Draw and Explain the Lewis Structure for the Azide Ion Showing All Its Possible Resonance Structures. Give the Name of the Molecular Geometry and the Hybridization Around the Central Atom.

The azide ion, represented as N₃^-1, consists of three nitrogen atoms. To draw the Lewis structure, we start by calculating the total number of valence electrons available. Each nitrogen atom has five valence electrons, giving us a total of 15 electrons. Since the azide ion carries a negative charge, we add one additional electron, making it 16 valence electrons in total.

We can arrange the nitrogen atoms linearly as follows:

N = N = N

The central nitrogen atom is bonded to both terminal nitrogen atoms. To distribute the electrons, we form double bonds between the central nitrogen and each terminal nitrogen. This makes the initial structure:

    :N:  
    ||
N = N = N

Now, we have used 12 electrons in bonds and 4 electrons in lone pairs on the terminal nitrogen. The formal charge on the terminal nitrogens is 0, and the central nitrogen has a formal charge of +1. Therefore, we can create resonance structures to reduce the formal charges on the molecule. The possible resonance structures for the azide ion are as follows:

Structure 1:    Structure 2:    Structure 3:  
    :N:            :N:            :N:  
    ||             ||             ||  
N = N = N   ↔   N = N ≡ N  ↔   N ≡ N = N  

In these structures, we can see that the double bond and triple bond are shuffled between the nitrogen atoms. The resonance structures indicate that the real structure is a hybrid of these forms.

Regarding molecular geometry, the azide ion adopts a linear shape due to the arrangement of three nitrogen atoms. The central nitrogen atom is in the sp hybridization state since it is involved in two sigma bonds with terminal nitrogens as well as showing a lone pair. Therefore, the azide ion’s molecular geometry is linear, and the hybridization around the central nitrogen atom is sp.