AP CHEMISTRY CHAPTER 9 BONDING 1

Hybridization 2

When drawing Lewis structures to explain bonding, we have been using the Localized Electron Model of bonding. This assumes that the electrons stay with (or close to) the atoms from which they originated and that bonds are formed by the overlap of atomic orbitals. This model needs to be developed a little further to explain experimental data. 3

In methane, CH 4, carbon has one 2s and three 2p orbitals available for bonding. Hydrogen has one 1s orbital available. We could imagine that one of the bonds in methane would be formed from the overlap of a hydrogen 1s orbital and a carbon 2s orbital and the other three bonds would be formed from the overlap of a hydrogen 1s orbital and a carbon 2p orbital. This would make 2 different kinds of bonds. Experimental data shows that all four bonds are identical. 4

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Hybridization is an addition to the localized electron model that explains this. Carbon needs 4 identical orbitals to form 4 identical bonds. Imagine throwing the one 2s orbital and the three 2p orbitals from carbon into a blender. Whrrrrr!! Out comes four new orbitals that are a homogenous blend of the four that went in!! 6

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The new orbitals have properties of the s and of the p orbitals. Since they were made from one s and three p orbitals, we can call them sp 3 orbitals. We put four atomic orbitals in and got four hybrid orbitals out. The hybrid sp 3 orbitals each overlap with the 1s orbitals from each hydrogen atom and make four identical bonds having o bond angles. 8

Atoms with four effective electron pairs have sp 3 hybridization, even if they have unshared pairs as in H 2 O and NH 3. The electron pairs have a tetrahedral arrangement. 9

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We can look at the hybridization in the bonding of ethene (C 2 H 4 ). H H C = C H H 12

Each carbon atom has three effective pairs and needs three equal orbitals. To get three hybrid orbitals we throw one s and two p orbitals into the blender. Whrrrr!!! Out come three identical orbitals made from one s and two p orbitals. The three new orbitals are called sp 2. This type of hybridization results in trigonal planar electron pair arrangements and 120 o bond angles. 13

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The overlap of the 1s orbitals from each hydrogen with the sp 2 hybrid orbitals of carbon results in sigma bonds. In a sigma bond, the orbitals overlap head-on and the electrons are in the region between the two nuclei. All single bonds are sigma bonds. Each double bond and each triple bond contains one sigma bond. 15

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What happened to the p orbital that wasn’t hybridized? The remaining p orbitals from each carbon atom overlap above and below the plane of the nuclei resulting in a sideways overlap that is called a pi bond. 17

A pi bond can form only if there is also a sigma bond between the same two atoms. Because the electrons in a pi bond are not directly between the nuclei, a pi bond is weaker than a sigma bond. A double bond contains one sigma plus one pi bond and is thus stronger and shorter than a single bond. 18

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Let’s look at the hybridization and bonding in ethyne (acetylene), C 2 H 2. H-C  C-H 21

Each carbon in ethyne has two effective electron pairs. We can throw one s and one p orbital from carbon into the blender. Whrrrr!!! Out come two sp orbitals. The s orbital from each hydrogen atom overlaps with one of the sp orbitals to form a sigma bond. The other sp orbital overlaps with one of the sp orbital on the other carbon to form a sigma bond between the two C atoms. 22

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The remaining two p orbitals from each carbon atom overlap outside the plane of the nuclei to form two pi bonds. A triple bond is one sigma and two pi bonds and is thus stronger and shorter than either a single or a double bond. Linear molecules and 180 o bond angles result from sp hybridization. 24

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We learned that certain elements can exceed the octet rule by utilizing unfilled d orbitals. These d orbitals are also involved in hybridization. For example, PF 5 has five effective electron pairs. These five pairs require five equal bonds. We must throw one s, three p and 1 d orbital into the blender to get five dsp 3 (or sp 3 d) orbitals. This form of hybridization results in trigonal bipyramidal shapes and 90 o and 120 o bond angles. 26

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With SF 6 we have six effective electron pairs. We must hybridize one s, three p, and two d orbitals to get six d 2 sp 3 or sp 3 d 2 orbitals. This results in the octahedral shape with 90 o bond angles. 28

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Sigma bonds- formed by the end-to-end overlap of atomic orbitals between two nuclei Pi bonds- formed by the side-to-side overlap of unhybridized p orbitals outside the plane of two nuclei 30

