IB DP Chemistry Structure 1.3: Electron configurations – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions test Structure 1.3, Electron configurations, as defined in the IB Chemistry guide for the Diploma Programme (published February 2023; first assessment 2025). Questions 1 to 7 and question 11 cover understandings 1.3.1 to 1.3.5, which are SL and HL. Questions 8, 9 and 10 cover 1.3.6 and 1.3.7, which are additional higher level and are labelled HL only. The set applies to the course as first assessed in 2025.

Revise first with the study guide and the revision notes. A broader set for all of Structure 1 mixes this unit with the rest of the component. Course pages: the chemistry hub and a checklist to print.

Questions

1. Three sources emit radiation: a microwave oven, a green laser pointer and an X-ray tube.

(a) State how the energy of a photon changes as its wavelength increases. [1] (b) List the three sources in order of increasing frequency of the radiation they emit. [1]

2. Light from a filament bulb and light from a neon discharge tube are each passed through a prism. Distinguish between the two spectra seen, and explain why the neon tube gives the spectrum it does. [3]

3. (a) Deduce how many electrons, at most, the fifth main energy level can hold. [1] (b) State the number of orbitals in a 4f sublevel. [1]

4. Describe the shape of an s orbital and the shapes and relative orientation of the three p orbitals in the same main energy level. [3]

5. Zinc has atomic number 30.

(a) Write the full electron configuration of a zinc atom. [1] (b) Write the condensed electron configuration of the Zn²⁺ ion. [1] (c) Explain why forming Zn²⁺ leaves the 3d sublevel unchanged. [1]

6. Chromium has atomic number 24.

(a) Draw an orbital diagram for the 3d and 4s sublevels of a chromium atom. [1] (b) State the number of unpaired electrons in the atom. [1] (c) Explain why the configuration of chromium differs from the one predicted by the Aufbau order. [2]

7. Use these values for the energy levels of the hydrogen atom, in kJ mol⁻¹:

n = 1: −1312     n = 2: −328     n = 3: −145.8    n = 4: −82.0
n = 5: −52.5     n = 6: −36.4    n = ∞: 0

(a) Calculate the energy released, in kJ mol⁻¹, when electrons fall from n = 5 to n = 3, and state the region of the electromagnetic spectrum in which this line appears. [2] (b) Of the drops n = 2 → 1, n = 6 → 2 and n = 4 → 3, identify the one that gives the line of shortest wavelength. Give a reason. [2] (c) Use the data to explain why the lines in each family converge at higher frequency. [2] (d) Explain how the hydrogen emission spectrum shows that the electron occupies discrete energy levels. [1]

8. HL only. Lines in the emission spectrum of potassium that end on its ground state converge at a wavelength of 286 nm. Calculate the first ionization energy of potassium in kJ mol⁻¹. Constants: Planck constant h = 6.63 × 10⁻³⁴ J s; speed of light c = 3.00 × 10⁸ m s⁻¹; Avogadro constant L = 6.02 × 10²³ mol⁻¹. [4]

9. HL only. Use these first ionization energies, in kJ mol⁻¹:

F 1681    Cl 1251    Br 1140    I 1008
Zn 906    Ga 579     As 947     Se 941

(a) Explain the trend from fluorine to iodine. [2] (b) Explain why the value for gallium is lower than the value for zinc. [2] (c) Explain why the value for selenium is slightly lower than the value for arsenic. [2]

10. HL only. The first seven successive ionization energies of an element in period 3 are, in kJ mol⁻¹: 1000, 2252, 3357, 4556, 7004, 8496, 27 107.

(a) Explain why the values keep rising. [1] (b) Deduce the group of the element, showing how you used the data. [2] (c) State the main energy level from which the seventh electron is removed. [1]

11. Cobalt has atomic number 27. It forms the ions Co²⁺ and Co³⁺.

(a) Write the condensed electron configuration of a cobalt atom. [1] (b) Write the condensed electron configurations of Co²⁺ and Co³⁺. [2] (c) Draw the 3d orbital diagram for Co³⁺ and state how many unpaired electrons it has. [2] (d) State the number of electrons in the third main energy level of a cobalt atom, and the maximum this level could hold. [2] (e) State which of the three species, Co, Co²⁺ or Co³⁺, has electrons in the fourth main energy level. [1]

Answers

1. (a) Energy decreases as wavelength increases [1]. [1] (b) Microwave oven < green laser < X-ray tube [1]. [1] Examiner insight: An order is only correct if every item is in the right place, so check it against the spectrum sequence rather than guessing.

2. The bulb gives a continuous spectrum containing all visible wavelengths with no gaps; the neon tube gives a line spectrum of separate coloured lines on a dark background [1]. Electrons in neon atoms are excited to higher energy levels and emit photons as they fall back to lower levels [1]. Only certain energy gaps exist between levels, so only certain frequencies are emitted [1]. [3] Examiner insight: “Distinguish” needs both spectra described in the same sentence or pair of sentences; describing only the line spectrum leaves the comparison incomplete.

3. (a) 2n² = 2 × 5² = 50 [1]. [1] (b) Seven [1]. [1] Examiner insight: An answer of 14 for (b) gives the electron capacity of 4f, not its number of orbitals; read whether the question asks for orbitals or electrons.

4. An s orbital is spherical, centred on the nucleus [1]. Each p orbital is shaped like a dumbbell: two lobes, one either side of the nucleus [1]. The three p orbitals lie along the x, y and z axes, at right angles to each other [1]. [3] Examiner insight: “Circle” and “figure of eight” describe flat shapes; use “spherical” and “dumbbell”, and state the 90° orientation for the third point.

