Question

The azide ion, \(\mathrm{N}_{3}^{-}\), is linear with two \(\mathrm{N}-\mathrm{N}\) bonds of equal length, \(1.16 \AA\). (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 \(\mathrm{N}_{3}^{-}\). (e) It is often observed that \(\sigma\) bonds that involve an \(s p\) 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 the observed N-N bond length of 1.16 Å being consistent with a double bond. The central Nitrogen is hybridized as \(sp\) and terminal Nitrogens as \(sp^2\). Sigma bonds involve the overlap of the central Nitrogen's \(sp\) orbitals and the terminal Nitrogens' \(sp^2\) orbitals, while pi bonds involve the unhybridized \(p\) orbitals. The bond lengths cannot be directly compared to the influence of \(sp\) versus \(sp^2\) or \(sp^3\) hybridization, as every bond involves an \(sp^2\) orbital from the terminal Nitrogen and an \(sp\) orbital from the central Nitrogen.

Step-by-step solution

Drawing the Lewis Structure

To determine the Lewis structure, take the total number of valence electrons from the atoms and the ion charge into consideration. Nitrogen has 5 valence electrons and there are 3 Nitrogens, so the total is 15. The azide ion carries an extra electron (because of its -1 charge), bringing the total to 16 electrons. Start by connecting all atoms with single bonds. This uses 4 electrons (2 per bond), leaving 12. Then, fill the octet of the terminal Nitrogens, which uses the remaining 12, leaving 0. The resulting structure is \( N = N - N^- \), it does not have formal charge.

Checking Bond Length Consistency

Table 8.5 depicts commonly found bond lengths. The bond length observed is 1.16 Å, which is expected for a N-N double bond. It means that one of the N-N bonds in the azide ion is consistent for a double bond as per the Lewis structure.

Identifying the Hybridization Scheme

A Nitrogen atom in a double bond usually has a hybridization of \(sp^2\). The central Nitrogen atom forms two sigma bonds with the surrounding Nitrogens and does not have any lone pairs. Therefore, it should be \(sp\). The terminal Nitrogens both have one sigma bond to the central Nitrogen and two lone pairs. According to that, both of these Nitrogens should be \(sp^2\) hybridized.

Identifying Orbitals involved in Sigma and Pi bonds

The sigma bonds in the molecule are formed by head-on overlap of the hybrid orbitals of the Nitrogens. Therefore, the sigma bonds involve the \(sp\) hybrids of the central Nitrogen and the \(sp^2\) hybrids of the terminal Nitrogens. The pi bonds are formed by the unhybridized \(p\) orbitals of the central and terminal Nitrogens, and they overlap sideways.

Evaluating Bond Lengths influenced by Orbital Types

\(sp\) hybrid orbitals have more \(s\) character, which means the electrons are held closer to the nucleus. As a result, \(sp\) hybrids form stronger and shorter bonds compared to \(sp^2\) or \(sp^3\). However, every bond in the azide ion involves an \(sp^2\) orbital from the terminal Nitrogen and an \(sp\) orbital from the central Nitrogen. Thus, the influence of \(sp\) versus \(sp^2\) or \(sp^3\) hybridization on bond length is not directly applicable here.

Key concepts

Lewis Structure

Understanding the Lewis structure is a foundation for predicting the molecular geometry and reactivity of a molecule. In the case of the azide ion (3]), the Lewis structure is drawn by accounting for the total number of valence electrons. Nitrogen, with 5 valence electrons, adds up to 15 for the three nitrogen atoms, plus an extra electron for the negative charge, totaling 16.

The Lewis structure shows each nitrogen atom bonded sequentially with the central nitrogen forming two bonds and carrying no formal charges. This linear arrangement reflects the actual geometry of the azide ion and indicates that we have a resonance structure, where the double bond character is distributed between the two terminal nitrogens and the central one.

Bond Length

Bond length is the distance between the nuclei of two bonded atoms. For the azide ion, both N-N bonds are equal in length, measured at 1.16 Å, typical for a N-N double bond. As per the Lewis structure, this bond length suggests partial double bond character due to resonance structures. In other words, although we draw single and double bonds in a Lewis structure, the actual molecule has an equal distribution of double bond character across both bonds.

The observed bond length implies that the electron density is shared between the three atoms in a manner that the two terminal nitrogen atoms do not entirely possess a full double bond, but rather, a bond order close to 1.5, consistent with the resonating structure of the azide ion.

Hybridization

Hybridization is a concept that describes the merging of atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in molecules. The azide ion shows variation in hybridization among its nitrogen atoms. The central nitrogen atom employs an hybridization, facilitating the formation of two sigma bonds with no lone pairs. On the other hand, the terminal nitrogens exhibit 2 hybridization each, having one sigma bond and two lone pairs.

Interestingly, the lone pairs on the terminal nitrogen atoms do not participate in bond formation, but they affect the molecule's shape, keeping it linear. Hybridization is a crucial factor in understanding the arrangement of atoms and the type of bonds (sigma and pi) that will be present in a given molecule.

Sigma and Pi Bonds

The sigma () and pi () bonds are types of covalent bonds. Sigma bonds are characterized by the head-on overlap of orbitals, which allows them to be rotationally symmetric around the bond axis. In 3]^-, the sigma bonds are formed by the overlap of the hybrid orbital of the central nitrogen with the 2 hybrid orbitals of the terminal nitrogens.

Pi bonds, on the other hand, are formed due to the sideways overlap of p orbitals, and they are found in areas above and below the sigma bond axis. Within the azide ion, the pi bond forms from the overlap of unhybridized p orbitals on both the central and the terminal nitrogen atoms. The combination of sigma and pi bonds gives azide its unique linear structure and contributes to the overall bond length consistency observed.