S4 Chemistry – Unit 2: Electronic Configuration of Atoms & Ions

2.1

Bohr's Atomic Model & Energy Levels

Limitations of Rutherford's Model

Rutherford's nuclear model could not explain why electrons don't radiate energy and spiral into the nucleus, nor why atoms emit light only at specific wavelengths. Niels Bohr (1913) proposed a modified model to address these issues.

Bohr's Postulates

Electrons orbit the nucleus in fixed circular paths called shells (energy levels), labelled n = 1, 2, 3, 4…
While in an allowed orbit, an electron does not radiate energy — it stays at constant energy.
Each shell has a fixed energy. Higher n = higher energy = further from nucleus.
An electron moves between shells by absorbing or emitting a photon of energy exactly equal to the energy difference: ΔE = hf.

E = hf = hc/λ h = 6.626 × 10⁻³⁴ J s c = 3.00 × 10⁸ m s⁻¹ λ = wavelength (m)

ℹ️

Ground state vs excited stateGround state = lowest energy arrangement (electrons in lowest available shells). Excited state = electron has absorbed energy and jumped to a higher shell. It rapidly falls back, emitting a photon.

⚠️

Limitation of Bohr's modelWorks well for hydrogen but fails for multi-electron atoms. Replaced by the quantum mechanical orbital model, but the concept of quantised energy levels remains valid.

Section 2.1 Quick Quiz

Bohr's Atomic Model & Energy Levels

10 Questions

Q1

In Bohr's model of the hydrogen atom, electrons:

Can occupy any energy level Orbit the nucleus only in allowed (quantised) energy levels without emitting radiation Move randomly in the electron cloud Are embedded in the nucleus

Q2

When an electron moves from a higher to a lower energy level, it:

Absorbs a photon of light Emits a photon of light with energy = ΔE between levels Gains kinetic energy Is lost from the atom

Q3

What does it mean to say that atomic energy levels are 'quantised'?

Atoms contain equal numbers of protons and electrons Energy levels can only have specific discrete values — not a continuous range All electrons have the same energy Energy levels are infinitely large

Q4

Which scientist proposed that electrons move in fixed circular orbits around the nucleus?

Thomson Rutherford Bohr Schrödinger

Q5

The ground state of an atom is defined as:

The state with maximum energy The lowest energy state — all electrons in their lowest possible energy levels The state where electrons are in the nucleus The ionised state

Q6

When an electron absorbs energy and moves to a higher energy level, the atom is described as:

Ionised In the ground state In an excited state Discharged

Q7

The energy levels in hydrogen are labelled n=1, 2, 3... Where is the electron in the ground state?

n=3 n=2 n=1 n=∞

Q8

Why did Bohr's model fail for atoms with more than one electron?

Bohr didn't know about neutrons Bohr's model only accounted for electron-nucleus attraction, ignoring electron-electron repulsion in multi-electron atoms The model only worked for gases Bohr used incorrect values for the speed of light

Q9

The energy of an emitted photon depends on:

The mass of the electron only The difference in energy between the two levels involved in the transition (ΔE = hf) The temperature of the atom The atomic number of the element

Q10

Bohr's model successfully explained:

The spectra of all atoms Only the line spectrum of hydrogen (and hydrogen-like ions) The shapes of atomic orbitals The reactivity of metals

2.2

Hydrogen Spectrum & Spectral Lines

Emission Spectrum of Hydrogen

When hydrogen is excited (electrical discharge or heat), electrons are promoted to higher levels. As they fall back, they emit photons of specific frequencies producing a line emission spectrum — bright coloured lines on a dark background. Each line corresponds to electrons falling from n₂ to n₁.

Series	Transition	Region
Lyman	n ≥ 2 → n=1	Ultraviolet (UV)
Balmer	n ≥ 3 → n=2	Visible light
Paschen	n ≥ 4 → n=3	Infrared (IR)
Brackett	n ≥ 5 → n=4	Infrared (IR)

Convergence and Ionisation Energy

As n increases, energy levels get closer together — spectral lines converge. At the convergence limit (n → ∞), the electron is completely removed. The frequency at convergence gives the ionisation energy directly:

IE₁(H) = hf_convergence = hc/λ_convergence

Section 2.2 Quick Quiz

Hydrogen Spectrum & Spectral Lines

10 Questions

Q1

The Lyman series in the hydrogen spectrum lies in the:

Visible region Ultraviolet (UV) region Infrared (IR) region X-ray region

Q2

The Balmer series corresponds to electron transitions:

From any level to n=1 From any higher level to n=2 From any higher level to n=3 From n=2 to any higher level

Q3

Which spectral line in the Balmer series has the longest wavelength (lowest energy)?

n=6→n=2 n=5→n=2 n=4→n=2 n=3→n=2

Q4

The line spectrum of hydrogen is evidence that:

Atoms are solid spheres Electrons can only occupy discrete energy levels (quantised) Electrons orbit in circular paths Hydrogen has only one electron

Q5

What happens to spectral lines as n increases in the Balmer series?

