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How is the atom structured, and how do mass spectrometry and electron configurations explain its behaviour?

Sub-atomic particles, isotopes and relative masses, the mass spectrometer and relative atomic mass calculations, electron configuration in s, p and d sub-shells, and ionisation energy evidence.

A CCEA A-Level Chemistry answer on sub-atomic particles, isotopes and relative masses, how the mass spectrometer works and is used to find relative atomic mass, electron configuration in s, p and d sub-shells, and the ionisation energy evidence for shells and sub-shells.

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  1. What this dot point is asking
  2. Sub-atomic particles and isotopes
  3. The mass spectrometer
  4. Electron configuration
  5. Common traps
  6. Examples in context
  7. Try this

What this dot point is asking

CCEA wants you to describe the three sub-atomic particles and their relative masses and charges, define isotopes and relative atomic mass, explain how a mass spectrometer works and use spectra to calculate relative atomic mass, write electron configurations using s, p and d sub-shells, and use successive ionisation energies as evidence for shells and sub-shells.

Sub-atomic particles and isotopes

The relative atomic mass (ArA_r) is the weighted mean mass of an atom of an element relative to 112\frac{1}{12} the mass of a carbon-12 atom.

The mass spectrometer

A modern time-of-flight mass spectrometer works in four stages: ionisation (the sample is ionised, for example by electron impact or electrospray), acceleration (ions are accelerated by an electric field to the same kinetic energy), flight or deflection (ions are separated by their mass-to-charge ratio m/zm/z), and detection (ions strike a detector, producing a signal proportional to abundance).

To find relative atomic mass from a spectrum, multiply each isotope mass by its percentage abundance, add the products, and divide by 100:

Ar=(isotope mass×% abundance)100A_r = \frac{\sum (\text{isotope mass} \times \% \text{ abundance})}{100}

For chlorine with 75%75\% chlorine-35 and 25%25\% chlorine-37, Ar=(35×75)+(37×25)100=35.5A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = 35.5.

Note that the molecular ion in a spectrum can carry a 2+2+ charge, so a peak at m/z=31.5m/z = 31.5 would correspond to a Cu2+\text{Cu}^{2+} species of mass 63. The detector reads the mass-to-charge ratio, not mass directly.

Electron configuration

Electrons occupy shells divided into sub-shells (s, p, d) that hold a maximum of 2, 6 and 10 electrons. Sub-shells fill in order of increasing energy: 1s2s2p3s3p4s3d4p1s\,2s\,2p\,3s\,3p\,4s\,3d\,4p, because the 4s4s sub-shell is slightly lower in energy than 3d3d. For example, iron is 1s22s22p63s23p63d64s21s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^6\,4s^2.

Two stability quirks are tested at CCEA: chromium is [Ar]3d54s1[\text{Ar}]\,3d^5\,4s^1 and copper is [Ar]3d104s1[\text{Ar}]\,3d^{10}\,4s^1, because a half-filled or fully filled 3d3d sub-shell is more stable than the expected 3d44s23d^4\,4s^2 or 3d94s23d^9\,4s^2. When writing ion configurations, remove the 4s4s electrons first: Fe2+\text{Fe}^{2+} is [Ar]3d6[\text{Ar}]\,3d^6 and Fe3+\text{Fe}^{3+} is [Ar]3d5[\text{Ar}]\,3d^5.

Common traps

Examples in context

Example 1. Carbon-14 dating and the constancy of chemical behaviour. Carbon-12, carbon-13 and carbon-14 are isotopes with 6 protons but 6, 7 and 8 neutrons. Because they share the same electron configuration, 1s22s22p21s^2\,2s^2\,2p^2, they react identically, which is why living organisms take up all three in the same ratio as the atmosphere. Only when the organism dies does the radioactive carbon-14 decay, and its lower abundance is read on a mass spectrometer to date the sample. This is a direct application of the CCEA point that isotopes differ in mass but not in chemistry.

Example 2. Chlorine in a mass spectrum of a chloroalkane. When chloromethane CH3Cl\text{CH}_3\text{Cl} is analysed, the molecular ion region shows two peaks two mass units apart in a 3:1 ratio, mirroring the 75%75\% chlorine-35 to 25%25\% chlorine-37 abundance. Recognising this 3:1 doublet is how chemists confirm a single chlorine atom is present, and a 9:6:1 triplet would signal two chlorines. The same weighted-mean idea that gives Ar(Cl)=35.5A_r(\text{Cl}) = 35.5 explains the relative peak heights.

Try this

Q1. Define the term isotopes. [2 marks]

  • Cue. Atoms of the same element with the same number of protons but different numbers of neutrons.

Q2. A sample of boron is 20%20\% boron-10 and 80%80\% boron-11. Calculate its relative atomic mass. [2 marks]

  • Cue. (10×20)+(11×80)100=10.8\frac{(10 \times 20) + (11 \times 80)}{100} = 10.8.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20194 marksA sample of magnesium contains three isotopes. Magnesium-24 has an abundance of 79.0 percent, magnesium-25 has 10.0 percent and magnesium-26 has 11.0 percent. Calculate the relative atomic mass of this sample, giving your answer to one decimal place.
Show worked answer →

This is the standard weighted-mean calculation, and CCEA wants the working set out clearly.

Multiply each isotope mass by its percentage abundance and sum:

(24×79.0)+(25×10.0)+(26×11.0)=1896+250+286=2432(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0) = 1896 + 250 + 286 = 2432

Divide by the total abundance (100100):

Ar=2432100=24.3A_r = \dfrac{2432}{100} = 24.3

Markers award one mark for the correct method (mass times abundance, summed), one for the numerator, one for dividing by 100, and one for the answer quoted to one decimal place. A frequent error is dividing by 3 (the number of isotopes) instead of by the total percentage.

CCEA 20213 marksExplain how the successive ionisation energies of magnesium provide evidence for the existence of three electron shells.
Show worked answer →

Magnesium is 1s22s22p63s21s^2\,2s^2\,2p^6\,3s^2, so its 12 successive ionisation energies fall into groups of 2, 8 and 2.

State that each ionisation energy rises as electrons are removed because the remaining electrons feel a greater effective nuclear charge.

Then identify the large jumps: between the 2nd and 3rd electrons (now removing from the inner n=2n=2 shell) and between the 10th and 11th (removing from the n=1n=1 shell). Each large jump marks an electron taken from a shell closer to the nucleus, more strongly attracted, giving evidence for three distinct shells of 2, 8 and 2 electrons. Markers reward linking each jump to a change of shell and to distance from the nucleus.

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