Chapter1McMurry

Information about Chapter1McMurry

Published on October 13, 2007

Author: Arkwright26

Source: authorstream.com

Content

Chapter 1: Structure and Bonding:  Chapter 1: Structure and Bonding Coverage: 1. Electron Configurations. 2. Lewis Structures 3. Covalent and Ionic bonds 4. Atomic and Molecular Orbitals 5. Hybridization Goals: Write the electron configuration for any element in the periodic table. Write Lewis line bond structures for simple organic molecules. Know how to calculate formal charges on atoms. Know the definitions of a sigma and pi bonds. Draw energy diagrams for formation of pi and sigma bonds. 5 Predict the number of sigma and pi bonds in a molecule. 6. Predict the hybridization of an atom in a molecule. Predict the approximate bond angles in a molecule. Slide2:  Linus Pauling 1901-1994 Nobel Prize Chemistry 1954 Nobel Prize Peace 1963 Chemical bonding, medicine, biology For more see: http://www.chemheritage.org/EducationalServices/chemach/home.html Both independently recogized that carbon atoms link together to form carbon chains. Kekule’ proposes that carbon is tetravalent Slide3:  The first step in drawing Lewis structures is to determine the number of electrons to be used to connect the atoms. This is done by simply adding up the number of valence electrons of the atoms in the molecule. Consider carbon dioxide CO2 Carbon (C) has four valence electrons x 1 carbon = 4 e- Oxygen (O) has six valence electrons x 2 oxygens = 12 e- There are a total of 16 e- to be placed in the Lewis structure. Connect the central atom to the other atoms in the molecule with single bonds. Carbon is the central atom, the two oxygens are bound to it and electrons are added to fulfill the octets of the outer atoms. Complete the valence shell of the outer atoms in the molecule. Lewis Structures Slide4:  4. Place any remaining electrons on the central atom. There are no more electrons available in this example. If the valence shell of the central atom is complete, you have drawn an acceptable Lewis structure. Carbon is electron deficient - it only has four electrons around it. This is not an acceptable Lewis structure. If the valence shell of the central atom is not complete, use a lone pair on one of the outer atoms to form a double bond between that outer atom and the central atom. Continue this process of making multiple bonds between the outer atoms and the central atom until the valence shell of the central atom is complete. becomes   Slide5:  The central atom is still electron deficient, so share another pair. becomes 5. Double check to make sure that you have used the correct number of electrons in the Lewis structure and that no atom that cannot exceed its valence shell, does not. The best Lewis structure that can be drawn for carbon dioxide is:   Slide6:  Formal Charge Formal charge is an accounting procedure. It allows chemists to determine the location of charge in a molecule as well as compare how good a Lewis structure might be. The formula for calculating formal charge is shown below: Consider the molecule H2CO2. There are two possible Lewis structures for this molecule. Each has the same number of bonds. We can determine which is better by determining which has the least formal charge. It takes energy to get a separation of charge in the molecule (as indicated by the formal charge) so the structure with the least formal charge should be lower in energy and thereby be the better Lewis structure. Slide8:  The two possible Lewis structures are shown below. They are connected by a double headed arrow and placed in brackets. The non-zero formal charge on any atoms in the molecule have been written near the atom. Slide9:  Orbitals and Wave Functions Bohr and Quantum Mechanical Method de Broglie – electrons have wave properties Y Wave Function that describes orbital Y2 Electron Density around Nucleus Analogy: Guituar String + and – indicate mathematical sign of wave, not charges! 1s orbital 2s 2p 1s Slide10:  2p orbital See HyperChem 3-D Model Slide11:  Atomic and Molecular Orbitals Guidelines for Combining Atomic Orbitals 1. Atomic Orbitals (AOs) on different atoms combine to produce Molecular Orbitals (Mos) that are bonding or antibonding. This is referred to as a Linear Combination of Atomic Orbitals (LCAO). MO1 = aAO1 + bAO2 + c AO3 where a,b,c are coefficients. Each AO used results in an MO. For example, if three AOs are used to construct the MOs, then three MOs must result. 3. When AOs on the same atom combine they produce hybrid AOs. Slide12:  H2 Molecule Antibonding MO Bonding MO 1s AO 1s AO In the Ground State, the Antibonding MO is empty. In the Ground State, the Bonding MO has two electrons + + + - + or - + or - . . . . . . s s* Slide14:  Sigma Bond (s) – covalent bond that is symmetric about the bond axis. Slide15:  Pi (p) Bond Antibonding MO Bonding MO 2p AO 2p AO In the Ground State, the Antibonding MO is empty. In the Ground State, the Bonding MO has two electrons . . . . p p* Slide16:  Two atomic p orbitals Pi (p) Bond 3D view of Pi (p) Bond (pi) Bond – overlap of two p orbitals oriented perpendicular to the line connecting the nuclei. Hybridization of Atomic Orbitals:  Hybridization of Atomic Orbitals AOs on a single atom mix to form new, hybrid orbitals. These hybrid orbitals have characteristics of both s and p orbitals. Provides a means of explaining observed bond angles in organic molecules. Acetylene 180o sp sp E __ s __ __ __ p __ __ __ __ __ __ p __ __ sp hydrid promote e- Mix orbitals D:\rw32b2a.exe Slide18:  s + p = sp + sp Two sp orbitals Remember, there are two p orbitals leftover and these would be located on the y and z axes. y z x Each sp orbital possesses 50% s character and 50% p character. Orbital Picture of Acetylene:  Orbital Picture of Acetylene D:\rw32b2a.exe Slide20:  Ethylene sp2 sp2 The carbon atom is sp2 hybridized to obtain trigonal planar geometry E __ s __ __ __ p __ __ __ __ __ p __ __ __ sp2 hydrid Promote e- Mix orbitals Slide21:  s + 2 p = 3 sp2 orbitals sp2 Hybridization There is one p orbital left over, and it would be along the z axis. Slide22:  Orbital Picture of Ethylene D:\rw32b2a.exe Slide23:  Methane sp3 The carbon atom is sp3 hybridized to obtain tetrahedral geometry. All bond angles are equal at 109.50. E __ s __ __ __ p __ __ __ __ __ __ __ __sp3 Promote e- Mix orbitals Slide24:  s + 3 p = 4 sp3 orbitals D:\rw32b2a.exe Slide25:  Predicting Hybridization of Atoms in Molecules Determine the number of other atoms (X) bonded to the atom of interest (A). Determine the number of nonbonded pairs of electrons (E) on the atom. If X + E = 2 then hybridization is sp. System Hybridization Example Lewis Structure AX2 sp CO2 :O=C=O: AXE sp CO If X + E = 3 then hybridization is sp2 AX3 sp2 H2C=O AX2E sp2 H2C=NH AXE2 sp2 H2C=O Slide26:  If X + E = 4 then hybridization is sp3 AX4 sp3 CH4 AX3E sp3 CH3NH2 AX2E2 sp3 CH3OCH3

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