Question

The azide ion, \(\left[\mathrm{N}_{3}\right]^{-},\) is linear with equal \(\mathrm{N}-\mathrm{N}\) bond lengths. Give a description of the bonding in \(\left[\mathrm{N}_{3}\right]^{-}\) in terms of valence bond theory.

Short answer

In azide ion, nitrogen atoms undergo \(sp\) hybridization, forming sigma bonds, with resonance giving delocalized pi bonds across the molecule.

Step-by-step solution

Understand the structure of azide ion

The azide ion, \([\mathrm{N}_{3}^-]\), has a linear structure. It consists of three nitrogen atoms aligned in a straight line with equal bond lengths between each adjacent pair of nitrogen atoms. This symmetry suggests resonance in its bonding.

Determine the hybridization of nitrogen atoms

In valence bond theory, we consider the hybridization of atomic orbitals to describe bonding. Each nitrogen atom in the azide ion undergoes \(sp\) hybridization because this allows the linear shape and formation of two sigma bonds – one with each adjacent nitrogen.

Describe the sigma bond formation

The \(sp\) hybrid orbitals of the nitrogen atoms overlap to form sigma bonds. There is a sigma bond between the first nitrogen and the central nitrogen, and another sigma bond between the central nitrogen and the third nitrogen atom.

Discuss the pi bond formation through pi-orbital overlap

Aside from sigma bonds, valence bond theory suggests that \(p\)-orbitals that are not involved in hybridization can overlap laterally to form pi bonds. The alignment and overlap of unhybridized \(p\) orbitals on each nitrogen allow for the formation of delocalized pi bonds over the molecule.

Explain resonance

The equal \(\mathrm{N}-\mathrm{N}\) bond lengths and linear configuration indicate resonance. Multiple resonant structures can be drawn where pi-electrons are delocalized, further stabilizing the ion by distributing electron density evenly among the nitrogen atoms.

Key concepts

Hybridization

Hybridization is a core concept in valence bond theory that helps explain the geometry and bonding in molecules. In the case of the azide ion \(_3^-\), each nitrogen atom undergoes \(sp\) hybridization. Hybridization involves the mixing of atomic orbitals to create new orbitals that are suitable for forming bonds.
The process of \(sp\) hybridization combines one s orbital and one p orbital from a nitrogen atom to form two equivalent sp hybrid orbitals. These hybrid orbitals are arranged linearly. This is why the azide ion is linear and can effectively form two sigma bonds, one connecting each adjacent nitrogen atom.
This type of hybridization is particularly conducive to maintaining the linear structure of the azide ion, allowing it to have equal \(N-N\) bond lengths and resembling a straight line.

Sigma Bonds

Sigma bonds are the strongest type of covalent chemical bond and form via the direct overlap of orbitals. Within the azide ion \(_3^-\), sigma bonds allow for stable connections between atoms. These sigma bonds are formed through the overlap of \(sp\) hybrid orbitals from the nitrogen atoms.
The geometry dictated by this overlap maintains the linear arrangement of the azide ion. The direct end-to-end overlap provides a strong, stable framework that ensures equal bond lengths between the nitrogen atoms, helping in maintaining resonance.
This alignment results in the azide ion having two sigma bonds, contributing significantly to its linear and symmetrical structure, while playing a pivotal role in maintaining the integrity of the molecule.

Pi Bonds

Pi bonds arise from the lateral overlap of unhybridized p orbitals. In the azide ion \(\text{N}_3^-\), pi bonds form after the initial sigma bonds have been established with the help of \(sp\) hybridization.
Non-hybridized p orbitals on each nitrogen atom can laterally overlap with those on neighboring nitrogen atoms to create pi bonds. These pi bonds are crucial for the concept of resonance, as they allow for electron delocalization.

• The lateral p orbital overlap leads to the delocalization of electrons.
• The presence of pi bonds increases the stability of the azide ion by spreading out the electron density.

A key feature is that these pi bonds are not fixed between two atoms but can extend across the molecule, contributing to the resonance observed in the azide ion.

Resonance

Resonance is an important concept that helps illustrate the stability and symmetry of molecules like the azide ion \(\text{N}_3^-\). Despite the ion having a linear structure, the bonding can be represented by multiple resonance structures.
In these structures, the pi bonds are depicted as being delocalized across the nitrogen atoms.

• Multiple resonance structures show the movement of electrons between different nitrogen atoms.
• This delocalization helps in stabilizing the azide ion by evenly distributing electronic charge.

As a result, the azide ion does not have fixed double or single bonds but rather equivalent bonds across the structure. This is why all \(N-N\) bond lengths are equal, offering additional stabilization to the ion's structure.

Azide Ion

The azide ion \(\text{N}_3^-\) is a fascinating molecular ion that consists of three nitrogen atoms connected in a straightforward, linear fashion. To understand this ion using valence bond theory, it’s important to consider several key bonding aspects.
As covered, various concepts such as hybridization, sigma bonds, pi bonds, and resonance play significant roles in understanding the azide ion's structure:

• The linear arrangement is a result of \(sp\) hybridization, allowing for even bond dispersion.
• The shared equal-length bonds are due to resonance, where pi electrons are delocalized across the entire ion.

Such features amalgamate to give the azide ion a stable and symmetric configuration, crucial for the chemistry applications it participates in, including its use as a precursor in various organic and inorganic syntheses.