DP Chemistry HL S1.3 Electron Configurations

DP Chemistry HL S1.3 Electron Configurations - Full notes

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DP Chemistry HL | S1.3 Electron Configurations

DP CHEMISTRY · STRUCTURE

S1.3

Electron Configurations

Full notes for HL · IB-style

Contents

1	Electromagnetic radiation and the spectrum
2	Emission spectra and energy levels
3	The hydrogen spectrum
4	Shells, sub-levels and orbitals
5	Writing electron configurations
6	Orbital diagrams, Hund's rule and exceptions
7	Ionization energy and the convergence limit (HL only)
8	Successive ionization energies and trends (HL only)
9	Electromagnetic radiation
10	Continuous and line spectra
11	The hydrogen emission spectrum in detail
12	Shells, sub-levels and orbitals
13	Writing electron configurations
14	Orbital diagrams
15	Configurations of ions
16	Blocks of the periodic table
17	Energy level diagrams
18	Flame tests and emission
19	Ionization energy: an introduction
20	Calculations: energy of a transition
21	Ionization energy from the convergence limit
22	Valence electrons and chemical properties
23	Mixed review: worked problems
24	Additional exam-style practice
25	Common misconceptions and links
26	Exam-style question with markscheme
27	Check your understanding
28	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the electromagnetic spectrum and relate frequency, wavelength and energy
explain how emission spectra provide evidence for discrete energy levels
explain the hydrogen emission spectrum and convergence of lines
describe s, p, d and f sub-levels and orbital shapes
write full and condensed electron configurations for the first 36 elements, including ions
apply the Aufbau principle, Hund's rule and the Pauli exclusion principle, including Cr and Cu exceptions
calculate ionization energy from the convergence limit
explain trends in first ionization energies and deduce group from successive ionization energies

Sections marked HL only (Sections 7 and 8) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How can we model the energy states of electrons in atoms? How do emission spectra provide evidence for these models?

How this topic is examined

S1.3 is examined through electron configuration questions, orbital diagrams and interpretation of emission spectra. HL questions include ionization energy graphs and calculations from the convergence limit.

Write configurations carefully in order of energy, and remember that 4s fills before 3d but empties first when ions form.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Electromagnetic radiation and the spectrum

Light is energy that travels as waves.

Think about it

Fireworks glow red, green or blue depending on the metal salts they contain. What could the colour of light tell us about atoms?

Electromagnetic radiation includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. All travel at the speed of light, c = 3.00 × 108 m s−1, and differ in wavelength (λ) and frequency (f): c = fλ.

Radiation also behaves as packets of energy called photons. The energy of a photon is proportional to its frequency: E = hf, where h is Planck's constant (6.63 × 10−34 J s). So high frequency means short wavelength and high energy. Ultraviolet photons have more energy than visible photons, which have more energy than infrared photons.

Visible light ranges from about 400 nm (violet) to 700 nm (red). A continuous spectrum (as from white light passed through a prism) contains all wavelengths; a line spectrum contains only specific wavelengths.

WORKED EXAMPLE 1 Energy of a photon

Calculate the energy of a photon of red light with wavelength 656 nm.

Solution

f = c ÷ λ = 3.00 × 108 ÷ 656 × 10−9 = 4.57 × 1014 HzM1

E = hf = 6.63 × 10−34 × 4.57 × 1014 = 3.03 × 10−19 JA1

Quick check

1. What is the relationship between frequency and wavelength?

2. Which has more energy: UV or infrared photons?

Answers: 1. c = fλ: inversely proportional 2. ultraviolet

2 Emission spectra and energy levels

Electrons can only have certain energies.

When atoms of an element are heated or an electric discharge is passed through a gas, they emit light. Passing this light through a prism gives a line emission spectrum: a series of bright coloured lines on a dark background. Each element has a unique line spectrum, like a fingerprint, which is used to identify elements (for example in flame tests and in the light from stars).

