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

My office hours next week

• Wednesday Dec 8: 10-12 AND 2-4

• Friday Dec 10: 10-12 AND 2-4

Planck Postulated

“forced to have only certain discrete values”

“Energy can only be transferred in discrete quantities.”

hE is the frequency of the energy

h is Planck’s constant, 6.626 x 10-34 J s.

Energy is not continuous

Planck…….

Energy is quantized

THE PHOTOELECTRIC EFFECT

light electron

metal

KE

of

elec

tro

n

Frequency of light ()0

When <0, no electrons are ejected at any light intensity.

KE of the ejected electrons depends only on the

light’s frequency

When >0, the number of electrons is proportional to the light intensity.

This lead Einstein to use Planck’s idea of quanta

EINSTEIN POSTULATED

“Electromagnetic radiation can be viewed as a stream of particle-like units called photons.”

Energy of a Photon: hE

hc

hE

The energy of the photon depends upon the frequency

c

Diffraction of an electron beam….

ph

mh v

We can relate these spacings to the electronwavelength

De Broglie:

SPECTROSCOPYEMISSION

Sample heated.

Many excited states populated

SPECTROSCOPYEMISSION

Sample heated.

Many excited states populated

n = 1 Ground state

n = 2

n = 3n = 4n = Ion8

Excited states

...

En

erg

y

The spectrum…..

• We will describe atoms and molecules using wavefunctions, which we will give symbols … like this:

• These wavefunctions contain all the information about the item we are trying to understand

• Since they are waves, they will have wave properties: amplitude, frequency, wavelength, phase, etc.

• We obtain the energy by performing the “energy operation” on the wavefunction – the result is a constant (the energy) times the wavefunction

H

• This equation is called the Schrodinger wave equation (SWE)

• Let’s see how this might work

• So H = KE operator + PE operator

H =

• H

2(h2 / 8m) + V

2(h2 / 8m) + V{ }

PARTICLE IN A BOX

ENERGY

0 L

(0) = 0 (L) = 0

BOUNDARY CONDITION

x

(x) A kx sin

PARTICLE IN A BOXThe SCHRODINGER WAVE EQUATION for a

h

m

d

dxE

2

2

2

28

for a 1-D particle with no potential acting on it!

THE HEISENBERG UNCERTAINTY PRINCIPLE

p is the uncertainty in the particle’s momentum

x is the uncertainty in the particle’s position

For a particle like an electron, we cannot know

both the position and velocity

to any meaningful precision simultaneously.

x ph

4

Electrons have both wavelike and particle like properties.

Because of the wavelike character of electron

we CANNOT say that an electron

WILL be found at certain point in an atom!

WAVE-PARTICLE DUALITY

What does Mean?????

The answer lies in

PARTICLE IN A BOX

n 1 2 3, , .......

The probability of finding the particle on a segment of the x-axis of length dx surrounding the point x is….

2dxIf we sum all of these infinitesimal probabilities, thetotal must be equal to one, since there must be some finite probability of finding the particle somewhere..

n(x) A

n

Lx sin

1

2

3

0 L

ENERGY OF A PARTICLE IN A BOX

n =1

n = 2

n = 3

EN

ER

GY

HALF-WAVELENGTHS

INTEGRAL NUMBER OF

2L

n

n =1

n = 2

n = 3

ENERGY

12

22

PROBABILITIES

2

ZERO

32

NO CHANCE OF FINDING ELECTRON

AND….

x

Proton at (0,0,0)

y

z

Electron at (r

2(h2 / 8m) + V{ }

Where now has terms in {d2/dr2 ; d2/d2 ; d2/d2}

and V = Ze2 / r

2

After the transformation we stillhave the Schrodinger equation

• The result of solving the Schrodinger equation this way is that we can split the hydrogen wavefunction into two:

(x,y,z) (r,) = R(r) x Y()

Depends on r onlyDepends on angular variables

• The solutions have the same features we have seen already:– Energy is quantized

• En = R Z2 / n2

= 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …]

– Wavefunctions have shapes which depend on the quantum numbers

– There are (n-1) nodes in the wavefunctions

• Because we have 3 spatial dimensions, we end up with 3 quantum numbers:

n, l, ml

• n = 1,2,3, …; l = 0,1,2 … (n1); ml = l, l+1, …0…l1, l

• n is the principal quantum number – gives energy and level

• l is the orbital angular momentum quantum number – it gives the shape of the wavefunction

• ml is the magnetic quantum number – it distinguishes the various degenerate wavefunctions with the same n and l

n l ml

1 0 (s) 0

2 0 (s) 01 (p) -1, 0, 1

3 0 (s) 01 (p) -1, 0, 12 (d) -2, -1, 0, 1, 2

The result (after a lot of math!)

