Organic Chemistry: Structure and Function
8th Edition
ISBN: 9781319079451
Author: K. Peter C. Vollhardt, Neil E. Schore
Publisher: W. H. Freeman
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Chapter 1.7, Problem 1.15TIY
Interpretation Introduction
Interpretation:Themolecular orbital and energy-splitting diagram for the bonding in
Concept Introduction:Molecular orbital theory explained the bonding, magnetic and spectral properties of molecule. It is based on the formation of molecular orbitals by the combination of atomic orbitals. On the basis of energy and stability these molecular orbitals can be further classified in three types:
- Bonding molecular orbitals (BMO): They have lesser energy than atomic orbital therefore more stable compare to atomic orbital.
- Antibonding molecular orbitals (ABMO): They have higher energy than atomic orbital therefore less stable compare to atomic orbital.
- Non-bonding molecular orbitals (NBMO): They have same energy as atomic orbital.
Molecular orbital diagrams represents the distribution of electrons in different molecular orbitals in increasing order of their energy. Hence lower energy molecular orbitals occupy first then only electron moves in higher energy orbitals.
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44. The molecule H3* has long been speculated to exist. The interest here is that the addition
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a) Draw what you think would be the most stable structure for H3* and show all
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b)
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Chapter 1 Solutions
Organic Chemistry: Structure and Function
Ch. 1.3 - Prob. 1.1ECh. 1.3 - Prob. 1.2ECh. 1.3 - Prob. 1.3ECh. 1.3 - Prob. 1.4ECh. 1.3 - Prob. 1.5ECh. 1.4 - Prob. 1.6ECh. 1.4 - Prob. 1.8TIYCh. 1.5 - Prob. 1.9ECh. 1.5 - Prob. 1.11TIYCh. 1.6 - Prob. 1.12E
Ch. 1.6 - Prob. 1.13ECh. 1.7 - Prob. 1.15TIYCh. 1.8 - Prob. 1.16ECh. 1.8 - Prob. 1.18TIYCh. 1.9 - Prob. 1.19ECh. 1.9 - Prob. 1.20ECh. 1.9 - Prob. 1.21ECh. 1.9 - Prob. 1.22ECh. 1 - Prob. 25PCh. 1 - Prob. 26PCh. 1 - Prob. 27PCh. 1 - Prob. 28PCh. 1 - Prob. 29PCh. 1 - Prob. 30PCh. 1 - Prob. 31PCh. 1 - Prob. 32PCh. 1 - Prob. 33PCh. 1 - Prob. 34PCh. 1 - Prob. 35PCh. 1 - Prob. 36PCh. 1 - Prob. 37PCh. 1 - Prob. 38PCh. 1 - Prob. 39PCh. 1 - Prob. 40PCh. 1 - Prob. 41PCh. 1 - Prob. 42PCh. 1 - Prob. 43PCh. 1 - Prob. 44PCh. 1 - Prob. 45PCh. 1 - Prob. 46PCh. 1 - Prob. 47PCh. 1 - Prob. 48PCh. 1 - Prob. 49PCh. 1 - Prob. 50PCh. 1 - Prob. 51PCh. 1 - Prob. 52PCh. 1 - Prob. 53PCh. 1 - Prob. 54PCh. 1 - Prob. 55PCh. 1 - Prob. 56PCh. 1 - Prob. 57PCh. 1 - Prob. 58P
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- Compare and contrast the MO model with the local electron model. When is each useful?arrow_forwardThe diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be 97.06 pm and 424.7 kJ/mol, respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy pz, orbital from oxygen with the higher-energy ls orbital of hydrogen (the OH bond lies along the z-axis). a. Which of the two molecular orbitals will have the greater hydrogen 1s character? b. Can the 2px orbital of oxygen form molecular orbitals with the 1s orbital of hydrogen? Explain. c. Knowing that only the 2p orbitals of oxygen will interact significantly with the 1s orbital of hydrogen, complete the molecular orbital energy-level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of OH+ will be greater than, less than, or the same as that of OH. Explain.arrow_forwardThe B2 molecule is paramagnetic; show how this indicates that the energy ordering of the orbitals in this molecule is given by Figure 6.18a rather than 6.18b.arrow_forward
