Problem 68 - Chemistry The Science in Context

Question

The Lewis structure of \(\mathrm{N}_{4} \mathrm{O},\) with the skeletal structure \(\mathrm{O}-\mathrm{N}-\mathrm{N}-\mathrm{N}-\mathrm{N},\) contains one \(\mathrm{N}-\mathrm{N}\) single bond, one \(\mathrm{N}=\mathrm{N}\) double bond, and a \(\mathrm{N} \equiv \mathrm{N}\) triple bond. Is the hybridization of all the nitrogen atoms the same?

Step-by-Step Solution

Answer

Answer: The hybridization states of the nitrogen atoms in N4O are as follows: N1 is sp² hybridized, while N2, N3, and N4 are all sp hybridized.

Step 1: Draw the Lewis structure

Start by drawing the Lewis structure of the N4O molecule, based on the given skeletal structure O-N-N-N-N, which includes one N-N single bond, one N=N double bond, and one N≡N triple bond. Make sure to include lone pairs of electrons on atoms where needed to complete their octets.

Step 2: Identify the number of sigma bonds and lone pairs around each nitrogen atom

For each nitrogen atom, identify the number of sigma bonds and lone pairs surrounding it in the structure. - N1: has one sigma bond with O, one sigma bond with N2, and one lone pair. - N2: has one sigma bond with N1, one sigma bond and one pi bond with N3 (forming the N=N double bond). - N3: has one sigma bond and one pi bond with N2, and one sigma bond and two pi bonds with N4 (forming the N≡N triple bond). - N4: has one sigma bond and two pi bonds with N3 (forming the N≡N triple bond), and one lone pair.

Step 3: Determine the hybridization states

Based on the number of sigma bonds and lone pairs around each nitrogen atom, apply the hybridization rules to find the hybridization state of each atom. - N1: 2 sigma bonds + 1 lone pair = 3 electron domains → sp² hybridization. - N2: 2 sigma bonds + 0 lone pairs = 2 electron domains → sp hybridization. - N3: 2 sigma bonds + 0 lone pairs = 2 electron domains → sp hybridization. - N4: 1 sigma bond + 1 lone pair = 2 electron domains → sp hybridization. So, the hybridization states of nitrogen atoms in the N4O molecule are not the same. Only the first nitrogen atom (N1) has a different hybridization state (sp²) compared to the other three nitrogen atoms (N2, N3, and N4), which all are sp hybridized.

Key Concepts

Chemical Bonding

Chemical bonding is the force that holds atoms together in molecules. Atoms bond to achieve a more stable electron configuration, which usually involves filling their outer shell with the maximum number of electrons, known as an octet. There are three primary types of chemical bonds:

• Ionic Bonds: Formed when electrons are transferred from one atom to another, resulting in positively and negatively charged ions that attract each other.
• Covalent Bonds: Occur when atoms share pairs of electrons. They can be single, double, or triple bonds, with single bonds sharing one pair of electrons, double bonds sharing two pairs, and triple bonds sharing three pairs.
• Metallic Bonds: These are found in metals, where electrons are free to move around a lattice of positively charged metal ions, creating a strong bond.

In the case of the N4O molecule from our exercise, the different types of N-N bonds (single, double, and triple) are examples of covalent bonding, where the nitrogen atoms share electrons to achieve stable electron configurations. The type of bond affects the molecule's properties and the hybridization state of the atoms involved.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The spatial configuration is determined by the number of electron pairs (bonding and lone pairs) surrounding the central atom, and the shape is crucial for understanding the molecule's physical properties and reactivity.The VSEPR (Valence Shell Electron Pair Repulsion) theory is used to predict molecular geometry. This theory states that electron pairs around a central atom tend to repel each other and will therefore arrange themselves as far apart as possible to minimize repulsion. This arrangement dictates the molecule's geometry.For instance, in the exercise, the nitrogen molecules adopt different hybridizations, which influence the molecular geometry around them. The first nitrogen (N1), with sp² hybridization, has a trigonal planar geometry, while the others (N2, N3, N4), with sp hybridization, have linear geometries due to the number and types of electron domains (sigma bonds and lone pairs) they have.

Electron Pair Repulsion Theory

The Electron Pair Repulsion Theory, often known as VSEPR theory, plays a pivotal role in determining the shape of molecules. The fundamental premise of this theory is that electron pairs, both bonding and non-bonding, will repel each other and, as a result, prefer to be as far apart as possible. This repulsion impacts molecular geometry by establishing the angles between adjacent bonds.In our exercise illustrating the N4O molecule, we must consider the electron pairs when deciding on the shape of the molecule. For example, the first nitrogen atom (N1) has a lone pair of electrons in addition to two sigma bonds, leading to a bent shape due to the repulsion between the electron pairs. The remaining nitrogen atoms (N2, N3, and N4) do not have lone pairs and form linear shapes, displaying the influence of electron pair repulsion on the molecular geometry.By applying the principles of the VSEPR theory, we gain insights into the molecular structures that significantly affect the chemical behavior and properties of substances. Understanding how these repulsions dictate the final arrangement of atoms in a molecule is critical for students venturing into the study of chemical bonding and molecular geometry.