OH⁻ Lewis Structure: Electrons, Lone Pairs, and Charge

OH⁻ Lewis structure

$$\left[\mathrm{H}-\underset{\cdot\cdot}{\overset{\cdot\cdot}{\mathrm{O}}}\mathord{:}\right]^{-}$$

The oxygen has three lone pairs: one above, one below, and one to the right. H–O is a single bond. The brackets enclose the entire hydroxide ion and the superscript gives its −1 overall charge.

Include the extra electron

Oxygen contributes six valence electrons and hydrogen contributes one. Add one more for the negative charge:

$$6+1+1=8\text{ valence electrons}.$$

Place two electrons in the O–H bond and the remaining six on oxygen as three lone pairs. Oxygen then has eight electrons around it, while hydrogen has the two-electron shell it requires. Hydrogen must not receive an octet or an extra lone pair.

Verify where the formal charge belongs

The formal charges add to −1. This allocation identifies oxygen as the formally negative atom; it does not turn the covalent O–H bond into a separate pair of monatomic ions.

Avoid confusing hydroxide with the hydroxyl radical

Removing the charge from the formula changes the electron count. Neutral OH has seven valence electrons and an unpaired electron, whereas OH⁻ has eight, all paired in this Lewis structure. An answer with only two oxygen lone pairs, or without the enclosing charge, does not represent the requested ion.