Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 9, Problem 57IL
Draw the two resonance structures that describe the bonding in SO2. Then describe the bonding in this compound using MO theory. How does MO theory rationalize the bond order of 1.5 for the two S—O bonds in this compound?
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Draw two resonance structures that describe the bonding in SO2. Then describe the bonding in this compound using MO theory. How does MO theory rationalize the bond order of 1.5 for the two S-O bonds in this compound?
1) The photoelectron spectrum for water has four bands, not including ionizations from the 1s electrons of oxygen. Three of the bands have vibrational fine structure. The fourth band, which is from the least energetic ionization, shows no such fine structure. Explain these results in terms of your MO scheme.
2) If we added two more electron pairs to the MO scheme for BeH2, this would describe the bonding in water if water were a linear molecule. Compare your MO scheme for bent water with this hypothetical scheme for linear water, and comment on why water is bent instead of linear.
2. Use molecular orbital (MO) theory in description of the following molecules:
Which of the following species has the greater bond enthalpy? These chemical species have the similar orbital structure to that of N2.
CO, CO+
Chapter 9 Solutions
Chemistry & Chemical Reactivity
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- It is possible to write a simple Lewis structure for the SO42- ion, involving only single bonds, which follows the octet rule. However, Linus Pauling and others have suggested an alternative structure, involving double bonds, in which the sulfur atom is surrounded by six electron pairs. (a) Draw the two Lewis structures. (b) What geometries are predicted for the two structures? (c) What is the hybridization of sulfur in each case? (d) What are the formal charges of the atoms in the two structures?arrow_forwardUse the molecular orbital model to fully describe the bonding in O2+, O2, O2-, and O22-.Determine which of the following statements are true and which are false. True or false? : Bond length increases with increasing bond order while bond energy decreases.The number of unpaired electrons in O2- and O2 is, respectively, 1 and 2. The electron configuration of O2 is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)1. The electron configuration of O2- is (σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3. The bond order in O2- and O22- is, respectively, 2.5 and 1. The bond lengths increase in the order: O22- < O2- < O2 < O2+.arrow_forwardConsider a heteronuclear diatomic molecule, AB, where A and B are period 2 elements. Using the following pieces of information, provide details on the molecular orbital diagram of AB: Clue 1: The MO diagram of AB does not contain non- bonding electrons. Clue 2: The s and p orbital energies of both atoms are quite close to one another that interactions between different orbital types occur. Clue 3: AB" is more stable than AB, but AB2 is less stable than AB". Clue 4: The potential energies of the atomic orbitals of A are higher than the potential energies of the atomic orbitals of B. Table 1. Atomic orbital energies of period 2 elements. Atom Orbital energy Orbital energy (2s) (2p) Li -5.45 Ве -9.30 B -14.0 -8.30 C -19.5 -10.7 -25.5 -13.1 -32.3 -15.9 F -46.4 -18.7 Ne -48.5 -21.5arrow_forward
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