Question

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}^{-} .\) (a) Draw the Lewis structure of the azide ion that minizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central Natom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

Short answer

a) The Lewis structure of the azide ion is either N≡N=N⁻ or N⁻=N≡N, and its molecular geometry is linear. b) The hybridization of the central nitrogen atom in the azide ion is sp. c) The central nitrogen atom forms 2 sigma (σ) bonds and 2 pi (π) bonds in the azide ion.

Step-by-step solution

Calculate the total number of valence electrons

In the azide ion, N3-, there are three nitrogen atoms, and since nitrogen has 5 valence electrons, the total valence electrons are 3 * 5 = 15. Additionally, as there is a negative charge on the ion, one extra electron must be considered, resulting in a total of 16 valence electrons.

Draw the skeleton Lewis structure

Place the least electronegative atom (central nitrogen) in the center and connect the other two nitrogen atoms to it: N - N - N

Distribute the remaining valence electrons to minimize formal charge

In this step, we will place the remaining valence electrons on the nitrogen atoms as lone pairs to minimize formal charges while considering the octet rule. N≡N=N⁻ ↔ N⁻=N≡N The structure on the left has a formal charge of -1 on the left nitrogen atom, +1 on the central nitrogen, and 0 on the right nitrogen atom. The structure on the right has a formal charge of 0 on the left nitrogen atom, +1 on the central nitrogen, and -1 on the right nitrogen atom. As both structures have the same overall formal charge, either structure can represent the azide ion. So, the final Lewis structure for the azide ion is either: N≡N=N⁻ or N⁻=N≡N

Determine the molecular geometry

The central nitrogen atom has two atoms directly attached and no lone pairs. Hence, it is linear.

Determine the hybridization of the central nitrogen atom

The central nitrogen atom is involved in a single bond and a triple bond; therefore, it has three sigma bonds in total. According to the VSEPR theory, the hybridization on the central nitrogen atom is sp, as it has two electron domains.

Identify the number of sigma and pi bonds for the central nitrogen atom

Since the central nitrogen is involved in a single bond (1 sigma bond) and a triple bond (1 sigma and 2 pi bonds), there is a total of: -2 sigma (σ) bonds: 1 from the single bond and 1 from the triple bond -2 pi (π) bonds: These are the remaining 2 bonds from the triple bond To summarize: a) The Lewis structure of the azide ion is either N≡N=N⁻ or N⁻=N≡N, and its molecular geometry is linear. b) The hybridization of the central nitrogen atom in the azide ion is sp. c) The central nitrogen atom forms 2 sigma (σ) bonds and 2 pi (π) bonds in the azide ion.

Key concepts

Lewis Structure

Understanding the Lewis structure of a molecule is a foundation in studying its chemical behavior. In the Lewis structure, electrons are represented as dots, and bonds between atoms are represented by lines. For the azide ion, (N_3^{-}), the process of drawing the structure begins with accounting for the valence electrons. Nitrogen, with five valence electrons, contributes a total of fifteen for three of its atoms, and an additional electron comes from the negative charge, making up sixteen valence electrons.

To minimize formal charges, which helps in predicting the most stable structure, we have two resonance forms: N≡N=N⁻ and N⁻=N≡N. The negative charge can be on the terminal nitrogen atoms, not on the central nitrogen, to ensure each nitrogen atom has a complete octet. By fulfilling the octet rule and reducing formal charges, the Lewis structure gives a depiction of stable electron distribution in the azide ion.

Molecular Geometry

The molecular geometry is determined by the spatial arrangement of atoms in a molecule or ion. The VSEPR (Valence Shell Electron Pair Repulsion) theory is invaluable in predicting these structures. For the azide ion, we observe that the central nitrogen atom has two bonding domains and no lone pairs, leading to a linear shape. This linear geometry is typical for species with a hybridization of sp, where two electron domains are present. A linear structure also implies a bond angle of 180 degrees, a characteristic trait of molecules with a 'straight line' appearance, which includes the azide ion.

Sigma and Pi Bonds

In chemical bonding, we differentiate between sigma (σ) and pi (π) bonds. Sigma bonds are the first bonds formed between two atoms and involve head-on overlap of orbitals. Pi bonds, on the other hand, occur when parallel orbital overlap happens and always accompany a sigma bond in multiple bonds.

For the central nitrogen in the azide ion, there are two sigma bonds and two pi bonds. One sigma bond originates from the end-to-end overlap with each of the neighboring nitrogen atoms. The triple bond with one of the terminal nitrogen atoms consists of one sigma and two pi bonds. This configuration is vital for understanding the chemical reactivity and physical properties of the azide ion.

Nitrogen Hybridization

The concept of hybridization explains how atomic orbitals mix to form new hybrid orbitals, affecting both molecular geometry and bonding. For nitrogen, an element in the second period of the periodic table, sp hybridization occurs when one s orbital and one p orbital blend together. This results in two sp-hybrid orbitals.

In the azide ion's central nitrogen atom, sp hybridization is anticipated due to the combination of one single bond and one triple bond, leading to two regions of electron density—consistent with the linear geometry of the molecule. Understanding hybridization helps in predicting the electron distribution in a molecule, which in turn helps in anticipating molecular reactivity.