Write the ground state configuration for Cu, Fe3+, Ti3+ and S4+. Cu+ _____ Ti3+ __________ Fe2+
The Correct Answer and Explanation is:
Correct Answers:
- Cu⁺ (Copper ion):
[Ar] 3d¹⁰ - Ti³⁺ (Titanium ion):
[Ar] 3d¹ - Fe²⁺ (Iron ion):
[Ar] 3d⁶ - S⁴⁺ (Sulfur ion):
1s² 2s² 2p⁶ (or simply [Ne])
Explanation
To understand the ground-state electron configurations of ions like Cu⁺, Ti³⁺, Fe²⁺, and S⁴⁺, we begin with their neutral atom configurations and remove electrons according to the ion’s charge.
1. Copper (Cu⁺):
- Neutral Cu has atomic number 29.
- Its ground state: [Ar] 4s¹ 3d¹⁰
- When Cu loses one electron to form Cu⁺, it is removed from the 4s orbital first, not the 3d.
- So, Cu⁺ = [Ar] 3d¹⁰
This fully filled d-subshell (3d¹⁰) makes Cu⁺ more stable due to electron configuration symmetry and energy considerations.
2. Titanium (Ti³⁺):
- Atomic number of Ti is 22.
- Neutral configuration: [Ar] 4s² 3d²
- Losing three electrons for Ti³⁺: remove the two 4s and one 3d electron.
- So, Ti³⁺ = [Ar] 3d¹
This is a common oxidation state for titanium in coordination complexes, and it results in one unpaired d-electron.
3. Iron (Fe²⁺):
- Fe atomic number: 26.
- Neutral: [Ar] 4s² 3d⁶
- To get Fe²⁺, remove two electrons. The 4s electrons are removed first.
- Fe²⁺ = [Ar] 3d⁶
This configuration has four unpaired electrons and is commonly seen in biological systems like hemoglobin.
4. Sulfur (S⁴⁺):
- Atomic number of sulfur: 16.
- Neutral configuration: 1s² 2s² 2p⁶ 3s² 3p⁴
- S⁴⁺ loses four electrons: two from 3p and two from 3s.
- Remaining configuration: 1s² 2s² 2p⁶, or simply [Ne]
This noble gas configuration is especially stable and occurs in compounds like SO₂.
Understanding which orbitals lose electrons first is key in writing ion configurations correctly.