The lines are explained by the idea that electrons in atoms can only occupy discrete energy levels. When an atom absorbs energy, an electron is excited to a higher energy level. When it falls back to a lower level, it emits a photon whose energy equals exactly the difference in energy between the two levels: ΔE = hf. Because the energy levels have fixed values, only certain frequencies of light are emitted, giving lines rather than a continuous spectrum.

An absorption spectrum is the reverse: dark lines appear in a continuous spectrum at the frequencies absorbed when electrons are excited.

Key point

Lines, not a continuous spectrum, mean that electron energy is quantized: electrons occupy discrete energy levels.

In the real world

Flame tests use emission spectra: sodium gives yellow-orange, potassium lilac, lithium red, copper blue-green and calcium orange-red. Fireworks use the same metal salts.

3 The hydrogen spectrum

Lines converge at high energy.

Hydrogen, with one electron, has the simplest spectrum. Energy levels are numbered by the principal quantum number n = 1, 2, 3, ... with n = 1 the lowest (ground state). The levels get closer together as n increases, converging towards n = ∞, where the electron is no longer attracted to the nucleus.

Transitions to n = 1 emit the highest-energy photons, in the ultraviolet (the Lyman series). Transitions to n = 2 emit visible light (the Balmer series: red, blue-green, blue and violet lines). Transitions to n = 3 emit infrared.

Within each series, the lines get closer together at higher frequency (they converge), because the energy levels themselves converge. The point at which the lines merge is the convergence limit, corresponding to the electron being removed completely.

Figure 1. Origin of the hydrogen emission spectrum.

Common mistake

Saying the lines are produced when electrons are excited. Emission lines come from electrons falling to lower levels.

Quick check

1. Which transitions produce visible lines in hydrogen?

2. Why do the lines converge?

Answers: 1. to n = 2 2. energy levels get closer together at higher n

4 Shells, sub-levels and orbitals

Energy levels are split into sub-levels made of orbitals.

Each main energy level (shell) n contains sub-levels: s, p, d and f. Level 1 has an s sub-level; level 2 has s and p; level 3 has s, p and d; level 4 has s, p, d and f. The maximum number of electrons in shell n is 2n².

Each sub-level consists of orbitals: regions of space where there is a high probability of finding an electron. Each orbital holds a maximum of two electrons, with opposite spins (the Pauli exclusion principle). An s sub-level has 1 orbital (2 electrons); p has 3 (6 electrons); d has 5 (10 electrons); f has 7 (14 electrons).

Shapes: s orbitals are spherical; p orbitals are dumbbell-shaped, with three arranged at right angles along the x, y and z axes (px, py, pz).

Shell (n)	Sub-levels	Orbitals	Maximum electrons
1	1s	1	2
2	2s, 2p	4	8
3	3s, 3p, 3d	9	18
4	4s, 4p, 4d, 4f	16	32

5 Writing electron configurations

Fill orbitals in order of increasing energy.

The Aufbau principle: electrons fill sub-levels in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p. Note that 4s fills before 3d, because 4s is slightly lower in energy when empty.

A full electron configuration lists each sub-level with the number of electrons as a superscript: sodium (11 electrons) is 1s2 2s2 2p6 3s1. A condensed configuration uses the previous noble gas in square brackets: Na is [Ne] 3s1; iron (26) is [Ar] 4s2 3d6.

Ions: for anions, add electrons; for cations, remove electrons from the highest-energy occupied level. For transition metals, 4s electrons are removed before 3d: Fe is [Ar] 4s2 3d6, so Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5.

The electron configuration determines the position of an element in the periodic table: the s-block (groups 1 and 2), p-block (groups 13 to 18), d-block (transition elements) and f-block (lanthanoids and actinoids) are named after the sub-level being filled.

WORKED EXAMPLE 2 Electron configurations

Write full configurations for (a) S, (b) S2−, (c) Ti, and (d) Ti2+.

Solution

(a) S (16): 1s2 2s2 2p6 3s2 3p4A1

(b) S2− (18 electrons): 1s2 2s2 2p6 3s2 3p6A1

(d) Ti2+: remove 4s first: [Ar] 3d2A1

Common mistake

Removing 3d electrons before 4s when forming transition metal ions. The 4s electrons are lost first.

