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***Test Dates: Syllabus onCLEW***

Test Dates: There will be no make-uptests or exams!

Test 1:October 4 - during class time

Test 2: November 6 - during class time

Final Exam: Saturday, Dec. 13, 12:00

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Atomic Structure

The hydrogen atom, wavefunctionsand orbitals, many electron atoms,

spin, Slaters rules and effectivenuclear charge.

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Lecture 1, lets start at1H!

Hydrogen economy old news?

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Making and Moving H2 Nearly all made from

hydrocarbons Most used on site

Ways of the futuremake for good mediacoverage Well see

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Start out with a Big Bang

Universe is 75% hydrogen and 23% helium. Irony? Has been since about since about 3 minutes after big bang

Surprised? Not really reflected by Earths composition.

BIGBANG

proton (p+)

neutron (n)

and

electron (e-)

soup

n + p+ 2H+ Deuterium

n + 2D+ 3H+ Tritium

p+ + 2D+ 3He2+ 3-Helium

2D + 2D 4He2+

3

H

+

+ p+

4

He

2+

3He2+ + n 4He2+4-Helium

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Abundance of Elements

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Stars at Work

Our sun is young, and too small to make heavy nuclei Big old stars convert He to larger nuclei At around Fe, these processes are no longer favourable Our solar system is recycled stardust.

p+ + p+ 2H+e

+

e

2H+ + p+ 3He2+

3He2+ + 3He2+ 4He2+ + 2 p+

YOUNG STARS

3 4He 12C

BIGGER STARS

etc up to 56Fe

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Quantum Prep:Particle in a box

States defined by asingle quantumnumbern.

Energy is neverzero

Energy dependson n2

8

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Back to Hydrogen

Expt: Ionize H2 into H atoms.Populate excited states andwatch emissions.

This is in the visible region What is the pattern?

(nm)

656 red

486 blue-green434 blue

410 violet

397 ultraviolet

22

2

256.364

m

m

Where m is an integer > 2.

Balmer series (1855)

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Bohr Model 1. Rydberg, 1890, slightly

different presentation to

Balmer

22

11

mnhvE

H

Bohr foundBalmers/Rydbergs

numericalsolution in 1913,

and arrived at a planetarymodel

Bohr was well aware of the issues; a circulating electronshould emit radiation and spiral into nucleus.

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Bohr Model 2. Bohrs wild guess: certain

orbits are stable - restrictangular momentum L.

Can equate electrostaticattraction to centrifugalforce to figure out moreabout these orbits.

2

hnmvrL

De Broglie would later interpret this as astanding wave condition for an electron(1923 thesis lead to Nobel Prize)

m = electron massv= electron velocityr= radius of orbit

n = principle quantum numberZ = nuclear chargeE

Zn

Rn

Rn

2

2 2

1radius(n) = n2a0a0 = Bohr radius= 0.529

R

e

2

4

2 4

0

2 2

h = 13.6 eV

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H E

= Rnl(r)Y

lml(,)

Rnl(r) radial function

Ylml

(,) angular function

polar coordinates

= wave function2 = probability (electron) density

3-D: Wave functions andOrbitals (Hydrogen 1925)

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The Solution to the Hydrogen AtomSchrodinger Equation

Three quantum numbers describe theorbitals that are essentially standing wavesolutions

n= principle quantum number (1, 2, 3, 4)l= orbital quantum number (0, 1, 2 n-1)

ml= magnetic quantum number (valuesbetweenland +l)

n comes strictly from radial component offunction, and is analogous to Bohrs model

The land mlquantum numbers come from

the angular components of the function

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Principle Quantum Number (n)Determines energy of orbital

NOTE: Only true for hydrogen;

energy of 2s =2p, etc.n= 1, 2, 3 n = 1 : ground staten = 2 : first excited state

n = 3 : second excited state

En

n

13 6

1

2.