They get further apart They get closer together (converge) towards a series limit They stay equally spaced They disappear

Q6

The series limit of the Lyman series corresponds to:

n=1→n=2 transition n=∞→n=1 transition (ionisation from n=1) n=1→n=∞ transition (ionisation from n=1) n=2→n=1 transition

Q7

An emission spectrum is produced when:

Atoms absorb photons of specific wavelengths Atoms are cooled below their boiling point Electrons fall from higher to lower energy levels, releasing photons Electrons are added to atoms

Q8

In the hydrogen spectrum, the visible region contains lines from the Balmer series because:

Hydrogen only exists in the visible range Transitions from n≥3 to n=2 produce energy differences corresponding to visible light photons (400-700 nm) Hydrogen has only 2 electron energy levels The Balmer series is the only one for hydrogen

Q9

The ionisation energy of hydrogen can be determined from its spectrum by:

Measuring the number of spectral lines Finding the series limit: the frequency at which lines converge — this corresponds to complete ionisation Counting the visible lines Measuring the colour of the brightest line

Q10

How does the emission spectrum of helium differ from that of hydrogen?

It has fewer lines because He has only 2 electrons He has more complex spectrum because 2 electrons create more possible transitions They are identical — both noble elements He has only one spectral line

2.3

Atomic Spectra

Emission SpectrumExcited atoms release energy — electrons fall to lower levels, emitting photons. Bright coloured lines on a dark background. Unique to each element ("fingerprint").

Absorption SpectrumWhite light passes through a cool gas. Atoms absorb photons at exact frequencies to promote electrons. Dark lines on a continuous spectrum — same positions as emission lines.

Evidence for Quantised Energy Levels

Discrete spectral lines (not a continuous spectrum) are direct evidence that electrons can only exist at specific, quantised energy levels. If energy were continuous, a continuous spectrum would result. Only certain wavelengths being emitted proves energy levels are discrete.

💡

Flame tests and emissionMetal salts heated in a flame emit characteristic colours: Li = crimson, Na = yellow, K = lilac, Ca = orange-red, Ba = green, Cu = blue-green. Each corresponds to specific emission lines in the visible region.

Section 2.3 Quick Quiz

Atomic Spectra

10 Questions

Q1

An absorption spectrum is produced when:

Atoms emit light of specific wavelengths White light passes through a cool gas — atoms absorb specific wavelengths, leaving dark lines Atoms are heated in a flame Electrons are removed from atoms

Q2

The dark lines in the absorption spectrum of the sun are called:

Balmer lines Fraunhofer lines Bohr lines Planck lines

Q3

Which statement correctly compares emission and absorption spectra of the same element?

They have no relation to each other The bright lines in the emission spectrum are at the same wavelengths as the dark lines in the absorption spectrum Emission spectra have more lines Absorption spectra are always in the UV region

Q4

Atomic emission spectroscopy is used to identify elements in stars because:

Stars are close to Earth Each element has a unique spectral fingerprint — its set of emission lines matches no other element Stars emit visible light only The mass spectrometer doesn't work in space

Q5

The flame test for sodium produces a persistent yellow-orange flame because:

Sodium burns in air Na⁺ ions are easily reduced in the flame Electrons in excited sodium atoms emit photons at 589 nm (yellow-orange) when returning to ground state Sodium has a yellow colour

Q6

What is meant by 'continuous spectrum'?

A spectrum with only a few lines A spectrum containing all wavelengths of light without gaps (like a rainbow) A spectrum that can be seen continuously A spectrum measured over a long time

Q7

The emission spectrum of hydrogen shows the Paschen series in the infrared because:

Hydrogen is a gas Transitions from n≥4 to n=3 release photons with lower energy than visible light → wavelengths >700 nm (IR) Infrared detectors are more sensitive Hydrogen has 3 electron shells

Q8

Line spectra prove that atomic energy is quantised because:

Atoms contain a fixed number of electrons If energy were continuous, all wavelengths would be emitted — the fact that only specific wavelengths appear shows only specific ΔE values are possible Line spectra are easier to measure Atoms can only exist in two states

Q9

The frequency of a spectral line is related to the energy of the photon by:

E = mc² E = hf (Planck's equation, h = 6.63×10⁻³⁴ J s) E = ½mv² E = qV

Q10

Why do different elements produce different colours in flame tests?

