Draw and explain the Lewis structure for Na2O

The Lewis structure for sodium oxide (Na₂O) can be drawn by following these steps:

1. Determine the total number of valence electrons: Sodium (Na) is in Group 1 of the periodic table, contributing 1 valence electron per atom. Oxygen (O) is in Group 16, contributing 6 valence electrons. Therefore, for Na₂O:
1. Total from sodium: 2 Na × 1 = 2 electrons
2. Total from oxygen: 1 O × 6 = 6 electrons
3. Total: 2 + 6 = 8 valence electrons
2. Sketch the skeletal structure: In Na₂O, sodium will lose one electron, while oxygen will gain two electrons to achieve a stable configuration. Therefore, the sodium atoms can be drawn as being bonded to the oxygen atom.
3. Distribute the electrons: Each sodium atom donates one electron to the oxygen atom. Consequently, the Lewis structure shows each Na atom connected to the O atom:

  Na   Na
   |    |
   O  

In this structure, the oxygen atom should have three lone pairs of electrons, since it originally has six valence electrons, and it will share two with the sodium atoms.

     ..
    :O:
     ..
    / \
   Na  Na

Final Explanation: In the Lewis structure of Na₂O, the oxygen atom has two single bonds with two sodium atoms and three lone pairs. This arrangement satisfies the octet rule for the oxygen atom and leaves each sodium atom with a stable electron configuration. Since sodium loses its one valence electron and becomes Na⁺, while oxygen gains two and becomes O^2-, this compound is ionic in nature.