Question

Nitryl fluoride \(\left(\mathrm{FNO}_{2}\right)\) is very reactive chemically. The fluorine and oxygen atoms are bonded to the nitrogen atom. (a) Write a Lewis structure for \(\mathrm{FNO}_{2}\). (b) Indicate the hybridization of the nitrogen atom. (c) Describe the bonding in terms of molecular orbital theory. Where would you expect delocalized molecular orbitals to form?

Short answer

(a) Lewis: :O=N-F-:O:. (b) N is \(\text{sp}^2\) hybridized. (c) Delocalized orbitals form over N and O atoms involving \(\pi\)-bonds.

Step-by-step solution

Understand Lewis Structure Notation

A Lewis structure represents the arrangement of electrons in a molecule. We need to show how the valence electrons are distributed among the atoms. In \(\text{FNO}_2\), we have one N, one F, and two O atoms.

Calculate Total Valence Electrons

Determine the total valence electrons by adding those from each atom: N (5 valence electrons), O (6 valence electrons each), F (7 valence electrons). Total: \(5 + 6 \times 2 + 7 = 24\) valence electrons.

Build the Basic Skeleton

Place N in the center, as it’s less electronegative than O and F. Attach O and F atoms to N: \( \text{O} - \text{N} - \text{O}, \text{F} - \text{N} \). This accounts for basic bonding.

Distribute Remaining Electrons

Distribute the remaining electrons to satisfy the octet rule. Connect N with double bonds to each O atom and a single bond to F, using lone pairs to complete the octet for each atom.

Draw Lewis Structure

Finalize the structure with N double-bonded to each O and single-bonded to F. Lone pairs are placed on O and F for stability. Lewis structure: \(:\text{O}=\text{N}-\text{F}-\text{:O}:\).

Determine Hybridization of N

The nitrogen atom forms three sigma bonds and has no lone pairs, leading to \(\text{sp}^2\) hybridization.

Describe Bonding with Molecular Orbital Theory

In \(\text{FNO}_2\), molecular orbitals extend over multiple atoms. The \(\pi\)-electron system formed by p orbitals on N and O is delocalized, contributing to resonance stability.

Identify Delocalized Molecular Orbitals

Delocalization occurs in the \(\pi\)-system involving the nitrogen and oxygen atoms, due to the resonance generated by electrons in \(\pi\)-bonds being shared across multiple atoms.

Key concepts

Lewis structure

The Lewis structure is a straightforward way to represent the arrangement of electrons in a molecular compound. In the case of nitryl fluoride, \(\mathrm{FNO}_{2}\), it's crucial to understand how to distribute the available valence electrons among the atoms. Let's explore this further.To construct the Lewis structure, you first calculate the total number of valence electrons. For \(\mathrm{FNO}_{2}\), consider:

• Nitrogen (N) contributes 5 valence electrons.
• Each Oxygen (O) provides 6, resulting in 12 from both oxygen atoms.
• Fluorine (F) adds 7 valence electrons.

This gives a total of \(24\) valence electrons to work with.Next, arrange these atoms. You place nitrogen in the center because it's less electronegative compared to oxygen and fluorine. Then, form sigma bonds connecting each oxygen and the fluorine directly to the nitrogen. After establishing these bonds, distribute the remaining electrons to satisfy the octet rule.The final Lewis structure for \(\mathrm{FNO}_2\) appears as \(:\text{O}=\text{N}-\text{F}-\text{:O}:\). Nitrogen is double-bonded to each oxygen and single-bonded to fluorine, with lone pairs on the oxygen and fluorine atoms.

Hybridization

Hybridization describes the process of mixing atomic orbitals to create new hybrid orbitals, which are used for bonding atoms. In nitryl fluoride \(\mathrm{FNO}_{2}\), we need to assess the hybridization state of the nitrogen atom, the central atom in the molecule.In the \(\mathrm{FNO}_{2}\) molecule, nitrogen forms three sigma bonds: one with fluorine and two with the oxygen atoms. To form these sigma bonds, nitrogen undergoes \(\text{sp}^2\) hybridization. This hybridization involves the mixing of one s orbital and two p orbitals from nitrogen, creating three equivalent \(\text{sp}^2\) hybrid orbitals.As a result of \(\text{sp}^2\) hybridization:

• Nitrogen forms a planar trigonal arrangement, fixing the molecules in a specific geometry.
• There are no lone pairs of electrons on the nitrogen atom, ensuring all hybrid orbitals are involved in bonding.

The \(\text{sp}^2\) hybridization is crucial for understanding the planar structure and angles within \(\mathrm{FNO}_2\), leading to an optimal arrangement for resonance and molecular stability.

Molecular orbital theory

Molecular Orbital (MO) Theory offers a more in-depth perspective on bonding than the traditional Lewis structures. This theory explains how atomic orbitals combine to form molecular orbitals, influencing the molecule's electronic properties.In nitryl fluoride, \(\mathrm{FNO}_{2}\), molecular orbitals extend over several atoms within the molecule. Specifically, the \(\pi\) electron system in \(\mathrm{FNO}_2\) plays a significant role. The p orbitals on nitrogen and the two oxygen atoms overlap laterally to form a set of delocalized \(\pi\) molecular orbitals.These \(\pi\) orbitals can:

• Spread over more than two atoms, giving rise to a delocalized electron cloud.
• Contribute to resonance stability by allowing the \(\pi\) electrons to be shared among the nitrogen and oxygen atoms.

Such delocalization in \(\pi\)-bonds strengthens the compound by stabilizing the energy levels of the involved atoms, contributing to the molecule's reactivity and stability.

Valence electrons

Valence electrons are the outermost electrons of an atom and play a fundamental role in chemical bonding. They are crucial for determining how atoms bond together and the structure of the resulting molecule.In nitryl fluoride \(\mathrm{FNO}_2\), each atom contributes a specific number of valence electrons:

• Nitrogen contributes 5 valence electrons.
• Each Oxygen atom contributes 6, totaling 12 from all oxygen atoms.
• Fluorine offers 7 valence electrons.

The sum of these valence electrons facilitates the formation of bonds that hold the molecule together, leading to a total of \(24\) valence electrons in \(\mathrm{FNO}_2\).Valence electrons are responsible for forming the bonds in the \(\mathrm{FNO}_2\) molecule:

• Single bonds are formed as the nitrogen atom shares one pair of electrons each with fluorine and oxygen.
• Double bonds consume more valence electrons but stabilize the molecule by forming stronger bonds between nitrogen and each oxygen atom.

Understanding the distribution and count of valence electrons is essential for predicting molecule form, stability, and reactivity.