Quick check

1. Which fills first, 4s or 3d?

2. Write the condensed configuration of Ca.

3. Write the configuration of Fe3+.

Answers: 1. 4s 2. [Ar] 4s2 3. [Ar] 3d5

6 Orbital diagrams, Hund's rule and exceptions

Electrons spread out before they pair up.

In orbital (box) diagrams, each orbital is drawn as a box and electrons as arrows (↑ and ↓ for opposite spins). Hund's rule: electrons occupy orbitals of the same sub-level singly first, with parallel spins, before pairing up, because this minimizes repulsion. Nitrogen's 2p sub-level therefore has three unpaired electrons: ↑ | ↑ | ↑.

Two important exceptions in the first row of the d-block: chromium is [Ar] 4s1 3d5 (not 4s2 3d4) and copper is [Ar] 4s1 3d10 (not 4s2 3d9). A half-filled or completely filled d sub-level gives extra stability.

Element	2p orbital diagram	Unpaired electrons
Carbon (2p2)	↑ \| ↑ \|	2
Nitrogen (2p3)	↑ \| ↑ \| ↑	3
Oxygen (2p4)	↑↓ \| ↑ \| ↑	2

Quick check

1. State Hund's rule.

2. Write the configuration of Cu.

Answers: 1. electrons fill orbitals singly with parallel spins before pairing 2. [Ar] 4s1 3d10

7 Ionization energy and the convergence limit (HL only)

Removing an electron completely.

HL only

This section is Additional Higher Level content. SL students do not need it.

The first ionization energy is the minimum energy required to remove one mole of electrons from one mole of gaseous atoms in their ground state: X(g) → X+(g) + e−.

For hydrogen, the convergence limit of the Lyman series (transitions to n = 1) corresponds to the electron moving from n = ∞ to n = 1; the energy at this frequency equals the energy needed to remove the electron from the ground state. So the ionization energy can be calculated from the convergence frequency or wavelength: E = hf per atom, multiplied by the Avogadro constant for one mole.

WORKED EXAMPLE 3 Ionization energy from the convergence limit

The Lyman series of hydrogen converges at a wavelength of 91.2 nm. Calculate the ionization energy of hydrogen in kJ mol−1.

Solution

f = c ÷ λ = 3.00 × 108 ÷ 91.2 × 10−9 = 3.29 × 1015 HzM1

E per atom = hf = 6.63 × 10−34 × 3.29 × 1015 = 2.18 × 10−18 JM1

Per mole: × 6.02 × 1023 = 1.31 × 106 J = 1310 kJ mol−1A1

8 Successive ionization energies and trends (HL only)

Big jumps reveal electron shells.

HL only

This section is Additional Higher Level content. SL students do not need it.

Successive ionization energies remove electrons one at a time. They always increase, because each electron is removed from an increasingly positive ion. Large jumps occur when an electron is removed from a shell closer to the nucleus. The position of the first big jump shows the number of valence electrons and therefore the group: for example, if the jump comes after the third electron, the element is in group 13.

Trends in first ionization energy: across a period, first IE generally increases, because nuclear charge increases while electrons are added to the same shell, so attraction increases. Down a group it decreases, because outer electrons are further from the nucleus and more shielded.

Discontinuities across period 3: IE drops from Mg to Al, because Al's outer electron is in a 3p orbital, higher in energy and slightly further from the nucleus than 3s. IE drops from P to S, because in S one 3p orbital contains a pair of electrons, and repulsion between them makes one easier to remove.

Figure 2. Successive ionization energies of aluminium: a large jump after the 3rd electron (group 13) and another after the 11th (inner shell).

Figure 3. First ionization energies across period 3, with dips at Al and S.

Quick check

1. Why do successive ionization energies increase?

2. Why is the first IE of Al lower than Mg?

Answers: 1. electrons removed from increasingly positive ions 2. Al's electron is in higher-energy 3p

9 Electromagnetic radiation

Wavelength, frequency and energy.

Light is electromagnetic radiation. Its speed in a vacuum is c = 3.00 × 108 m s−1, and c = fλ (frequency f, wavelength λ). The energy of a photon is E = hf, where h = 6.63 × 10−34 J s. Higher frequency (shorter wavelength) means higher energy.

