Problem 58 - Chemistry: The Central Science in SI Units, Global Edition

Question

The oxygen atoms in \(\mathrm{O}_{2}\) participate in multiple bonding, whereas those in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the oxygen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{O}-\mathrm{O}\) bond?

Step-by-Step Solution

Answer

(a) In \(\mathrm{O}_{2}\), there is a double bond between the oxygen atoms, whereas in \(\mathrm{H}_{2}\mathrm{O}_{2}\), there is a single O-O bond and each oxygen atom is bonded to one hydrogen atom. (b) The oxygen atoms in both molecules have sp³ hybridization. (c) The \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) is stronger than the bond in \(\mathrm{H}_{2}\mathrm{O}_{2}\).

Step 1: Drawing Lewis structures for \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\)

For \(\mathrm{O}_{2}\) molecule, each oxygen atom has 6 valence electrons. In order to achieve the full octet for both atoms, they will form a double bond by sharing two pairs of electrons. In case of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (hydrogen peroxide), it is a bit more complicated. Each oxygen atom has 6 valence electrons, and hydrogen atom has 1 valence electron. For the oxygen atoms to have a full octet, they will share one pair of electrons to form an oxygen-oxygen single bond. The remaining two pairs of electrons from each oxygen atom will form bonds with hydrogen atoms.

Step 2: Determining the hybridization of oxygen atoms in both molecules

We can determine the hybridization of an atom by counting its number of sigma bonds and lone pairs; summing these values gives the steric number of the atom. The oxygen in the \(\mathrm{O}_{2}\) molecule has two sigma bonds (one double bond) and two lone pairs, so it has a steric number of 4 and hybridization of sp³. In \(\mathrm{H}_{2} \mathrm{O}_{2}\), each oxygen atom has one sigma bond with the other oxygen atom, two sigma bonds with hydrogen atoms, and one lone pair of electrons. So, each oxygen atom has a steric number of 4 and hybridization of sp³, too.

Step 3: Comparing the strength of O-O bond in both molecules

In the \(\mathrm{O}_{2}\) molecule, there is a double bond between the oxygen atoms, whereas in \(\mathrm{H}_{2} \mathrm{O}_{2}\) there is a single bond. In general, double bonds are stronger than single bonds because of the greater electron density in the bonding region. This means the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) is stronger than that in \(\mathrm{H}_{2} \mathrm{O}_{2}\). So, to summarize: (a) \(\mathrm{O}_{2}\) has a double bond between the oxygen atoms, and \(\mathrm{H}_{2} \mathrm{O}_{2}\) has a single O-O bond and each oxygen atom bonded to one hydrogen atom. (b) The oxygen atoms in both molecules have sp³ hybridization. (c) The \(\mathrm{O}-\mathrm{O}\) bond in the \(\mathrm{O}_{2}\) molecule is stronger than the bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\).

Key Concepts

Molecular Hybridization

Molecular hybridization is a concept used to explain how atoms form bonds by combining their atomic orbitals. Hybridization involves the mixing of atomic orbitals to create new hybrid orbitals, which contain different energies and shapes. These hybrid orbitals can effectively overlap with orbitals from other atoms to form stable chemical bonds.

For example, when considering oxygen in both \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\), we need to identify the number of sigma bonds and lone pairs of electrons to determine the hybridization state. Each oxygen atom is involved in forming sigma bonds with neighboring atoms, which dictates the arrangement and type of hybrid orbitals.

In both \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\), the oxygen atoms exhibit \(\mathrm{sp}^{3}\) hybridization. This is because the steric number, which is the sum of sigma bonds and lone pairs, is 4 for each oxygen atom:

Chemical Bonding

Chemical bonding is a key aspect of how atoms join to form molecules. It involves the attraction between atoms, allowing them to share or transfer electrons. There are various types of chemical bonds, including covalent bonds, ionic bonds, and metallic bonds. Each type of bond involves a different mechanism of electron interaction.

In the case of \(\mathrm{O}_{2}\), the oxygen atoms form a double bond between them. Double bonds involve sharing two pairs of electrons, resulting in a strong and stable link between the atoms. This kind of bond increases electron density, making it stronger than a single bond.

Conversely, \(\mathrm{H}_{2} \mathrm{O}_{2}\) exhibits a single bond between the oxygen atoms, and each oxygen atom forms additional single bonds with hydrogen atoms. Single bonds share only one pair of electrons, making them weaker than double bonds. Understanding these bonding patterns is crucial when predicting the behavior of molecules in chemical reactions.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a vital role in chemical bonding. These electrons are involved in the formation of bonds with other atoms. The number of valence electrons determines an atom's chemical properties, such as its ability to form bonds and its reactivity with other elements.

For oxygen, which is in group 16 of the periodic table, there are six valence electrons available for bonding. In \(\mathrm{O}_{2}\), each oxygen atom shares two pairs of these electrons to achieve a full octet, satisfied by a double bond. Consequently, the molecule becomes stable with a complete set of valence electrons around each oxygen.

In \(\mathrm{H}_{2} \mathrm{O}_{2}\), each oxygen atom also starts with six valence electrons. By forming single bonds with another oxygen and two hydrogen atoms, each oxygen achieves a stable electron configuration. This demonstrates the role of valence electrons and how they determine the type and number of bonds an atom can form.