Question

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \mathrm{~A}\) (a) Draw a Lewis structure for the azide ion. (b) With reference to Table \(8.5\), is the observed \(\mathrm{N}-\mathrm{N}\) bond length consistent with your Lewis structure? (c) What hybridization scheme would you expect at each of the nitrogen atoms in \(\mathrm{N}_{3}^{-} ?\) (d) Show which hybridized and unhybridized orbitals are involved in the formation of \(\sigma\) and \(\pi\) bonds in \(\mathrm{N}_{3}^{-} .(\mathrm{e}) \mathrm{It}\) is often observed that \(\sigma\) bonds that involve an sp hybrid orbital are shorter than those that involve only \(s p^{2}\) or \(s p^{3}\) hybrid orbitals. Can you propose a reason for this? Is this observation applicable to the observed bond lengths in \(\mathrm{N}_{3}^{-}\) ?

Short answer

The azide ion, N₃⁻, has a Lewis structure of N≡N-N with a formal negative charge on the terminal nitrogen atom. The given bond length of 1.16 A is consistent with the presence of both triple and single N-N bonds. The terminal nitrogen atom has sp³ hybridization, while the central nitrogen atom has sp hybridization. The σ bond is formed by the overlap of hybrid orbitals, and the π bonds are formed by the overlap of unhybridized p orbitals. Although sp hybrid orbitals generally form shorter σ bonds due to their higher s character, the observed bond length in azide may be influenced by resonance and charge distribution.

Step-by-step solution

Drawing the Lewis structure of N3-

To draw the Lewis structure for N3-, first determine the total number of valence electrons. Each nitrogen atom has 5 valence electrons, so it will be a total of 15 electrons. Since the ion has a negative charge, add one more electron, making a total of 16 electrons. Start by connecting the three nitrogen atoms in a linear arrangement with single bonds. This uses up 4 electrons. Distribute the remaining 12 electrons as lone pairs around the nitrogen atoms, ensuring that each nitrogen satisfies the octet rule. The two terminal nitrogen atoms will have three lone pairs, while the central nitrogen atom will have one lone pair. Thus, the Lewis structure of N3- is: N≡N-N with a formal negative charge on the terminal nitrogen atom.

Checking the given N-N bond length with the Lewis structure

The given N-N bond length is 1.16 A. In the Lewis structure, there is a single bond and a triple bond between the nitrogen atoms. According to Table 8.5, the average bond length for a single N-N bond is 1.45 A, and for a triple N-N bond is 1.10 A. The observed bond length of 1.16 A is between these two values. So, the observed bond length is consistent with the Lewis structure.

Determining the hybridization scheme for each nitrogen atom

To determine the hybridization scheme, count the electron domains around each nitrogen atom (bonds + lone pairs). The terminal nitrogen atom has one triple bond (one electron domain) and three lone pairs (three electron domains), totaling four electron domains. It has sp³ hybridization. The central nitrogen atom has one triple bond (one electron domain) and one single bond (one electron domain), totaling two electron domains. It has sp hybridization.

Identifying orbitals involved in σ and π bond formation

In N3-, the σ bonds are formed by the overlap of hybrid orbitals. The terminal nitrogen atom forms one σ bond with its sp³ hybrid orbital overlapping with the sp hybrid orbital of the central nitrogen atom. The π bonds are formed by the sideways overlap of unhybridized p orbitals. Two pairs of unhybridized p orbitals from the terminal and central nitrogen atoms overlap to form two π bonds, making a total of three bonds between these atoms (one σ and two π bonds).