Different elements have different masses Each element has unique atomic energy levels → unique spectral lines → unique colours All elements produce the same spectral lines Colour depends on the concentration

2.4

Orbitals & Quantum Numbers

Quantum Mechanical Model

Bohr's circular orbits were replaced by the quantum mechanical model. Electrons do not follow defined paths — instead we describe regions of space where there is high probability of finding an electron: atomic orbitals. Each orbital holds a maximum of 2 electrons (opposite spins).

Quantum Number	Symbol	Describes	Values
Principal	n	Shell (energy level), distance from nucleus	1, 2, 3, 4…
Angular momentum	l	Subshell (orbital shape)	0 to (n−1)
Magnetic	mₗ	Orbital orientation in space	−l to +l
Spin	mₛ	Electron spin direction	+½ or −½

l value	Subshell	Shape	Orbitals	Max e⁻
0	s	Spherical	1	2
1	p	Dumbbell (3 orientations)	3	6
2	d	Double dumbbell (5 orientations)	5	10
3	f	Complex (7 orientations)	7	14

Shell (n)	Subshells	Total orbitals	Max electrons (2n²)
n=1	1s	1	2
n=2	2s, 2p	4	8
n=3	3s, 3p, 3d	9	18
n=4	4s, 4p, 4d, 4f	16	32

Three Rules for Filling Orbitals

1. Aufbau Principle — fill in order of increasing energy:

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p…

2. Pauli Exclusion Principle — no two electrons in the same atom can have identical quantum numbers. Each orbital holds max 2 electrons with opposite spins (↑↓).

3. Hund's Rule — electrons fill degenerate orbitals singly with parallel spins before any pairing occurs. Minimises repulsion.

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4s fills before 3dIn neutral atoms, 4s has lower energy than 3d due to penetration. So K = [Ar]4s¹, Ca = [Ar]4s². When transition metal ions form, the 4s electrons are removed first.

Section 2.4 Quick Quiz

Orbitals & Quantum Numbers

10 Questions

Q1

An atomic orbital is best described as:

A fixed circular path for an electron A 3D region of space where there is a high probability (>90%) of finding an electron The nucleus of an atom The electron itself

Q2

The principal quantum number (n) determines:

The shape of the orbital The orientation of the orbital The main energy level (shell) — larger n = higher energy and further from nucleus The spin of the electron

Q3

The angular momentum quantum number (l) for a p subshell has the value:

0 1 2 3

Q4

How many orbitals are in the 3d subshell?

1 3 5 7

Q5

The spin quantum number (ms) for an electron can only be:

0 or +1 −½ or +½ 1 or 2 +1 or −1

Q6

The Pauli exclusion principle states that:

Electrons always pair up before occupying new orbitals No two electrons in the same atom can have the same set of four quantum numbers Electrons in degenerate orbitals must have parallel spins Orbitals fill in order of increasing energy

Q7

Hund's rule states that in degenerate (equal energy) orbitals:

Electrons always pair before filling new orbitals Electrons fill each degenerate orbital singly with parallel spins before any pairing occurs Orbitals fill in order of increasing energy Each orbital must have 2 electrons

Q8

The aufbau principle states that:

Orbitals must be filled with paired electrons Electrons fill orbitals in order of increasing energy (lowest first) All electrons have the same spin p orbitals are always filled before s orbitals

Q9

How many electrons can the n=3 shell hold in total?

6 8 18 32

Q10

The shape of a p orbital is best described as:

Spherical Dumbbell (figure-8) shaped, with one lobe on each side of the nucleus Cloverleaf shaped Toroidal (donut shaped)

2.5

Electronic Configuration of Atoms & Ions

Writing Electron Configurations

List each subshell and its electron count: e.g., 1s²2s²2p⁶. Shorthand uses the previous noble gas: Na = [Ne]3s¹.

Element	Z	Configuration	Shorthand
H	1	1s¹	—
He	2	1s²	—
Li	3	1s²2s¹	[He]2s¹
C	6	1s²2s²2p²	[He]2s²2p²
N	7	1s²2s²2p³	[He]2s²2p³
O	8	1s²2s²2p⁴	[He]2s²2p⁴
Ne	10	1s²2s²2p⁶	—
Na	11	1s²2s²2p⁶3s¹	[Ne]3s¹
Mg	12	1s²2s²2p⁶3s²	[Ne]3s²
Al	13	1s²2s²2p⁶3s²3p¹	[Ne]3s²3p¹
P	15	1s²2s²2p⁶3s²3p³	[Ne]3s²3p³
S	16	1s²2s²2p⁶3s²3p⁴	[Ne]3s²3p⁴
Ar	18	1s²2s²2p⁶3s²3p⁶	—
K	19	…3p⁶4s¹	[Ar]4s¹
Ca	20	…3p⁶4s²	[Ar]4s²
Fe	26	…3d⁶4s²	[Ar]3d⁶4s²
Cr*	24	…3d⁵4s¹	[Ar]3d⁵4s¹
Cu*	29	…3d¹⁰4s¹	[Ar]3d¹⁰4s¹
Zn	30	…3d¹⁰4s²	[Ar]3d¹⁰4s²

