To draw the Lewis structure for sodium hydroxide (NaOH), we begin by identifying the components involved in the compound. Sodium (Na) is a metal that exists as a cation with a charge of +1, while hydroxide (OH–) is a polyatomic ion consisting of oxygen and hydrogen.
1. **Identify Valence Electrons**: Sodium has 1 valence electron and contributes it to form a bond, while oxygen has 6 valence electrons, and hydrogen has 1 valence electron, totaling 8 valence electrons in OH–. This gives us:
- Na: 1 electron
- O: 6 electrons
- H: 1 electron
- Total: 1 + 6 + 1 = 8 electrons
2. **Construct the Structure**: In NaOH, sodium donates its electron to the hydroxide ion. The hydroxide ion itself has a typical Lewis structure with oxygen at the center, bonded to hydrogen, and possessing 2 lone pairs of electrons around the oxygen atom. So, we represent it as:
H | :O: | Na+
3. **Add Lone Pairs**: The oxygen atom has 6 valence electrons total, and two of these are used in the bond with hydrogen, leaving 4 electrons to form 2 lone pairs on oxygen. Therefore, the structure now looks like:
H | :O: .. | Na+
4. **Formal Charges**: To determine the formal charges, we use the formula:
Formal Charge = Valence Electrons – (Non-bonding Electrons + 0.5 x Bonding Electrons)
- For Na: 1 – 0 = +1
- For O: 6 – (4 + 1) = 1
- For H: 1 – 0 = 0
So, oxygen carries a formal charge of -1, making the overall charge of the hydroxide ion -1, and sodium carries a +1 charge, maintaining charge neutrality for the molecule. The final drawing of the Lewis structure for sodium hydroxide, with formal charges indicated, shows sodium donating its electron to the hydroxide ion, stabilizing the molecule:
H | :O: - .. | Na+
This structure effectively represents sodium hydroxide with the proper lone pairs and formal charges accounted for.