EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393630817
Author: KARTY
Publisher: W.W.NORTON+CO. (CC)
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Chapter 3, Problem 3.44P
Interpretation Introduction

Interpretation:

The HOMO and LUMO of the uncharged CH3CH2 molecule is to be determined.

Concept introduction:

According to VSEPR theory, atoms that are bonded to two or more other atoms are considered to be hybridized. Such atoms contribute their valence orbitals for mixing. The new orbitals are formed by mixing just two AOs. The hybridization of an atom can be determined based on the electron geometry about an atom. The electron geometry about an atom describes the orientation of group of electrons around it. MOs are constructed from the AOs of different atoms. The two atomic orbitals, on mixing along bonding axes, form two molecular orbitals – one is σ, and the other is σ*. The σ is a bonding MO, and σ* is an antibonding MO. The σ MOs have lower energy and greater stability than the corresponding σ* MOs. The highest occupied molecular orbital, i.e., the highest energy MO that contains an electron is HOMO. The lowest unoccupied molecular orbital, i.e., lowest energy MO having no electron is LUMO.

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Determine a molecular formula, e.g. CH4, from the line structure below. Specify elements in the following order: C, H, others(in alphabetical order). Example: C4H,CIOS OH Molecular formula
a. CH3CHO(g) + O2(9) CH3COOH(1) b. AgCl(s) + Br2(1) AgBr(s) + c. C12H22011(s) + O2(g) Given: Ethanal, CH3CHO(g) Oxygen, O2(g) Ethanoic Acid, CH3COOH (1) Silver Chloride, AgCl(s) Bromine gas, Br2(1) J/Kmol 3 sig. figs. J/Kmol 3 sig. figs. CO₂(g) + H₂O(1) (+) J/Kmol 4 sig. figs. Chlorine gas,Cl2(g) Silver Bromide, AgBr (s) Sucrose, C12H22011(s) Carbon Dioxide, CO2(g) Water H₂0(1) Cl₂(g) Sm(J/K.mol) 250.3 205.138 159.8 96.2 152.23 223.07 107.1 360.2 213.74 69 91
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