⚠️

Chromium & Copper exceptions (*)Cr: [Ar]3d⁵4s¹ (not 3d⁴4s²) — half-filled 3d has extra stability. Cu: [Ar]3d¹⁰4s¹ (not 3d⁹4s²) — fully-filled 3d has extra stability. One 4s electron promotes to 3d in each case.

Ion Electron Configurations

Cations: remove electrons from the highest energy subshell first. For transition metals, remove 4s before 3d.
Anions: add electrons to the next available subshell.

WORKED EXAMPLE

Ion Configurations

Fe (Z=26): [Ar]3d⁶4s²
Fe²⁺: remove 2×4s → [Ar]3d⁶
Fe³⁺: remove 2×4s + 1×3d → [Ar]3d⁵

O (Z=8): 1s²2s²2p⁴
O²⁻: add 2e⁻ to 2p → 1s²2s²2p⁶ (isoelectronic with Ne)

Cu (Z=29): [Ar]3d¹⁰4s¹
Cu²⁺: remove 4s¹ + one 3d → [Ar]3d⁹

Box Notation (Orbital Diagrams)

Arrows (↑ = spin up, ↓ = spin down) in boxes represent electrons in orbitals. Demonstrates Hund's rule clearly.

Carbon (1s²2s²2p²): 1s[↑↓] 2s[↑↓] 2p[↑][ ↑][ ] ← Hund: unpaired in separate orbitals Nitrogen (1s²2s²2p³): 1s[↑↓] 2s[↑↓] 2p[↑][↑][↑] ← all 3 p orbitals singly occupied

Section 2.5 Quick Quiz

Electronic Configuration of Atoms & Ions

10 Questions

Q1

What is the electronic configuration of sodium (Z=11)?

1s²2s²2p⁶ 1s²2s²2p⁶3s¹ 1s²2s²2p⁵3s² 1s²2s²2p⁶3s²

Q2

The electronic configuration of a chloride ion (Cl⁻, Z=17) is:

1s²2s²2p⁶3s²3p⁵ 1s²2s²2p⁶3s²3p⁶ 1s²2s²2p⁶3s²3p⁴ 1s²2s²2p⁶3s²3p⁷

Q3

Why does chromium (Z=24) have the configuration [Ar]3d⁵4s¹ and not [Ar]3d⁴4s²?

There are only 5 d orbitals The half-filled 3d⁵ configuration with one 4s¹ electron is more stable due to exchange energy Chromium is a metal 4s orbital is full before 3d

Q4

When transition metals form ions, electrons are removed from:

The d subshell first The s subshell first (4s electrons are lost before 3d) The p subshell The f subshell

Q5

The electronic configuration [Ar]3d¹⁰4s²4p⁶ corresponds to:

Zinc (Zn) Krypton (Kr) Calcium (Ca) Selenium (Se)

Q6

How many unpaired electrons does oxygen (Z=8) have in its ground state?

0 1 2 4

Q7

What is the configuration of Fe³⁺ (Z=26)?

[Ar]3d⁵4s¹ [Ar]3d⁵ [Ar]3d⁴4s¹ [Ar]3d⁶

Q8

Which species is isoelectronic with Ar (Z=18)?

Na⁺ K⁺ Cl⁻ Both Na⁺ and Cl⁻

Q9

The block of an element in the periodic table is determined by:

Its group number Its period number The subshell being filled by the last electron (s-block, p-block, d-block, f-block) Its atomic mass

Q10

An atom has configuration 1s²2s²2p⁶3s²3p¹. Which element is this?

Magnesium Aluminium Silicon Phosphorus

2.6

Ionisation Energy, Energy Levels & Influencing Factors

First Ionisation Energy (IE₁)Energy to remove 1 mol of electrons from 1 mol of gaseous atoms in their ground state.
X(g) → X⁺(g) + e⁻ ΔH = IE₁ (always positive — endothermic)

Successive Ionisation EnergiesEnergy to remove the 1st, 2nd, 3rd… electrons in sequence. Each successive IE is always greater than the previous (removing from a more positive ion).

