Problem 100 - Chemistry: The Central Science in SI Units, Global Edition

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 minimizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central \(\mathrm{N}\) atom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

Step-by-Step Solution

Answer

The azide ion has a linear shape with a Lewis structure of \[ \dddot{N} \equiv \stackrel{.}{\ddot N} - \stackrel{.}{\ddot N} \]. The hybridization of the central nitrogen atom is sp, and it forms 2 sigma bonds (σ) and 2 pi bonds (π).

Step 1: Draw the Lewis Structure of the Azide Ion

To draw the Lewis structure, start by considering the number of valence electrons. Nitrogen has 5 valence electrons, so three nitrogen atoms contribute a total of 15 valence electrons (5 × 3 = 15). In the azide ion, two of the nitrogen atoms are bonded through a single bond, and the central nitrogen atom is bonded to the other terminal nitrogen atom via a triple bond: N ≡ N - N To show this linear configuration, we draw the azide ion as: \[ \dddot{N} \equiv \stackrel{.}{\ddot N} - \stackrel{.}{\ddot N} \] All three nitrogen atoms have a formal charge of zero, which minimizes the formal charge.

Step 2: Determine Linear or Bent Shape

Since there are no lone pair electrons on the central nitrogen atom and it has two bonding areas (one single bonded terminal nitrogen and one triple bonded terminal nitrogen), the azide ion has a linear shape.

Step 3: Identify Hybridization of Central Nitrogen Atom

To determine the hybridization of the central nitrogen atom, count its bonding regions and lone pairs. The central nitrogen atom has two bonding regions and no lone pairs. For two effective electron pairs around the central atom, the hybridization will be sp, as it includes one s and one p orbital. So, the hybridization of the central nitrogen atom in the azide ion is sp.

Step 4: Calculate Sigma and Pi Bonds

With the azide ion's Lewis structure, we can now determine the number of sigma (σ) bonds and pi (π) bonds made by the central nitrogen atom. For each single bond, there is one sigma bond, and for every double or triple bond, there is only one sigma bond. In the azide ion, the central nitrogen atom is connected with one single bond and one triple bond. Thus, it has 1 single bond (1 σ bond) and 1 triple bond (1 σ bond + 2 π bonds). Therefore, the central nitrogen atom in the azide ion makes 2 sigma bonds (σ) and 2 pi-bonds (π).

Key Concepts

Formal Charge

When considering the Lewis structure of a molecule, one of the key elements is the formal charge. The formal charge helps determine the most stable Lewis structure by indicating the charge each atom would have if all the atoms in the molecule were sharing electrons equally. To calculate the formal charge, use the formula:

• Formal charge = (Valence electrons of the atom) – (Non-bonding electrons) – (1/2 * Bonding electrons)

In the case of the azide ion \( ext{N}_3^- \), each nitrogen atom strives for a formal charge of zero to achieve stability.

• The first nitrogen atom in the azide structure is attached to one triple bond, incorporating no lone pairs, and it maintains a formal charge of zero.
• The central nitrogen, also achieving a formal charge of zero, participates in one single and one triple bond with the other nitrogen atoms.
• The third nitrogen with a single bond also aims for a formal charge of zero by having appropriate lone pairs.

Minimizing the formal charges is crucial for determining the most favorable structure, which in this case, results in a linear molecular geometry.

Hybridization

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals. These new orbitals are designed to accommodate electron pairs in bond formation, essentially explaining molecular geometry.
For the central nitrogen atom in the azide ion, which is represented as \(N\equiv N-N\), the hybridization is sp.

• This sp hybridization arises because the central nitrogen atom has two regions of electron density around it—one associated with the triple bond and one with the single bond.
• To accommodate these two bonding regions, the central nitrogen atom uses one s orbital and one p orbital to form two sp hybrid orbitals.

Sp hybridization results in a linear structure as the angle between the hybrid orbitals is 180 degrees. This linearity aligns with the actual structure of the azide ion, demonstrating how hybridization helps predict molecular shape.

Sigma and Pi Bonds

When examining the covalent bonds within a molecule, they are typically characterized as either sigma \((\sigma)\) or pi \((\pi)\) bonds, based on their formation.

• Sigma bonds occur when orbitals overlap head-to-head. These bonds are strong and allow free rotation of bonded atoms. Every single bond contains one sigma bond.
• Pi bonds, by contrast, are formed when orbitals overlap side-to-side. These are weaker than sigma bonds and restrict rotation due to their electron cloud placement above and below the bond axis.

In the azide ion \( (\text{N}_3^-) \):

• The central nitrogen atom engages in a single bond with the terminal nitrogen, which includes one sigma bond.
• In addition to the single bond, it participates in a triple bond with the other terminal nitrogen atom, consisting of one sigma bond and two pi bonds.
• Overall, the central nitrogen atom forms two sigma bonds and two pi bonds, illustrating how its bonding arrangement supports the ion's stability and linear structure.