Textbook Question
Use the MO energy diagram in Figure 8.22b to describe thebonding in O2+, O2, and O2-. Which of the three is likely tobe stable? What is the bond order of each? Which containunpaired electrons?
Ch.8 - Covalent Compounds: Bonding Theories and Molecular Structure
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McMurry 8th EditionCh.8 - Covalent Compounds: Bonding Theories and Molecular StructureProblem 105
McMurry 8th EditionCh.8 - Covalent Compounds: Bonding Theories and Molecular StructureProblem 105Chapter 8, Problem 105
Calcium carbide, CaC2, reacts with water to produce acetylene,C2H2, and is sometimes used as a convenient source ofthat substance. Use the MO energy diagram in Figure 8.22ato describe the bonding in the carbide anion, C22-. What isits bond order?
Verified step by step guidance1
Step 1: The carbide anion, C22-, is a diatomic molecule. In order to understand its bonding, we need to fill in the molecular orbital (MO) energy diagram. The MO energy diagram for a diatomic molecule like C22- consists of a series of energy levels representing bonding and antibonding molecular orbitals. The energy levels are filled with electrons starting from the lowest energy level.
Step 2: Each carbon atom has 6 electrons, so the C22- anion, which has two extra electrons, has a total of 14 electrons. Start filling in the MO energy diagram from the lowest energy level, which is the sigma 1s orbital, and continue to the higher energy levels. Each orbital can hold a maximum of 2 electrons.
Step 3: After filling in the sigma 1s, sigma* 1s, sigma 2s, and sigma* 2s orbitals, you will have 8 electrons left. These will fill the pi 2p and sigma 2p orbitals. The pi 2p orbitals can hold 4 electrons and the sigma 2p orbital can hold 2 electrons.
Step 4: To calculate the bond order, use the formula: Bond Order = 0.5 * (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals). In this case, the bonding orbitals are sigma 1s, sigma 2s, pi 2p, and sigma 2p, and the antibonding orbitals are sigma* 1s and sigma* 2s.
Step 5: Plug the number of electrons in the bonding and antibonding orbitals into the bond order formula to calculate the bond order of the C22- anion. A higher bond order indicates a stronger, more stable bond.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Molecular Orbital Theory
Molecular Orbital (MO) Theory explains how atomic orbitals combine to form molecular orbitals, which can be occupied by electrons. In this theory, bonding and antibonding orbitals are formed, and the distribution of electrons in these orbitals determines the stability and properties of the molecule. Understanding MO diagrams is crucial for analyzing the bonding characteristics of complex ions like the carbide anion.
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Molecular Orbital Theory
Bond Order
Bond order is a measure of the number of chemical bonds between a pair of atoms, calculated as the difference between the number of bonding and antibonding electrons divided by two. It provides insight into the strength and stability of a bond; a higher bond order indicates a stronger bond. In the case of the carbide anion, determining the bond order helps assess the nature of the bonding between carbon atoms.
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Carbide Anion Structure
The carbide anion, C2^2-, consists of two carbon atoms sharing electrons, resulting in a unique bonding arrangement. This anion can be analyzed using MO theory to understand its electronic configuration and stability. The structure and bonding characteristics of the carbide anion are essential for predicting its reactivity and the products formed during its reactions, such as with water.
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Related Practice
Textbook Question
The C2 molecule can be represented by an MO diagram similar to that in Figure 8.22a. (b) To increase the bond order of C2, should you add or remove an electron?
Textbook Question
Look at the MO diagrams of corresponding neutral diatomic species in Figure 8.22, and predict whether each of the following ions is diamagnetic or paramagnetic. Diagrams for Li2 and C2 are similar to N2; Cl2 is similar to F2. (c) F2-
Textbook Question
At high temperatures, sulfur vapor is predominantly in the form of S2(g) molecules. (a) Assuming that the molecular orbitals for third-row diatomic molecules are analogous to those for second-row molecules, construct an MO diagram for the valence orbitals of S2(g).
Textbook Question
At high temperatures, sulfur vapor is predominantly in the form of S2(g) molecules. (d) When two electrons are added to S2, the disulfide ion S22- is formed. Is the bond length in S22- likely to be shorter or longer than the bond length in S2? Explain.
Textbook Question
Carbon monoxide is produced by incomplete combustion of fossil fuels. (a) Give the electron configuration for the valence molecular orbitals of CO. The orbitals have the same energy order as those of the N2 molecule.
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