Factor	Effect	Explanation
Nuclear charge (Z)	↑Z → ↑IE	More protons attract electrons more strongly.
Distance from nucleus	↑distance → ↓IE	Outer electrons further away feel weaker attraction.
Shielding	↑shielding → ↓IE	Inner electrons reduce effective nuclear charge felt by outer electrons.
Paired electron repulsion	Paired → ↓IE	Electrons paired in the same orbital repel each other — easier to remove one.

Trends in First Ionisation Energy

Across a period (left → right): IE₁ generally increases — Z increases while shielding stays roughly constant → greater effective nuclear charge → stronger hold on outer electrons.

Anomalous dips across Period 2/3:

Group 2 → Group 13 dip (Be→B, Mg→Al): Group 13 outer electron is in a higher-energy p subshell, shielded by the s electrons → easier to remove despite higher Z.
Group 15 → Group 16 dip (N→O, P→S): Group 16 has a paired p electron; repulsion between paired electrons makes it easier to remove one → IE₁(O) < IE₁(N).

Down a group: IE₁ decreases — each new period adds a shell, increasing distance and shielding → outer electron easier to remove.

Successive IEs and Shell Structure

A plot of log(IE) vs ionisation number shows large jumps when removal crosses into an inner shell. These jumps confirm the number of electrons in each shell.

Sodium (Na, Z=11, config 2,8,1):

IE₁ low: remove 3s¹ (lone outer electron) IE₂–IE⁹ moderate, rising: removing n=2 electrons (8 electrons) Large jump IE⁹ → IE₁₀: crossing into n=1 shell (core) IE₁₀, IE₁₁ very high: n=1 electrons, closest to nucleus

WORKED EXAMPLE

Identifying an Element from Successive IEs

Successive IEs (kJ mol⁻¹): 738, 1451, 7733, 10540, 13630, 17995, 21703. Identify element X and its group.

Large jump between IE₂ (1451) and IE₃ (7733) — a ~5× increase.

2 electrons removed easily before jump → 2 outer shell electrons → Group 2.

IE₁ = 738 kJ mol⁻¹ → Magnesium (Mg, Z=12).

Section 2.6 Quick Quiz

Ionisation Energy, Energy Levels & Influencing Factors

10 Questions

Q1

The first ionisation energy (IE₁) is defined as the energy required to:

Remove all electrons from one mole of atoms Remove one electron from each atom in one mole of gaseous atoms: X(g) → X⁺(g) + e⁻ Remove the most loosely bound electron from a solid Add one electron to one mole of gaseous atoms

Q2

Which factor increases the first ionisation energy?

Larger atomic radius More electron shielding Higher nuclear charge (Z) with same shielding More electrons

Q3

Across Period 3 (Na→Ar), the general trend in first ionisation energy is:

Decreasing Increasing, with two notable exceptions (Mg→Al and P→S are drops) Constant First increasing then decreasing

Q4

Down Group 1 (Li→Cs), first ionisation energy:

Increases Decreases Stays the same First increases then decreases

Q5

The successive ionisation energies of sodium show a large jump between:

IE₁ and IE₂ IE₂ and IE₃ IE₁₀ and IE₁₁ IE₅ and IE₆

Q6

The difference in IE₁ between Mg (738 kJ/mol) and Al (578 kJ/mol) is because:

Mg has more protons Al's outermost electron (3p¹) is in a higher energy subshell than Mg's (3s²) and is more easily removed Al is larger than Mg Mg has more electrons

Q7

Why is IE₁(S) < IE₁(P) despite S having higher Z?

S is larger than P S (3p⁴) has a paired electron in one 3p orbital — electron-electron repulsion makes this electron easier to remove than P's half-filled 3p³ (all unpaired, more stable) P has fewer electrons S is in a different period

Q8

Electron shielding refers to:

Electrons blocking X-rays Inner shell electrons reducing the effective nuclear charge felt by outer electrons The nucleus repelling electrons Electrons absorbing energy from the nucleus

Q9

The second ionisation energy (IE₂) is always greater than IE₁ because:

The second electron is in a different subshell After removing one electron, the remaining electrons are held more tightly by the same nuclear charge — higher Zeff per electron The second electron is heavier IE₂ involves a different element

Q10

High first ionisation energy is characteristic of:

Alkali metals (Group 1) Transition metals Noble gases (Group 18) Halogens (Group 17)

