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# 1 Polyelectronic atoms Many electrons systems. 2 Two-electron atoms Electron 1 Electron 2 Nucleus...

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1 Polyelectronic atoms Many electrons systems
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1

Polyelectronic atoms

Many electrons systems

2

Two-electron atomsélectron 2

noyau +Ze

électron 1

r1r2

r12Electron 1

Electron 2

Nucleus

+Ze

- e

- e

Write the Schrödinger Equation for He or Li+

3

Two-electron atomsélectron 2

noyau +Ze

électron 1

r1r2

r12Electron 1

Electron 2

Nucleus

+Ze

- e

- e

TVE

4

T is positive

V contains positive and negative contributions

E is negative (for bounded states)

Attraction: negative

repulsion: positive

Attraction: negative

5

T1 acts on e1 and does not on e2

V contains positive and negative contributions

Ve1,e2 acts on both

T2 acts on e2 and does not on e1

Vn,e1 acts on e1 and does not on e2

Vn,e2 acts on e2 and does not on e1

This is the difficult part that couples electrons.

The coupling of the electron is called “the electronic correlation”. Each electron depends on the position of the other electron (not only on its average distribution).

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First approach: Complete neglect of Ve1,e2

Consequence: We neglect a repulsion overestimation of the atom stability

Atomic orbitals are solutions for one electron: 1s

Let try the product of two orbitals: 1s2 = 1s(e1) 1s(e2) Sum of monoelectronic operators

Sum

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Conclusions

•The atomic wave function is a 2-electron function.

•The product of two orbitals (1-e function) is a solution •The corresponding energy is the sum of the energies.

1s2s (e1,e2) = 1s(e1) 2s (e2) E = E1s + E2s

1s2s is an atomic configuration.

Its energy is higher than that for 1s2

1s2 is the ground state; 1s2s is an excited state

Orbitalar Approximation

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Ionization Potential

He He+ + e- IP = 24.5 eVHe+ He2+ + e- IP = 54.4 eV

The second IP is that of the hydrogenoid: Z2(-13.6) eV = 54.4 eV no approximation

The first IP is wrong ! Electron Affinity

A- A + e- EA = IP of the negative ionDefined to be positive when the anion is stableH- H + e- = 0.77 eV not equal to 13.6 eV

2-e repulsion differs EA may be positive or negative

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Second approach: Ve1,e2 replaced by it mean value: -5/4 Z (E1sH)

Replacing the repulsion by a constant still allows the orbitalar approximation

-5/4Z(13.6) is a constant and do not depend on the electron position.

The first IP = 20.4 eV (24.5 eV) experimentallyThe second IP is again that of the hydrogenoid

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Third approach: The Slater Model

Each monoelectronic operator is the hamiltonian for the hydrogenoid,

replacing Z by Z*=Z-2 (Z-)2 E1s(H) = - 77.69 eVIP = 23.29 eV assuming = 0.31

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The atom with many electrons

Orbitalar approximation : Every wave function describing a polyelectronic atom will be expressed as a product of atomic orbitals.The expression 1s(1) 1s(2) 2s(3) 2s(4) 2p(5) 2p(6)… describes an atomic configuration.

We neglect the electronic correlation. Electrons are not coupled.

We neglect part of the antisymmetry that should respect the polyelectronic wave function: 1s(1) 1s(2) - 1s(2) 1s(1) The exchange of two electrons should givethe same expression changing sign.A requirement for the Pauli Principle

12

Four rules to determine the atomic configurations for the ground state

•The Pauli principle: fundamental principle of physics = always verified•The principle of stability. Just commonsense. Obvious to have the ground state•The Klechkovsky rule. Practical. Necessary to order the atomic levels. Many exceptions •The Hund rule (s). To remind that for high spin states, one of them is lower in energy

Pauli principle: Two different electrons cannot be in the same state (Two electrons cannot have the same four quantum numbers). This imposes the maximum occupancy for orbitals and spin-orbitals.One orbital is occupied by no more than 2 electrons (with opposite spin). When occupied a spin-orbital has only one electron.

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The Slater model

Generalization of the method used for 2-electrons

Requires defining the screen factors, Allows respecting the Pauli principle and the

principle of stability

It provides its own ordering of atomic levels.

Z* = Z -

Sum over the n-1 electrons screening the one that we consider

John Clarke Slater (1900-1976)

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Screening factorsScreening Electron i are classified in families: I1sI2spI3spI3dI4spI4dI…

Electron j in which we are interested

If i is inside (closer) it is screeningIf i is outside, it has no influence (Gauss theorem) ranges from 1 to 0.

No distinction between s and p

ii

j

15

Electron j in which we are interested

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We thus calculate atomic orbitals, orbital energies and orbital radii.

