Predict the ground-state electron configuration of each ion. Use the abbreviated noble gas notation. Ru2+. W3+ B UX2
The correct answer and explanation is :
Electron Configuration of Ions Using Noble Gas Notation
To determine the ground-state electron configuration of each ion, we must first write the electron configuration of the neutral atom and then account for the loss or gain of electrons due to ion formation.
1. Ruthenium (Ru²⁺)
- Atomic number: 44
- Neutral electron configuration: [Kr] 5s² 4d⁶
- For Ru²⁺: Removal of two electrons occurs from the 5s orbital first, followed by the 4d orbital.
- Ru²⁺ configuration: [Kr] 4d⁶
2. Tungsten (W³⁺)
- Atomic number: 74
- Neutral electron configuration: [Xe] 6s² 4f¹⁴ 5d⁴
- For W³⁺: Remove three electrons starting with the 6s orbital, followed by the 5d orbital.
- W³⁺ configuration: [Xe] 4f¹⁴ 5d¹
3. Boron (B)
- Atomic number: 5
- Neutral electron configuration: [He] 2s² 2p¹
- Boron typically forms B³⁺ in some compounds, removing three electrons.
- B³⁺ configuration: [He] (since all valence electrons are removed)
4. UX₂ (Assuming U = Uranium)
- Uranium atomic number: 92
- Neutral electron configuration: [Rn] 7s² 5f⁴ 6d¹
- The oxidation state depends on X (a halogen or ligand), but common uranium oxidation states include U⁴⁺ and U⁶⁺.
- If UX₂ suggests U⁴⁺, remove four electrons.
- U⁴⁺ configuration: [Rn] 5f²
- If UX₂ suggests U⁶⁺, remove six electrons.
- U⁶⁺ configuration: [Rn] 5f⁰
Explanation
- Electron removal follows energy levels: Electrons are removed first from the outermost orbitals (highest principal quantum number) before inner orbitals.
- Transition metals lose s-electrons before d-electrons: Even though 5s fills before 4d (Ru case), it is lost first when forming cations.
- Lanthanide contraction affects configurations: W³⁺ retains 4f¹⁴, showing the stability of filled f orbitals.
- Actinide contraction in uranium: Removal of electrons favors the more stable 5f² (U⁴⁺) or empty 5f⁰ (U⁶⁺) state.