Unit 1 · Spec 3.1.1

Atomic Structure & TOF Mass Spectrometry

Subatomic particles, isotopes, TOF mass spectrometry, relative atomic mass from mass spectra, electron configurations (Cr/Cu anomalies), and first and successive ionisation energies.

1. Fundamental Subatomic Particles

Every atom consists of a dense central nucleus surrounded by an electron cloud. Three fundamental particles define atomic architecture:

Subatomic Particle Relative Charge Relative Mass Physical Location
Proton (\(p\)) +1 1 Nucleus
Neutron (\(n\)) 0 1 Nucleus
Electron (\(e^-\)) -1 1 / 1840 (0.00054) Orbitals outside nucleus

Atomic Number (\(Z\))

The number of protons in the nucleus of an atom. This quantity defines the chemical identity of the element.

Mass Number (\(A\))

The total number of nucleons (protons plus neutrons) in the nucleus: \[ A = Z + N \]

Isotopes

Atoms of the same element having the same number of protons (same atomic number \(Z\)), but different numbers of neutrons (different mass number \(A\)). Isotopes exhibit identical chemical properties because they have identical electron configurations, but possess slightly differing physical properties (such as density and rate of diffusion).

Worked Example: Counting Particles in Atoms and Ions
Give the numbers of protons, neutrons and electrons in ³⁷Cl⁻ and in ²⁴Mg²⁺.
  • ³⁷Cl⁻: Z = 17, so 17 protons. Neutrons = 37 - 17 = 20. The 1- charge means one extra electron: 18 electrons.
  • ²⁴Mg²⁺: Z = 12, so 12 protons. Neutrons = 24 - 12 = 12. The 2+ charge means two electrons lost: 10 electrons.

How our model of the atom changed

Ideas about atoms have changed over time as new evidence was found:

  • Dalton: atoms are tiny solid spheres that cannot be divided.
  • Thomson: the plum pudding model, with electrons in a ball of positive charge.
  • Rutherford: most of the mass and all the positive charge are in a tiny central nucleus.
  • Bohr: electrons orbit the nucleus in fixed shells (energy levels).

2. Time of Flight (TOF) Mass Spectrometry

Time of Flight mass spectrometry determines the mass-to-charge ratio (\(m/z\)) and abundance of isotopes and molecules. The operation proceeds through five distinct stages:

Time of flight mass spectrometer e⁻ + + + − + + heavy ion: slower light ion: faster vacuum, no electric field sample (g) 1 Ionise 2 Accelerate negative plate 3 Drift (flight tube, length d) 4 Detect electron impact X(g) + e⁻ → X⁺(g) + 2e⁻ same kinetic energy KE = ½mv² lighter ions arrive first t = d / v = d√(m / 2KE) ions gain e⁻ current shows abundance

Stage 1: Ionisation

The sample must be converted into positive ions. The usual method is electron impact:

  • Electron Impact (EI): The vaporised sample is bombarded by high-energy electrons fired from an electron gun. A high-energy electron knocks out an outer shell electron: \[ \text{X}(g) \rightarrow \text{X}^+(g) + e^- \] This technique creates molecular ions with high internal energy, causing extensive fragmentation. It is used for elements and low-mass organic molecules.

Beyond the OxfordAQA specification

Electrospray ionisation is used for large molecules in research. 9620 does not require it.

Stage 2: Acceleration

Positive ions are accelerated towards a negatively charged electric plate. All ions acquire the same kinetic energy (KE):

\[ \text{KE} = \frac{1}{2} m v^2 \]

Stage 3: Ion Drift

Ions pass through a flight tube of length \(d\) under high vacuum with no electric field. Their velocity is inversely proportional to the square root of their mass:

\[ v = \sqrt{\frac{2\text{KE}}{m}} \quad \implies \quad t = \frac{d}{v} = d \sqrt{\frac{m}{2\text{KE}}} \]

Because kinetic energy and tube length are constant, lighter ions travel at higher velocities and arrive at the detector in a shorter flight time.

Stage 4: Detection

Positive ions strike the negatively charged detector plate. Each ion accepts one or more electrons from the plate, neutralising its charge. This electron flow produces a tiny electric current. The size of the current is directly proportional to the abundance of that ion.

Stage 5: Data Analysis

The detector signal is processed by a computer to give a mass spectrum of \(m/z\) against relative abundance.

Worked Example: TOF Flight Time Calculation
In a TOF mass spectrometer with a 1.50 m flight tube, an ion of mass 1.16 × 10⁻²⁵ kg arrives at the detector in 2.80 × 10⁻⁵ s. Calculate the kinetic energy of the ion, and determine the flight time of an ion of mass 1.45 × 10⁻²⁵ kg under identical acceleration.

