Atomic
Structure

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

A beam of protons, neutrons and electrons passes between two charged plates. Which way does each one bend, and which bends the most?

2

Chlorine has a relative atomic mass of 35.5, yet no chlorine atom has a mass of 35.5. Why not?

3

If an electron does not orbit the nucleus like a tiny planet, where is it, and what shape is the space it occupies?

4

Why does potassium put its 19th electron into the 4s orbital rather than a 3d orbital?

5

Why is the second electron of sodium about nine times harder to remove than the first?

Recall: what you know from O levels

1A sodium atom has 11 protons and 12 neutrons. How many electrons does it have?

11. An atom is neutral, so the number of electrons equals the number of protons.[1]

2What is the charge on a sodium ion? Why?

+1. Na (2.8.1) loses its one valence electron to form Na⁺ (2.8): 11 protons but only 10 electrons.[1]

3Write the electronic configuration of chlorine (17 electrons) as you did at O levels.

2.8.7[1]In this topic we look inside each shell: 2.8.7 becomes 1s² 2s² 2p⁶ 3s² 3p⁵.
1

Sub-atomic Particles

1.1Protons, neutrons and electrons

ParticleRelative chargeRelative mass
pproton+11
nneutron01
eelectron−11/1836
  • Protons and neutrons reside together in the nucleus, which holds almost all the mass and all the positive charge.
  • Electrons are found in regions of space around the nucleus, and are attracted electrostatically by the protons.
NoteIf the nucleus were a 1 cm marble, the whole atom would be about 1 km across. The atom is mostly empty space.
1

Sub-atomic Particles

3517Cl
17 p,  35 − 17 = 18 n,  17 e⁻

1.2Proton number and nucleon number

DefinitionNucleon number (mass number): total number of protons and neutrons in the nucleus of an atom.
DefinitionProton number (atomic number): number of protons in the nucleus of an atom. It determines the identity of the element.
DefinitionNuclide: a specific combination of protons and neutrons in a nucleus. Neutrons = nucleon number − proton number.
Atomno. of p = no. of e⁻
Cationloses e⁻: p > e⁻
Aniongains e⁻: e⁻ > p
1

Sub-atomic Particles

Your turnComplete the table. The first row is done for you.

SpeciesProtonsNeutronsElectrons
23Na111211
23Na⁺111210
32S²⁻161618
56Fe³⁺263023
Exam tipCharge changes only the electrons: no. of e⁻ = proton number − charge. Protons and neutrons never change in chemical reactions.
1

Sub-atomic Particles

1.3Isotopes

DefinitionIsotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
  • Same electronic configuration → same chemical properties: chemical reactions involve electrons, not the nucleus.
  • Different masses → different physical properties, e.g. density, rate of diffusion.

Name an isotope with its nucleon number, e.g. chlorine-35, 35Cl, or 2H. In every stable isotope except 1H and 3He, neutrons ≥ protons.

1

Sub-atomic Particles

Quick checkWhich statement about 35Cl and 37Cl atoms is correct?

AThey have the same number of neutrons.
BThey have the same nucleon number.
CThey have the same chemical properties.
DThey have the same density.
NoteC: both have 17 electrons arranged 1s² 2s² 2p⁶ 3s² 3p⁵, so they react the same way. They differ in neutrons (18, 20), nucleon number (35, 37), and therefore mass and density.
1

Sub-atomic Particles

1.4Beams of p, n and e⁻ in an electric field

PredictWhich way does each beam bend? Which bends the most?
  • Neutrons have no charge: undeflected.
  • Protons are deflected towards the negatively charged plate.
  • Electrons are deflected towards the positively charged plate, to a much larger extent: they are far lighter.
1

Sub-atomic Particles

Why does the electron bend more?

