Atomic Structure Problem Solving

This lesson brings together everything from the previous nine lessons and focuses on the types of multi-step problems you will encounter in Edexcel A-Level exams. We will work through exam-style calculations, data interpretation, and questions that require you to link different concepts together.

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Type 1: Relative Atomic Mass from Mass Spectra

These questions give you isotopic data (either as percentages or as peak heights from a mass spectrum) and ask you to calculate Ar.

Strategy

flowchart TD
    A[Read m/z values and abundances from spectrum] --> B{Are abundances percentages?}
    B -- Yes --> C["Ar = Sum(mass x %) / 100"]
    B -- No --> D["Calculate total abundance"]
    D --> E["Ar = Sum(mass x abundance) / total"]

Worked Example 1: Standard Calculation

A mass spectrum of rubidium shows two peaks:

m/z	Relative Abundance
85	72.2
87	27.8

Ar = (85 x 72.2 + 87 x 27.8) / (72.2 + 27.8) Ar = (6137.0 + 2418.6) / 100.0 Ar = 8555.6 / 100.0 Ar = 85.6

Worked Example 2: Reverse Calculation (Finding Abundances)

Bromine has two isotopes, 79-Br and 81-Br, with Ar = 79.9. Calculate the percentage abundance of each isotope.

Let x = % abundance of 79-Br. Then (100 - x) = % abundance of 81-Br.

79.9 = (79x + 81(100 - x)) / 100 7990 = 79x + 8100 - 81x 7990 = 8100 - 2x 2x = 110 x = 55

Therefore: 79-Br = 55.0% and 81-Br = 45.0%.

Worked Example 3: Three Isotopes with Non-Percentage Abundances

A mass spectrum shows peaks at m/z = 24 (height 7.9), 25 (height 1.0), 26 (height 1.1).

Total = 7.9 + 1.0 + 1.1 = 10.0

Ar = (24 x 7.9 + 25 x 1.0 + 26 x 1.1) / 10.0 Ar = (189.6 + 25.0 + 28.6) / 10.0 Ar = 243.2 / 10.0 Ar = 24.3 (magnesium)

Quick Check Method

The Ar should always lie between the lightest and heaviest isotope masses, and should be closer to the mass of the most abundant isotope. If your answer falls outside this range, you have made an error.

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Type 2: Ionisation Energy Data Interpretation

These questions present successive ionisation energy data and ask you to deduce electronic structure or identify an element.

Strategy

1. Look for large proportional jumps between consecutive ionisation energies
2. Calculate ratios: IE(n+1) / IE(n)
3. A ratio significantly above 2 usually indicates a change from one shell to the next
4. Count electrons before the first big jump -> number of outer shell electrons -> group number

Worked Example 1: Identifying Group Number

The first five ionisation energies of an element are (kJ/mol): 786, 1577, 3232, 4356, 16091.

Analysis:

• 1st to 2nd: 786 -> 1577 (ratio = 2.0) -- moderate increase
• 2nd to 3rd: 1577 -> 3232 (ratio = 2.0) -- moderate increase
• 3rd to 4th: 3232 -> 4356 (ratio = 1.3) -- moderate increase
• 4th to 5th: 4356 -> 16091 (ratio = 3.7) -- large jump

The big jump is between the 4th and 5th IE, so the element has 4 electrons in its outer shell -> Group 14.

If the element is in Period 3, it is silicon.

Worked Example 2: Subshell Evidence Within a Shell

Within a shell, you may notice smaller jumps that correspond to moving between subshells.

Consider removing electrons from aluminium ([Ne] 3s2 3p1):

IE	Electron Removed	Subshell	Ratio to Previous
1st (578)	3p1	p	--
2nd (1817)	3s2	s	3.1
3rd (2745)	3s1	s	1.5
4th (11578)	2p6	NEW SHELL	4.2

The jump from 1st to 2nd (ratio 3.1) is larger than 2nd to 3rd (ratio 1.5) because removing the 3s electron after the 3p electron involves a change from the higher-energy p subshell to the lower-energy s subshell within the same shell.

Worked Example 3: Using IE Data to Determine Electronic Structure

An element has the following IEs (kJ/mol): 1000, 2252, 3389, 4540, 7004, 8496, 27107

• Ratios: 2.25, 1.50, 1.34, 1.54, 1.21, 3.19
• Wait -- the ratio 1.54 between 4th and 5th looks notable but is not huge
• The really big jump is between 6th (8496) and 7th (27107) with ratio 3.19
• This suggests 6 outer electrons -> Group 16

Actually, we need to look more carefully. The jump from 4th to 5th (4540 -> 7004, ratio 1.54) is larger than the surrounding ratios (1.34 and 1.21). This could indicate that the first 4 electrons are from the p subshell, and the 5th is from the s subshell within the same shell. The biggest shell-change jump is between 6th and 7th.

This is consistent with a Group 16 element (e.g., sulfur: 3s2 3p4).

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Type 3: Electron Configurations of Ions

These questions require you to write or interpret electron configurations for ions, often transition metal ions.