Step 1: Calculate velocity of the first ion:

\[ v_1 = \frac{d}{t_1} = \frac{1.50}{2.80 \times 10^{-5}} = 53571.4\text{ m s}^{-1} \]

Step 2: Calculate kinetic energy:

\[ \text{KE} = \frac{1}{2} m_1 v_1^2 = 0.5 \times (1.16 \times 10^{-25}) \times (53571.4)^2 = 1.6645 \times 10^{-16}\text{ J} \]

Step 3: Calculate flight time for the second ion:

Since both ions possess identical kinetic energy:

\[ \frac{t_2}{t_1} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{1.45 \times 10^{-25}}{1.16 \times 10^{-25}}} = \sqrt{1.25} = 1.11803 \]

\[ t_2 = 2.80 \times 10^{-5} \times 1.11803 = \mathbf{3.13 \times 10^{-5}\text{ s}} \]

3. Using Mass Spectra

Reading the mass spectrum of an element

  • Each peak is one isotope. For 1+ ions, the m/z value of a peak is the relative isotopic mass.
  • The peak height gives the relative abundance of that isotope.
  • The m/z values and abundances are used to work out Ar. The Ar identifies the element.
Mass Spectrum of Neon 0 25 50 75 100 19 20 21 22 23 90.9 0.3 8.8 m/z Relative abundance / %

Relative atomic mass from isotopic abundance

Ar is the weighted mean of the isotopic masses. In 9620, these calculations are limited to mononuclear ions: single-atom ions from an element, such as Ne⁺.

Ar = Σ(isotopic mass × abundance) ÷ Σ(abundance)

Worked Example: Relative Atomic Mass of Neon
The mass spectrum of neon shows ²⁰Ne 90.9%, ²¹Ne 0.3% and ²²Ne 8.8%. Calculate the relative atomic mass of neon.

Ar = (20 × 90.9 + 21 × 0.3 + 22 × 8.8) / 100

Ar = 2017.9 / 100 = 20.18, so Ar = 20.2 (3 significant figures).

Worked Example: Finding an Unknown Isotope (Magnesium)
Magnesium has Ar = 24.31. Its spectrum shows ²⁴Mg 78.8% and ²⁵Mg 11.7%. The third isotope makes up 9.5%. Find the mass number of the third isotope.

Step 1: 24.31 × 100 = 24 × 78.8 + 25 × 11.7 + 9.5x

Step 2: 2431 = 1891.2 + 292.5 + 9.5x

Step 3: 9.5x = 247.3, so x = 26.0

Answer: the third isotope is ²⁶Mg.

Finding Mr by mass spectrometry

A molecular substance gives a molecular ion, M⁺. The molecular ion peak is the peak with the highest m/z value (ignore any tiny peak at M+1). Its m/z value gives Mr.

High resolution mass spectrometry is covered in Unit 2: see Organic Analysis.

4. Electronic Configuration

Electrons reside in principal energy levels (\(n = 1, 2, 3, 4\)) divided into subshells (\(s, p, d, f\)):

  • \(s\)-subshell: 1 orbital (holds maximum 2 electrons).
  • \(p\)-subshell: 3 degenerate orbitals (holds maximum 6 electrons).
  • \(d\)-subshell: 5 degenerate orbitals (holds maximum 10 electrons).
  • \(f\)-subshell: 7 degenerate orbitals (holds maximum 14 electrons).

Transition Metal Anomalies

Chromium (\(Z = 24\)) and Copper (\(Z = 29\))

Chromium and copper do not follow the standard Aufbau filling sequence due to the extra stability of half-filled and fully-filled \(3d\) subshells:

  • Chromium (Cr): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\) (or \([\text{Ar}] 3d^5 4s^1\)), NOT \(3d^4 4s^2\).
  • Copper (Cu): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\) (or \([\text{Ar}] 3d^{10} 4s^1\)), NOT \(3d^9 4s^2\).

Ion Formation Rule: When transition metals form positive ions, electrons are lost from the \(4s\) orbital before the \(3d\) orbitals. For example, Fe²⁺ is \([\text{Ar}] 3d^6\), having lost both \(4s\) electrons.

5. First Ionisation Energy Trends

First Ionisation Energy

The enthalpy change required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1⁺ ions under standard conditions: \[ \text{X}(g) \rightarrow \text{X}^+(g) + e^- \]

First Ionisation Energy Trend Across Period 3 Period 3 Elements First IE / kJ mol⁻¹ 0 500 1000 1500 Na Mg Al Si P S Cl Ar Al dip: 3p¹ electron higher energy, shielded by 3s² S dip: 3p⁴ spin-pairing paired 3p electrons repel

Period 3 General Increase & Anomalies

Across Period 3 (Na to Ar), first ionisation energy exhibits a general increase because nuclear charge increases (more protons) while electrons are added to the same energy shell with similar shielding, drawing valence electrons closer.