  • Between the plates, each particle feels a sideways force that depends on its charge. A proton and an electron feel forces of the same size.
  • The same force changes the path of a light particle far more than a heavy one. A proton is 1836 times heavier.
angle of deflection ∝ charge / mass
Particlecharge / massDeflection
proton+1 / 1 = +1small, towards −
neutron0 / 1 = 0none
electron−1 / (1/1836) = −1836large, towards +
1

Sub-atomic Particles

Worked exampleA beam of 241Am+ ions is deflected by +2°. Under the same conditions, find the angle of deflection of 32S− ions.

Step 1Write the relationship with a constant:  θ = k × chargemass

Step 2Use the ion you know to find k:  Am⁺:  +2 = k × +1241 → k = 482

Step 3Apply k to the new ion:  S⁻:  θ = 482 × −132 = −15.1°

NoteThe negative sign means S⁻ bends the opposite way to Am⁺. It bends more because its charge / mass is larger in size: 1/32 against 1/241.
1

Sub-atomic Particles

Your turnUnder certain conditions, a beam of 1H⁺ is deflected by +12°. Under the same conditions, find the angle of deflection of (a) 2H⁺, (b) 4He²⁺ and (c) 16O²⁻.

For ¹H⁺: +12 = k × +11, so k = 12
(a)
²H⁺: 12 × +12 = +6°[1]
(b)
⁴He²⁺: 12 × +24 = +6°[1]
(c)
16O²⁻: 12 × −216 = −1.5°[1]
Note2H⁺ and 4He²⁺ have the same charge / mass (1/2), so they are deflected by the same angle.
2

Relative Masses of Atoms

2.1Relative isotopic mass

DefinitionRelative isotopic mass: the mass of one mole of atoms of an isotope relative to 1/12 the mass of one mole of 12C atoms.

The ruler is carbon-12: one twelfth of a 12C atom is 1 unit.

ExampleA 37Cl atom weighs as much as 37 of those units, so its relative isotopic mass is 37 (more precisely 36.97). Relative masses have no units.
2

Relative Masses of Atoms

2.2Relative abundance

A natural sample of an element is a mixture of its isotopes. To find its average mass (Ar, next) we need to know how much of each isotope there is.

Relative abundance: the percentage of atoms of each isotope in a naturally occurring sample of the element.
ExampleChlorine: in every 100 atoms, 75 are 35Cl and 25 are 37Cl, so the relative abundances are 75% and 25%.
2

Relative Masses of Atoms

2.3Relative atomic mass, Ar

DefinitionRelative atomic mass: the weighted average mass of one mole of atoms of an element relative to 1/12 the mass of one mole of 12C atoms.

A simple average of 35 and 37 is 36, but there are three times as many 35Cl atoms, so the average is pulled towards 35.

Ar = Σ (relative isotopic mass × relative abundance)total abundance (100)

Ar(Cl) = (35 × 75) + (37 × 25)100 = 35.5

2

Relative Masses of Atoms

Worked exampleCopper is 69.2% 63Cu and 30.8% 65Cu. Find Ar(Cu).

Ar = (63 × 69.2) + (65 × 30.8)100

Ar = 4359.6 + 2002100 = 63.6

Your turnMagnesium is 79.0% 24Mg, x% 25Mg and the rest 26Mg. Ar(Mg) = 24.32. Find x.

% 26Mg = 100 − 79.0 − x = 21.0 − x
Ar = 24(79.0) + 25x + 26(21.0 − x)100 = 24.32[1]
1896 + 25x + 546 − 26x = 2432
x = 10.0[1]
2

Relative Masses of Atoms

ChallengeChlorine is 75% 35Cl and 25% 37Cl. Cl₂ molecules contain two Cl atoms picked at random. Find the relative abundances of the three kinds of Cl₂ molecule, of masses 70, 72 and 74.

