1.3 Exam Practice
Exam-style practice questions on Electron Configurations
Section B: Data Analysis (Paper 1B Style)
Calculator and Data Booklet permitted. Show all working clearly.
Question 1: Successive Ionisation Energies Deduce HL
5 marksThe successive ionisation energies (in kJ mol\(^{-1}\)) for an unknown element X are shown below:
| IE | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
|---|---|---|---|---|---|---|---|
| kJ mol\(^{-1}\) | 578 | 1817 | 2745 | 11578 | 14831 | 18378 | 23293 |
(a) Define the term "first ionisation energy." [2]
(b) Deduce the group number of element X. Justify your answer using the data. [2]
(c) Explain why there is a large increase between the 3rd and 4th ionisation energies. [1]
Show Mark Scheme
(a) The minimum energy required to remove one electron [1]
from one mole of gaseous atoms in their ground state [1]
(b) Group 13 / Group 3 (in older notation) [1]
There is a large jump between the 3rd and 4th IE, indicating that the 4th electron is removed from a different (inner) shell, so X has 3 outer-shell electrons [1]
(c) The 4th electron is being removed from a lower energy level / inner shell which is closer to the nucleus, so it experiences a much stronger nuclear attraction [1]
Section C: Structured Questions (Paper 2 Style)
Show all working. State answers with appropriate significant figures and units.
Question 2: Electron Configurations of Atoms and Ions State
4 marks(a) Write the full electron configuration for a ground-state atom of vanadium (Z = 23). [1]
(b) Write the electron configuration for the V\(^{3+}\) ion and state how many unpaired electrons it contains. [2]
(c) Explain why transition metal ions lose their 4s electrons before their 3d electrons when forming cations. [1]
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(a) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³ [1]
(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² [1]
2 unpaired electrons [1]
(c) In the ion / when occupied, the 4s subshell has a higher energy than 3d, so 4s electrons are removed first [1]
Question 3: Aufbau Principle and Hund's Rule Explain
4 marks(a) State the Aufbau principle. [1]
(b) State Hund's rule. [1]
(c) Using the orbital box notation, show the arrangement of electrons in the 2p subshell of a nitrogen atom. [1]
(d) Explain why the electron configuration of copper is [Ar] 3d\(^{10}\) 4s\(^1\) rather than [Ar] 3d\(^9\) 4s\(^2\). [1]
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(a) Electrons fill orbitals starting from the lowest energy level first [1]
(b) Every orbital in a subshell is singly occupied with parallel spin before any orbital is doubly occupied [1]
(c) Three boxes each containing one up-arrow: [↑] [↑] [↑] [1] (Accept any clear representation showing 3 singly-occupied orbitals with parallel spins)
(d) A completely filled 3d subshell (3d\(^{10}\)) provides additional stability compared to 3d\(^9\) / the fully filled d-subshell has lower energy [1]
Question 4: Trends in Ionisation Energy Discuss HL
4 marks(a) Explain the general trend in first ionisation energy across Period 3 (Na to Ar). [2]
(b) Explain why the first ionisation energy of aluminium is lower than that of magnesium, despite aluminium having a greater nuclear charge. [2]
Show Mark Scheme
(a) The general trend is an increase in first IE across the period [1]
Because nuclear charge increases while shielding remains approximately constant / the effective nuclear charge increases, so the outer electrons are held more tightly [1]
(b) Aluminium's outer electron is in a 3p orbital, which is higher in energy than the 3s orbital of magnesium [1]
The 3p electron is slightly further from the nucleus and more shielded by the 3s electrons, so it is easier to remove [1]
Question 5: Convergence Limit and Ionisation Energy Calculate HL
5 marksThe hydrogen emission spectrum shows a series of lines that converge towards a limit. For the Lyman series (transitions to n = 1), the convergence limit occurs at a frequency of \(3.28 \times 10^{15}\) Hz.
(a) Explain the significance of the convergence limit in an emission spectrum. [2]
(b) Calculate the first ionisation energy of hydrogen using this data. Give your answer in kJ mol\(^{-1}\). [2]
\(h = 6.626 \times 10^{-34}\) J s \(L = 6.022 \times 10^{23}\) mol\(^{-1}\)
(c) Explain why the lines in the Lyman series converge as frequency increases. [1]
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(a) The convergence limit corresponds to the ionisation of the atom / removal of the electron completely [1]
At this point the electron is being removed from the atom from the ground state to an infinite distance [1]
(b) \(E = hf = 6.626 \times 10^{-34} \times 3.28 \times 10^{15} = 2.173 \times 10^{-18}\) J [1]
IE = \(2.173 \times 10^{-18} \times 6.022 \times 10^{23} = 1.309 \times 10^{6}\) J mol\(^{-1}\) = 1309 kJ mol\(^{-1}\) [1]
(c) The energy levels become closer together as they get further from the nucleus, so the energy difference (and therefore frequency) between successive transitions becomes smaller, causing lines to converge [1]