Which of the following is the ground-state electron configuration of the Fe^{3+} ion?

• A. 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6
• B. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5
• C. 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5
• D. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3

Answer: (C)

Verified step by step guidance

1. Step 1: Identify the atomic number of iron (Fe), which is 26. This means a neutral Fe atom has 26 electrons.
2. Step 2: Write the electron configuration for neutral Fe. The order of filling orbitals is 1s, 2s, 2p, 3s, 3p, 4s, then 3d. So, the configuration is \$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$.
3. Step 3: Understand that Fe^{3+} means the iron ion has lost 3 electrons. Electrons are removed first from the highest energy level orbitals, which for transition metals is the 4s orbital before the 3d orbital.
4. Step 4: Remove 2 electrons from the 4s orbital and 1 electron from the 3d orbital to account for the 3+ charge. This results in the configuration \$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$.
5. Step 5: Confirm that this configuration corresponds to a half-filled 3d subshell, which is known to be particularly stable, supporting that \$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5$ is the ground-state electron configuration for Fe^{3+}.