Problem 64 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) Write a single Lewis structure for \(\mathrm{N}_{2} \mathrm{O},\) and determine the hybridization of the central \(\mathrm{N}\) atom. (b) Are there other possible Lewis structures for the molecule? (c) Would you expect \(\mathrm{N}_{2} \mathrm{O}\) to exhibit delocalized \(\pi\) bonding?

Step-by-Step Solution

Answer

(a) The single Lewis structure for N2O is O = N = N, with the central N atom having an sp hybridization. (b) There is another possible Lewis structure (N - N ≡ O), but it is less stable due to formal charges, so O = N = N is more favorable. (c) N2O does not exhibit delocalized π bonding as there are no resonance structures involving π bond rearrangement.

Step 1: Determine the total number of valence electrons

First, we need to count the total number of valence electrons in the N2O molecule. Nitrogen (N) has 5 valence electrons, and Oxygen (O) has 6 valence electrons. So, the total number of valence electrons in N2O is 5 + 5 + 6 = 16.

Step 2: Write a single Lewis structure for N2O

Now, we can draw a single Lewis structure for N2O: 1. Connect the atoms with a single bond. 2. Complete the octets of the outer atoms (Oxygen in this case) by adding lone pairs. 3. Place any remaining electrons on the central atom (Nitrogen in this case). By following these steps, we get the following Lewis structure for N2O: O = N = N

Step 3: Determine the hybridization of the central N atom

To determine the hybridization of an atom, we use the formula: Hybridization = (Number of Bonding Electron Regions + Number of Lone Pairs) In the N2O Lewis structure that we drew, the central N atom has two bonding electron regions and no lone pairs. Therefore, the hybridization of the central N atom is: Hybridization = (2 + 0) = 2 Since the number 2 corresponds to sp hybridization, the hybridization of the central N atom is sp.

Step 4: Discuss possible alternative Lewis structures

There is another possible Lewis structure for N2O: N - N ≡ O However, this structure would result in a less stable molecule due to the positive formal charge on the central N atom and a negative formal charge on the terminal N atom. Therefore, the structure O = N = N is more favorable.

Step 5: Determine if N2O exhibits delocalized π bonding

A molecule exhibits delocalized π bonding when it has multiple resonance structures involving π bond rearrangement. In the case of N2O, there is no resonance due to the lack of alternative stable structures involving π bond rearrangement. In both O = N = N and N - N ≡ O structures, the π bonds do not move between different atoms. Therefore, N2O does not exhibit delocalized π bonding.

Key Concepts

Valence Electrons

Valence electrons are the electrons found in the outermost shell of an atom. They play a critical role in chemical bonding and determining how atoms interact to form molecules. For many students, understanding how to count valence electrons accurately is crucial when drawing Lewis structures. Each element on the periodic table has a specific number of valence electrons based on its group number. For example:

• Nitrogen (N), which belongs to Group 15, has 5 valence electrons.
• Oxygen (O), from Group 16, has 6 valence electrons.

When creating Lewis structures, first calculate the total number of valence electrons available for bonding. For \\( ext{N}_2 ext{O}\), you add:

• 5 electrons from each nitrogen atom (since there are two nitrogen atoms, this gives \(5 \times 2 = 10\)).
• 6 electrons from the oxygen atom.

The total valence electrons for \\( ext{N}_2 ext{O}\) is thus 16 (\(10 + 6 = 16\)). With this total, you can proceed to arrange electrons in bonds and lone pairs to complete the octets, ensuring that each atom achieves a stable electronic arrangement.

Hybridization

Hybridization is a concept that describes the mixing of atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds. This concept is crucial to understanding the geometry of molecules. In simpler terms, it explains how atoms, especially central atoms in a molecule, arrange their electrons to create optimal geometry for bonding. For the molecule \\( ext{N}_2 ext{O}\), the focus is on the central nitrogen atom:

• Assess the number of electron regions around the central atom considering both bonding and lone pairs.
• Use the formula: Hybridization = Number of Bonding Regions + Number of Lone Pairs.

In \\( ext{N}_2 ext{O}\), the central nitrogen atom is surrounded by two bonding regions (one double bond with oxygen and one with the other nitrogen) and zero lone pairs. Hence, the calculation gives:

• Hybridization = \(2 + 0 = 2\).

A value of 2 corresponds to \(sp\) hybridization, indicating that the central nitrogen uses one \(s\) and one \(p\) orbital to form two linear hybrid orbitals.

Delocalized π Bonding

Delocalized \(\pi\) bonding involves the spreading of \(\pi\) electrons across adjacent atoms in a molecule, rather than being confined to a specific atom or bond. This gives rise to resonance, which enhances stability as electrons are essentially shared among multiple structures. Typically, resonance structures are different possible arrangements of \(\pi\) electrons that maintain the fixed positions of atoms. However, for \\( ext{N}_2 ext{O}\), delocalized \(\pi\) bonding is not present. Here's why:

• In the primary structure, \(O = N = N\), the \(\pi\) bonds are fixed and do not transfer between atoms.
• Even the alternative Lewis structure \(N - N \equiv O\), while chemically plausible, remains less stable due to unwelcome charges on atoms.
• Without additional stable resonance forms involving \(\pi\) electrons, the delocalization that typically implies multiple resonance structures does not occur.

Thus, the \\( ext{N}_2 ext{O}\) molecule lacks the flexible \(\pi\) electron distribution necessary for delocalization, rendering any potential resonance unstably ineffective.