Region	Wavelength	Energy
ultraviolet	10–400 nm	high
visible	400–700 nm	medium
infrared	700 nm–1 mm	low

WORKED EXAMPLE 4 Photon energy (standard)

Calculate the frequency and energy of a photon of wavelength 656 nm.

Solution

f = 3.00 × 108/(656 × 10−9) = 4.57 × 1014 HzM1 A1

E = 6.63 × 10−34 × 4.57 × 1014 = 3.03 × 10−19 JA1

Quick check

1. Which has more energy: a red or a blue photon?

2. Calculate the wavelength of radiation with frequency 1.00 × 1015 Hz.

Answers: 1. blue (shorter wavelength, higher frequency) 2. 300 nm (ultraviolet)

10 Continuous and line spectra

White light versus excited atoms.

A continuous spectrum contains all wavelengths (white light through a prism). An emission (line) spectrum contains only certain wavelengths, seen as bright lines on a dark background. An absorption spectrum shows dark lines on a continuous background where atoms absorb specific wavelengths.

WORKED EXAMPLE 5 Interpreting lines (standard)

Explain why the emission spectrum of hydrogen consists of discrete lines.

Solution

Electrons can only occupy fixed energy levelsR1

Transitions between levels emit photons of specific energies, so specific frequenciesR1

In the real world

Astronomers identify the elements in stars from the dark absorption lines in their spectra; helium was discovered in the Sun's spectrum before it was found on Earth.

Quick check

1. What is the difference between an emission and an absorption spectrum?

Answers: 1. emission: bright lines on dark; absorption: dark lines on a continuous spectrum, at the same wavelengths

11 The hydrogen emission spectrum in detail

Series and convergence.

Transitions ending at n = 1 give the Lyman series (ultraviolet); ending at n = 2, the Balmer series (visible); ending at n = 3, the Paschen series (infrared). Within each series, lines get closer together at higher energy and converge, because the energy levels themselves converge at higher n.

WORKED EXAMPLE 6 Identifying a series (standard)

Which series contains the transition n = 4 → n = 2, and in which region is it?

Solution

Ends at n = 2: Balmer seriesA1

Visible (blue-green, 486 nm)A1

Common mistake

Thinking that emission lines come from electrons absorbing energy. Emission lines are produced when excited electrons fall to lower energy levels.

Quick check

1. Why do lines in a series get closer together?

2. Which transition in hydrogen releases the most energy: 2 → 1, 3 → 1 or 3 → 2?

Answers: 1. energy levels converge at higher n 2. 3 → 1

12 Shells, sub-levels and orbitals

The quantum mechanical model.

Energy levels (shells, n = 1, 2, 3, ...) contain sub-levels: s, p, d and f. Each sub-level contains orbitals, each holding up to two electrons with opposite spins. Level n holds a maximum of 2n2 electrons.

Sub-level	Number of orbitals	Maximum electrons
s	1	2
p	3	6
d	5	10
f	7	14

WORKED EXAMPLE 7 Capacity (basic)

How many electrons can the third shell hold, and which sub-levels does it contain?

Solution

2(3)2 = 18A1

3s, 3p, 3dA1

Quick check

1. How many orbitals are in the n = 4 shell?

2. What is the shape of an s orbital and a p orbital?

Answers: 1. 16 (4s, 4p, 4d, 4f: 1 + 3 + 5 + 7) 2. s spherical; p dumbbell

13 Writing electron configurations

Aufbau order.

Fill sub-levels in order of increasing energy: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p. Use the Pauli exclusion principle (two electrons per orbital, opposite spins) and Hund's rule (orbitals in a sub-level are singly occupied before pairing). A condensed configuration uses a noble gas core, such as [Ne]3s23p3 for phosphorus.

WORKED EXAMPLE 8 Full and condensed (standard)

Write the full and condensed configurations of iron (Z = 26).