Single bonds= 1 sigma bond Double bonds = 1 sigma bond + 1 pi bond Triple bonds = 1 sigma bond + 2 pi bonds 31

Practice: For each of the following, determine: a)hybridization of each atom b)bond angles c)molecular shape d)# of sigma bonds and # of pi bonds 32

SbCl 5 Cl Cl Cl Sb Cl Cl 33

NO 3 - O O N O 34

CHCl 3 H Cl - C - Cl Cl 35

:N  C C  N: C = C :N  C C  N: 36

MOLECULAR ORBITAL BONDING THEORY (MO) 37

The MO model is more complex than the localized electron model but works better in many instances. 38

A molecular orbital (MO) is formed from valence atomic orbitals. Two atomic orbitals combine to form two MOs. One MO has a higher energy than the atomic orbitals and is called an antibonding MO. The other MO has a lower energy (more stable) and is called a bonding MO. 39

In a bonding orbital, electrons have the greatest probability of being between the nuclei. In an antibonding orbital, electrons are mainly outside the space between the nuclei. We can have  and  MOs. Orbitals are conserved. The number of MOs will be the same as the number of AOs used to construct them. 40

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When H 2 is formed from two H atoms, its 1s orbitals combine to form molecular orbitals as follows: 42

H 2 antibonding  1s * Energy 1s 1 1s 1 bonding  1s MO configuration for H 2 = (  1s 2 ) Since both electrons are in the bonding MO, H 2 is very stable. 43

He 2 antibonding  1s * Energy 1s 2 1s 2 bonding  1s MO configuration for He 2 = (  1s 2 ) (  * 1s 2 ) Not stable! He 2 doesn’t exist. 44

For diatomic molecules with valence electrons in the first energy level, the molecular orbitals are:  * 1s ___  1s ____ 45

Figure 9.34 The Molecular Orbitals from p Atomic Orbitals 46

For diatomic molecules with valence electrons in the second energy level, the molecular orbitals are:  * 2p ____  * 2p ____ ____  2p ____  2p ____ ____  * 2s ____  2s ____ There are 2 pi orbitals, so  2p and  * 2p can each hold 4 electrons. 47

Figure 9.39 Molecular Orbital Summary of Second Row Diatomics 48

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C 2 Carbon has 4 valence electrons, so C 2 has 8. MO configuration = (  2s 2 )(  * 2s 2 )(  2p 4 ) All electrons are paired so C 2 is diamagnetic. 50

Bond order- an indication of bond strength. Bond order = # bonding electrons- # antibonding electrons 2 For C 2 = (6-2)/2 = 2 The larger the bond order, the stronger the bond. A bond order of zero means that the molecule or ion is unstable and would not exist. 51

O 2 Oxygen has 6 valence electrons. O 2 would have 12. MO configuration= (  2s 2 )(  * 2s 2 )(  2p 2 )(  2p 4 )(  * 2p 2 ) The last two electrons are unpaired, this explains the experimental results that show that oxygen is paramagnetic. Bond order = (8-4)/2 = 2 52

Figure 9.43 A Partial Molecular Orbital Energy- Level Diagram for the HF Molecule 53

PARAMAGNETISM AND DIAMAGNETISM 54

Substances that contain unpaired electrons are weakly attracted into magnetic fields and are said to be paramagnetic. Substances in which all of the electrons are paired are very weakly repelled by a magnetic field and are said to be diamagnetic. Paramagnetism is much weaker than ferromagnetism (Fe, Co, Ni) 55

Figure 9.37 Diagram of the Kind of Apparatus Used to Measure the Paramagnetism of a Sample 56

The Lewis structure of a molecule does not always correctly predict paramagnetism. The oxygen molecule is an example of this. O=O Experimental data shows that oxygen is paramagnetic (has unpaired electrons). The more complex molecular orbital model will correctly predict oxygen’s paramagnetism. 57

Paramagnetic or Diamagnetic? NO 2 17 electrons O = N - O 58

Resonance- To eliminate the need for resonance we can use the localized electron model to describe the  bonding and the MO model to describe the  bonding. In benzene, each C is sp 2 hybridized. The 6 additional p orbitals perpendicular to the ring are used to form  MOs. The electrons in these  MOs are delocalized above and below the plane of the ring. 59

Figure 9.48 The Pi System for Benzene 60