Determine the bond order and ground state configuration of C22+, N2-, and F23+. 2. The ionization energy of the 2p-orbitals of carbon, nitrogen, and oxygen are: Carbon: 10.7 eV Nitrogen: 13.1 eV Oxygen: 15.9 eV Using this information, construct an MO diagram for the pi-bond of C=C, C=N, and C=O. 3. In nucleophilic “attack”, a nucleophile donates a pair of electrons to an electrophile. Using your MO from part b), predict which atom the nucleophile will be bound to after the reaction.

Determine the bond order and ground state configuration of C22+, N2-, and F23+. 2. The ionization energy of the 2p-orbitals of carbon, nitrogen, and oxygen are: Carbon: 10.7 eV Nitrogen: 13.1 eV Oxygen: 15.9 eV Using this information, construct an MO diagram for the pi-bond of C=C, C=N, and C=O. 3. In nucleophilic “attack”, a nucleophile donates a pair of electrons to an electrophile. Using your MO from part b), predict which atom the nucleophile will be bound to after the reaction.

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter6: Quantum Mechanics And Molecular Structure

Section: Chapter Questions

Problem 39P: The photoelectron spectrum of HBr has two main groups of peaks. The first has ionization energy...

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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.
By using FMO (orbital coefficients and delta EHOMO-LUMO), ALSO sizes of molecules, explain the regioselectivity .
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Using the molecular orbital method and its qualitative explanations, justify that the bond formation energy in the B2 molecule is equal to -297 kJ mol-1, while in the ionized N2+ molecule it is equal to -842 kJ mol-1. Explain whether these molecules are diamagnetic or paramagnetic.
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1. 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.
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Use the data given below to calculate the bond energies (Do) (in eV) of 'H³°CI and 'H®l Br molecules. Which bond is stronger? Molecule ve (cm=1) De((cm¯1) HCI 2991 53194 HBr 2649 34570
Photoelectron spectroscopy studies have determined the orbital energies for fluorine atoms to be 15 -689 ev 25 -34 ev 2p -12 ev Estimate the value of Zff for F in each of these orbitals.
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