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Unformatted text preview: Hybridization Part 2 Reading: Gray: (47) OGC: (6.4) and (6.5) XI1 MOLCAO works great for diatomic molecules! But... The Story so far: We can use hybridization of the central atom and MOLCAO together to describe small polyatomic molecules. Does hybridization / MOLCAO enable us to describe more complicated molecules? XI2 Watson and Crick s Original DNA Model XI3 We ll take the challenge: C 2 H 6 From VSEPR and Lewis dot structures, we know it looks like this: H H H H H c c C C H H H H H H XI4 C C H H H H H H C 2 H 6 Look at the left C . Notice it has a steric number of four: What other molecule had a steric number of four? CH 4 This suggests we should hybridize our current C the same way. Let s analyze one C at a time. H H H H H H 1 2 3 4 C H C XI5 C 2 H 6 sp 3 hybridize the carbon: Add 3 hydrogens (using MOLCAO) sp 3 sp 3 sp 3 sp 3 sp 3 sp 3 sp 3 H H H C 1s 1s 1s sp 3 Notice the tetrahedral shape. C C H H H H H H XI6 C 2 H 6 Do the same to the other side: Bring the two halves together: C C H H H H H H sp 3 sp 3 sp 3 sp 3 H H H C H C H H XI7 C 2 H 6 Notice that all the bonds formed are bonds. Each is rotationally symmetrical about its axis. H H H C H C H H C C H H H H H H XI8 Let s try another: C 2 H 4 From VSEPR and Lewis dot structures, we know it looks like this: C C H H H H C H H C C C H H H H XI9 Look at the left C . Notice it has a steric number of three: What other molecule had a steric number of three? BH 3 This suggests we should hybridize our current C the same way. Let s analyze one C at a time. C C H H H H C 2 H 4 B H H H C 1 2 3 C H H XI10 C 2 H 4 sp 2 hybridize the carbon: Notice there is an extra p orbital that is not involved in hybridization: sp 2 sp 2 sp 2 p sp 2 sp 2 sp 2 p x p x p y p z p x C C H H H H sp 2 sp 2 sp 2 p XI11 H C C H H H H C 2 H 4 For now, we will deemphasize the p orbital. sp 2 sp 2 sp 2 p p Add hydrogen: H Do the same for the other C: H p H XI12 C C H H H H C 2 H 4 H p H H p H Now bond the two C together: Notice that the two sp 2 orbitals form a orbital that is rotationally symmetric. H p H H p H 2sp 2 * b C 2sp 2 C XI13 C 2 H 4 p p Count the bonding e around each carbon: left C 1 1 1 3 + right C 1 1 1 3 + But each C has 4 e ! What do we do with the extras? e e left over: 1 left over: 1 C C H H H H XI14 C 2 H 4 H p H H p H Let s reemphasize the p orbitals: The p orbitals can combine, forming a bond: H H H H H p H H p H 2p z * b C C 2p z C C H H H H XI15 bonds 2p z * b C C 2p z node Notice that...
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This note was uploaded on 10/27/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.
 Fall '08
 Lewis
 Chemistry, Atom, Mole

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