Question

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

Short answer

The best choice of hybridization scheme for the atoms in ozone is sp2 hybridization for all oxygen atoms. In the first resonance form, the central oxygen atom uses its sp2 hybrid orbitals to form one sigma bond with each terminal oxygen atom and holds a lone pair of electrons, while its unhybridized p orbital contains one π electron. Terminal oxygen atoms use their sp2 hybrid orbitals to form sigma bonds with the central oxygen atom and hold their lone pairs of electrons. The unhybridized p orbitals of all three oxygen atoms are used for delocalizing the π electrons. There are a total of 2 delocalized π electrons in the π system of ozone.

Step-by-step solution

Determine the hybridization scheme of ozone

To determine the hybridization of ozone, let's consider the Lewis structure of O3. Ozone has two resonance structures, where the central oxygen atom forms a double bond with one terminal oxygen atom and a single bond with the other terminal oxygen atom. The two resonance structures can be represented as: O = O - O ↔ O - O = O The central oxygen atom has three electron domains (one double bond, one single bond, and one lone pair), so it adopts the sp2 hybridization, forming a trigonal planar geometry. The terminal oxygen atoms have two bonding electron domains (single or double bond), and two lone pairs, so they also adopt sp2 hybridization. Thus, the best choice of hybridization scheme for ozone is sp2 hybridization for all oxygen atoms.

Identify orbitals used for bond formation and nonbonding pairs in a resonance form

Let's consider the first resonance form of ozone, where the central oxygen atom is double-bonded to the left terminal oxygen atom and single-bonded to the right terminal oxygen atom: O = O - O In this resonance form: - Central Oxygen Atom: One of the sp2 hybrid orbitals forms a sigma bond with each terminal oxygen atom, while the remaining sp2 hybrid orbital holds a lone pair of electrons. The unhybridized p orbital contains one π electron involved in the double bond. - Terminal Oxygen Atoms: The sp2 hybrid orbitals are used to form sigma bonds with the central oxygen atom, and each terminal oxygen atom has two sp2 hybrid orbitals occupied by the lone pairs of electrons.

Identify the orbitals used for delocalizing the π electrons

The unhybridized p orbitals of the central oxygen atom and both terminal oxygen atoms can overlap to form delocalized π bonds in ozone. Therefore, these are the orbitals that can be used for the delocalization of π electrons.

Determine the number of electrons delocalized in the π system of ozone

In both resonance forms of ozone, the central oxygen atom contributes one π electron while the double bonded terminal oxygen atom contributes another π electron to the delocalized π system. As these two π electrons are shared among all three oxygen atoms, there are a total of 2 delocalized π electrons in the π system of ozone.

Key concepts

Hybridization in Molecules

In the study of molecules like ozone (\(\text{O}_3\)), hybridization helps us understand how atomic orbitals mix to form new hybrid orbitals. This concept is crucial as it describes the arrangement of electrons around atoms within a molecule. In ozone, each oxygen atom uses \(sp^2\) hybridization. Here's how it works:

• The central oxygen atom has three electron domains: one double bond, one single bond, and one lone pair.
• This leads to a trigonal planar geometry, characteristic of \(sp^2\) hybridization.
• Each terminal oxygen atom also has two bonding domains (single or double bonds) and two lone pairs.

Understanding hybridization aids in predicting the molecule's shape and bonding characteristics, ensuring students grasp how atoms connect within ozone.

Resonance Structures

Resonance structures are multiple ways to depict the electron arrangement in molecules with the same spatial configuration. In the case of ozone:

• The molecule showcases two resonance structures:
• In one, the central oxygen atom forms a double bond with one terminal oxygen and a single bond with the other.
• The roles of the double and single bonds switch in the second structure.

These structures illustrate that electrons are not fixed but can move between positions, thereby stabilizing the molecule. Resonance structures emphasize the concept that the true form is a hybrid of all structures, central in understanding molecular stability and reactivity.

Delocalization of Electrons

Electron delocalization refers to electrons spread over several atoms, rather than localized between two. In ozone, this concept is pivotal in its stability:

• The unhybridized p orbitals in ozone contribute to delocalized electrons.
• These orbitals allow \(\pi\) electrons to move freely across the oxygen atoms.
• This "sharing" of electrons reduces energy and increases molecule stability.

Delocalization is vital in this context as it highlights a key feature of compounds with resonance—it leads to greater stabilization, as seen with ozone’s shared electrons ensuring its structural integrity.

Pi Bonds

Pi bonds form when unhybridized p orbitals overlap, commonly found in double bonds. In ozone (\(\text{O}_3\)), these bonds play a critical role:

• In \(sp^2\) hybridization, oxygen atoms have unhybridized p orbitals that can create \(\pi\) bonds.
• These \(\pi\) bonds in ozone are involved in delocalization.
• Two \(\pi\) electrons are distributed across the molecule.

Pi bonds are essential for understanding delocalization and resonance in ozone. They illustrate how molecules achieve stability and play a role in chemical reactivity due to energy-efficient overlaps.