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✏️

Exercises

State Bohr's three postulates about electron behaviour. What evidence from atomic spectra supports quantised energy levels?
(1) Electrons orbit in fixed shells (n=1,2,3…) without radiating energy. (2) Each shell has a fixed energy; higher n = higher energy. (3) Electrons move between levels by absorbing/emitting photons of energy ΔE=hf. Evidence: Only specific wavelengths are observed (line spectrum, not continuous). This proves electrons occupy discrete energy levels — only certain transitions are possible, each producing one specific photon frequency.
Write the full electron configurations for: (a) S (Z=16), (b) Fe (Z=26), (c) Br (Z=35).
(a) S: 1s²2s²2p⁶3s²3p⁴
(b) Fe: 1s²2s²2p⁶3s²3p⁶3d⁶4s²
(c) Br: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵
Write the electron configurations for: (a) Fe²⁺, (b) Fe³⁺, (c) Cl⁻, (d) Cu²⁺.
(a) Fe²⁺: [Ar]3d⁶ (b) Fe³⁺: [Ar]3d⁵
(c) Cl⁻: 1s²2s²2p⁶3s²3p⁶ (isoelectronic with Ar)
(d) Cu²⁺: [Ar]3d⁹
State and explain why IE₁(B) < IE₁(Be) and IE₁(O) < IE₁(N).
B < Be: B has its outer electron in 2p (higher energy, shielded by 2s electrons). Be's outer electron is in 2s. The 2p electron is easier to remove despite higher Z. O < N: O (2p⁴) has one doubly-occupied 2p orbital. Electron-electron repulsion in the paired orbital makes it easier to remove one electron, lowering IE₁ below N (2p³, all singly occupied, no such repulsion).
The successive IEs (kJ mol⁻¹) of element Y are: 578, 1817, 2745, 11578, 14831, 18378. Determine Y's group and suggest its identity.
Large jump between IE₃ (2745) and IE₄ (11578). Three outer electrons → Group 13. IE₁ = 578 kJ mol⁻¹ → Aluminium (Al, Z=13).
Draw the orbital (box) diagrams for (a) carbon and (b) phosphorus, showing Hund's rule.
(a) C (1s²2s²2p²): 1s[↑↓] 2s[↑↓] 2p[↑][↑][ ] — two unpaired electrons in separate 2p orbitals.
(b) P (1s²2s²2p⁶3s²3p³): …3s[↑↓] 3p[↑][↑][↑] — all three 3p orbitals singly occupied, all spins parallel.

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Multiple Choice Quiz — 25 Questions

📝

Unit Test — 50 Marks

Section A — Short Answer

20 marks

Q1 [4 marks]

State the three rules used to determine the electron configuration of an atom. Name each rule and state what it requires, with one example. [4]

(1) Aufbau Principle: electrons fill orbitals in order of increasing energy (1s→2s→2p→3s→3p→4s→3d…). [1] (2) Pauli Exclusion Principle: no two electrons can share all four quantum numbers; each orbital holds max 2 electrons with opposite spins. [1] (3) Hund's Rule: in degenerate orbitals, electrons occupy each singly with parallel spins before pairing to minimise repulsion. [1] Example: N (2p³) — the three 2p electrons occupy three separate p orbitals, all with the same spin direction, before any pairing. [1]

Q2 [4 marks]

Write electron configurations for: (a) Cl⁻ [1]; (b) Mn (Z=25) [1]; (c) Mn²⁺ [1]; (d) Zn²⁺ (Z=30) [1].

(a) Cl⁻: 1s²2s²2p⁶3s²3p⁶ (18e, isoelectronic with Ar). [1] (b) Mn: [Ar]3d⁵4s². [1] (c) Mn²⁺: remove 2×4s → [Ar]3d⁵. [1] (d) Zn: [Ar]3d¹⁰4s²; Zn²⁺: remove 4s² → [Ar]3d¹⁰. [1]

Q3 [4 marks]

Explain why: (a) IE₁(B) < IE₁(Be) [2]; (b) IE₁(S) < IE₁(P) [2].

(a) Be: outer electron in 2s. B: outer electron in 2p, which is higher energy and shielded by 2s electrons → requires less energy to remove despite higher Z. IE₁(B) < IE₁(Be). [2] (b) P: 3p³ — all three 3p orbitals singly occupied, no pairing. S: 3p⁴ — one 3p orbital doubly occupied. Repulsion between paired electrons makes one easier to remove. IE₁(S) < IE₁(P). [2]

Q4 [4 marks]

Successive IEs (kJ mol⁻¹) of element Z: 590, 1145, 4912, 6474, 8144, 10496, 12270… (a) How many outer shell electrons does Z have? [1] (b) State Z's group. [1] (c) Suggest the identity of Z. [1] (d) Explain why successive IEs always increase. [1]

