Orbitals and covalent bonding

advertisement

1

Chapter 9

Orbitals and Covalent Bond

2

Atomic Orbitals Don’t Work

 to explain molecular geometry.

 In methane, CH

4 tetrahedral.

, the shape s

 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

3

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.

 sp

3 hybridization has tetrahedral geometry.

4

2s

In terms of energy

2p

Hybridization sp

3

5

How we get to hybridization

We know the geometry from experiment.

 We know the orbitals of the atom

 hybridizing atomic orbitals can explain the geometry.

 So if the geometry requires a tetrahedral shape, it is sp

3 hybridized

 This includes bent and trigonal pyramidal molecules because one of the sp 3 lobes holds the lone pair.

6 sp 2 hybridization

C

2

H

4

Double bond acts as one pair.

 trigonal planar

 Have to end up with three blended orbitals.

 Use one s and two p orbitals to make sp

2 orbitals.

 Leaves one p orbital perpendicular.

7

8

9

2s

In terms of energy

2p

Hybridization

2p sp

2

10

Where is the P orbital?

 Perpendicular

 The overlap of orbitals makes a sigma bond ( s bond)

11

Two types of Bonds

 Sigma bonds from overlap of orbitals.

Between the atoms.

Pi bond ( p bond) above and below atoms

 Between adjacent p orbitals.

 The two bonds of a double bond.

12

H

H

C C

H

H

13 sp 2 hybridization

 When three things come off atom.

 trigonal planar

120º on s one p bond

14

What about two

 When two things come off.

 One s and one p hybridize.

 linear

15 sp hybridization

End up with two lobes 180º apart.

 p orbitals are at right angles

Makes room for two p bonds and two sigma bonds.

 A triple bond or two double bonds.

16

2s

In terms of energy

2p

Hybridization

2p sp

17

CO

2

C can make two s and two p

O can make one s and one p

O C O

18

Breaking the octet

 PCl

5

 The model predicts that we must use

 the d orbitals.

dsp

3 hybridization

 There is some controversy about how involved the d orbitals are.

19

 Trigonal

 bipyrimidal can only s bond.

can’t p bond.

 basic shape for five things.

dsp 3

PCl

5

Can’t tell the hybridization of Cl

Assume sp

3 to minimize repulsion of electron pairs.

20

21 d 2 sp 3

 gets us to six things around

 octahedral

Molecular Orbital Model

 Localized Model we have learned explains much about bonding.

It doesn’t deal well with the ideal of resonance, unpaired electrons, and bond energy.

 The MO model is a parallel of the atomic orbital, using quantum mechanics.

 Each MO can hold two electrons with

22

 opposite spins

Square of wave function tells probability

23

What do you get?

Solve the equations for H

2

H

A

H

B

 get two orbitals

 MO

1

= 1s

A

- 1s

B

 MO

2

= 1s

A

+ 1s

B

24

The Molecular Orbital Model

The molecular orbitals are centered on a line through the nuclei

– MO

1 the greatest probability is between the nuclei

– MO

2 it is on either side of the nuclei

– this shape is called a sigma molecular orbital

25

The Molecular Orbital Model

In the molecule only the molecular orbitals exist, the atomic orbitals are gone

MO

1 is lower in energy than the 1s orbitals they came from.

– This favors molecule formation

– Called an bonding orbital

MO

2 is higher in energy

– This goes against bonding

– antibonding orbital

26

The Molecular Orbital Model

MO

2

1 s

1 s

MO

1

27

The Molecular Orbital Model

We use labels to indicate shapes, and whether the MO’s are bonding or antibonding.

– MO

1

– MO

2

= s

1s

= s

1s

* (* indicates antibonding)

Can write them the same way as atomic orbitals

– H

2

= s

1s

2

28

The Molecular Orbital Model

Each MO can hold two electrons, but they must have opposite spins

Orbitals are conserved. The number of molecular orbitals must equal the number atomic orbitals that are used to make them.

29

1 s

H

2

s

1s

*

1 s s

1s

30

Bond Order

 The difference between the number of bonding electrons and the number of antibonding electrons divided by two

Bond Order =

# bonding-#antibonding

2

Download