+$ K $^+$ Li $^+$
2	$2+$	Mg $^{2+}$ Ca $^{2+}$
3	$3+$	Al $^{3+}$
5	$3-$	N $^{3-}$
6	$2-$	O $^{2-}$ S $^{2-}$
7	$1-$	Cl $^-$ Br $^-$ I $^-$

The pattern is clear: elements in groups 1—3 tend to lose electrons (forming positive ions whose Charge equals the group number), while elements in groups 5—7 tend to gain electrons (forming Negative ions whose charge equals $8 - \mathrm{group number$ ).

Derivation of the charge rule: For groups 1—3, the atom loses all of its outer electrons to Achieve a full outer shell. The charge equals the number of electrons lost, which equals the group Number. For groups 5—7, the atom gains $(8 - \mathrm{group number)$ electrons to fill its outer Shell. The charge is the negative of this number: $-(8 - \mathrm{group number)$ .

7. Higher Tier Content

7.1 Ionisation Energy

The first ionisation energy is the energy required to remove one electron from each atom in a Mole of gaseous atoms.

$\mathrm{X(g) \to \mathrm{X^+(g) + e^-$

Trends across a period: Ionisation energy generally increases because the nuclear charge Increases while the shielding remains roughly constant. The outer electron is more strongly Attracted, so more energy is needed to remove it.

Trends down a group: Ionisation energy decreases because the outer electron is further from the Nucleus and is shielded by additional inner shells.

Exceptions:

Between Group 2 and Group 3 (e.g. Mg to Al): the Group 3 electron is in a p-subshell which is slightly higher in energy than the s-subshell, so it is easier to remove.
Between Group 5 and Group 6 (e.g. P to S): the Group 5 element has a half-filled p-subshell which is relatively stable; in Group 6, pairing begins in the p-subshell, and the repulsion between paired electrons makes one of them easier to remove.

7.2 Successive Ionisation Energies

Each successive ionisation energy is larger than the previous one because the remaining electrons Are held more tightly by an increasingly positive ion. Large jumps occur when an electron is removed From a new, inner shell.

Worked Example. The first five ionisation energies of an element are: 578, 1817, 2745, 11578, 14842 kJ/mol. Identify the group.

The large jump occurs between the third and fourth ionisation energies. This means the first three Electrons are in the outer shell, and the fourth electron is in a new inner shell. The element is in Group 3 (three outer electrons).

7.3 Using Mass Spectrometry to Determine $A_r$

Mass spectrometry can measure the relative abundances and masses of isotopes directly. The resulting Spectrum shows peaks at different mass-to-charge ratios ( $m/z$ ), and the height of each peak is Proportional to the abundance of that isotope.

This is a more precise method than simple calculation from percentage abundances, and it is the way That relative atomic masses are measured in practice.

7.4 Mass Spectrum Interpretation

Worked Example. The mass spectrum of an element shows peaks at $m/z = 63$ (69%) and $m/z = 65$ (31%). Calculate the relative atomic mass and identify the element.

$A_r = \frac{(63 \times 69) + (65 \times 31)}{100} = \frac{4347 + 2015}{100} = \frac{6362}{100} = 63.62$

This matches copper ( $A_r \approx 63.5$ ). The isotopes are Cu-63 and Cu-65.

Common Pitfalls

Confusing atomic number and mass number. Atomic number = protons; mass number = protons + neutrons. Remember: $A = Z + N$ .
Forgetting that ions have different numbers of protons and electrons. A positive ion has fewer electrons than protons. For Na $^+$ There are 11 protons and 10 electrons.
Stating that noble gases do not react at all. They are very unreactive but not completely inert (xenon and krypton do form some compounds).
Confusing the mass number of an ion with its atomic number. An ion has the same atomic number as its atom but may have gained or lost electrons. The mass number does not change when an ion forms because electrons have negligible mass.
Assuming all isotopes are radioactive. Most elements have at least one stable isotope; only some isotopes are radioactive. Carbon-12 and carbon-13 are stable; carbon-14 is radioactive.
Writing electron configurations incorrectly. Always fill lower shells first; do not exceed the maximum for each shell. A common error is writing phosphorus as 2, 4, 9 instead of 2, 8, 5.
Misidentifying trends in reactivity. Group 1 reactivity increases down the group; Group 7 reactivity decreases down the group. The reasons are different (easier to lose vs. Harder to gain electrons).
Using the wrong half-life formula. Remember that after $n$ half-lives, the remaining amount is $\frac{1}{2^n}$ of the original, not $\frac{1}{2n}$ .
Confusing relative atomic mass with mass number. $A_r$ is a weighted average of all isotopes; mass number is the total of protons and neutrons in a specific atom.
Forgetting that the mass number does not change when an ion forms. Ions lose or gain electrons, which have negligible mass. Only the charge changes.

