Transcript AP Ch 09 apchapt9r

Chapter 9
Orbitals and Covalent Bond
1
Atomic Orbitals Don’t Work
to explain molecular geometry.
 In methane, CH4, the shape is
tetrahedral.
 The valence electrons of carbon should
be two in s, and two in p.
 The p orbitals would have to be at right
angles.
 The atomic orbitals change when
making a molecule.
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9.1 Hybridization
We blend the s and p orbitals of the
valence electrons and end up with the
tetrahedral geometry.
 We combine one s orbital and 3 p
orbitals.
 The atoms are responding as needed to
give the minimum energy for the
molecule.
 sp3 hybridization has tetrahedral
geometry.
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In terms of energy
2p
Energy
Hybridization
2s
4
sp3
How we get to hybridization - CH4
We know the geometry from experiment.
Four bonds of equal length and strength.
 We know the orbitals of the central atom.
Hybridizing atomic orbitals can explain
the geometry.
 So if the geometry requires a tetrahedral
shape, it is sp3 hybridized.
 This includes bent and trigonal pyramidal
molecules because one of the sp3 lobes
holds the lone pair.
5
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(YDVD)
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2
sp
hybridization
C2H4 Trigonal planar. 120°
 Double bond acts as one pair. This
results in 3 effective pairs surrounding
the carbon atoms.
 One s and two p orbitals hybridize into
3 identical orbitals of equal length and
energy to make sp2 orbitals.
 This leaves one p orbital unhybridized.
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In terms of energy
2p
Energy
Hybridization
2s
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2p
sp2
Two types of Bonds
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Sigma bonds () form from the overlap of
orbitals along the internclear axis.
Pi bond () occupies the space above and below
internclear axis.
Between adjacent unhybridized p orbitals.
The double bond always consists of one  bond
and one  bond.
C-C double bond (BDVD)
 and  bonds
(YDVD)
H

H
C C
H
H
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10
sp hybridization
CO2
 Each carbon has two hybridized orbitals
180º apart. Also 2 unhybridized p
orbitals.
 p orbitals are at right angles (Fig. 9.17)
 Makes room for two p bonds and two
sigma bonds.
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In terms of energy
2p
Energy
Hybridization
2s
12
2p
sp
(YDVD)
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13
CO2
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C can make two s and two p
O can make one s and one p
(Fig. 9.19)
O
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C O
3
dsp
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PCl5
Five pairs of electrons around the central
atom. Trigonal bypyramidal. Only  bonds
no  bonds.
The model predicts that we must use the d
orbitals.
Five electron pairs require dsp3
hybridization. (Fig. 9.21)
There is some controversy about how
involved the d orbitals are.
2
3
d sp
SF6
 Six pairs of electrons around the central
atom.
 Octahedral shape. (Fig. 9.23)
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How do we figure this out?
Use the Localized Electron Model.
 Draw the Lewis structure(s).
 Determine the arrangement of electron
pairs (VSEPR model).
 Specify the necessary hybrid orbitals
based upon the pairs of electrons
around the central atom.
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