5. (a) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² [1]. [1] (b) [Ar] 3d¹⁰ [1]. [1] (c) The two electrons lost come from 4s, which is higher in energy than 3d once 3d is occupied, so 3d keeps all ten electrons [1]. [1] Examiner insight: For (a) the question asks for the full configuration, so a noble-gas core is not what was asked; always match the form the command requests.

6. (a)

4s  [↑ ]
3d  [↑ ][↑ ][↑ ][↑ ][↑ ]

Five singly occupied 3d orbitals and one 4s electron, all with parallel spins [1]. [1] (b) Six [1]. [1] (c) The 3d and 4s sublevels are very close in energy [1]. Moving one electron from 4s to 3d gives a half-filled 3d⁵ set, which has lower overall energy than 3d⁴ 4s² [1]. [2] Examiner insight: For (c), “half-filled is more stable” earns more when you also say the 3d and 4s energies are close; the closeness is what lets the electron move.

7. (a) −52.5 − (−145.8) = 93.3 kJ mol⁻¹ [1]; the drop ends on n = 3, so the line is in the infrared [1]. [2] (b) n = 2 → 1 [1]. It releases the most energy (984 kJ mol⁻¹, compared with 291.6 and 63.8), so it has the highest frequency and shortest wavelength [1]. [2] (c) The gaps between neighbouring levels shrink as n increases: n = 4 to 5 is 29.5 kJ mol⁻¹ but n = 5 to 6 is only 16.1 kJ mol⁻¹ [1]. Lines from successive higher levels therefore differ less and less in energy, so they crowd together towards the high-frequency end [1]. [2] (d) Only specific frequencies appear, so only specific energy gaps (and levels) can exist [1]. [1] Examiner insight: In (c), quote two gap values from the data when the question says “use the data”; a purely descriptive answer does not show the data was used.

8. λ = 286 × 10⁻⁹ m [1] E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (286 × 10⁻⁹) = 6.955 × 10⁻¹⁹ J per atom [1] × L: 6.955 × 10⁻¹⁹ × 6.02 × 10²³ = 4.19 × 10⁵ J mol⁻¹ [1] First ionization energy = 419 kJ mol⁻¹ [1]. [4] Examiner insight: Show the per-atom energy before multiplying by L, and keep an extra figure in it; rounding to 6.95 × 10⁻¹⁹ J first gives 418 kJ mol⁻¹, and a visible intermediate step is the only place your method can be seen.

9. (a) First IE decreases from F to I [1]. Going down, the outer electron sits in a higher main energy level, at greater distance and with more inner electrons screening it, so it is attracted less strongly despite the larger nuclear charge [1]. [2] (b) Gallium’s outer electron is in 4p, while zinc’s is in 4s [1]. The 4p sublevel is higher in energy (and partly screened by the 4s pair), so gallium’s electron is easier to take away [1]. [2] (c) Arsenic has three unpaired 4p electrons; selenium has four 4p electrons, so one 4p orbital holds a pair [1]. Repulsion between the paired electrons makes one of them easier to remove [1]. [2] Examiner insight: Name the sublevels involved (4s, 4p) in (b) and (c); a general answer about “more shielding” or “further away” does not explain a drop within one period.

10. (a) After each removal the ion carries more positive charge, so the electrons left behind are attracted more strongly [1]. [1] (b) The largest increase is between the sixth and seventh values (8496 to 27 107) [1], so six electrons are in the outer level and the element is in group 16 [1]. [2] (c) The second main energy level (n = 2) [1]. [1] Examiner insight: Compare increases, not ratios alone: here the 6th-to-7th increase (18 611) is far larger than any other, and naming the two values that bracket the jump shows your reasoning.

11. (a) [Ar] 3d⁷ 4s² [1]. [1] (b) Co²⁺: [Ar] 3d⁷ [1]. Co³⁺: [Ar] 3d⁶ [1]. [2] (c)

3d  [↑↓][↑ ][↑ ][↑ ][↑ ]

One paired and four singly occupied orbitals, following Hund’s rule [1]; four unpaired electrons [1]. [2] (d) 3s² 3p⁶ 3d⁷ gives 15 electrons [1]; the maximum is 2 × 3² = 18 [1]. [2] (e) Co only, because both ions have lost their 4s electrons [1]. [1] Examiner insight: In (b), writing [Ar] 3d⁵ 4s² for Co²⁺ is a common slip; check the electron total (25) and that 4s is emptied first.

Where marks are usually lost

• Taking 3d electrons before 4s electrons when writing transition-element ions.
• Writing the predicted Cr or Cu configuration instead of the real one.
• Drawing paired electrons with the same spin, or pairing before all equal-energy orbitals hold one electron.
• Giving the number of electrons when the number of orbitals was asked for.
• Placing hydrogen lines for drops to n = 3 in the visible region.
• Saying lines converge at low frequency or long wavelength.
• (HL) Substituting nm into E = hc/λ, or stopping at the energy of one atom.
• (HL) Explaining the group 15 to 16 drop without paired-electron repulsion.
• (HL) Counting the group from the position of the largest value instead of the electrons removed before the jump.

Next steps

• All free 10-minute diagnostics
• Structure 1.3 revision notes for a quick recap
• Structure 1.3 study guide for full explanations
• Course hub for IB DP Chemistry
• Checklist to print and tick off
• Book a free trial class

Official syllabus

Chemistry guide, Diploma Programme, International Baccalaureate Organization: first assessment 2025, published February 2023. Sub-topic Structure 1.3, Electron configurations (1.3.1 to 1.3.5 SL and HL; 1.3.6 and 1.3.7 additional higher level).