After drawing the Lewis structure of the nitrite ion (NO_2^-), what is the hybridization of the central nitrogen atom?

• A. sp
• B. sp^3
• C. sp^3d
• D. sp^2

Answer: (D)

Verified step by step guidance

1. First, draw the Lewis structure of the nitrite ion (NO\_2\^-). Count the total valence electrons: nitrogen has 5, each oxygen has 6, and there is an extra electron due to the negative charge, giving a total of 18 valence electrons.
2. Arrange the atoms with nitrogen as the central atom bonded to two oxygen atoms. Use single bonds initially and then distribute the remaining electrons to satisfy the octet rule for each atom.
3. After placing lone pairs and forming double bonds if necessary, determine the number of regions of electron density (bonding and lone pairs) around the central nitrogen atom. This includes single bonds, double bonds, and lone pairs.
4. Use the steric number (number of electron density regions) to determine the hybridization. The steric number corresponds to hybridization as follows: 2 regions = sp, 3 regions = sp\^2, 4 regions = sp\^3, and so on.
5. Since the nitrogen in NO\_2\^- has three regions of electron density (two bonds and one lone pair), the hybridization is sp\^2.