Solution

1s22s22p63s23p64s23d6A1

[Ar]4s23d6A1

WORKED EXAMPLE 9 Exceptions (exam level)

Write the configurations of chromium (Z = 24) and copper (Z = 29).

Solution

Cr: [Ar]4s13d5 (half-filled d sub-level is more stable)A1

Cu: [Ar]4s13d10 (full d sub-level)A1

Quick check

1. Write the configuration of sulfur.

2. Write the condensed configuration of zinc.

3. Write the configuration of bromine.

Answers: 1. 1s22s22p63s23p4 2. [Ar]4s23d10 3. [Ar]4s23d104p5

14 Orbital diagrams

Arrows in boxes.

Orbital (box) diagrams show each orbital as a box and each electron as an arrow. Hund's rule means that, for example, nitrogen's 2p electrons occupy three separate boxes with parallel spins, giving three unpaired electrons.

Element	2s	2p	Unpaired electrons
C	↑↓	↑ ↑ _	2
N	↑↓	↑ ↑ ↑	3
O	↑↓	↑↓ ↑ ↑	2
F	↑↓	↑↓ ↑↓ ↑	1

WORKED EXAMPLE 10 Unpaired electrons (standard)

How many unpaired electrons does an iron atom have?

Solution

3d6: five orbitals, one paired and four singleM1

4 unpaired electronsA1

Quick check

1. How many unpaired electrons are in a phosphorus atom?

2. How many unpaired electrons are in Mn (Z = 25)?

Answers: 1. 3 2. 5

15 Configurations of ions

Remove from the highest n first.

For positive ions, remove electrons from the outermost shell first. For transition metals, 4s electrons are removed before 3d electrons. For negative ions, add electrons to the next available orbital.

WORKED EXAMPLE 11 Transition metal ions (exam level)

Write the configurations of Fe2+ and Fe3+.

Solution

Fe: [Ar]4s23d6A1

Fe2+: [Ar]3d6; Fe3+: [Ar]3d5A1 A1

Common mistake

Removing the 3d electrons before the 4s electrons when forming transition metal ions. The 4s electrons are lost first.

Quick check

1. Write the configuration of Cu2+.

2. Write the configuration of S2−.

3. Which noble gas configuration does Ca2+ have?

Answers: 1. [Ar]3d9 2. 1s22s22p63s23p6 3. argon

16 Blocks of the periodic table

s, p, d and f.

The periodic table is divided into blocks according to the sub-level being filled: groups 1–2 are the s-block, groups 13–18 the p-block, the transition elements the d-block, and the lanthanoids and actinoids the f-block. The period number equals the outer shell number.

WORKED EXAMPLE 12 Locating an element (standard)

An element has configuration [Kr]5s24d105p4. State its period, group and block.

Solution

Period 5A1

Group 16, p-block (tellurium)A1

Quick check

1. State the block of titanium.

2. An element has outer configuration 4s24p1. Identify it.

Answers: 1. d-block 2. gallium

17 Energy level diagrams

Visualizing order of filling.

An energy level diagram shows sub-levels as horizontal lines at increasing energy. 4s is slightly lower in energy than 3d in neutral atoms before filling, which is why 4s fills first; once 3d is occupied, it drops below 4s, which is why 4s empties first.

WORKED EXAMPLE 13 Order (standard)

Explain why potassium's 19th electron enters 4s rather than 3d.

Solution

4s is lower in energy than 3d for potassiumR1

Electrons occupy the lowest available energy levelR1

Quick check

1. List the order of filling from 3p to 5s.

Answers: 1. 3p, 4s, 3d, 4p, 5s

18 Flame tests and emission

Identifying metal ions.

When metal compounds are heated in a flame, electrons are excited and emit characteristic colours as they fall back. Each metal has a distinctive line spectrum, so flame colours identify metal ions.

Ion	Flame colour
Li+	red
Na+	yellow-orange
K+	lilac
Ca2+	brick red
Cu2+	blue-green

In the real world

Fireworks owe their colours to metal salts: strontium compounds give red, barium compounds green and sodium compounds yellow.

Quick check

1. Why do different metals give different flame colours?

Answers: 1. different energy level spacings give photons of different energies (wavelengths)

19 Ionization energy: an introduction

Energy to remove an electron.

