Question

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in $$: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}$$ be described in terms of orbitals?

Short answer

The correct arrangement for the $\mathrm{N}_{2} \mathrm{O}$ molecule is NNO, as it results in a stable molecule with zero formal charges. The Lewis structure is N≡N-O with no resonance forms, and formal charges of N (0), N (0), and O (0). The central nitrogen atom has sp2 hybridization. The triple bond between the nitrogen atoms consists of one sigma (σ) bond formed by the overlap of sp2 hybrid orbitals and two pi (π) bonds formed by the side-by-side overlap of unhybridized p orbitals.

Step-by-step solution

Determine the correct arrangement of the N2O molecule

To decide whether the arrangement is NNO or NON, we must check which arrangement results in a polar molecule, since the problem tells us that the molecule is linear and polar.

Check polarity of the arrangements

For the NNO arrangement: Since N has a higher electronegativity than O (3.44 for N and 3.44 for O), the N-O bond will be polar with the electron density being pulled towards the N atom. Overall, this arrangement results in a polar molecule. For the NON arrangement: The molecule has a central N atom, so the N-O bond would be polar in the opposite direction compared to the N-N bond (nonpolar). This results in a net polarity in the molecule. Since both arrangements result in a polar molecule, we need to check other factors to determine the correct arrangement.

Check formal charges and stability of the molecule

For the NNO arrangement: There would be a triple bond between N atoms and a single bond between N and O atoms. Formal charge on each atom: N (0), N (0), O (0). This arrangement results in a stable molecule with zero formal charges. For the NON arrangement: There would be a single bond between N and O atoms and a single bond between N and N atoms. Formal charge on each atom: N (1-), N (1+), O (0). This arrangement results in an unstable molecule with nonzero formal charges. Since the NNO arrangement leads to a stable molecule with zero formal charges, it is the correct arrangement.

Draw the Lewis structure and provide the formal charges

The correct Lewis structure for \(\mathrm{N}_{2} \mathrm{O}\) is N≡N-O, and there are no resonance forms. Formal charges for each atom are N (0), N (0), and O (0).

Determine the hybridization of the central atom

In the N≡N-O structure, the central nitrogen atom is connected to two other atoms (N and O) and has one lone pair of electrons, making a total of 3 hybrid orbitals. This corresponds to sp2 hybridization for the central nitrogen atom.

Describe the multiple bonding in terms of orbitals

The triple bond between the two nitrogen atoms consists of one sigma (σ) bond and two pi (π) bonds. The σ bond is formed by the overlap of sp2 hybrid orbitals from both nitrogen atoms, while the two π bonds are formed by the side-by-side overlap of unhybridized p orbitals from each nitrogen atom.

Key concepts

Molecular Polarity

Molecular polarity refers to how electric charge is distributed across a molecule. It determines if a molecule has distinct positive and negative ends, known as dipoles. In molecules like \(\mathrm{N}_{2} \mathrm{O}\), linear but polar characteristics mean the overall distribution of electrons results in uneven charge.

Here's how to analyze polarity:

• Electronegativity: Atoms differ in electronegativity. This is a measure of an atom's ability to attract shared electrons. Nitrogen and oxygen have distinct electronegativities, affecting how electrons are shared.
• Molecular Geometry: Despite the linear geometry, the electronegativity difference between nitrogen and oxygen creates polarity.
• Bond Polarity: Each bond within the molecule can be polar. In \(\mathrm{N}_{2} \mathrm{O}\), an asymmetrical distribution causes the entire molecule to be polar.

Understanding these elements helps predict molecule behavior, such as solubility and reactivity.

Formal Charge

Formal charge is a concept that helps determine how electrons are assigned within a molecule, assessing charge distribution over individual atoms. Calculating formal charge ensures molecules are stable:

• Formula: Formal charge = Valence electrons - (Non-bonding electrons + 0.5 * Bonding electrons)
• Stability Factor: Molecules are most stable when formal charge is minimized across atoms.
• Application: In \(\mathrm{N}_{2} \mathrm{O}\), the arrangement with zero formal charges on atoms is more stable (i.e. \(\text{N} \equiv \text{N} - \text{O}\)) resulting in the chosen correct structure.

Mastering this calculation will help decide the most probable molecular structure.

Hybridization

Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals, influencing molecular shape and bonding properties. This is important in understanding the geometry of molecules:

• Basic Concept: Hybridization involves combining s, p, and sometimes d orbitals, forming new orbitals like sp, sp\(^2\), sp\(^3\).
• Example for \(\mathrm{N}_{2} \mathrm{O}\): The central nitrogen has sp\(^2\) hybridization. This is because it's bonded to two other atoms and has a lone pair, forming three hybrid orbitals.
• Shapes: These hybrid orbitals form bonds leading to specific geometries, such as linearity in the case of \(\mathrm{N}_{2} \mathrm{O}\).

Grasping hybridization concepts supports understanding of molecular structure and interaction.

Resonance Structures

Resonance structures are different Lewis structures for the same molecule that illustrate delocalized electrons. They don’t exist individually but contribute to the actual form of the molecule:

• Purpose: Represent electron distribution possibilities, not actual shifting electrons.
• For \(\mathrm{N}_{2} \mathrm{O}\): Despite possible resonance, only the N≡N-O structure holds stability due to formal charge optimization.
• Implication: Resonance can affect properties like energy distribution and reactivity.

Understanding resonance helps predict molecular behavior beyond a single Lewis structure.