Strategy

1. Write the electron configuration of the neutral atom
2. For positive ions: remove electrons (4s before 3d for transition metals)
3. For negative ions: add electrons to the next available orbital
4. Check for special cases (Cr, Cu)

Worked Example 1: Transition Metal Ion

Write the electron configuration of Co3+ (cobalt, Z = 27).

Step 1: Co neutral = [Ar] 4s2 3d7 Step 2: Remove 4s2 first -> [Ar] 3d7 Step 3: Remove one 3d electron -> [Ar] 3d6

Co3+ = [Ar] 3d6

Note: Co3+ is isoelectronic with Fe2+.

Worked Example 2: Copper Ion

Write the configurations of Cu+ and Cu2+.

Cu neutral = [Ar] 4s1 3d10 (exception -- full d subshell)

Cu+: Remove 4s1 -> [Ar] 3d10 (full d subshell -- very stable) Cu2+: Remove 4s1, then one 3d -> [Ar] 3d9

Worked Example 3: Determining Unpaired Electrons

How many unpaired electrons are in Fe3+ ([Ar] 3d5)?

By Hund's rule, 5 electrons in 5 d orbitals means one electron per orbital, all with parallel spins:

d-orbital: [up] [up] [up] [up] [up]

5 unpaired electrons -- this is a half-filled subshell, which is particularly stable.

Complete Table: First-Row d-Block Ions

Ion	Config	d-electrons	Unpaired e-
Sc3+	[Ar] 3d0	0	0
Ti3+	[Ar] 3d1	1	1
V3+	[Ar] 3d2	2	2
Cr3+	[Ar] 3d3	3	3
Mn2+	[Ar] 3d5	5	5
Fe2+	[Ar] 3d6	6	4
Fe3+	[Ar] 3d5	5	5
Co2+	[Ar] 3d7	7	3
Ni2+	[Ar] 3d8	8	2
Cu2+	[Ar] 3d9	9	1
Zn2+	[Ar] 3d10	10	0

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Type 4: Predicting Properties from Periodic Position

These questions give you an element's position and ask you to predict its physical or chemical properties.

Strategy

1. Determine the block (s, p, d, or f)
2. Determine the group -> number of outer electrons -> typical ion formed
3. Consider the period -> size, number of shells
4. Apply periodic trends (IE, radius, EN, melting point, oxide character)

Worked Example

Element X is in Period 4, Group 2. Predict:

(a) Its electron configuration: [Ar] 4s2 -> This is calcium.

(b) The formula of its oxide: Ca forms Ca2+, oxygen forms O2- -> CaO

(c) Whether its oxide is acidic or basic: Group 2 oxide -> basic (CaO + H2O -> Ca(OH)2, which is alkaline)

(d) Its first ionisation energy relative to potassium (Group 1, Period 4): Calcium has higher nuclear charge with the same shielding -> higher IE than potassium (Ca: 590 kJ/mol vs K: 419 kJ/mol)

(e) Its atomic radius relative to potassium: Ca has higher nuclear charge pulling electrons closer -> smaller than potassium

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Type 5: Multi-Step Linking Problems

Exam questions often require you to connect several concepts. Here is a typical example:

Worked Example 1

An element Y has the following successive ionisation energies (kJ/mol): 1st: 1251, 2nd: 2298, 3rd: 3822, 4th: 5159, 5th: 6542, 6th: 9362, 7th: 11018

(a) Identify the group: Ratios: 1.84, 1.66, 1.35, 1.27, 1.43, 1.18. The jump from 5th to 6th (ratio 1.43) is the largest, but is it large enough? Compare: the jump between 6th and 7th is only 1.18. The biggest proportional jump is between 5th and 6th -> 5 outer electrons -> Group 15 (or Group 5 of the p-block).

(b) If Y is in Period 3, identify the element: Period 3, Group 15 -> Phosphorus (P)

(c) Write the electron configuration: 1s2 2s2 2p6 3s2 3p3

(d) Predict the formula of its oxide: P forms P4O10 (or P2O5 as the empirical formula)

(e) Predict the pH when this oxide is added to water: Acidic -> P4O10 + 6H2O -> 4H3PO4 -> pH approximately 1-2

(f) Explain why the first ionisation energy of Y (1012 kJ/mol for P) is higher than that of element Z in Group 16 (1000 kJ/mol for S): Phosphorus has a half-filled 3p3 subshell (one electron per orbital, all unpaired). Sulfur has 3p4 with one paired electron. The additional electron-electron repulsion in the paired orbital of sulfur makes it slightly easier to remove an electron, despite sulfur having a higher nuclear charge.

Worked Example 2: Combining Mass Spec and Periodic Trends

An element has a mass spectrum showing peaks at m/z = 63 (abundance 69.2) and m/z = 65 (abundance 30.8).