Practice Questions

An atom of element X has 16 protons, 16 neutrons, and 16 electrons. Another atom of element X has 16 protons, 18 neutrons, and 16 electrons. Explain the relationship between these two atoms.
Neon has three isotopes: Ne-20 (90.5%), Ne-21 (0.3%), and Ne-22 (9.2%). Calculate the relative atomic mass of neon.
Describe the plum pudding model and explain how Rutherford’s gold foil experiment disproved it.
Explain why potassium is more reactive than sodium, with reference to electron configuration.
Write the electron configuration of sulfur ( $Z = 16$ ). Explain what this tells you about its position in the periodic table.
Predict the products when fluorine is added to a solution of potassium chloride, and explain your reasoning.
An ion has 10 electrons and 8 protons. Identify the ion and state its charge.
Explain why Mendeleev’s periodic table was more successful than Newlands’ classification.
State three properties that are typical of transition metals and give an example of each.
Explain why argon is placed after potassium in the modern periodic table, even though argon has a smaller relative atomic mass.
A radioactive isotope has a half-life of 6 hours. If a sample initially contains 40 g, how much remains after 24 hours?
Explain why the first ionisation energy of aluminium is less than that of magnesium.
Write the electron configuration of the following ions: (a) O $^{2-}$ (b) Mg $^{2+}$ (c) Cl $^-$ .
A sample of boron has $A_r = 10.81$ . Boron has two isotopes, B-10 and B-11. Calculate the percentage abundance of each isotope.
Explain, in terms of electron configuration, why calcium reacts vigorously with water but copper does not.
An element has successive ionisation energies: 738, 1451, 7733, and 10540 kJ/mol. Which group does this element belong to? Explain your reasoning.
The mass spectrum of magnesium shows three peaks at $m/z = 24$ 25, and 26 with relative abundances 79%, 10%, and 11%. Calculate the relative atomic mass of magnesium.
A sample of a radioactive isotope has an initial mass of 20 g. After 2 hours, 2.5 g remains. Calculate the half-life of the isotope.
Explain why the boiling point of helium ( $-269^{\circ}\mathrm{C$ ) is much lower than that of radon ( $-62^{\circ}\mathrm{C$ ).
Write balanced equations for the reactions of: (a) lithium with water, (b) fluorine with potassium bromide solution, (c) calcium with oxygen.

Practice Problems

Question 1: Electron configuration and reactivity

Sodium has the electron configuration 2, 8, 1. Explain why sodium is highly reactive and describe what happens when sodium reacts with chlorine. Include the electron transfers in your answer.

Answer

Sodium has one electron in its outer shell. It is highly reactive because it can lose this single outer electron to achieve a stable noble gas electron configuration (2, 8). When sodium reacts with chlorine, sodium loses its outer electron to form a $\mathrm{Na^+$ ion, and chlorine gains this electron to form a $\mathrm{Cl^-$ ion. The electrostatic attraction between these oppositely charged ions forms ionic bonds in sodium chloride ( $\mathrm{NaCl$ ).

Question 2: Relative atomic mass calculation

Chlorine has two isotopes: $\mathrm{Cl$ -35 (75.8% abundance) and $\mathrm{Cl$ -37 (24.2% abundance). Calculate the relative atomic mass of chlorine.

Answer

$A_r = (35 \times 75.8 + 37 \times 24.2) / 100 = (2653 + 895.4) / 100 = 3548.4 / 100 = 35.48$ .

The relative atomic mass of chlorine is 35.5 (to 1 decimal place).

Question 3: Periodic trends

Explain why the melting point of sodium is higher than that of neon, but lower than that of magnesium. Refer to the types of bonding in your answer.

Answer

Neon is a noble gas with a full outer shell. It exists as individual atoms held together only by weak London forces, so it has a very low melting point. Sodium is a metal with metallic bonding (positive ions in a sea of delocalised electrons), which requires more energy to overcome. Magnesium has stronger metallic bonding than sodium because it has two delocalised electrons per atom (compared to one for sodium) and a smaller atomic radius, resulting in stronger electrostatic attraction and a higher melting point.

Question 4: Flame tests and emission spectra

Describe how a flame test can be used to identify metal ions. Explain why different metals produce different flame colours.

Answer

A clean platinum or nichrome wire is dipped into a sample and held in a Bunsen flame. The flame colour is observed and compared to known colours: lithium (crimson), sodium (yellow), potassium (lilac), calcium (orange-red), copper (blue-green).

Different metals produce different colours because when they are heated, their electrons absorb energy and move to higher energy levels. When the electrons fall back to lower levels, they emit light of specific wavelengths (colours) that are characteristic of that element. Each element has a unique arrangement of electron energy levels, so the wavelengths emitted are unique.

Question 5: Group 1 and Group 7 trends

Describe the trends in reactivity down Group 1 (alkali metals) and down Group 7 (halogens). Explain these trends in terms of atomic structure.

Answer

Group 1: Reactivity increases down the group. As you go down, the atomic radius increases and the outer electron is further from the nucleus, shielded by more inner shells. The attraction between the nucleus and the outer electron decreases, so it is easier to lose the outer electron. Cesium is more reactive than lithium.

Group 7: Reactivity decreases down the group. As you go down, the atomic radius increases. Although there is a stronger nuclear charge, the increased distance and shielding make it harder for the atom to attract an additional electron. Fluorine is more reactive than iodine.

Worked Examples

Example 1:

A typical exam question on Atomic Structure requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Atomic Structure often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the essential chemistry of atomic structure, including key reactions, underlying theories, and practical applications.

Key concepts include:

key chemical principles and theories
mathematical relationships in chemistry
practical techniques and apparatus
applications of chemistry in industry
environmental and ethical considerations

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

Mark this page as reviewed

Atomic Structure