Question

draw the lewis structure of nitrite (no₂⁻) and then determine the hybridization of the central atom.

Explanation:

Step1: Calculate the number of valence electrons

Nitrogen has 5 valence electrons, each oxygen has 6 valence electrons, and there is 1 negative charge (extra electron). So the total number of valence electrons is \(5 + 2\times6+ 1=18\).

Step2: Draw the Lewis structure

The central atom is \(N\). We can draw two resonance structures: one with a double - bond to one \(O\) and a single - bond to the other \(O\) (with a negative charge on the single - bonded \(O\)), and vice - versa. The general Lewis structure has \(N\) in the center, bonded to two \(O\) atoms.

Step3: Determine the electron - pair geometry

Using the formula \(VSEPR\) (Valence - Shell Electron - Pair Repulsion) theory, the number of electron - pairs around \(N\) is calculated. The formula for the number of electron - pairs \(=\frac{\text{valence electrons of central atom}+\text{number of monovalent atoms}-\text{charge (if any)}}{2}\). For \(NO_{2}^{-}\), the number of electron - pairs \(=\frac{5 + 0+ 1}{2}=3\) (where \(0\) is for non - monovalent \(O\) atoms). The electron - pair geometry is trigonal planar.

Step4: Determine the hybridization

For trigonal planar electron - pair geometry, the hybridization of the central atom (\(N\)) is \(sp^{2}\).

Answer:

The Lewis structure of \(NO_{2}^{-}\) has \(N\) as the central atom, bonded to two \(O\) atoms (with resonance between a single bond and a double bond). The hybridization of the central \(N\) atom is \(sp^{2}\) (Option E).