(a) Large jump between IE₂ (1145) and IE₃ (4912) → 2 outer electrons. [1] (b) Group 2. [1] (c) IE₁=590 → Calcium (Ca, Z=20). [1] (d) After each ionisation, the ion has fewer electrons but the same nuclear charge → higher effective nuclear charge per electron → each subsequent electron is more strongly attracted and requires more energy to remove. [1]

Q5 [4 marks]

Describe the Lyman, Balmer, and Paschen series of the hydrogen emission spectrum (transitions + spectral regions) [3]. Explain how the Lyman convergence limit gives the ionisation energy of hydrogen. [1]

Lyman: n≥2 → n=1; UV. [1] Balmer: n≥3 → n=2; visible. [1] Paschen: n≥4 → n=3; IR. [1] Lyman convergence: as n→∞, the transition n=∞→n=1 corresponds to complete removal of the electron from n=1 (the ground state). The frequency at convergence gives E=hf = IE₁ of hydrogen directly, since it represents the energy to fully remove the electron from the atom. [1]

Section B — Extended Answer

30 marks

Q6 [8 marks]

Explain what is meant by an emission spectrum. Using Bohr's model, explain how the line emission spectrum of hydrogen is produced. Why does the spectrum provide evidence for quantised energy levels? Describe the Balmer series including why lines converge. [8]

Emission spectrum [1]: Produced when excited atoms release energy as light. Electrons fall from higher to lower energy levels, emitting photons. Appears as bright coloured lines on a dark background.
Hydrogen emission by Bohr model [3]: At ground state, H’s electron is in n=1. When energy is supplied (electrical discharge), the electron is promoted to n=2,3,4… (excited state). The excited state is unstable — the electron rapidly falls to a lower level, emitting a photon of energy ΔE = E_upper − E_lower = hf. Since only specific energy levels are allowed, only specific photon frequencies are emitted → line spectrum.
Evidence for quantisation [2]: A continuous range of energies would give a continuous spectrum (rainbow). Only discrete wavelengths are observed → electrons can only exist at specific, discrete energy levels. Each line proves one specific transition — direct evidence for quantised energy levels.
Balmer series [2]: Transitions n≥3 → n=2; produces 4 visible lines (red n=3→2, blue-green n=4→2, violet n=5→2, etc.). Lines converge at shorter wavelengths/higher frequencies because energy levels get closer together as n increases — the energy gap between consecutive levels decreases, so photons emitted become more similar in energy/frequency until they merge at the convergence limit (n=∞→n=2).

Q7 [8 marks]

Define quantum number and describe all four quantum numbers. Explain Aufbau, Pauli, and Hund's principles. Write full orbital box diagrams for (a) oxygen and (b) silicon. [8]

Quantum numbers [2]: Numbers describing the state of an electron in an atom. (1) n (principal): shell number (1,2,3…); energy and distance from nucleus. (2) l (angular momentum, 0 to n−1): subshell and orbital shape (0=s sphere, 1=p dumbbell, 2=d, 3=f). (3) mₗ (−l to +l): orbital orientation in space. (4) mₛ (+½ or −½): spin direction of electron.
Three principles [3]: Aufbau: fill orbitals starting from lowest energy (1s→2s→2p→3s→3p→4s→3d…). Pauli: no two electrons share all four quantum numbers; max 2 per orbital with opposite spins. Hund: fill degenerate orbitals singly with parallel spins before pairing to minimise electron repulsion.
Orbital diagrams [3]:
(a) O (Z=8, 1s²2s²2p⁴):
1s[↑↓] 2s[↑↓] 2p[↑↓][↑][↑] — first two 2p electrons fill separately (Hund), then third 2p must pair up because all three orbitals are singly filled. [1.5]
(b) Si (Z=14, 1s²2s²2p⁶3s²3p²):
1s[↑↓] 2s[↑↓] 2p[↑↓][↑↓][↑↓] 3s[↑↓] 3p[↑][↑][ ] — two 3p electrons fill separately (Hund's rule). [1.5]

Q8 [6 marks]

Define first ionisation energy. State the four main factors that influence IE. Use these to explain the general increase in IE₁ across Period 3 and the two anomalous dips. [6]

IE₁ [1]: Energy to remove 1 mol of electrons from 1 mol of gaseous atoms in their ground state: X(g) → X⁺(g) + e⁻. Always endothermic.
Four factors [2]: (1) Nuclear charge Z: higher Z → greater attraction → higher IE. (2) Distance/shell: outer electrons in higher shells → weaker attraction → lower IE. (3) Shielding: inner electrons reduce effective nuclear charge felt by outer electrons; more inner electrons = more shielding = lower IE. (4) Paired electron repulsion: electrons paired in the same orbital repel each other, making one easier to remove → lower IE than expected.
General increase across Period 3 [2]: From Na to Ar, Z increases by 1 each step. All outer electrons are added to the same shell (n=3), so distance and shielding change only slightly. Increasing Z provides greater effective nuclear charge → stronger nuclear attraction on outer electrons → IE₁ generally increases.
Two dips [1]: Mg→Al: Al’s outer electron is in 3p (higher energy, shielded by 3s), while Mg’s is in 3s → easier to remove despite higher Z. P→S: S has a paired 3p electron causing repulsion → one electron is easier to remove; P (3p³) has no paired 3p electrons.