The first ionization energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms: X(g) → X+(g) + e−. It increases across a period (increasing nuclear charge, similar shielding) and decreases down a group (outer electron further from the nucleus, more shielding).

WORKED EXAMPLE 14 Trend (standard)

Explain why sodium has a lower first ionization energy than magnesium.

Solution

Mg has a greater nuclear chargeR1

Outer electrons in the same shell with similar shielding, so held more strongly in MgR1

Quick check

1. Write the equation for the first ionization of calcium.

2. Explain why potassium has a lower first ionization energy than sodium.

Answers: 1. Ca(g) → Ca+(g) + e− 2. outer electron further from the nucleus and more shielded

20 Calculations: energy of a transition

Linking spectra and energy levels.

The energy of an emitted photon equals the difference between two energy levels: ΔE = hf = hc/λ. Multiplying by the Avogadro constant gives the energy per mole.

WORKED EXAMPLE 15 Energy per mole (exam level)

The hydrogen line at 486 nm corresponds to n = 4 → n = 2. Calculate the energy difference in kJ mol−1.

Solution

E = hc/λ = (6.63 × 10−34 × 3.00 × 108)/(486 × 10−9) = 4.09 × 10−19 JM1 A1

× 6.02 × 1023 = 246 kJ mol−1A1

Quick check

1. Calculate the wavelength of a photon with energy 3.98 × 10−19 J.

Answers: 1. 500 nm

21 Ionization energy from the convergence limit

HL extension.

The convergence limit of the Lyman series corresponds to the electron moving from n = 1 to n = ∞: the ionization of hydrogen. Its frequency (3.29 × 1015 Hz) gives the ionization energy: E = hf × L = 1312 kJ mol−1.

WORKED EXAMPLE 16 Convergence limit (exam level)

The convergence limit of the Lyman series is at 91.2 nm. Calculate the ionization energy of hydrogen in kJ mol−1.

Solution

E = hc/λ = 2.18 × 10−18 JM1 A1

× L = 1313 kJ mol−1A1

22 Valence electrons and chemical properties

Why configuration matters.

The outer (valence) electrons determine how an element reacts. Elements in the same group have the same outer configuration (for example, ns1 for group 1), which explains their similar chemical properties. Noble gases have full outer s and p sub-levels (ns2np6), which makes them very unreactive.

Group	Outer configuration	Typical ion
1	ns1	+1
2	ns2	+2
16	ns2np4	−2
17	ns2np5	−1
18	ns2np6	none

WORKED EXAMPLE 17 Predicting reactivity (standard)

Use configurations to explain why sodium and potassium react similarly with water.

Solution

Both have one electron in the outer s sub-level (3s1, 4s1)R1

Both lose this electron easily to form +1 ionsR1

Quick check

1. State the outer configuration of the halogens.

2. Why are noble gases unreactive?

Answers: 1. ns2np5 2. full outer s and p sub-levels; high ionization energy

23 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 18 Mixed problem (exam level)

Element Q is in period 4 and its ion Q3+ has the configuration [Ar]3d3. (a) Identify Q. (b) Write the configuration of a Q atom. (c) State the number of unpaired electrons in a Q atom.

Solution

(a) Q3+ has 21 electrons, so Q has 24: chromiumM1 A1

(b) [Ar]4s13d5A1

Quick check

1. Which ion has the configuration [Ne]: Mg2+, Cl−, K+?

Answers: 1. Mg2+

24 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 19 Paper 1 question (basic)

How many electrons are in the p sub-levels of a silicon atom? A. 2 B. 6 C. 8 D. 14

Solution

2p6 + 3p2 = 8 (C)A1

WORKED EXAMPLE 20 Paper 2 question (exam level)

(a) Write the electron configuration of a manganese atom. (b) Draw the orbital diagram of its 3d and 4s sub-levels. (c) Write the configuration of Mn2+.