E1 = -13.6 eV

E2 = -3.40 eV

E3 = -1.51 eV

E4 = -0.85 eV

E5 = -0.54 eVE = 0 eV

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0

-500

-1000

-1500Energ

y(kJ/mol)

1s

2s3s4s

2p3p4p

3d4d 4f

Quantum Numberll= azimuthal, orbital angular momentum

degeneracy, shape SUBSHELLl= 0, 1, 2 n1 degeneracy = 2 l+ 1

Degenerate states have the same energy

l 0 1 2 3 4, 5, 6

type s p d f g, h, i degeneracy 1 3 5 7 9, 11, 13

s = sharpp = principald = diffuse

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ml= magnetictype, orientation in space

ml= 0, 1, 2 l

# of orbitals in a subshell = 2 l+ 1

Magnetic Quantum Number (ml)

l ml orbital

0 0 s

1 0 pz

1 px1 py

2 0 d2z2-x2 -y2 = dz2

1 dxz1 dyz2 dx2y2

2 dxy

2p orbitals

y

x

z

y

x

z

y

x

z

px py pz

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The 1s Orbital n = 1, l= 0, m

l= 0

No nodes

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The 2s and 3s Orbitals: Nodes n = 2, l= 0, m

l= 0

n = 3, l= 0, ml= 0

Number of nodes = n 1Number of radial nodes = nl1Number of angular nodes = l

Nodes are surfaces where the values of the function is zero

The s orbitals only have radial nodes

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The p Orbitals n = 2, l= 1, m

l= -1,0,+1

Has one node throughnucleus (2p)

y

xz

y

xz

y

xz

px py pz

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The 3p Orbital n = 3, l= 1, m

l= -1, 0, +1

Seems like 3 nodes, butactually 2 (n-1 = 2) 1 radial, 1 angular node

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3dz2

z-axis is blue

The 3d Orbitals

n = 3, l= 2,

ml= -2,-1, 0, +1, +2

4 look the same, 1looks different (dz2)

2 nodes (n-1)=2

z

xy

z

xy

z

xy

dxz

dz2dx2-y2

z

xy

dxy

z

xy

dyz

3dxz

y-axis is green

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Many Electron Atoms

Something is wrong with our model!

+

--

EZ

nR

n

2

2

E1

1

R22

2

He, Z = 2

Predict: E1 = -54.4 eV

Actual: E1 = -24.6 eV

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Electron spin (ms)

ms = spin electron spinms = (- = ) (+ = )

Pauli exclusion principle:

Each electron must have a unique set of quantum numbers.Two electrons in the same orbital must have opposite spins.

Electron spin is a purely quantum mechanical concept.

N

S

H

E L l Di f H

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Energy Level Diagram for He.Electron Configuration: 1s2

N

S

H

N

S

He

paramagnetic one (more) unpaired electrons

diamagnetic all paired electrons

Energy

1

2

3

0 1 2

n

l

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Effective Nuclear Charge, Z*

The presence of other electrons around a nucleus screens

an electron from the full charge of the nucleus.

We can approximate the energy of the electrons bymodifying the Bohr equation to account for the lower

effective nuclear charge:

EZ

nR

n

*2

2

Z* = Z-

Z* is the effective nuclear chargeZis the atomic number

is the shielding or screening constant

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Helium Z = 2

Predicted: E1 = -54.4 eV

Actual: E1 = -24.6 eV- 24.6 -13.6

1

2

2

Z *

Z*.

.

24 6 1

13 6

2

Z* = 1.341.34 = 2 -

= 0.66

+

--

EZ

nR

n

*2

2

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Lithium Z = 3

Predicted: E2 = -30.6 eV

Actual: E2 = -5.4 eV-5.4 -13.6

2

2

2

Z*

Z*.

.

