Lewis Structure - Example

Example

The formula of the nitrite ion is NO₂−.

1. Nitrogen is the least electronegative atom, so it is the central atom by multiple criteria.
2. Count valence electrons. Nitrogen has 5 valence electrons; each oxygen has 6, for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron, so the total number of electrons is 18.
3. Place ion pairs. Each oxygen must be bonded to the nitrogen, which uses four electrons — two in each bond. The 14 remaining electrons should initially be placed as 7 lone pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.
4. Satisfy the octet rule. Both oxygen atoms currently have 8 electrons assigned to them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an oxygen atom must form a double bond, but either atom will work equally well. We therefore must have a resonance structure.
5. Tie up loose ends. Two Lewis structures must be drawn: Each structure has one of the two oxygen atoms double-bonded to the nitrogen atom. The second oxygen atom in each structure will be single-bonded to the nitrogen atom. Place brackets around each structure, and add the charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the two resonance forms.