Q9 [8 marks]

Successive IEs (kJ mol⁻¹) of Q: 738, 1451, 7733, 10540, 13630, 17995, 21703, 25656. (a) Identify Q's group and element. [2] (b) Sketch a log(IE) vs ionisation number graph, marking the large jump. [2] (c) Explain what successive IEs reveal about Q's electronic structure. [2] (d) Write equations for the first three IEs of Q. [2]

(a) Large jump between IE₂ (1451) and IE₃ (7733) → 2 outer electrons → Group 2. IE₁=738 kJ mol⁻¹ → Magnesium (Mg, Z=12). [2]
(b) x-axis: ionisation number 1–8; y-axis: log(IE). Two low points (IE₁, IE₂) at left, large upward step between points 2 and 3 marked clearly, then gradually rising plateau for IE₃–IE₁⁰ (n=2 shell), then another very large jump at IE₁₁/IE₁₂ (n=1 shell). [2]
(c) IE₁–IE₂ relatively low and close: 2 electrons in n=3 outer shell (3s²) — well-shielded, far from nucleus. IE₃–IE₁⁰ much higher, rise gradually: 8 electrons in n=2 shell (closer to nucleus, stronger attraction). IE₁₁–IE₁₂ extremely high: 2 electrons in n=1 (maximum attraction, no shielding). The two jumps confirm Mg’s electron structure: 2,8,2 across shells n=1,2,3. [2]
(d) Mg(g) → Mg⁺(g) + e⁻ ΔH = +738 kJ mol⁻¹ [1]
Mg⁺(g) → Mg²⁺(g) + e⁻ ΔH = +1451 kJ mol⁻¹ [0.5]
Mg²⁺(g) → Mg³⁺(g) + e⁻ ΔH = +7733 kJ mol⁻¹ [0.5]

← Unit 1: Structure of Atom Unit 3: Ionic & Metallic Bonds →

Electronic Configuration
of Atoms & Ions

Bohr's Atomic Model & Energy Levels

Limitations of Rutherford's Model

Bohr's Postulates

Section 2.1 — Shells & Subshells

Hydrogen Spectrum & Spectral Lines

Emission Spectrum of Hydrogen

Convergence and Ionisation Energy

Section 2.2 — Electronic Configurations & The Periodic Table

Atomic Spectra

Evidence for Quantised Energy Levels

Orbitals & Quantum Numbers

Quantum Mechanical Model

Three Rules for Filling Orbitals

Electronic Configuration of Atoms & Ions

Writing Electron Configurations

Ion Electron Configurations

Ion Configurations

Box Notation (Orbital Diagrams)

Ionisation Energy, Energy Levels & Influencing Factors

Trends in First Ionisation Energy

Successive IEs and Shell Structure

Identifying an Element from Successive IEs

Section 2.3 — Ionisation Energies

Exercises

Multiple Choice Quiz — 25 Questions

Unit 2 Quiz

Unit Test — 50 Marks

Section A — Short Answer

Section B — Extended Answer

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Electronic Configurationof Atoms & Ions

Bohr's Atomic Model & Energy Levels

Limitations of Rutherford's Model

Bohr's Postulates

Section 2.1 — Shells & Subshells

Hydrogen Spectrum & Spectral Lines

Emission Spectrum of Hydrogen

Convergence and Ionisation Energy

Section 2.2 — Electronic Configurations & The Periodic Table

Atomic Spectra

Evidence for Quantised Energy Levels

Orbitals & Quantum Numbers

Quantum Mechanical Model

Three Rules for Filling Orbitals

Electronic Configuration of Atoms & Ions

Writing Electron Configurations

Ion Electron Configurations

Ion Configurations

Box Notation (Orbital Diagrams)

Ionisation Energy, Energy Levels & Influencing Factors

Trends in First Ionisation Energy

Successive IEs and Shell Structure

Identifying an Element from Successive IEs

Section 2.3 — Ionisation Energies

Exercises

Multiple Choice Quiz — 25 Questions

Unit 2 Quiz

Unit Test — 50 Marks

Section A — Short Answer

Section B — Extended Answer

💳 Pay to Download

Electronic Configuration
of Atoms & Ions