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Chapter 10: Problem 42

What is the hybridization state of the central \(\mathrm{N}\) atom in the azide ion, \(\mathrm{N}_{3}^{-} ?\) (Arrangement of atoms: NNN.)

Short Answer

Expert verified
The hybridization state of the central N atom in the azide ion, \(N_{3}^{-}\), is sp.

Step by step solution

01

Draw the Lewis Structure

Before determining hybridization, the Lewis structure of the molecule has to be drawn. With azide ion (\(N_3^-\)), all three nitrogen atoms are bonded linearly. The central atom, \(N\), forms a triple bond with one \(N\) atom and a double bond with the other \(N\) atom.
02

Calculate the Total Number of Electron Groups

The groups of electrons around the central nitrogen atom determine its hybridization. Electron groups include both the bonds (single, double, triple) and lone pairs of electrons. For the central nitrogen atom, there is one double bond and one triple bond, with no lone pairs.
03

Determine the Hybridization

The number of electron groups determines the hybridization of the atom. If an atom has two groups, it is sp hybridized; if it has three groups, it is sp2 hybridized; if it has four groups, it is sp3 hybridized. In our situation, since there are two electron groups around the central \(N\) atom, it means that this atom is 'sp' hybridized.

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