Each atom is 35Cl with probability 0.75 and 37Cl with probability 0.25.
Mass 70 (35Cl–35Cl): 0.75 × 0.75 = 0.5625
Mass 72 (35Cl–37Cl or 37Cl–35Cl): 2 × 0.75 × 0.25 = 0.375
Mass 74 (37Cl–37Cl): 0.25 × 0.25 = 0.0625[1]
Relative abundances 70 : 72 : 74 = 9 : 6 : 1[1]
3

Shells, Subshells and Orbitals

3.1Principal quantum shells, subshells and orbitals

Atomthe hotel
→
Shellsn = 1, 2, 3 …floors
→
Subshellss, p, d, ftypes of room
→
Orbitalss: 1, p: 3, d: 5rooms
→
2 e⁻ maxper orbital2 guests per room
  • A larger n gives a larger, more diffused orbital, further from the nucleus, and hence a higher energy: higher floors cost more.
  • Shell n has n types of room: shell 3 has 3s, 3p and 3d.
3

Shells, Subshells and Orbitals

AnalogyGuests (electrons) check in one at a time. Every guest wants to pay as little as possible.

  • Rooms on lower floors are cheaper, and within a floor s < p < d.
  • Guests take the cheapest room available.
  • Nobody wants to share with a stranger, but if the next empty room costs much more, sharing is the better deal.
  • Rooms of the same price: each guest takes an empty one before anyone shares.

These are the three filling rules in Section 4.

3

Shells, Subshells and Orbitals

What is an orbital?

We cannot know exactly where an electron is, and it does not circle the nucleus like a planet. Instead, imagine taking a snapshot of its position again and again, and marking a dot each time.

The dots build up a cloud: dense near the nucleus, thinning out further away.

DefinitionOrbital: the region of space where there is a high probability (95%) of finding an electron. Each orbital can accommodate 2 electrons.
3

Shells, Subshells and Orbitals

nSubshellsOrbitals in eachTotal orbitals (n²)Max e⁻ (2n²)
11s112
22s, 2p1, 348
33s, 3p, 3d1, 3, 5918
44s, 4p, 4d, (4f)1, 3, 5, (7)1632
PredictHow many orbitals, and how many electrons, can shell 3 hold?

4f is not in the syllabus. The 3 p orbitals are px, py, pz, the 5 d orbitals are dxy, dxz, dyz, dx²−y², dz².

3

Shells, Subshells and Orbitals

3.2Relative energies of the orbitals

  • Energy rises with n: 1s < 2s < 3s < 4s.
  • Within a shell: s < p < d < f, as an s electron is more likely to be found near the nucleus than a p electron, and a p than a d. Orbitals of the same subshell have the same energy: they are degenerate.
  • The levels get closer together as n increases, until they overlap. The first overlap: 4s is lower in energy than 3d, so 4s fills first.
3

Shells, Subshells and Orbitals

3.3Shapes of orbitals

  • s: spherical. p: dumb-bell, one along each axis.
  • d: dx²−y² and dz² point along the axes, dxy, dxz and dyz point between the axes.
4

Electronic Configuration

Two ways to represent it

Electrons-in-boxes

Each orbital is a box, each electron is a half arrow. Two electrons in one box point in opposite directions (opposite spins).

Argon. Label boxes with the subshell only (2p, not 2p⁶).

4

Electronic Configuration

4.1Rules for filling orbitals

Aufbau principle
Electrons fill the orbitals of lowest energy first. After 3p, 4s is filled before 3d.
Pauli exclusion principle
An orbital holds a maximum of two electrons, of opposite spins.
Hund's rule
Orbitals of a subshell are filled singly first, with parallel spins, before pairing. This gives less inter-electronic repulsion.

Ground state: all electrons in the lowest energy levels. Excited state: an electron promoted to a higher level, e.g. O: 1s² 2s² 2p³ 3s¹.

4

Electronic Configuration

Order of filling

Filling order and the Periodic Table

  • The block tells you the subshell that is filled last.
  • The period number = n of the valence shell. A new period starts when electrons enter a new principal quantum shell.
  • The position in the block tells you how many electrons are in that subshell.
  • Elements in the same Group have similar valence shell configurations.
4

Electronic Configuration

4.2Writing configurations of atoms

Na: Period 3, 1st in s-blockends 3s¹: 1s² 2s² 2p⁶ 3s¹
P: Period 3, 3rd in p-blockends 3p³: [Ne] 3s² 3p³
Mn: Period 4, 5th in d-blockends 3d⁵: [Ar] 3d⁵ 4s² (d is one shell behind: 3d in Period 4)
Se: Period 4, 4th in p-blockends 4p⁴: [Ar] 3d10 4s² 4p⁴
NoteCr and Cu break the pattern (4.3). Always check those two.
4

Electronic Configuration

Worked exampleWrite the full electronic configuration of iron, Fe (26 electrons).

