Outline for second exam period

• Molecular orbitals in diatomics, triatomics and higher• Bonding in metals – crystal lattices and unit cells• Bonding in ionic solids

– Electrostatic model of ionic bond– Lattice energy– Thermochemical (Born-Haber) cycles

• Intermolecular forces (weak bonding interactions)• Solubilities of ionic compounds

Second Exam – Wednesday, February 25

The Lewis structure for molecular oxygen

● Lewis structures of the superoxide and peroxide ions suggest that each has a single bond

Experiment indicates that superoxide has a bond order of 1.5

O O O O 2-O O1-

● Predicts an oxygen-oxygen double bond and no unpaired electrons.

Experiment indicates that the oxygen-oxygen bond order is 2, but that there are two unpaired electrons.

● Molecular oxygen can be reduced by one electron, to superoxideion, and by two electrons to peroxide ion. The former has one unpaired electron.

Molecular orbitals from two H 1s orbitals

E H(1) H(2)

N1 + N2

N1 -N2

F1s(F1)

F1s*(F2)

Electron densities for MO’s (from Coulson, 2nd ed., p. 84)

Contour maps for MO’s (from Coulson)

N1 + N2

N1 - N2

H2+ σ1 0.5 1.05 256

H2 σ2 1 0.74 432

B.O. D/D E, kJ/mol

MO’s for homonuclear 2nd period diatomics

N1 + N2

N1 - N2

N1 - N2

N1 + N2

ss

pp

E

E

N1 + N2

N1 - N2

s s

p p

N1 - N2

N1 + N2

s & p orbital energies vs Z

-60

-50

-40

-30

-20

-10

0

Li Be B C N O F

E/eV

sp

Molecular orbitals from p orbitals

N1 - N2

N1 + N2

N1 + N2

N1 - N2

s s

p p

σs

σs*

σp

σp*

π

π*

σsp

ss

pp

σsp*

σps

σps*

π

π*

B2 KK4σ2σ*2π2 1 1.59 291

C2 KK4σ2σ*2π4 2 1.24 599

N2 KK4σ2σ*2π4σ2 3 1.10 942

O2 KK4σ2σ*2σ2 π4π*2 2 1.21 494

F2 KK4σ2σ*2σ2 π4π*4 1 1.41 155

O2+ KK4σ2σ*2σ2 π4π*1 2.5 1.12 643

O2- KK4σ2σ*2σ2 π4π*3 1.5 1.35 395

O22- KK4σ2σ*2σ2 π4π*4 1 1.49 126

Li2 KK4σ2 1 2.67 101

B.O. D/D E, kJ/mol

Prediction of molecular properties

● Magnetic properties● Ionization energy relative to the constituent

atoms● Electron affinity relative to the constituent

atoms

s s

p p

σs

σs*

σp

σp*

π

π*

σsp

ss

pp

σsp*

σps

σps*

π

π*

E E

0.28B0.4512.1O2

>1.3B2

1.4613.6O

EA, eVIE, eV

Orbital occupancies for electronic states of molecular oxygen

E

B*

B

3Eg- (ground state)

1)g (0.98 eV above ground state) andB

B*

1Eg+ (1.63 eV above ground state)

B

B*

Singlet oxygen: formation and properties

• Photosensitization with dyes• Electric discharge• Chemical reactions

• High reactivity, especially with double bonds• Luminescence (t1/2 for 1Δg = 14 min at atm pressure; 45 min at

low pressure)

• Photodynamic therapy as a treatment for cancer involves use of a photosensitizer and light of the appropriate wavelength to generate singlet oxygen, which can cause tumor cell death.http://cis.nci.nih.gov/fact/7_7.htm

Chemiluminescence from singlet O2

http://www.chem.leeds.ac.uk/delights/photos/singlet_O2

Singlet oxygen can be generated by passing Cl2 through a basic solution of H2O2

Cl2 + 2 OH– = OCl– + Cl– + H2OH2O2 + OCl– = ClOO– + H2OClOO– = 1O2 + Cl–

2 1O2 = 2 3O2 + h< (634 nm)

