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 hybridization scheme for ozone is sp3 for the central oxygen atom and sp2 for the two end oxygen atoms. In one resonance form, the end oxygen atoms use one sp2 orbital for sigma bonding while the remaining sp2 orbitals and the unhybridized p orbital hold nonbonding pairs. The unhybridized p orbitals of the end oxygen atoms and the central oxygen atom are involved in the delocalization of the \(\pi\) electrons, resulting in a total of 4 delocalized \(\pi\) electrons.

Step-by-step solution

1. Lewis structure of ozone

First, let's draw the Lewis structure for ozone. Ozone has a total of 18 valence electrons as each oxygen atom contributes 6 electrons. In ozone, the two end oxygen atoms are equivalent and double bonded to the central oxygen atom with a single bond between the central and end atoms. We will have: O = O - O

2. Hybridization of central oxygen atom

Now, let's determine the hybridization of the central oxygen atom in the ozone molecule. The central oxygen atom is bonded to two other oxygen atoms, thus it has two sigma bonds. Additionally, it has two lone pairs of electrons. Therefore, the central oxygen atom requires 4 hybrid orbitals. The hybridization that results in 4 hybrid orbitals is sp3 hybridization.

3. Hybridization of end oxygen atoms

The two end oxygen atoms are double bonded to the central oxygen atom. Each end oxygen atom has one sigma bond and one \(\pi\) bond to the central atom, as well as, two lone pairs of electrons. The end oxygen atoms require three hybrid orbitals to accommodate the sigma bond and the two lone pairs. Hence, the hybridization of the end oxygen atoms is sp2 hybridization.

4. Orbitals used for bonding and nonbonding pairs in one resonance form

In one of the resonance forms of the ozone molecule, the end oxygen atoms use one of their sp2 hybrid orbitals to form a sigma bond with the central oxygen atom. The remaining two sp2 hybrid orbitals and the unhybridized p orbital are used to hold the nonbonding pairs of electrons.

5. Orbitals involved in delocalizing \(\pi\) electrons

The delocalization of the \(\pi\) electrons occurs between the end oxygen atoms and the central oxygen atom. The unhybridized p orbitals of the end oxygen atoms and one of the unhybridized p orbitals of the central oxygen atom (with sp3 hybridization) can form \(\pi\) bonds, resulting in the delocalization of the \(\pi\) electrons.

6. Number of electrons delocalized in the \(\pi\) system

As mentioned previously, the end oxygen atoms form a \(\pi\) bond with the central oxygen atom using their unhybridized p orbitals. Each of these \(\pi\) bonds contains 2 electrons, resulting in a total of 4 \(\pi\) electrons that are delocalized in the ozone molecule. In summary, the best hybridization scheme for ozone is the central oxygen atom with sp3 hybridization, and the two end oxygen atoms with sp2 hybridization. In one of the resonance forms, the orbitals used for bonding are one sp2 hybrid orbital of each end oxygen atom and the sp3 hybrid orbitals of the central oxygen atom, while the remaining sp2 hybrid orbitals hold nonbonding pairs. The unhybridized p orbitals allow delocalization of the \(\pi\) electrons, and 4 electrons are delocalized in the \(\pi\) system.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and they play a crucial role in the formation of chemical bonds. In the ozone molecule,

• Each oxygen atom contributes 6 valence electrons.
• There are three oxygen atoms in ozone, giving a total of 18 valence electrons.

These electrons are arranged to form bonds and nonbonding pairs.
In ozone's Lewis structure:

• The central oxygen atom is connected by single bonds to the end oxygen atoms.
• Double bonds alternate between the central and one of the end oxygen atoms.
• This results in resonance structures, where electrons appear to "move" or are shared between bonds to stabilize the molecule.

Understanding valence electrons is key to predicting a molecule's geometry and reactivity, as they determine how atoms connect in a structure.

Pi Electron Delocalization

Pi electron delocalization occurs when

• The \(\pi\)-electrons in a molecule are spread across adjacent atoms, rather than being localized to a single bond.
• In ozone, this involves the central oxygen and the two end oxygen atoms.

The delocalization of \(\pi\) electrons helps stabilize the ozone molecule by

• allowing electrons to be shared across all three oxygen atoms.

In one of the resonance forms, the \(\pi\) bonds are formed by overlapping the p orbitals of the central and end oxygen atoms. In resonance:

• The apparent position of the double bond shifts, signifying that \(\pi\) electrons are moving between allowed energetic states.

This sharing of electrons reduces the potential energy of the molecule, adding to its stability and influencing its chemical properties.

sp3 and sp2 Hybridization

Hybridization is a process where atomic orbitals mix to form new equivalent hybrid orbitals, which help explain molecular geometry and bonding. In the ozone molecule:

• The central oxygen atom undergoes \(sp^3\) hybridization.
• It forms four hybrid orbitals to accommodate two sigma bonds and two lone pairs of electrons.

For the end oxygen atoms,

• They exhibit \(sp^2\) hybridization.
• Each forms one sigma bond with the central oxygen and uses the remaining hybrid orbitals to house lone pairs.
• The unhybridized p orbital in \(sp^2\) oxygen overlaps with p orbitals from other atoms to create \(\pi\) bonds.

This hybridization scheme in ozone is vital as it

• Ensures the correct geometry and bonding seen in its resonance forms.
• Allows efficient \(\pi\) electron delocalization, contributing to the molecule's overall energy minimization and stability.

Understanding hybridization helps predict and explain how atoms form bonds in different molecular environments.