Anomaly 1: Aluminium Dip (\(3p^1\))

Aluminium (578 kJ mol⁻¹) has a lower first ionisation energy than Magnesium (738 kJ mol⁻¹).

Reason: In Mg, the electron is removed from the \(3s\) subshell. In Al, the electron is removed from the higher energy \(3p\) subshell, which is shielded by the inner \(3s^2\) electrons.

Anomaly 2: Sulfur Dip (\(3p^4\))

Sulfur (1000 kJ mol⁻¹) has a lower first ionisation energy than Phosphorus (1012 kJ mol⁻¹).

Reason: In P, all three \(3p\) orbitals contain a single electron (\(3p_x^1 3p_y^1 3p_z^1\)). In S, one \(3p\) orbital contains a paired set of electrons (\(3p_x^2 3p_y^1 3p_z^1\)). Spin-pair repulsion between the two electrons in the same orbital makes that electron easier to remove.

Trend Down Group 2

First ionisation energy decreases down Group 2 (Be to Ba). Although nuclear charge increases down the group, atomic radius increases significantly and there are more intervening electron shells. The increased distance and electron shielding weaken electrostatic attraction to the nucleus.

  • Each element down the group has one more shell of electrons.
  • The outer electron is further from the nucleus and more shielded.
  • So it is less strongly attracted, even though the nuclear charge is greater.

This trend is evidence for shells of electrons. The Period 3 pattern above, with its dips at Al and S, is evidence for sub-shells. The Periodicity page uses the same Period 3 data.

6. Successive Ionisation Energies

An element has as many ionisation energies as it has electrons. The second ionisation energy removes an electron from a 1+ ion:

  • 1st: Na(g) → Na⁺(g) + e⁻
  • 2nd: Na⁺(g) → Na²⁺(g) + e⁻
  • Each successive ionisation energy is larger. The electron is removed from an increasingly positive ion, so it is more strongly attracted.
  • A big jump shows that the electron is being removed from a new shell, closer to the nucleus and less shielded.
  • Count the electrons removed before the first big jump. This is the number of outer electrons, so it gives the group.
Successive Ionisation Energies of Aluminium (log scale) 100 1 000 10 000 100 000 1 000 000 IE / kJ mol⁻¹ (log scale) 1 2 3 4 5 6 7 8 9 10 11 12 13 Electron removed (1st to 13th ionisation) Shell 3: 3 electrons Shell 2: 8 electrons Shell 1: 2

Aluminium has a large jump between the 3rd and 4th ionisation energies. So it has 3 outer electrons and is in Group 3 (13). The second large jump, after the 11th, shows the 8 electrons of the second shell. Successive ionisation energies are evidence for shells.

Worked Example: Finding the Group
An element has successive ionisation energies (kJ mol⁻¹) of 738, 1451, 7733 and 10541. Which group is it in?

The big jump is after the 2nd ionisation energy (1451 to 7733). Two electrons are in the outer shell.

Answer: Group 2 (this is magnesium).

Go deeper

These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.

7. Practice Questions

Practice Problem (3 Marks)
State and explain the general trend in first ionisation energy across Period 3 from sodium to argon, and write an equation including state symbols for the first ionisation energy of sodium.
Show Answer & Mark Scheme

Equation: \(\text{Na}(g) \rightarrow \text{Na}^+(g) + e^-\) (1 mark)

Trend & Explanation:

  • General increase across the period. (1 mark)
  • Nuclear charge increases (more protons) while electron shielding remains approximately constant because electrons enter the same third quantum shell. The stronger attraction pulls valence electrons closer, requiring more energy to remove an electron. (1 mark)
Practice Problem 2 (2 Marks)
A sample of boron contains 19.9% ¹⁰B and 80.1% ¹¹B. Calculate the relative atomic mass of boron. Give your answer to 3 significant figures.
Show Answer & Mark Scheme

Ar = (10 × 19.9 + 11 × 80.1) / 100 = (199 + 881.1) / 100 (1 mark)

Ar = 1080.1 / 100 = 10.801, so 10.8 (1 mark)

Practice Problem 3 (3 Marks)
The first four ionisation energies of an element X are 738, 1451, 7733 and 10541 kJ mol⁻¹. Write an equation for the second ionisation energy of X. State the group of X and explain your answer.
Show Answer & Mark Scheme
  • X⁺(g) → X²⁺(g) + e⁻ (1 mark)
  • Group 2 (1 mark)
  • There is a big jump between the 2nd and 3rd ionisation energies, because the 3rd electron is removed from a shell closer to the nucleus (1 mark)
Practice Problem 4 (2 Marks)
Explain why the first ionisation energy of barium is lower than that of magnesium.
Show Answer & Mark Scheme
  • The outer electron in barium is in a shell further from the nucleus and is more shielded by inner electrons (1 mark)
  • So it is less strongly attracted to the nucleus, despite the greater nuclear charge (1 mark)
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