MO’s in heteronuclear diatomic molecules

aN1 - bN2

bN1 + aN2

aN1 - bN2

bN1 + aN2

E

B

A

χA>χB

• MO’s are formed from linear combinations of atomic orbitals

• Because the orbitals are not of the same energy the contribution of each atomic orbital will not be the same

• The AO of the most electronegative atom is lowest in energy

• The atomic orbital that is closest in energy to a MO will make the greatest contribution

• This leads naturally to the concept of unequal distribution of bonding electron density, to bond polarity, and an ionic contribution to the bond energy

• Bond energy is the sum of the covalent and ionic contributions

MO’s in H-X molecules

s

p

s

F

F*

n

n

aN1 - bN2

bN1 + aN2

x

z

B*

FE

s

s

p

p

F

F

MO’s when both atoms have p orbitals

B

F*

CN CN- CO NO

Surface contours of MO’s in homo- and heteronuclear diatomic molecules

MO’s in linear triatomic molecules – H3+

• Begin by taking linear combinations of terminal atom s orbitals

• Determine how these combinations of terminal atom orbitals can overlap with atoms of the central atom

• Construct energy level diagram

• Distribute electrons starting with the lowest energy orbital

• Energy of molecule is sum of the one-electron energies less electron-electron repulsions

sE sN1 + N2

N1 - N2

F1

F2

F3

F1

F2

F3

σ4σ3

σ2σ1

πn

MO’s in linear triatomic molecules – BeH2

p

sE

ssN1 + N2

N1 - N2

F1

F2

F3

F4

• The electronegativity of hydrogen is greater than that of beryllium so the hydrogen orbital are lower in energy.

• Take linear combinations of terminal atom orbitals and determine how they can overlap with orbitals on the central atom

• Construct energy level diagram

• Distribute electrons starting with the lowest energy orbital

• What is bond order of the individual Be-H bonds?

• BeH2 is strongly (Lewis) acidic as a result of its electron deficiency (vacant p orbitals)

x

zBn

x

z

sp

sE

MO’s in bent triatomic molecules – OH2

N1 - N2

N1 + N2

x

z

x

z

OH H

zyx

3D contours of MO’s for water

x

z

Evidence for nonequivalent lone pairs

x

zFour ionizations in the PES spectrum confirms the presence of non-equivalent lone pairs

Photoelectron spectroscopy (PES)h< = IE + KE (ejected electron)

MO’s of triatomics with p orbitals on all atoms

As molecules get more complex the number of atomic orbitals and the number of molecular orbitals increases rapidly. Energy level diagrams become very complex and therefore less useful, but drawings of 3D contours for MO’s are still very useful.

On the following three pages are representations of MO’s for N3-, a linear

homonuclear triatomic, CO2, a linear heteronuclear triatomic, and NO2-, a

bent heteronuclear triatomic. Two views of the MO’s for nitrite ion are provided; one view is perpendicular to the plane of the ion, the other is in the plane of the ion toward the nitrogen atom (along the z axis).

Each of these species has 12 atomic orbitals and therefore 12 molecular orbitals.

Azide ion and carbon dioxide each have 16 electrons, whereas nitrite ion has 18.

Nitrite ion might be considered as a model for the product of addition of a base (electron pair donor) to CO2 or the addition of two electrons to NO2

+. In each case the addition of an electron pair to the LUMO results in bending of the molecule.

3D Contours for MO’s of Azide Ion

F

B

F

F

F

F* F*

B

B*

3D Contours for MO’s of carbon dioxide

Two views of 3D contours of MO’s for NO2-

z to planez to plane In-plane along z In-plane along z

-36.8

-21.2

-32.7

-16.3

-15.9

-15.9

-14.7

-14.3

-12.3

-10.8

9.8

17.6

E, evE, ev

1

4

3

2

5

6

11

12

10

9

8

7