- The photoelectron spectrum of HBr has two main groups of peaks. The first has ionization energy 11.88 eV. The next peak has ionization energy 15.2 eV, and it is followed by a long progression of peaks with higher ionization energies. Identify the molecular orbitals corresponding to these two groups of peaks.arrow_forward6. For heteronuclear diatomic molecules, the energy-level diagrams of the molecular orbitals need to be modified via calculation. For both NO and CN', the order of molecular orbitals are similar to that of O,. Write electron configurations of NO and CN including core electrons, starting with o1,'01,*². Which bond is stronger? (hint: bond order)arrow_forwardDraw the molecular orbital energy diagram (like we did in class, not the fancy general chemistry diagrams) for the bond between the nitrogen atom labeled a and the carbon atom labeled b in the following molecule. Clearly indicate which atomic orbitals are interacting, which resulting orbital the electrons are in, and label HOMO, LUMO, bonding and antibonding orbitals, and sigma (o), sigma* (o*), pi (n), and pi* (7*) where appropriate.arrow_forward
- Given that the following MO-energy-level diagram applies to the the diatomic molecular cation BO+, match the letter label of each energy level with the corresponding orbital descriptor. (See attached image!) O2pσ2pσ2s*π2pσ2p*B+2sπ2p*σ2sO2sB+2p Does this diagram display the effect of s-p orbital mixing? YES or NO? Which kind of magnetism would be observed for this cation? paramagnetism diamagnetism What is the bond order for this cation? 0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 2.0arrow_forward1. Derive the following expression for the bonding and antibonding orbitals of a heteronuclear diatomic molecule consisting of two electrons in a molecular orbital of the form y = CAA + CBB with atomic orbitals A and B. Let aa=HAA, aB=HBB, B=HAB=HBA, SAA-SBB=1, and, as a simplification, let SAB=SBA=S=0 (zero-overlap approximation). Note that the assumption of SAB=SBA=S=0 is used simply to get a more transparent expression. Show that the secular determinant is given by a,-E - E and energies are given by 1/2 2B E̟ = }(a, +a,)±(a,-a,) For B<0, E+ is the lower energy solution.arrow_forward4. Molecular Orbital Theory Let's construct an MO diagram for KrBr*, the krypton bromide cation. Let's focus only on the n=4 valence shell in for both species. a) Which of the two atoms, Kr or Br, has a 4p electrons that are lower in energy? Justify your answer in terms of periodic trends. Also provide experimental evidence for your answer in terms of ionization energies (available online). b) Based on part (a), draw an MO diagram for KrBr*, reflecting the appropriate relative energies of the Kr/Br valence electrons. Connect the orbitals that mix/split with dashed lines. c) Populate the MO diagram with the appropriate number of valence electrons. What is the expected bond order?arrow_forward
- 1. Draw the MO diagram for HBr. The orbital potential energies for the 4s and 4p of Br are -18.65 ev and -12.49ev, respectively. Recall that the H 1s is -13.61ev.arrow_forwardGiven that the following MO-energy-level diagram (see attached image) applies to the the diatomic molecular cation BO+, match the letter label of each energy level with the corresponding orbital descriptor. Match the letter on the daigram (a, b, c, d, etc) to the given below: O2p σ2p σ2s* π2p σ2p* B+2s π2p* σ2s O2s B+2parrow_forwardIn reference to the following figure, which of the following statements is true? a. An energy level diagram can be made of the two overlapping 1s orbitals, and the Aufbau process used to determine the electronic configurations of H2 and He2. b. In-phase overlap of electron waves represented by the two 1s orbitals results in a new orbital having lower energy than either of the s orbitals. This new bonding orbital concentrates the electron probability between the two nuclei. c. Out-of-phase overlap between two 1s orbitals results in a new orbital having higher energy than either of the s orbitals. This new antibonding orbital would place most of the electron probability to the left and right of the two nuclei. d. The energy-lowering of the bonding molecular orbital and energy-raising of the antibonding molecular orbital with respect to the atomic orbitals will result in equal energy splitting with no net bonding between helium atoms. e. All of the above statements are true.arrow_forward
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