What’s coming up???• Oct 25 The atmosphere, part 1 Ch. 8• Oct 27 Midterm … No lecture• Oct 29 The atmosphere, part 2 Ch. 8• Nov 1 Light, blackbodies, Bohr Ch. 9• Nov 3,5 Postulates of QM, p-in-a-box Ch. 9• Nov 8,10Hydrogen and multi – e atoms Ch. 9• Nov 12 Multi-electron atoms Ch.9,10• Nov 15 Periodic properties Ch. 10• Nov 17 Periodic properties Ch. 10• Nov 19 Valence-bond; Lewis structures Ch. 11• Nov 22 VSEPR Ch. 11• Nov 24 Hybrid orbitals; VSEPR Ch. 11, 12• Nov 26 Hybrid orbitals; MO theory Ch. 12• Nov 29 MO theory Ch. 12• Dec 1 bonding wrapup Ch. 11,12• Dec 2 Review for exam

The Final Exam

• December 13 (Monday) • 9:00 – 12:00

• Cumulative (covers everything!!)• Worth 50% of total mark

• Multiple choice

The Final Exam

• From my portion, you are responsible for:

– Chapter 8 … material from my lecture notes– Chapter 9 … everything– Chapter 10 … everything– Chapter 11 … everything– Chapter 12 … everything except 12.7

The Final Exam

• You will need to remember

– Relationship between photon energy and frequency / wavelength

– De Broglie AND Heisenberg relationships– Equations for energies of a particle-in-a-box

AND of the hydrogen atom– VSEPR shapes AND hybribizations which

give them

COMBINATION OF ORBITALS

1sA + 1sB = MO1

builds up electron density between nuclei.

Remember, when we take linear combinations of orbitals we get out as many as we put in.

Here, the sum of the 2 orbitals

90% probability

COMBINATION OF ORBITALS

1sA + 1sB = MO1

builds up electron density between nuclei.

1sA – 1sB = MO2

results in low electron density between nuclei

THE MO’s FORMED BY TWO 1s ORBITALS

EEnergy of a 1s orbital in a free atom

Energy of a 1s orbital in a free atom

A B

ADDITION gives an….

Energy more negative than average of original orbitals

Energy more positive than average of original orbitalsSUBTRACTION gives an….

1s

1s*

E1s

1s*

1s

1s

H HH2

The bonding in H2

E1s

1s*

1s

1s

H2: (1s)2

H HH2

E1s

1s*

1s

1s

He2: (1s)2(1s*)2

He HeHe2

The bonding effect of the (1s)2 is cancelled by the

antibonding effect of (1s*)2

The He2 molecule is not

a stable species.

BOND ORDER

= {

A high bond order indicates high bond energy and short bond length.

# of bonding electrons(nb)

# of antibonding electrons (na)

– 1/2 }

A measure of bond strength and molecular stability.

If # of bonding electrons > # of antibonding electrons

Bondorder

the molecule is predicted to be stable

Consider H2+,H2,He2

+,He2……….

= 1/2 (nb - na)

1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

First row diatomic molecules and ions

H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2

—

0

—

—

H2

Dia-

1

436

74

E

1s

1s*

1s

1s

2s

2s*

2s

2s Put the electrons in the MO’s

Li2

ELECTRONS FOR DILITHIUM

E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

(1s)2(1s*)2(2s)2

Li2

Bond Order = 1/2 (nb - na)

= 1/2(4 - 2) =1

A single bond.

E2s

2s*

2s

2s

Li2 (2s)2

Li LiLi2

(1s)2(1s*)2 assumed

Only valence orbitals contribute to molecular bonding

E2s

2s*

2s

2s

Be2Be BeBe2

Electron configuration for DIBERYLLIUM

Configuration: (2s)2(2s*)2 Bond order?

B2

The Boron atomic configuration is

1s22s22p1

form molecular orbitals.

So we expect B to use 2p orbitals to

How do we do that???

Combine them by addition and subtraction

BUT … remember there are 3 sets of p-orbitals to combine

molecular orbitals

2p* antibonding

2p bonding

+-

-+ +

-

-

ADD

SUBTRACT

The molecular orbitals.

2p* antibonding

2p bonding

+ -

+-

-

+ADD

SUBTRACT

The molecular orbitals.

