To draw the Lewis structure for a hydroxide (OH⁻) ion, follow these steps:
- Count the total number of valence electrons: Oxygen has 6 valence electrons, and hydrogen has 1. Since the hydroxide ion has a negative charge, add one extra electron. This gives a total of 6 (from oxygen) + 1 (from hydrogen) + 1 (for the charge) = 8 valence electrons.
- Place the least electronegative atom in the center: In this case, oxygen is more electronegative than hydrogen, so hydrogen will be placed on the outside.
- Draw a single bond between the oxygen and hydrogen atoms: This uses 2 of the 8 valence electrons.
- Place the remaining electrons around the oxygen atom: After forming the single bond, you have 6 electrons left. Place these as lone pairs around the oxygen atom. Oxygen will have 3 lone pairs (6 electrons).
- Check the octet rule: Oxygen now has 8 electrons around it (2 from the bond and 6 from the lone pairs), satisfying the octet rule. Hydrogen has 2 electrons, which is also stable.
- Add the negative charge: The extra electron is placed on the oxygen atom, giving it a negative charge.
Here is the final Lewis structure for the hydroxide (OH⁻) ion:
[O]⁻ | H
In this structure, the oxygen atom is surrounded by 3 lone pairs and one single bond to the hydrogen atom, with a negative charge on the oxygen.