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What is inside an atom, and how do mass spectra and electron arrangements explain chemical behaviour?

Subatomic particles and isotopes, relative atomic mass from mass spectra, the principles of time-of-flight mass spectrometry, and electron configuration in shells, sub-shells and orbitals including ionisation energy evidence.

An Eduqas A-Level Chemistry C1.2 answer on subatomic particles, isotopes, relative atomic mass from mass spectra, time-of-flight mass spectrometry and electron configuration with ionisation energy evidence.

Generated by Claude Opus 4.811 min answer

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  1. What this topic is asking
  2. Subatomic particles and isotopes
  3. Relative masses
  4. Time-of-flight mass spectrometry
  5. Electron configuration
  6. Ionisation energy as evidence
  7. Examples in context
  8. Try this

What this topic is asking

Eduqas topic C1.2 covers the structure of the atom, isotopes, how mass spectrometry measures relative masses, and how electrons are arranged in shells, sub-shells and orbitals. Ionisation energies provide the experimental evidence for that arrangement, so the topic links the model of the atom to measurable data.

Subatomic particles and isotopes

Relative masses

The relative atomic mass ArA_r is the weighted mean mass of an atom relative to one twelfth the mass of a carbon-12 atom. It is calculated from the abundances of the isotopes.

Time-of-flight mass spectrometry

Eduqas requires the principle of the modern time-of-flight (TOF) mass spectrometer. A sample is ionised (usually to 1+1+ ions), the ions are accelerated by an electric field to the same kinetic energy, they drift along a flight tube, and lighter ions arrive at the detector first. The time of flight gives the mass-to-charge ratio m/zm/z, and the relative size of each peak gives the abundance of that isotope.

Electron configuration

Electrons occupy shells (principal quantum number nn) divided into sub-shells (s,p,d\text{s}, \text{p}, \text{d}) made of orbitals, each holding two electrons of opposite spin. They fill in order of increasing energy, which puts 4s4\text{s} below 3d3\text{d}:

12233434p1\text{s}\ 2\text{s}\ 2\text{p}\ 3\text{s}\ 3\text{p}\ 4\text{s}\ 3\text{d}\ 4\text{p}

So iron is 1s22s22p63s23p63d64s21\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2, often written [Ar]3d64s2[\text{Ar}]3\text{d}^6 4\text{s}^2. Chromium and copper are exceptions, taking 3d54s13\text{d}^5 4\text{s}^1 and 3d104s13\text{d}^{10} 4\text{s}^1 to gain a more stable half-full or full 3d3\text{d} sub-shell.

Ionisation energy as evidence

The first ionisation energy is the energy to remove one mole of electrons from one mole of gaseous atoms: X(g)X+(g)+e\text{X}(\text{g}) \rightarrow \text{X}^+(\text{g}) + \text{e}^-. Successive ionisation energies always increase, because each electron is pulled from an increasingly positive ion. Large jumps appear when removal moves to a shell closer to the nucleus, revealing the shell structure and the group number.

Examples in context

Example 1. Radioactive dating uses isotope ratios. Carbon-14 and carbon-12 are chemically identical, so living tissue takes them up in the same proportion as the atmosphere; after death the carbon-14 decays, and the changed isotope ratio (a mass-spectrometry measurement) dates the sample.

Example 2. The successive-ionisation pattern for sodium. Sodium shows a small first ionisation energy, then a jump to the second (removing from the full n=2n = 2 shell), then a much larger jump after the ninth electron (removing from the n=1n = 1 shell). The pattern 2,8,12, 8, 1 falls straight out of the data, confirming sodium is in Group 1.

Try this

Q1. Write the full electron configuration of a sulfur atom (atomic number 16). [1 mark]

  • Cue. 1s22s22p63s23p41\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^4.

Q2. Explain why the first ionisation energy of magnesium is higher than that of sodium. [2 marks]

  • Cue. Magnesium has one more proton, so a greater nuclear charge attracts the outer electrons more strongly, and its atomic radius is slightly smaller, so more energy is needed to remove the outer electron.

Exam-style practice questions

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

Eduqas 20184 marksBoron exists as two isotopes. A sample contains 19.9%19.9\% of boron-10 and 80.1%80.1\% of boron-11. (a) Calculate the relative atomic mass of boron to 1 decimal place. (b) State the number of protons, neutrons and electrons in an atom of boron-11.
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(a) Ar=(10×19.9)+(11×80.1)100=199+881.1100=10.8A_r = \dfrac{(10 \times 19.9) + (11 \times 80.1)}{100} = \dfrac{199 + 881.1}{100} = 10.8 (2: 1 for the weighted-mean method, 1 for the answer).

(b) Boron has atomic number 55, so 55 protons and 55 electrons; mass number 1111 gives 115=611 - 5 = 6 neutrons (2).

Markers reward the weighted-mean calculation and the correct particle count from atomic and mass number.

Eduqas 20203 marksThe first four ionisation energies of an element (in kJ mol1\text{kJ mol}^{-1}) are 738738, 14511451, 77337733 and 1054110541. Deduce which group of the periodic table the element is in and explain your reasoning.
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The element is in Group 2 (1).

There is a large jump between the second and third ionisation energies (14511451 to 77337733) (1). This shows that the first two electrons are removed relatively easily from the outer shell, but the third must be removed from an inner shell much closer to the nucleus, so the element has two electrons in its outer shell and is in Group 2 (1).

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