A more interesting way to look at things is by using the radial probability distribution, which gives probabilities of finding the electron within an annulus at distance r (think of onion skins)

max. away from nucleus

The Radial Probability Distribution for the 3s, 3p, and 3d Orbitals

Another quantum number!

Electrons are influenced by a magnetic field as though they were spinning charges.They are not really, but we think of them as having “spin up” or “spin down” levels.These are labeled by the 4th quantum number: ms, which can take 2 values.

1s

E

2s2p

3s3p

3d4s

4p5s

4d

Remember the energies are < 0

THE MULTI-ELECTRON ATOM ENERGY LEVEL DIAGRAM

THE PAULI PRINCIPLE

An orbital is described by three quantum numbers,

each orbital may contain a maximum of two

electrons, and they must have opposite

spins.

No two electrons in the same atom can havethe same set of four quantum numbers (n, l, ml , ms).

Then each electron in a given orbital

must have a different ms

HOW MANY ELECTRONS IN AN ORBITAL?

1s 2s 2p

Ne: 1s22s22p6

1s 2s 2p

F: 1s22s22p5

1s 2s 2p

O: 1s22s22p4

1s 2s 2p

N: 1s22s22p3

Nitrogen

Oxygen

Fluorine

Neon

THE ELECTRON CONFIGURATIONS FOR NITROGEN TO NEON

0

0.1

0.2

0.3

0.4

0.5

0.6

0 5 10 15 20 25 30

SCREENING AND PENETRATION

THE 1s close to the nucleus

PENETRATES WELL SEES A CHARGE OF Z=2

EFFECTIVE NUCLEAR CHARGE…...

Is

3p

THE 3p DOES NOT SCREEN THE NUCLEUS

0

0.02

0.04

0.06

0.08

0.1

0.12

0 5 10 15 20 25 30

THE 3s orbital penetrates better than 3p orbital

3s

3p

3d

The 3p orbital penetrates better than 3d orbital

Zeff(s) > Zeff(p) > Zeff(d)

the metals that fill the d orbitals in their valence

shell.

TRANSITION METALS

HUND’S RULE OBEYED FOR ALL EXCEPT

Cr and Cu

Cr: [Ar] 4s23d4EXPECTED

OBSERVED Cr: [Ar] 4s13d5 …. The d-shell is ½ filled this way; all spin up

WHEN n=3

FOR COPPER

1

2 13 14 15 16 17

18

1

2

3

4

5

6

7

1s 1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

6d

3d

4d

5d

4f

5f

La

Ac

Zeff INCREASES

RADIUS DECREASES

DOWN GROUP...

1

2 13 14 15 16 17

18

1

2

3

4

5

6

7

1s 1s

2s

3s

4s

5s

6s

7s

2p

3p

4p

5p

6p

6d

3d

4d

5d

4f

5f

La

Ac

Zeff D

EC

RE

AS

ES

RA

DIU

S IN

CR

EA

SE

S

TRENDS IN EA

ELECTRON AFFINITYMORE NEGATIVE

EL

EC

TR

ON

AF

FIN

ITY

MO

RE

NE

GA

TIV

E

EL

EC

TR

ON

AF

FIN

ITY

MO

RE

NE

GA

TIV

E

TRENDS IN FIRST IE

Electrons closer to nucleus more tightly held

Zeff D

EC

RE

AS

ES

Zeff INCREASES UP THE GROUP ION

IZA

TIO

N E

NE

RG

Y

ION

IZA

TIO

N E

NE

RG

YFirst Ionization

energies decrease down the group

TRENDS IN FIRST IE

Greater effective nuclear charge across period

Poor shielding by electrons added

Zeff INCREASESI E A

Z

neff. . 2

2

IONIZATION ENERGY

ION

IZA

TIO

N E

NE

RG

Y

Based on the idea that all electron pairs repel

each other.

we can find the molecular shape!

Lets see how it works…...

VALENCE SHELL ELECTRON PAIR REPULSION:

VSEPR

The bonding and lone pairs push apart as far as possible……..

This means that atoms bound to a central atom are as far apart as possible…….

COMBINING ORBITALS TO FORM HYBRIDS

HYBRIDIZATION :

the combination of two or more “native” atomic

orbitals on an atom to produce “hybrid” orbitals

RULE: the number of atomic orbitals that are combined must equal the number which are formed

All resulting hybrid orbitals are identical.

The bondingframe work of lactic acid

LACTIC ACID

C C H

H

H

H

C

O

H

O

OH

HH

H

C

C

O

HH

C

O

O

LACTIC ACID

H

THE MO’s FORMED BY TWO 1s ORBITALS

The molecular orbitals.

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

2s

2s*

2s

2s

E

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

Double bonds involve interacting p orbitals, outside of the bonding line

p- bondingspread overwhole molecule

p- antibondingp- non-bonding

Benzene - aromatic molecules

Figure 9.48The Pi System for Benzene