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(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) Do you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{H}_{2} \mathrm{O}_{2}\) to be longer or shorter than the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) ?

Short Answer

Expert verified
(a) The Lewis structure for hydrogen peroxide (H2O2) is: H | O-O : : H (b) The O-O bond in H2O2 is a single bond, while the O-O bond in O2 is a double bond. Since double bonds are generally shorter than single bonds, we expect the O-O bond in H2O2 to be longer than the O-O bond in O2.

Step by step solution

01

Determine the total number of electrons in the molecule

To construct a Lewis structure for H2O2, we first need to find the total number of valence electrons in the molecule. For hydrogen (H), there is 1 valence electron per atom, and for oxygen (O), there are 6 valence electrons per atom. In H2O2, there are 2 hydrogen atoms and 2 oxygen atoms, so the total number of electrons is: Total electrons = (1 x 2) + (6 x 2) = 2 + 12 = 14
02

Write the skeleton structure for H2O2

The skeleton structure shows the covalent bonds between atoms in the molecule. Hydrogen forms one bond, whereas oxygen can form two bonds. So, one possible arrangement can be to place hydrogen atoms singly bonded to each oxygen atom, and a bond between the two oxygen atoms: O-H | O-H
03

Distribute the remaining electrons

We have used 4 electrons (2 sharing each bond) in the skeleton structure. We have 10 more electrons to distribute. We'll start by completing the octet for the oxygen atoms by adding lone pairs. Each oxygen atom needs 6 more electrons to complete its octet. We can add 3 lone pairs around each oxygen atom: O-H : : O-H
04

Finalize the Lewis structure

Now that all atoms have a complete octet, the Lewis structure of hydrogen peroxide is: H | O-O : : H (a) The Lewis structure for hydrogen peroxide is presented above. (b)
05

Compare the bond lengths in H2O2 and O2

The O-O bond in H2O2 is a single bond, whereas the O-O bond in O2 is a double bond. Single bonds are generally longer than double bonds because double bonds involve more electron density between the nuclei, causing a greater attractive force and resulting in shorter bond lengths. Therefore, we expect the O-O bond in H2O2 to be longer than the O-O bond in O2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Peroxide
Hydrogen peroxide, commonly known by its chemical formula \( \text{H}_2\text{O}_2 \), is a simple compound made up of two hydrogen atoms and two oxygen atoms. It appears as a pale blue liquid in its pure form and is commonly used as a disinfectant or bleaching agent. In the world of chemistry, hydrogen peroxide is interesting for its relatively unstable nature, which makes it a reactive species. Due to this, it decomposes readily into water and oxygen, especially in the presence of light.

Hydrogen peroxide is crucial to understand in the realm of Lewis structures. A Lewis structure helps to visualize the arrangement of electrons around atoms in a molecule, providing insights into its chemical behavior. By focusing on hydrogen peroxide, students learn about electron sharing (covalent bonding), molecular geometry, and the principle of achieving a complete valence electron shell (the octet rule) through bonding.
Valence Electrons
Valence electrons play a pivotal role in determining how atoms bond with one another. They are the outermost electrons of an atom and are responsible for an atom's chemical properties. For hydrogen peroxide, understanding valence electrons is essential for constructing its Lewis structure.
  • Each hydrogen atom has 1 valence electron.
  • Each oxygen atom possesses 6 valence electrons.
In constructing the Lewis structure for \( \text{H}_2\text{O}_2 \), the sum of valence electrons is crucial. The total is 14 valence electrons—2 from the hydrogen atoms and 12 from the oxygen atoms. This total guides the placement of bonds and lone pairs in the Lewis structure.

Remember, the goal in the Lewis structure is to have all oxygen atoms achieving an octet configuration, which means having eight electrons in their valence shell through shared pairs (bonds) and lone pairs.
Oxygen-Oxygen Bond
The bond between the two oxygen atoms in hydrogen peroxide is a significant aspect of its structure. In \( \text{H}_2\text{O}_2 \), the oxygen-oxygen bond is a single covalent bond. This means there is one shared pair of electrons between the two oxygen atoms, which contributes to the overall stability of the molecule.

When comparing this bond with the oxygen-oxygen bond in \( \text{O}_2 \), which is a double bond, we note a key difference. Double bonds involve two shared pairs of electrons, making them shorter and stronger due to increased electron density. Consequently, the \( \text{O}-\text{O} \) bond in \( \text{H}_2\text{O}_2 \), being a single bond, is longer and weaker. This aspect is crucial for predicting molecular behavior and reactivity, especially when comparing molecular oxygen (\( \text{O}_2 \)) with hydrogen peroxide.
Covalent Bonds
Covalent bonds are a fundamental concept in chemistry, characterized by the sharing of electron pairs between atoms. They are what hold the atoms together in a molecule like hydrogen peroxide.

In hydrogen peroxide, the covalent bonds are established as follows:
  • Each hydrogen forms one single covalent bond with an oxygen atom. This involves sharing one pair of electrons between hydrogen and oxygen.
  • The oxygen atoms share an additional pair of electrons between them, creating a single covalent bond between the oxygen atoms.
Understanding covalent bonds is vital because they determine the molecule’s shape, reactivity, and properties. They also help in predicting how the molecule might interact with other substances or break down into simpler products, such as in the decomposition of hydrogen peroxide into water and oxygen.

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Most popular questions from this chapter

In the vapor phase, \(\mathrm{BeCl}_{2}\) exists as a discrete molecule. (a) Draw the Lewis structure of this molecule, using only single bonds. Does this Lewis structure satisfy the octet rule? (b) What other resonance forms are possible that satisfy the octet rule? (c) Using formal charges, select the resonance form from among all the Lewis structures that is most important in describing \(\mathrm{BeCl}_{2}\).

You and a partner are asked to complete a lab entitled "Fluorides of Croup \(6 \mathrm{~B}\) Metals" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a colorless liquid and the other a green powder. You also find the following notes in your partner's notebook-Compound 1: \(47.7 \% \mathrm{Cr}\) and \(52.3 \% \mathrm{~F}\) (by mass), Compound 2: \(45.7 \% \mathrm{Mo}\) and \(54.3 \% \mathrm{~F}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds you find that the colorless liquid solidifies at \(18^{\circ} \mathrm{C}\), while the green powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the colorless liquid? What is the identity of the green powder? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

(a) Describe the molecule chlorine dioxide, \(\mathrm{ClO}_{2}\), using three possible resonance structures. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? Why or why not? (c) Using formal charges, select the resonance structure(s) that is (are) most important.

Write the electron configuration for phosphorus. Identify the valence electrons in this configuration and the nonvalence electrons. From the standpoint of chemical reactivity, what is the important difference between them?

Average bond enthalpies are generally defined for gasphase molecules. Many substances are liquids in their standard state. \(\mathrm{OB}\) (Section 5.7) By using appropriate thermochemical data from Appendix C, calculate average bond enthalpies in the liquid state for the following bonds, and compare these values to the gas-phase values given in Table 8.4: (a) \(\mathrm{Br}-\mathrm{Br}\), from \(\mathrm{Br}_{2}(l) ;\) (b) \(\mathrm{C}-\mathrm{Cl}\), from \(\mathrm{CCl}_{4}(l) ;\) (c) \(\mathrm{O}-\mathrm{O}\), from \(\mathrm{H}_{2} \mathrm{O}_{2}(l)\) (assume that the \(\mathrm{O}-\mathrm{H}\) bond enthalpy is the same as in the gas phase). (d) What can you conclude about the process of breaking bonds in the liquid as compared to the gas phase? Explain the difference in the \(\Delta H\) values between the two phases.

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