The M.O.’s formed by p orbitals

2p*

2p

2p

2p*

The do not split as much as the because of weaker overlap.

E2p 2p

Combine this with the s-orbitals…..

E

Expected orbital splitting:

2s

2s*

2s

2s

2p

2p*

2p

2p

2p

2p*

The do not split as much because of weaker overlap.

But the s and p along the internuclear axis DO interact

This pushes the 2p up..

E

MODIFIED ENERGY LEVEL DIAGRAM

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p* interaction

Notice that the 2p and 2p

have changed places!!!!

Now look at B2...

E

2s

2s*

2s

2s

Electron configuration for B2

2p

2p*

2p2p

2p

2p*B is [He] 2s22p1

E

Electron configuration for B2:

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

(2s)2(2s*)2(2p)2

Abbreviated configuration

(1s)2(1s*)2(2s)2(2s*)2(2p)2

E

2s

2s*

2s

2s

Bond order

2p

2p*

2p2p

2p

2p*Molecule is predicted to be stable and paramagnetic.

1/2(nb - na)

= 1/2(4 - 2) =1

SECOND ROW DIATOMICS

B2 C2 N2 O2 F2

E

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

Li2

O O

Back to Oxygen

2p*

2p*

2p

2p

2s*

2s

E

12 valence electrons

BO = 2 but PARAMAGNETIC

BUT REMEMBER …THE LEWIS STRUCTURE WAS DIAMAGNETIC

2p*

2p*

2p or 2p

2por 2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2

Para-

1

290

159

C2

Dia-

2

620

131

N2

Dia-

3

942

110

O2

Para-

2

495

121

F2

Dia-

1

154

143

E

NOTE SWITCH OF LABELS

Example: Give the electron configuration and bond order

for O2, O2+ , O2

- & O22-. Place them in order of bond

strength and describe their magnetic properties.

Step 1:Determine the number of valence electrons in each:

O2+ : 6 + 6 - 1 = 11

O2– : 6 + 6 + 1 = 13

O22- : 6 + 6 + 2 = 14

O2 : 6 + 6 = 12

Step 2: Determine the valence electrons configurations:

O2 : (2s)2(2s*)2 (2p)2(2p)4 (2p*)2

O2+ :

O2– :

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

O2 : (2s)2(2s*)2 (2p)2 (2p)4(2p*)2

O2+ : (2s)2(2s*)2 (2p)2 (2p)4(2p*)1

O2– : (2s)2(2s*)2 (2p)2 (2p)4(2p*)3

O22- : (2s)2(2s*)2 (2p)2 (2p)4(2p*)4

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

O2 : B.O. = (8 - 4)/2 = 2

O2+ : B.O. = (8 - 3)/2 = 2.5

O2– : B.O. = (8 - 5)/2 = 1.5

O22- : B.O. = (8 - 6)/2 =

1

Step 3: Determine the bond orders of each species:

2s

2s*

2s

2s

E

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

HETERONUCLEAR DIATOMICS

2p*

2p*

2p

2p

2s*

2s

E

NITRIC OXIDE (NO)Number of valence electrons: 5 + 6 = 11

USE THE MO DIAGRAM FOR HOMONUCLEAR DIATOMIC MOLECULES WITH s-p INTERACTION AS AN APPROXIMATION FOR < 12 ELECTRONS

Put the electrons in…..

Molecule is stable and paramagnetic.

NITRIC OXIDE (NO)

2p*

2p*

2p

2p

2s*

2s

E

Bond order 522

38.

Experimental data agrees.

NO+ and CN-

Ions are both stable and diamagnetic.

NO+: Number of valence electrons: 5 + 6 - 1 = 10

CN–: Number of valence electrons: 4 + 5 + 1 = 10

ISOELECTRONIC

2p*

2p*

2p

2p

2s*

2s

EBond order 03

2

28.

Experimental data agrees.

TRIPLE BOND

CAN NeO EXIST?

How can we answer this question?

Check bond order……...

Therefore …. It could exist.

NeO: Number of valence electrons: 8 + 6 = 14

2p*

2p*

2p

2p

2s*

2s

E

Bond order 012

68.

SINGLE BOND