Electron Configuration: Writing and Interpreting Configurations
Practice writing electron configurations using the Aufbau principle, Hund's rule, and the Pauli exclusion principle for atoms and ions.
Problem Scenario
Write the full and abbreviated (noble gas core) electron configurations for the following: (a) Sulfur (S, Z = 16), (b) Iron (Fe, Z = 26), (c) Fe²⁺ ion, and (d) Fe³⁺ ion. Identify any that are paramagnetic and explain why.
Given Data
Requirements
- Write the full electron configuration for S
- Write the abbreviated configuration for Fe using [Ar] core
- Write configurations for Fe²⁺ and Fe³⁺ ions
- Determine which species are paramagnetic and why
Solution
Step 1:
Sulfur (Z = 16): Fill orbitals in order — 1s²2s²2p⁶3s²3p⁴. The 3p subshell has 4 electrons across 3 orbitals: by Hund's rule, three go one per orbital (↑ ↑ ↑), then the fourth pairs (↑↓ ↑ ↑). Abbreviated: [Ne]3s²3p⁴.
Step 2:
Iron (Z = 26): Full configuration — 1s²2s²2p⁶3s²3p⁶4s²3d⁶. Abbreviated: [Ar]4s²3d⁶. The 3d subshell has 6 electrons: five go one per orbital first (Hund's rule), then the sixth pairs one, giving 4 unpaired electrons.
Step 3:
Fe²⁺ (24 electrons): Remove 2 electrons from Fe. Important: electrons are removed from the highest principal quantum number first — the 4s electrons, not 3d. Configuration: [Ar]3d⁶. This has 4 unpaired electrons.
Step 4:
Fe³⁺ (23 electrons): Remove 3 electrons from Fe — both 4s electrons and one 3d electron. Configuration: [Ar]3d⁵. This is a half-filled d subshell with 5 unpaired electrons (extra stable).
Step 5:
Paramagnetism: All four species are paramagnetic because they all have unpaired electrons. S has 2 unpaired (in 3p), Fe has 4 unpaired (in 3d), Fe²⁺ has 4 unpaired (in 3d), and Fe³⁺ has 5 unpaired (in 3d). Fe³⁺ is the most paramagnetic.
Final Answer
S: [Ne]3s²3p⁴ (2 unpaired). Fe: [Ar]4s²3d⁶ (4 unpaired). Fe²⁺: [Ar]3d⁶ (4 unpaired). Fe³⁺: [Ar]3d⁵ (5 unpaired). All are paramagnetic. Electrons are removed from 4s before 3d when forming transition metal cations.
Key Takeaways
- ✓Aufbau principle: fill lowest energy orbitals first (1s → 2s → 2p → 3s → 3p → 4s → 3d)
- ✓Hund's rule: maximize unpaired electrons within a subshell
- ✓When forming cations, remove electrons from the highest n first (4s before 3d)
- ✓Half-filled and fully-filled subshells have extra stability
Common Errors to Avoid
- ✗Removing 3d electrons instead of 4s when forming transition metal ions
- ✗Forgetting Hund's rule — pairing electrons before filling all orbitals singly
- ✗Writing 3d before 4s in the neutral atom configuration (4s fills first but is removed first)
- ✗Confusing paramagnetic (unpaired electrons) with diamagnetic (all paired)
Practice More Problems with AI
Snap a photo of any problem and get instant explanations.
Download ChemistryIQFAQs
Common questions about this problem type
Although 4s fills before 3d in neutral atoms, once the 3d orbitals are occupied they become lower in energy than 4s due to greater effective nuclear charge. So 4s electrons are higher in energy and are removed first in cation formation.
Half-filled subshells (d⁵) have maximum exchange energy — every electron has the same spin, maximizing favorable electron-electron interactions. This extra stability explains anomalies like Cr being [Ar]4s¹3d⁵ instead of [Ar]4s²3d⁴.