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Chapter 8: Problem 36

(a) Construct a Lewis structure for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), in which each atom achieves an octet of electrons. (b) How many bonding electrons are between the two oxygen atoms? (c) 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 \(\mathrm{O}_{2}\) ? Explain.

Short Answer

Expert verified
In \(\mathrm{H}_{2}\mathrm{O}_{2}\), the \(\mathrm{O-O}\) bond is a single bond with 2 bonding electrons and is longer than the double bond in \(\mathrm{O}_{2}\).

Step by step solution

01

Count Valence Electrons

Identify the total number of valence electrons in the molecule. Each hydrogen atom contributes 1 electron, and each oxygen atom contributes 6 electrons. Therefore, for \( \mathrm{H}_{2}\mathrm{O}_{2} \), the total number of valence electrons is \( 2(1) + 2(6) = 14 \) electrons.
02

Setup Atom Framework

Arrange the atoms in the molecule, keeping in mind typical bonding patterns. Place two hydrogen atoms at each side of the structure and place the oxygen atoms in the center to connect them, forming a skeletal structure: \(\mathrm{H-O-O-H}\).
03

Connect with Single Bonds

Initially, connect the hydrogen and oxygen atoms using single bonds. Place a single bond between each hydrogen atom and its adjacent oxygen atom, and one between the two oxygen atoms. Each bond consists of 2 electrons, totaling 8 electrons used (4 bonds).
04

Distribute Remaining Electrons

Assign the remaining electrons as lone pairs to fulfill the octet rule for each atom wherever possible. After using 8 electrons for bonding, distribute the remaining 6 electrons as lone pairs on oxygen atoms such that each oxygen atom achieves the octet (8 electrons total, including the bonds).
05

Calculate Bonding Electrons for O-O Bond

To find the number of bonding electrons between the two oxygen atoms, note that a single bond was placed between them. A single bond contains 2 bonding electrons.
06

Compare O-O Bond Lengths

In \(\mathrm{H}_{2}\mathrm{O}_{2}\), the \(\mathrm{O-O}\) bond is a single bond, while in \(\mathrm{O}_{2}\) the bond is a double bond. A double bond generally has a shorter bond length than a single bond due to the increased electron sharing. Therefore, the \(\mathrm{O-O}\) bond in \(\mathrm{H}_{2}\mathrm{O}_{2}\) is expected to be longer than in \(\mathrm{O}_{2}\).

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Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom. They play a crucial role in chemical bonding and the formation of molecules. For the molecule \(\mathrm{H}_{2}\mathrm{O}_{2}\), each type of atom contributes a specific number of valence electrons:
  • Each hydrogen atom contributes 1 valence electron.
  • Each oxygen atom contributes 6 valence electrons.
This results in a total of 14 valence electrons for hydrogen peroxide, calculated as \(2 \times 1 + 2 \times 6 = 14\) electrons.
The valence electrons are essential because they determine how atoms will bond with each other. They are involved in forming chemical bonds, holding atoms together in the structure of a molecule.
Octet Rule
The octet rule is a fundamental concept in chemistry that suggests atoms tend to bond in such a way that they have eight electrons in their valence shell, achieving a noble gas-like configuration. For molecules, this rule helps guide the arrangement of electrons in Lewis structures.
In hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)), each oxygen atom aims to have a total of eight valence electrons:
  • Oxygen achieves this by sharing electrons with hydrogen atoms and another oxygen atom through single bonds.
  • Hydrogen atoms are the exception to the octet rule, as they are stable with just 2 electrons in their valence shell.
Once the skeletal structure is established with single bonds, the remaining electrons are distributed to satisfy the octet rule for the oxygen atoms. In this way, each oxygen atom can have its full complement of 8 electrons, including those involved in bonding.
Bond Length
Bond length is the average distance between the nuclei of two bonded atoms. It varies depending on the types of atoms and the number of shared bonding electrons or bond order.
In the case of \(\mathrm{H}_{2}\mathrm{O}_{2}\), the bond between the two oxygen atoms is a single bond, which usually results in a greater bond length than a double bond found in \(\mathrm{O}_{2}\). This is because:
  • A single bond shares 2 electrons, while a double bond shares 4 electrons, increasing the electron density between the atoms.
  • Higher electron density in a double bond pulls the atoms closer together, reducing bond length.
Thus, the \(\mathrm{O-O}\) bond in \(\mathrm{H}_{2}\mathrm{O}_{2}\) is expected to be longer compared to the double-bonded oxygen atoms in \(\mathrm{O}_{2}\).
Bonding Electrons
Bonding electrons are the pairs of electrons shared between atoms to form covalent bonds. In Lewis structures, each bond line represents a pair of bonding electrons.
For hydrogen peroxide, consider the bond between the two oxygen atoms:
  • A single line or a single covalent bond connects these oxygen atoms, representing 2 bonding electrons being shared.
  • These electrons help each oxygen atom meet the octet rule in coordination with the other bonds and lone pairs.
Understanding the role of bonding electrons explains how atoms complete their valence electron configurations through covalent bonding, ensuring molecule stability.

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Most popular questions from this chapter

By referring only to the periodic table, select \((\mathbf{a})\) the most electronegative element in group \(13 ;(\mathbf{b})\) the least electronegative element in the group As, Se and Br; (c) the most electronegative element in the group K, Mg, Al and In; (d) the element in the group \(\mathrm{Na}\), Be, \(\mathrm{Si}\), Ar, that is most likely to form an ionic compound with \(B \mathrm{r}\).

Which of the following statements about electronegativity is false? (a) Electronegativity is the ability of an atom in a molecule to attract electron density toward itself. (b) Electronegativity is the same thing as electron affinity. (c) The numerical values for electronegativity have no units. (d) Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" 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 and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound \(1: 40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \%\) M, \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of M? (b) What is the empirical formula for Compound 2 and the identity of M? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. \((\mathbf{e})\) Are compounds 1 and 2 ionic or molecular?

(a) Write the electron configuration for the element titanium, Ti. How many valence electrons does this atom possess? (b) Hafnium, Hf, is also found in group 4 . Write the electron configuration for Hf. (c) Ti and Hf behave as though they possess the same number of valence electrons. Which of the subshells in the electron configuration of Hf behave as valence orbitals? Which behave as core orbitals?

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: \((\mathbf{a}) \mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2},(\mathbf{b}) \mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3},(\mathbf{c}) \mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3} .\)
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