Propose a reason why sp hybrid orbitals form shorter σ bonds and if this applies to N3-

Sp hybrid orbitals have more s character (50% s character) than sp² (33.3% s character) or sp³ hybrid orbitals (25% s character). Since s orbitals are more close to the nucleus than p orbitals, the more s character an orbital has, the closer it is to the nucleus, and the shorter the bond it forms. This suggests that sp hybrid orbitals form shorter σ bonds than sp² or sp³ hybrid orbitals. However, in azide ion N3-, the observed N-N bond length is 1.16 A, which is not shorter than the average bond lengths of single and triple N-N bonds. This observation may not be directly applicable to the azide ion, as the effects of resonance and charge distribution may also play a role in its bond lengths.

Key concepts

Hybridization

Understanding hybridization is crucial for predicting molecular shapes and ensuring accurate Lewis structures. In the azide ion \( \text{N}_3^- \), hybridization occurs to allow the nitrogen atoms to form appropriate bonds. When atoms hybridize, they mix atomic orbitals (s and p orbitals, primarily) to create new hybrid orbitals. These hybrid orbitals accommodate bonded and lone pair electrons.

In the azide ion, consider each nitrogen atom separately:

• The central nitrogen in \( \text{N}_3^- \) employs \( sp \) hybridization. This creates two hybrid orbitals, aligning with the bond formation in a linear fashion.
• The terminal nitrogen atoms use \( sp^3 \) hybridization, although this is specific to accommodating lone pairs, and they participate in multiple bond types extending their bonding functionality.

Understanding these variations in hybridization aids in visualizing the linear shape and specific bonding interactions in azide.

Sigma and Pi Bonds

To grasp the structure of the azide ion, it's vital to differentiate between sigma (\( \sigma \)) and pi (\( \pi \)) bonds, which are key to chemical bonding theories.

• \( \sigma \) bonds are formed through the head-on overlapping of hybrid orbitals. In \( \text{N}_3^- \), for instance, a distal nitrogen atom using an \( sp^3 \) hybrid orbital overlaps directly with the \( sp \) hybrid orbital of the central nitrogen, forming a strong \( \sigma \) bond.
• \( \pi \) bonds, conversely, arise from the side-by-side overlap of unhybridized p-orbitals found in the nitrogen atoms. In the azide ion, \( \pi \) bonds form part of the multiple bonds between nitrogen atoms, specifically in the arrangement of one \( \sigma \) and two \( \pi \) bonds.

The combination of these bonds contributes to the azide's structural stability and remarkable chemical characteristics.

Azide Ion

The azide ion, \( \text{N}_3^- \), is a fascinating molecular ion with a linear shape characterized by its unique bonding and resonance structures. Azides are generally known for their explosive properties and are used in chemical synthesis and applications such as detonators.

In chemistry, the azide ion is crucial for understanding resonance and stability in bonding. The ion is a linear system composed of three nitrogen atoms. Its electronic structure must maintain a balance of charges, and often features a resonance hybrid where electrons are delocalized across the nitrogen atoms. For the azide ion, the best Lewis structure doesn’t emit a unique form but rather a blend that explains its equidistant \( \text{N}-\text{N} \) bonds. The assignment of formal charges further defines its reactivity and stability. The overall negative charge is settled on one of the terminal nitrogens in its resonance forms, affecting how it interacts with other ions and molecules.

Nitrogen Bond Lengths

Nitrogen-nitrogen bond lengths in the azide ion are pivotal to interpreting its chemical nature. Each bond's length gives insights into the atom interactions and stability within the molecule.

The azide ion has two \( \text{N}-\text{N} \) bonds, both uniformly measured at 1.16 Å. This distance is significant in providing clues about resonance effects and hybridization.

• This bond length is neither purely single nor triple; instead, it indicates resonance, where electron density is spread across the \( \text{N}_3^- \) framework, leading to bond equalization.
• The resonance forms of \( \text{N}_3^- \) suggest partial \( \sigma \) and \( \pi \) character blending, accounting for a shared bond length disparate from isolated single or triple bonds.

This understanding is critical in predicting its chemical behavior and interactions. Resonance, hybridization, and molecular geometry collectively affect these measured bond lengths, showcasing the intricate balance within azide ion structures.