Step 1Fill in energy order: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² (20 e⁻ so far)

Step 2The last 6 go into 3d: 3d⁶ (26 e⁻)

Step 3Write in order of increasing n: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Exam tipWrite the full configuration with 3d before 4s, and check that the superscripts add up to the number of electrons.

Your turnWrite the full electronic configurations of:

(a) K (19)

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹[1]

(b) Mn (25)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²[1]

(c) Br (35)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d10 4s² 4p⁵[1]
Noble gas configuration: show the inner electrons as the previous noble gas, e.g. Br: [Ar] 3d10 4s² 4p⁵.
4

Electronic Configuration

Anomalies: Cr and Cu

PredictUsing the Aufbau principle, what do you expect for Cr (24 e⁻) and Cu (29 e⁻)?

Cr expected [Ar] 3d⁴ 4s², actual [Ar] 3d⁵ 4s¹

Cu expected [Ar] 3d⁹ 4s², actual [Ar] 3d10 4s¹

A half-filled (d⁵) or fully-filled (d10) 3d subshell is more stable.

3d and 4s are very close in energy, so moving one electron costs little, and the more stable d⁵ or d10 arrangement wins.

4

Electronic Configuration

4.3Configurations of ions

Fe fills 4s before 3d.

PredictSo does Fe²⁺ lose its last-in electrons, two of the 3d?

Cation: electrons are removed from the outermost shell of the ground state configuration: 4s before 3d. Once 3d holds electrons, 4s (n = 4) is the outermost shell.
Anion: electrons are added to the empty or partially-filled valence orbitals, e.g. S²⁻: 1s² 2s² 2p⁶ 3s² 3p⁶.
4

Electronic Configuration

Quick checkWhat is the electronic configuration of Cu²⁺?

A[Ar] 3d⁹
B[Ar] 3d⁸ 4s¹
C[Ar] 3d⁷ 4s²
D[Ar] 3d10
NoteA: Cu is [Ar] 3d10 4s¹ (an anomaly). Remove the 4s electron first, then one 3d electron: [Ar] 3d⁹. Option C forgets the anomaly and removes 3d before 4s.
5

Ionisation Energy

DefinitionIonisation energy: the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms or ions.
X(g) → X⁺(g) + e⁻ΔH = 1st IE of X X⁺(g) → X²⁺(g) + e⁻ΔH = 2nd IE of X X(n−1)+(g) → Xn+(g) + e⁻ΔH = nth IE of X
  • All IEs are endothermic: energy must be put in to pull an electron away from the attraction of the nucleus.
  • Gaseous atoms are far apart, so the energy measured is only for removing the electron. Always show (g) in the equation.
5

Ionisation Energy

First IE of elements 1 to 20 / kJ mol⁻¹

DescribeDescribe the patterns you see.
  • Peaks at the noble gases, a sharp drop to each Group 1 metal.
  • Generally increases across a period.
  • Decreases down a group: He > Ne > Ar.
  • Two small dips in each period: B, Al and O, S.
5

Ionisation Energy

Shielding and effective nuclear charge

The outer electron is pulled by every proton, but the inner electrons sit between it and the nucleus and repel it, cancelling part of that pull: shielding.

Effective nuclear charge, Zeff = Z − shielding constant: the pull an outer electron actually feels.

AnalogyStanding in the back row of a concert: the crowd in front (inner electrons) blocks most of the stage, however big the stage (nucleus) is.

Electrons in the same shell shield each other very little.

