Electron Configuration Practice Questions with Full Answers

Electron configurations are a skill: you get fluent by doing lots of them. These thirty questions are grouped into five sets of increasing difficulty. Work through a set, check your answers, and move on only when you’re getting nearly all of them right. If you get stuck, the relevant guides are electron configuration rules, the Aufbau principle, noble gas shorthand and electron configurations of ions.

Filling order reminder: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p

Set A: Light elements (full configurations)

Write the full electron configuration.

1. Lithium (Z = 3)
2. Boron (Z = 5)
3. Neon (Z = 10)
4. Aluminium (Z = 13)
5. Chlorine (Z = 17)
6. Potassium (Z = 19)

Answers

1. 1s² 2s¹
2. 1s² 2s² 2p¹
3. 1s² 2s² 2p⁶
4. 1s² 2s² 2p⁶ 3s² 3p¹
5. 1s² 2s² 2p⁶ 3s² 3p⁵
6. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Check yourself: add the superscripts. They must equal the atomic number (for potassium: 2 + 2 + 6 + 2 + 6 + 1 = 19). Notice potassium’s last electron goes into 4s, not 3d, because 4s is lower in energy at this point.

Set B: Noble gas shorthand

Write the configuration in noble gas shorthand.

7. Magnesium (Z = 12)
8. Phosphorus (Z = 15)
9. Titanium (Z = 22)
10. Nickel (Z = 28)
11. Arsenic (Z = 33)
12. Barium (Z = 56)

Answers

7. [Ne] 3s²
8. [Ne] 3s² 3p³
9. [Ar] 3d² 4s²
10. [Ar] 3d⁸ 4s²
11. [Ar] 3d¹⁰ 4s² 4p³
12. [Xe] 6s²

Tip: always use the noble gas at the end of the previous period. For arsenic in period 4, that’s argon. Don’t forget the 3d¹⁰ when you cross the d-block to reach the p-block.

Set C: Ions

Write the configuration of each ion.

13. Na⁺
14. F⁻
15. S²⁻
16. Ca²⁺
17. Fe²⁺
18. Cu²⁺

Answers

13. 1s² 2s² 2p⁶ (= [Ne])
14. 1s² 2s² 2p⁶ (= [Ne])
15. [Ne] 3s² 3p⁶ (= [Ar])
16. [Ne] 3s² 3p⁶ (= [Ar])
17. [Ar] 3d⁶
18. [Ar] 3d⁹

The key trap: for Fe²⁺, the two electrons are removed from 4s first, not from 3d. Fe is [Ar] 3d⁶ 4s², so Fe²⁺ is [Ar] 3d⁶, not [Ar] 3d⁴ 4s². For Cu²⁺, start from copper’s actual configuration, [Ar] 3d¹⁰ 4s¹: remove the 4s electron and one 3d electron to give [Ar] 3d⁹.

Set D: Identify the element or ion

Each configuration belongs to a neutral atom unless a charge is stated. Identify it.

19. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
20. [Kr] 5s² 4d¹⁰ 5p²
21. [Ar] 3d⁵ 4s¹
22. A 2+ ion with configuration [Ar] 3d⁵
23. A 3− ion with configuration 1s² 2s² 2p⁶
24. An atom in period 3 with three unpaired electrons

Answers

19. 2 + 2 + 6 + 2 + 6 + 2 + 10 + 5 = 35 electrons → bromine
20. 36 + 2 + 10 + 2 = 50 → tin
21. 18 + 5 + 1 = 24 → chromium (one of the famous exceptions: 3d⁵ 4s¹ rather than 3d⁴ 4s²)
22. The ion has 23 electrons; a 2+ ion has 2 more protons than electrons, so Z = 25 → Mn²⁺
23. 10 electrons, 3 more than protons, so Z = 7 → N³⁻ (nitride)
24. Three unpaired electrons in period 3 means 3p³ → phosphorus

Set E: Explain and apply

25. Explain why the configuration of copper is [Ar] 3d¹⁰ 4s¹ rather than [Ar] 3d⁹ 4s².

Answer: In copper, the 3d and 4s orbitals are very close in energy. Moving one electron from 4s into 3d produces a completely filled 3d subshell, which has extra stability (lower electron repulsion and favourable exchange energy). Overall, [Ar] 3d¹⁰ 4s¹ is lower in energy, so it’s the ground state. See electron configuration exceptions.

26. Which of these configurations is an excited state rather than a ground state? (a) 1s² 2s² 2p¹; (b) 1s² 2s¹ 2p²; (c) 1s² 2s² 2p⁶ 3s¹

Answer: (b). A ground-state atom with 5 electrons would be 1s² 2s² 2p¹. In (b), one electron has been promoted from 2s to 2p, a higher-energy arrangement. Excited states like this can form when atoms absorb energy, for example in a flame.

27. Which configuration breaks the Pauli exclusion principle, and which breaks Hund’s rule? (a) 2p boxes ↑↓ ↑ _ for a 2p³ atom; (b) a 2s box containing ↑↑

Answer: (b) breaks the Pauli exclusion principle: two electrons in the same orbital must have opposite spins. (a) breaks Hund’s rule: the three 2p electrons should occupy separate orbitals, ↑ ↑ ↑. See Hund’s rule and the Pauli exclusion principle.

28. How many unpaired electrons does Co²⁺ (Z = 27) have?

Answer: Co is [Ar] 3d⁷ 4s². Co²⁺ loses the two 4s electrons: [Ar] 3d⁷. Five d orbitals, seven electrons: five go in singly, then two pair up, leaving 3 unpaired electrons. See how to count unpaired electrons.

29. Explain why sodium forms Na⁺ but not Na²⁺.

Answer: Na is [Ne] 3s¹. Removing the single 3s electron leaves the stable neon configuration. A second electron would have to come from the 2p subshell, which is much closer to the nucleus and far less shielded, so the second ionisation energy is enormous (about 4,560 kJ/mol compared with 496 kJ/mol for the first). The energy needed isn’t recovered in forming compounds. See ionization energy trend.

30. Use electron configurations to explain why chlorine and bromine have similar chemical properties.

Answer: Chlorine is [Ne] 3s² 3p⁵ and bromine is [Ar] 3d¹⁰ 4s² 4p⁵. Both have the outer configuration ns² np⁵: seven outer electrons, one short of a noble gas configuration. Chemical properties depend mainly on outer electrons, so both readily gain one electron to form X⁻ ions and behave similarly. See halogens.

Scoring and next steps

• 27–30 correct: you’re exam-ready on configurations. Try effective nuclear charge next to understand the trends behind them.
• 20–26: review the traps in Sets C and E, especially transition metal ions and exceptions.
• Under 20: rework Sets A and B with the electron configuration tool open to check each answer as you go, and draw orbital box diagrams for the ones you get wrong.

Common errors spotted in these questions

1. Filling 3d before 4s in neutral atoms, or removing 3d before 4s in ions.
2. Forgetting 3d¹⁰ (or 4f¹⁴) when jumping to the p-block of periods 4–6.
3. Using the wrong noble gas core.
4. Forgetting the exceptions (Cr, Cu, and heavier analogues such as Mo, Ag and Au).
5. Miscounting: always add the superscripts and compare with Z (minus the charge for ions).

Key takeaways

• Fill subshells in Aufbau order; check that superscripts add up to the number of electrons.
• In noble gas shorthand, use the previous noble gas and include any filled d or f subshells crossed.
• For ions, remove electrons from the highest n first (4s before 3d) and add electrons to the next available orbitals.
• Learn the common exceptions and be ready to explain them.
• Configurations explain chemistry: similar outer configurations mean similar properties.