Formation of Na2O
The formation of sodium oxide (Na₂O) involves the transfer of electrons between sodium (Na) and oxygen (O) atoms. This process can be described in detail as follows:
1. Electron Configuration
- Sodium (Na) has an atomic number of 11, with an electron configuration of 1s² 2s² 2p⁶ 3s¹.
- Oxygen (O) has an atomic number of 8, with an electron configuration of 1s² 2s² 2p⁴.
2. Electron Transfer
- Sodium has one electron in its outermost shell (3s¹) and tends to lose this electron to achieve a stable electron configuration similar to neon (Ne).
- Oxygen has six electrons in its outer shell (2s² 2p⁴) and needs two more electrons to complete its octet, achieving a stable configuration like neon.
3. Formation of Ions
- When sodium loses one electron, it becomes a positively charged ion (Na⁺):
- Na → Na⁺ + e⁻
- When oxygen gains two electrons, it becomes a negatively charged ion (O²⁻):
- O + 2e⁻ → O²⁻
4. Ionic Bond Formation
- The Na⁺ ions and O²⁻ ions attract each other due to their opposite charges, forming an ionic bond.
- Two sodium ions are required to balance the charge of one oxide ion (O²⁻), leading to the formula Na₂O.
5. Overall Reaction
- The overall reaction for the formation of sodium oxide can be summarized as:
- 4 Na + O₂ → 2 Na₂O
This reaction illustrates how the transfer of electrons results in the formation of an ionic compound, sodium oxide (Na₂O).