5

Ionisation Energy

5.1Across periods 2 and 3, 1st IE generally ↑

  1. Nuclear charge ↑, while there is a negligible ↑ in inner shell shielding effect, as e⁻ are added to the same shell
  2. → effective nuclear charge ↑, atomic radius ↓
  3. → valence e⁻ more strongly attracted by the nucleus → more energy required to remove them
5

Ionisation Energy

Down a group, 1st IE ↓

  1. Atomic radius ↑
  2. No. of principal quantum shells ↑ → shielding effect ↑, which largely cancels out the ↑ in nuclear charge
  3. → valence e⁻ less attracted by the nucleus → less energy required to remove them
Exam tipAlways state the effect on attraction and then on the energy required: these are the mark points.
5

Ionisation Energy

Dips: Be → B (also Mg → Al)

Be1s2 2s2B1s2 2s2 2p1

PredictThe 2p electron is higher in energy. Does that make it harder or easier to remove?

AnalogyThink of a deep well. Removing an electron means lifting it to the top (where it is free). A ball resting on a higher ledge needs a shorter lift.

The 2p electron removed from B is higher in energy than the 2s electron removed from Be.[1]
Hence, less energy is required to remove it, and the first ionisation energy of B is lower than that of Be.[1]
5

Ionisation Energy

Dips: N → O (also P → S)

AnalogyBack to the hotel: O's 8th guest had to share a 2p room with a stranger. Two guests in one room push each other, so one of them is easier to move out.

There is Coulombic repulsion between the paired 2p electrons in O, so it is easier to remove one of them than an unpaired 2p electron from N.[1]
Hence, less energy is required, and the first ionisation energy of O is lower than that of N.[1]
5

Ionisation Energy

5.2Successive IEs of fluorine, 1s² 2s² 2p⁵ (log scale)

Answer structureReading the graph

  1. General trend: successive IEs ↑ as each e⁻ is removed from an ion of increasing positive charge
  2. Gradual increase (1st to 7th IE): removal of e⁻ from the same principal quantum shell
  3. Sharp increase from 7th to 8th IE: 8th e⁻ removed from an inner principal quantum shell, nearer to the nucleus, with less shielding

Note: the small jump from 5th to 6th IE is an e⁻ from a lower energy subshell (2s after 2p) of the same shell.

5

Ionisation Energy

Worked exampleThe successive IEs of element X are shown below. Predict the Group of X.

IE / kJ mol⁻¹8701800300036005800700013000
  1. Successive IEs ↑ as each e⁻ is removed from an ion of increasing positive charge
  2. 1st to 6th IE ↑ gradually: removal of the outermost e⁻ only
  3. Sharp ↑ from 6th to 7th IE: 7th e⁻ removed from an inner principal quantum shell, nearer to the nucleus, with less shielding
  4. ∴ 6 e⁻ in the outermost shell (ns² np⁴) → X is in Group 16
  5. Each big jump marks an inner shell, so big jumps + 1 = occupied shells.
5

Ionisation Energy

Your turnElement Y is in Period 3. Its successive IEs / kJ mol⁻¹ are:

578181727451157714842

Find its Group and identify Y.

The large jump is between the 3rd and 4th ionisation energies, so Y has 3 electrons in its outermost shell. Group 13, and in Period 3, Y is aluminium.[1]

ChallengeElement Z has successive IEs / kJ mol⁻¹:

496456269129543

Which Group is Z in, and why is the 2nd IE so much larger?

Group 1: the large jump is after the 1st ionisation energy.[1]
The 2nd electron is removed from an inner principal quantum shell, nearer the nucleus and less shielded, so it is much more strongly attracted.[1]

Exit ticket

1How many neutrons and electrons are in 56Fe³⁺?

30 neutrons, 23 electrons[1]

2Write the full electronic configuration of Cr.

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹[1]

3Why is the first IE of O lower than that of N?

Repulsion between the paired 2p e⁻ in O makes one of them easier to remove.[1]

4Which is deflected more in an electric field, 2H⁺ or 3He²⁺?

3He²⁺: its charge / mass, 2/3, is larger than 1/2.[1]