(a) Calculate Ar: Ar = (63 x 69.2 + 65 x 30.8) / 100 = (4359.6 + 2002.0) / 100 = 63.6

(b) Identify the element: Ar = 63.6 corresponds to copper (Cu)

(c) Write the electron configuration: [Ar] 4s1 3d10 (exception)

(d) Write the configuration of Cu2+: [Ar] 3d9

(e) Is copper a transition metal? Yes -- Cu2+ has a partially filled d subshell (3d9)

(f) Predict two characteristic properties: Variable oxidation states (+1 and +2); forms coloured compounds (Cu2+ is blue)

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Exam Technique Tips

1. Show your working in calculations -- even if you make an arithmetic error, you can still gain method marks.

2. State the trend before explaining it. For example: "First ionisation energy generally increases across Period 3 (trend) because nuclear charge increases while shielding remains constant (explanation)."

3. Use precise language. Say "ionisation energy" not "energy." Say "electron shielding" not just "shielding." Say "nuclear charge increases" not "more protons pulling."

4. When explaining anomalies, always reference the specific subshell involved: "The drop from nitrogen to oxygen occurs because oxygen has a paired electron in one of its 2p orbitals, experiencing additional electron-electron repulsion."

5. For successive IE questions, calculate the ratio between consecutive values to find the biggest proportional jump -- this is more reliable than looking at absolute differences.

6. Remember the order of electron removal for transition metals: 4s electrons are removed before 3d electrons when forming positive ions.

7. Check your Ar calculations by verifying the answer falls between the lightest and heaviest isotope masses, and is closer to the most abundant isotope.

8. For oxide/chloride questions, always state the type of bonding first, then predict the acid-base character or reaction with water based on that bonding type.

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Quick Reference: Key Data Tables

First Ionisation Energies Across Period 3 (kJ/mol)

Na	Mg	Al	Si	P	S	Cl	Ar
496	738	578	789	1012	1000	1251	1521

Electronegativity Across Period 3 (Pauling)

Na	Mg	Al	Si	P	S	Cl
0.9	1.3	1.6	1.9	2.2	2.6	3.2

Atomic Radius Across Period 3 (pm)

Na	Mg	Al	Si	P	S	Cl
186	160	143	117	110	104	99

Oxide Character Across Period 3

Na2O	MgO	Al2O3	SiO2	P4O10	SO2/SO3
Strongly basic	Basic	Amphoteric	Weakly acidic	Acidic	Strongly acidic

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A-Level Deep Dive: Atomic Structure Problem Solving

Spec mapping

Edexcel 9CH0 specification Topic 1 (synoptic across sub-topics 1.1–1.6) integrates atomic models, electron configuration, ionisation energies, mass spectrometry, periodicity (atomic radius, electronegativity, m.p., conductivity, reactions, oxides/chlorides) and block classification into multi-part problems requiring fluent transfer between data, configuration and bonding/property prediction (refer to the official Pearson Edexcel specification document for exact wording). Examined synoptically in Paper 3 (General and Practical), where extended-response questions integrate Topic 1 with Topic 2 (Bonding), Topic 4 (Inorganic chemistry of Group 1, 2, 7) and Topic 5 (Energetics). The skill examined here is integration — using one piece of data (e.g. successive IE pattern) to deduce another (element identity), then a third (likely chemistry).

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Worked example with full mark scheme

Question (12 marks):

An unknown element X has the following properties:

• Successive ionisation energies (kJ mol⁻¹): 738, 1451, 7733, 10541, 13630, 17995.
• Mass spectrum shows three peaks at m/z = 24, 25, 26 with relative intensities 78.6, 10.1, 11.3.
• Reacts vigorously with cold water producing hydrogen gas and an alkaline solution.
• Forms a chloride that is white, melts at 714 °C and dissolves in water to give a near-neutral solution.

(a) Identify element X. Justify with at least two of the data sources. (4) (b) Calculate the relative atomic mass of X to 2 d.p. (3) (c) Predict the bonding type, structure and electrical conductivity of solid X. (2) (d) Predict the pH and write a balanced equation for the reaction of X with water. (3)

Solution with mark scheme:

(a) M1 — IE pattern: large jump from IE2 (1451) to IE3 (7733) → 2 outermost electrons → Group 2.

M1 — A_r calculation gives 24.32 (see (b)) → atomic mass of magnesium.

M1 — vigorous reaction with cold water → reactive metal, consistent with active s-block metal.

A1 — Element X is magnesium (Mg).

(b) M1 — A_r = (24 × 78.6 + 25 × 10.1 + 26 × 11.3) / 100.

M1 — = (1886.4 + 252.5 + 293.8) / 100 = 2432.7 / 100 = 24.327.

A1 — A_r = 24.33 to 2 d.p.

(c) B1 — Mg metal: giant metallic structure with Mg²⁺ cations and delocalised electrons.

B1 — high m.p. (650 °C), good electrical conductor (delocalised electrons mobile in lattice), malleable and ductile.

(d) B1 — equation: Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g) (very slow with cold water).

B1 — pH ≈ 9–10 (Mg(OH)₂ is moderately basic but limited by low solubility).

B1 — observation: slow effervescence from H₂ release, formation of cloudy white Mg(OH)₂. With steam the reaction is much faster: Mg(s) + H₂O(g) → MgO(s) + H₂(g).