5 4 2

13 6

2

Z* = 1.261.26 = 3 -

= 1.74

+-

-

-

EZ

nR

n

*2

2

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Predicting and Z*Slaters rules for the prediction of for an electron:

1. Group electron configuration as follows:(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p) etc.

2. Electrons to the right (in higher subshells and shells) of an electron donot shield it.

3. If the electron of interest is an ns ornp electron:a) each other electron in the same group contributes 0.35 (0.30 for 1s)b) each electron in an n-1 group contributes 0.85c) each electron in an n-2 or lower group contributes 1.00

4. If the electron of interest is an nd ornf electron:a) each other electron in the same group contributes 0.35b) each electron in a lower group (to the left) contributes 1.00

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Example: Oxygen Valence Shell

Z* = Z-

O, Z= 8

Electron configuration: 1s2

2s2

2p4

a) (1s2) (2s2 2p4)

b) = (2 * 0.85) + (5 * 0.35) = 3.451s 2s,2p

Z* = Z-

Z* = 8 3.45 = 4.55

This electron is actually held with about 57% of theforce that one would expect for a +8 nucleus.

E l ith T El t f

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Example with Two Electrons forNickel:

Z* = Z-

Ni, Z= 28

Electron configuration: 1s2

2s2

2p6

3s2

3p6

4s2

3d8

(1s2) (2s2 2p6) (3s2 3p6) (3d8) (4s2)

For a 3d electron: = (18 * 1.00) + (7 * 0.35) = 20.45

1s,2s,2p,3s,3p 3d

Z* = Z- Z* = 28 20.45 = 7.55

For a 4s electron: = (10 * 1.00) + (16 * 0.85) + (1 * 0.35) = 23.95

1s,2s,2p 3s,3p,3d 4s

Z* = Z- Z* = 28 23.95 = 4.05

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The basis of Slaters rules for

s and p orbitals have better penetration to the nucleus than

d (or f) orbitals for any given value ofni.e. there is a greater probability of s and p electrons beingnear the nucleus

This means:1. ns and np orbitals completely shield nd orbitals2. (n-1) s and p orbitals dont completely shield n s and p

orbitals

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Z* on valence electrons

For the elements shown here, the effective nuclearcharge on the valence electrons never exceeds 8.25,despite nuclear charges up to 86 (radon)!

H1.00

He1.65

Li1.30

Na2.20

K2.20

Rb2.20

Cs2.20 Ba2.85

Sr2.85

Ca2.85

Mg2.85

Be1.95

B2.60

Al3.50

Ga5.00

In5.00

Tl5.00 Pb5.65

Sn5.65

Ge5.65

Si4.15

C3.25

N3.90

P4.80

As6.30

Sb6.30

Bi6.30 Po6.95

Te6.95

Se6.95

S5.45

O4.55

F5.20

Cl6.10

Br7.60

I7.60

At7.60 Rn8.25

Xe8.25

Kr8.25

Ar6.75

Ne5.85

Shi ldi d Eff ti N l

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Shielding and Effective NuclearCharge

The energy of valence electrons in an atom/ionchanges with the loss of addition of an electron.

Slaters rules are only approximate and can give poor

predictions. For example:

They ignore the differences in penetration between s and porbitals. Real s and p orbitals do not have the same energy(except H).

They assume that all electrons in lower shells shield outerelectrons equally effectively.

Z* can be used to estimate ionization energy:H 13.6

ie

2

2

Z*

n

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Summary 4 quantum numbers describe orbitals: n, l, m

l,

ms

The s, p and d-orbitals, nodes, radial

distribution function, degeneracy

Effective nuclear charge and Slaters rules

NEXT LECTURE: Hunds rules, trends: atomicradii, ionization enthalpy

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What is an eV? Good question. A single atom is such a small thing that to talk

about its energy in joules would be inconvenient.But instead of taking a definite unit in the samesystem, like 1020 J, [physicists] have unfortunatelychosen, arbitrarily, a funny unit called anelectronvolt (eV) ... I am sorry that we do that, but

that's the way it is for the physicists.

Richard Feynman, in a 1961 recorded lecture.

1eV = 1.60217653(14)1019 joules

An electron volt is the kinetic energy gained by an electronwhen it is accelerated through an electrostatic potential of 1 V.