Problem 100 - Chemistry The Central Science In Si Units

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

(a) Linear; (b) \( sp \); (c) Two \( \sigma \) bonds and two \( \pi \) bonds.

Step 1: Drawing the Lewis Structure

To draw the Lewis structure of the azide ion \( \mathrm{N}_{3}^{-} \), start by counting the total number of valence electrons. Nitrogen has 5 valence electrons, and there are 3 nitrogen atoms, giving us \(5 \times 3 = 15\) electrons. Since the ion has a \(-1\) charge, we add one more electron, resulting in a total of 16 electrons. Create a linear arrangement of the nitrogen atoms (it does not form a triangle) and place the electrons:1. Place a nitrogen atom in the center with an adjacent nitrogen atom on each side.2. Distribute electrons to satisfy the octet rule, minimizing formal charges. The structure that minimizes formal charge has a double bond between each terminal nitrogen and the central nitrogen, and one additional lone pair of electrons on one terminal nitrogen. Assign formal charges, resulting in \(-1\) on one of the terminal nitrogen atoms and 0 on the others.

Step 2: Determine Molecular Geometry

The Lewis structure shows that the molecule is linear. All the nitrogen atoms are arranged in a straight line, and there are no lone pairs on the central nitrogen that would push the atoms into a bent shape. Thus, the azide ion is linear.

Step 3: Hybridization of the Central N Atom

To determine the hybridization of the central nitrogen atom, look at its bonding and electron geometry. The central nitrogen is forming two double bonds, which means it is involved in two \( \sigma \) bonds and contributes to two \( \pi \) bonds. This indicates \( sp \) hybridization because two hybrid orbitals are needed to form two \( \sigma \) bonds, and these orbitals come from one \( s \) orbital and one \( p \) orbital.

Step 4: Count \( \sigma \) and \( \pi \) Bonds

In the azide ion, the central nitrogen makes two \( \sigma \) bonds, each to one of the terminal nitrogen atoms. It also shares one \( \pi \) bond with each terminal nitrogen atom. Therefore, the central nitrogen makes two \( \sigma \) bonds and two \( \pi \) bonds.

Key Concepts

Formal Charge

Formal charge is a concept used to determine the distribution of electrons in a molecule and helps us identify the most stable Lewis structure. It helps us in predicting the most likely arrangement of atoms in a molecule. To calculate the formal charge for an atom:

• Start with the number of valence electrons in the isolated atom.
• Subtract the number of non-bonding electrons (lone pairs).
• Subtract half the number of bonding electrons (those shared in bonds).

For nitrogen in the azide ion (\(N_3^-\)), we can calculate formal charges to ensure that one nitrogen carries a \(-1\) charge and the others carry a charge of \(0\). This arrangement minimizes the formal charges across the molecule, leading to greater stability. Remember, the sum of all formal charges must equal the overall charge of the ion.

Hybridization

Hybridization is a concept that explains the mixing of atomic orbitals in order to form new hybrid orbitals. These hybrids are essential for forming chemical bonds and define the geometry of molecules. In the azide ion, the central nitrogen exhibits \(sp\) hybridization. This type of hybridization involves the mixing of one \(s\) orbital and one \(p\) orbital.

In \(sp\) hybridization, the two new orbitals are used to form \(\sigma\) bonds. The linear geometry of the azide ion is a direct consequence of this kind of hybridization. The other \(p\) orbitals remain unhybridized and participate in \(\pi\) bonding, which is crucial for the double bonds formed in the ion.

Molecular Geometry

Molecular geometry is the three-dimensional arrangement of atoms in a molecule. It determines many physical and chemical properties. In the azide ion (\(N_3^-\)), the molecular geometry is linear. This is because there are no lone pairs on the central nitrogen atom that can distort the shape by repelling bond pairs.

The azide ion adopts a linear geometry due to its \(sp\) hybridization, which perfectly aligns the participating hybrid orbitals into a straight line. This linear alignment is typical for molecules that exhibit \(sp\) hybridization, resulting in bond angles of \(180^\circ\) between the atoms.

Sigma Bonds

Sigma (\(\sigma\)) bonds are the strongest type of covalent chemical bonds. They form when electron orbitals overlap linearly, head-to-head. In the azide ion, the central nitrogen atom forms two \(\sigma\) bonds.

• These bonds occur between the central nitrogen and each terminal nitrogen.
• \(\sigma\) bonds are characterized by free rotation around the bond axis and are generally stronger than \(\pi\) bonds.

The \(\sigma\) bonds in the azide ion help maintain a stable linear structure by providing a strong bonding framework. The central nitrogen’s \(sp\) orbitals directly overlap with the \(p\) orbitals of the terminal nitrogen atoms to form these bonds.

Pi Bonds

Pi (\(\pi\)) bonds add further bonding between atoms and result from the side-by-side overlap of unhybridized \(p\) orbitals. In the azide ion, the central nitrogen is involved in forming two \(\pi\) bonds, one with each terminal nitrogen.

• The existence of these \(\pi\) bonds contributes to the presence of double bonds between the nitrogen atoms.
• \(\pi\) bonds prevent rotation around the bond axis, making them less flexible than \(\sigma\) bonds.

The formation of \(\pi\) bonds is fundamental for achieving the required bonding and electron sharing in certain configurations, such as in aromatic compounds and acids. However, in the azide ion, they play a pivotal role in maintaining its linear structure through additional electron sharing.