Exam Practice

Topic 3.1.1: Atomic Structure Exam Practice

Test your understanding of fundamental particles, TOF mass spectrometry, configurations, and ionisation trends with exam-style questions.

Atomic Structure Exam Practice

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📋 Structured Questions

Complete each question on paper, then check your answers against the mark scheme.

Question 1: TOF Mass Spectrometry

8 marks

(a) Describe the processes of electrospray ionisation and electron impact ionisation in a Time of Flight (TOF) mass spectrometer. Compare when each method is typically used. [4]

(b) An ion of 82Kr+ is accelerated in a TOF mass spectrometer. The kinetic energy of the ion is 4.82 x 10^-15 J. The length of the drift tube is 1.20 m. Calculate the time of flight of this ion. Give your answer to 3 significant figures. (Avogadro constant L = 6.022 x 10^23 mol^-1). [4]

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(a)

  • Electron impact: Sample is vaporised and high energy electrons from an electron gun are fired at it, knocking off an electron to form 1+ ions. [1] Used for elements or low molecular mass compounds. [1]
  • Electrospray: Sample is dissolved in a polar solvent and injected through a fine hypodermic needle connected to a high voltage supply. Each molecule gains a proton (H+) to form MH+ ions. [1] Used for high molecular mass compounds (e.g. proteins) to prevent fragmentation. [1]

(b)

  • Mass of one 82Kr+ ion in kg:
    m = 0.082 / (6.022 x 10^23) = 1.3617 x 10^-25 kg [1]
  • Using KE = 1/2 * m * v^2:
    v = sqrt(2 * KE / m) = sqrt(2 * 4.82 x 10^-15 / 1.3617 x 10^-25) = 2.6607 x 10^5 m s^-1 [1]
  • Using t = d / v:
    t = 1.20 / (2.6607 x 10^5) = 4.51 x 10^-6 s (or 4.51 microseconds) [2]
Examiner tip: For TOF calculations, remember that the mass must be converted to kg per single particle. Always convert the mass in grams per mole (82 g/mol = 0.082 kg/mol) by dividing by Avogadro's number. Do not round numbers in intermediate steps, or your final answer might be slightly off.

Question 2: Electron Configurations and Trends

7 marks

(a) Write the full electron configuration of the S2- ion and the Ni atom. [2]

(b) Explain why the second ionisation energy of sodium is much higher than its first ionisation energy. [2]

(c) Explain why the first ionisation energy of sulfur is lower than that of phosphorus, despite sulfur having a higher nuclear charge. [3]

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(a)

  • S2-: 1s2 2s2 2p6 3s2 3p6 [1]
  • Ni: 1s2 2s2 2p6 3s2 3p6 3d8 4s2 [1]

(b)

  • The second electron of sodium is removed from the 2p sub-shell, which is in a lower energy level / closer to the nucleus than the 3s sub-shell. [1]
  • This electron experiences less shielding, and a stronger electrostatic attraction from the nucleus. [1]

(c)

  • Phosphorus has three unpaired electrons, one in each of the three 3p orbitals (3p3). [1]
  • Sulfur has four electrons in the 3p sub-shell (3p4), meaning two electrons are paired in one 3p orbital. [1]
  • The repulsion between these paired electrons in the same orbital makes the outer electron in sulfur easier to remove. [1]
Examiner tip: For part (c), draw the orbital box diagram if you are unsure. Pairing electrons in the same orbital results in spin-pair repulsion. Make sure to name the orbital specifically (3p) and state that it is electron-electron repulsion.

Question 3: Mass Spectrometry and Ar

4 marks

(a) Define relative atomic mass (Ar). [2]

(b) A mass spectrum of a sample of bromine contains peaks at m/z = 79 and m/z = 81. The peak heights are 50.5% and 49.5% respectively. Calculate the relative atomic mass of the bromine in this sample. Give your answer to 2 decimal places. [2]

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(a)

  • The weighted average mass of an atom of an element [1]
  • relative to 1/12th of the mass of an atom of carbon-12. [1]

(b)

  • Ar = (79 * 50.5 + 81 * 49.5) / 100 [1]
  • Ar = 79.99 [1]
Examiner tip: The definition of Ar must mention "weighted average mass" and "relative to 1/12th of carbon-12". Do not say "average mass of isotopes" without mentioning "of an atom of an element".

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