Solution

(a) [Ar]4s23d5A1

(b) 5 single arrows in 3d; paired 4sA1

Quick check

1. Which sub-level is filled after 4s?

Answers: 1. 3d

25 Common misconceptions and links

Students often think…	Actually…
emission lines come from absorbing energy	they come from electrons falling to lower levels
3d fills before 4s	4s fills first but empties first in ions
electrons pair up before filling empty orbitals	Hund's rule: singly first
IE always increases across a period	dips at groups 13 and 16

How S1.3 connects to the rest of the course

Topic	Connection
S1.2 Nuclear atom	protons and electrons
S3.1 Periodic table	blocks and periodic trends
S2.1 Ionic model	ion formation
S2.2 Covalent model	valence electrons

26 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. When an electric discharge passes through hydrogen gas, a line spectrum is observed in the visible region.

(a) Write the full electron configuration of iron. [1]

(b) Write the configuration of Fe3+. [1]

(d) Explain why lines in the spectrum converge at higher frequency. [2]

(e) Draw an orbital diagram for the 2p sub-level of nitrogen and explain it using Hund's rule. [2]

(f) Explain why sulfur has a lower first ionization energy than phosphorus. [2]

Markscheme

Part	Answer	Marks
(a)	1s2 2s2 2p6 3s2 3p6 4s2 3d6	A1
(b)	[Ar] 3d5	A1
(c)	line spectrum: only certain frequencies emitted	R1
	photons emitted when electrons fall between levels	R1
	ΔE = hf fixed values, so levels are discrete	R1
(d)	energy levels get closer together	R1
	at higher n	R1
(e)	three singly occupied boxes with parallel spins	A1
	electrons occupy separate orbitals before pairing to minimize repulsion	R1
(f)	S has a paired 3p electron	R1
	electron-electron repulsion makes it easier to remove	R1

Examiner's comments. In (b), the most common error was removing 3d electrons before 4s.

27 Check your understanding

Section A: multiple choice

1. Which has the highest energy photons?

A. ultraviolet B. infrared C. visible red D. microwaves

2. Visible lines in the hydrogen spectrum result from transitions to

A. n = 1 B. n = 3 C. n = 2 D. n = ∞

3. A p sub-level can hold

A. 2 B. 6 electrons C. 10 D. 14

4. The configuration of Cr is

A. [Ar] 4s2 3d4 B. [Ar] 3d6 C. [Ar] 4s2 3d5 D. [Ar] 4s1 3d5

5. The shape of an s orbital is

A. dumbbell B. spherical C. cloverleaf D. linear

6. Maximum electrons in shell n = 3

A. 18 B. 8 C. 9 D. 32

7. The first IE generally

A. decreases across a period B. increases down a group C. is constant D. increases across a period

8. A big jump between the 2nd and 3rd ionization energies shows the element is in group

A. 1 B. 3 C. 2 D. 13

Section B: structured

9. Explain how emission spectra arise. [3]

10. Write configurations for K, Zn and Zn2+. [3]

11. State the Aufbau principle and Pauli exclusion principle. [2]

12. Explain the trend in first ionization energy down group 1. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Electrons excited; fall to lower levels; emit photons with ΔE = hf; discrete levels give lines.

10. K: [Ar] 4s1; Zn: [Ar] 4s2 3d10; Zn2+: [Ar] 3d10.

11. Fill lowest energy first; maximum two electrons per orbital with opposite spins.

12. Decreases: outer electron further from nucleus, more shielding, weaker attraction.

28 Summary and topic map

Idea	Remember
EM spectrum	c = fλ; E = hf
Emission spectra	discrete energy levels; ΔE = hf
Hydrogen	to n = 1 UV; to n = 2 visible; convergence
Orbitals	s (2), p (6), d (10), f (14)
Aufbau	1s 2s 2p 3s 3p 4s 3d 4p
Rules	Pauli, Hund; Cr and Cu exceptions
HL	IE from convergence limit; successive IEs; period 3 trends

Before the exam, make sure you…

•fill 4s before 3d; empty 4s first
•explain lines as electrons falling
•use orbital diagrams with Hund's rule
•remember Cr and Cu exceptions

Electron Configurations on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for Electron Configurations.

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