/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Use molecular orbital theory to ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 8

Use molecular orbital theory to predict the relative bond energies and bond lengths of a diatomic carbon molecule, \(\mathrm{C}_{2}\), and the acetylide ion, \(\mathrm{C}_{2}^{2-}\).

Short Answer

Expert verified
\( \mathrm{C}_2 \) has a bond order of 2; \( \mathrm{C}_2^{2-} \) has a bond order of 3.

Step by step solution

01

Determine Molecular Orbital Diagrams

To apply molecular orbital theory, first, we need to construct MO diagrams for both \( \mathrm{C}_2 \) and \( \mathrm{C}_2^{2-} \). \( \mathrm{C}_2 \) has 12 valence electrons (6 from each carbon atom), while \( \mathrm{C}_2^{2-} \) has 14 valence electrons due to the addition of 2 extra electrons.
02

Arrange Molecular Orbitals for \( \mathrm{C}_2 \)

For \( \mathrm{C}_2 \), the molecular orbitals fill in the following order up to \( \sigma_{2p_z} \): \( \sigma_{2s} \), \( \sigma_{2s}^* \), \( \pi_{2p_x} = \pi_{2p_y} \), \( \sigma_{2p_z} \). Since \( \mathrm{C}_2 \) has 12 electrons, they fill as: 2 in \( \sigma_{2s} \), 2 in \( \sigma_{2s}^* \), 4 in \( \pi_{2p_x} \) and \( \pi_{2p_y} \), and 4 in \( \sigma_{2p_z} \). This results in a bond order of 2.
03

Arrange Molecular Orbitals for \( \mathrm{C}_2^{2-} \)

For \( \mathrm{C}_2^{2-} \), the molecular orbitals fill upto \( \pi_{2p_x}^* = \pi_{2p_y}^* \). The 14 electrons within \( \mathrm{C}_2^{2-} \) fill as: 2 in \( \sigma_{2s} \), 2 in \( \sigma_{2s}^* \), 4 in \( \pi_{2p_x} \) and \( \pi_{2p_y} \), 2 in \( \sigma_{2p_z} \), followed by 2 more in \( \pi_{2p_x}^* = \pi_{2p_y}^* \). The bond order is 3.
04

Compare Bond Energies and Lengths

Bond order is directly related to bond energy and inversely related to bond length. \( \mathrm{C}_2 \) with a bond order of 2 has a lower bond energy and longer bond length compared to \( \mathrm{C}_2^{2-} \), which has a bond order of 3, indicating a higher bond energy and shorter bond length.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bond Order
Bond order gives us an idea of the strength and stability of a bond in a molecule. It is calculated using the formula:\[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]This tells us the number of chemical bonds between a pair of atoms. For example,
  • A bond order of 1 is typical for a single bond.
  • A bond order of 2 suggests a double bond.
  • A bond order of 3 indicates a triple bond.
In the case of diatomic molecules like
  • \( \mathrm{C}_2 \), the bond order is calculated to be 2, indicating a double bond.
  • \( \mathrm{C}_2^{2-} \) or acetylide ion has a bond order of 3, representing a triple bond, thus a more robust connection.
Higher bond order means more bonds holding the atoms together, generally leading to a more stable and stronger bond.
Bond Energy
Bond energy refers to the amount of energy required to break one mole of bonds in a molecule.
  • Bonds with higher bond orders generally have higher bond energies since more electrons are involved in bonding.
  • For instance, the acetylide ion \( \mathrm{C}_2^{2-} \), with a bond order of 3, will have a higher bond energy compared to
  • The diatomic carbon molecule \( \mathrm{C}_2 \), which has a bond order of 2.
This increased bond energy leads to greater stability in the acetylide ion. The closer the atoms, the stronger the bond, requiring more energy to break it. Therefore, to break a triple bond in \( \mathrm{C}_2^{2-} \), more energy is needed than for a double bond in \( \mathrm{C}_2 \). The larger the bond energy, the harder it is to separate the atoms.
Bond Lengths
Bond length is the distance between the centers of two bonded atoms. It inversely correlates with bond order.
  • Fewer bonds mean longer bond lengths.
  • Thus, the diatomic carbon \( \mathrm{C}_2 \) with a bond order of 2 has a longer bond length than
  • The acetylide ion \( \mathrm{C}_2^{2-} \), which has a bond order of 3.
The shorter length in \( \mathrm{C}_2^{2-} \) is due to the additional bonding, which pulls the atoms closer. A shorter bond length usually indicates a stronger bond because the atoms are held more tightly together. A simple analogy is to think of bond length like an elastic band: the shorter and tighter it is, the harder it is to snap (break).
Diatomic Molecules
Diatomic molecules consist of two atoms. They can be composed of identical or different atoms. Examples don't just include only gases such as
  • \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \),
  • but also molecules like the carbon molecule \( \mathrm{C}_2 \) which is diatomic but rarely found in nature.
In molecular orbital theory, these molecules are often analyzed to understand bonding behavior. It's straightforward to apply
  • Because it involves fewer atoms,
  • making it a useful model for understanding basic bonding principles.
They feature prominently in studies of chemical bonding as they provide a simplified model to explain more complex structures.
Acetylide Ion
The acetylide ion \( \mathrm{C}_2^{2-} \) is a molecule in which two carbon atoms are triple-bonded, with an extra two electrons present, giving the molecule a negative two charge. Molecular orbital theory helps to explain its high bond order of 3, signifying a robust triple bond.
  • Because of this triple bond,
  • The ion has significant stability and higher bond energy compared to its neutral counterpart, \( \mathrm{C}_2 \).
Its structural representation also means that the carbon atoms are much closer together, reflecting a shorter bond length. This unique bonding configuration makes the acetylide ion an important compound in organic chemistry and materials science. With improved stability and reactivity, it serves as an important building block in synthesizing other compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ketones are organic compounds with the general Lewis formula where \(\mathrm{R}\) and \(\mathrm{R}^{\prime}\) are alkyl groups (see previous problem) that may or may not be different. The simplest ketone is 2-propanone, commonly known as acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\), one of the most important solvents. Describe the bonding and shape of an acetone molecule.

How many valence electrons are there in a XeF \(_{2}\) molecule? Use hybrid orbitals to describe the bonding in a \(\mathrm{XeF}_{2}\) molecule.

Use molecular orbital theory to explain why the bond energy of a \(\mathrm{N}_{2}\) molecule is greater than that of a \(\mathrm{N}_{2}^{+}\) ion.

In our localized bond description of \(\sigma\) -bonding orbitals in molecules, we generally chose to ignore the presence of antibonding orbitals. Why are we able to do this?

Discuss the differences between the molecular orbital and the localized bond descriptions of a molecule. What can each theory tell us about a molecule? What are the advantages and disadvantages of each description?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.