Summing electron contributions, we calculate atomic properties

ApplicationIP for C

C → C+ + e-

IP = E(C+)-E(C) 1s22s22p1 1s22s22p2

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We thus calculate atomic orbitals, orbital energies and orbital radii.

Summing electron contributions, we calculate atomic properties

ApplicationIP for C

C → C+ + e-

IP = E(C+)-E(C) E2p

1s22s22p1 1s22s22p2

The atomic functions have the same shape but differ !

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C → C+ + e-

IP = E(C+)-E(C) 1s22s22p1 1s22s22p2

Here the 1s2 contribution is the same (modification in the valence shell only)

The 2sp in 2s22p1 energy differs from the 2sp one in 2s22p2

There have not the same Z*

Z characterizes the nucleus chargeNot the number of electronsIt never varies

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C → C+ + e-

IP = E(C+)-E(C) 1s22s22p1 1s22s22p2

Here the 1s2 contribution is the same (modification in the valence shell only)

The 2sp in 2s22p1 energy differs from the 2sp one in 2s22p2

There have not the same Z*

# of electron in the shellvaries

# of electron in the shellVaries minus the one considered - varies

20

Z characterizes the nucleus chargeAnd not the number of electronsNever varies

C and C+ orbitals

21

C and C+ orbitals

# of electron in the shellminus the one considered - varies

22

trendsZ* increases for negatively charged species

Z* increases for orbitals of the same shell

n → n+1 Z* → Z*+(1-)

Energy decrease with n*

Effective quantum number

n 1 2 3 4 5 6

n* 1 2 3 3.7 4 4.2

Slater values originate from an ab initio energy minimization (variation principle).

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The Klechkovsky rule

This rule provides a more reliable ordering of the atomic levels than using the Slater model. It helps building the Mendeleev table. It is not always satisfied.

The atomic orbitals are ordered according to n+l values

For equivalent n+l, they are ordered according to n values: first smaller n (larger l)

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The Klechkovsky rule

n s p d f

l = 0

l = 1

l = 2

l = 3

1 1

2 2 3

3 3 4 5

4 4 5 6 7

5 5 6 7 8

6 6 7 8 9Table of n + l values

1s

2s 2p

3s 3p 3d

4s 4p 4d

5s 5p 5d

4f

5f

6s6p

25

Exceptions to the Klechkovsky rule

46Pd IS 1s22s22p63s23p63d104s24p64d10

IS NOT 1s22s22p63s23p63d104s24p64d85s2.

Filled d band explains; however Ni is 3d84s2

and Pt is 5d96s1

Noble metal atoms: Cu is 3d104s1 Ag 4d105s1 and 5d106s1

Properties are close to alkali

26

Hund(s) ruleWhen the filling of a shell is incomplete, ground states are high spin states with the maximum number of electrons with the same spin.

C

2s 2p

No order among the p levels here

27

Paramagnetism - diamagnetismDiamagnetism = closed shell systems

Paramagnetism = open shell systems

Is an atom with an even number of electron necessarily diamagnetic?

Is an atom with an odd number of electron necessarily paramagnetic?

What is the (l, ml) values for Lithium?

Is Li dia or para? Why?

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Periodicity from IPs

Extraction of s orbitals marks a periodic discontinuity (see H, Li, Na, K, Cs and Rb).

We place on the same column these atoms. The number of electrons in the row (period) is the number of valence electrons. The electrons of the previous rows are core electrons

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lines

Isolobal compounds are placed in a same column

Column c: 1-2 3-12 13-18

Configuration: nsc ns2(n-1)dc-2 ns2(n-1)d10npc-12

Start a new period filling each new ns type orbital

columns

30

Building the Mendeleev table

1s2 2

2s22p6 +8 =10

3s23p6 +8 = 18

4s23d104p6 +18 = 36

5s24d105p6 +18 = 54

6s24f145d106p6 +32 = 86

7s25f146d107p6 +32 = 118

configuration number of e of the rare gas core electrons for the next period

Total number of electrons = core + valence

Z (for atoms) 2-10-18-36-54-86 n° of column

31http://www.webelements.com/

alkaline

Alkaline earth Halogens

Rare gases

Transition metals Noble

metals

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Where is Ga(Z=31) in the table?

Where is Hg(Z=80) in the table?

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Where is Ga(Z=31) in the table?

31 = 18 + 13

Where is Hg(Z=80) in the table?

80 = 54 + 14 + 12

Column 13

18<31<36Row 4

Column 12

54<80<86Row 6 f electrons

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Trends deduced from the Slater model

Electronegativity increases from left to rightZ* increases for orbitals of the same shell

n → n+1 Z* → Z*+(1-)

for Na: Z*=11-8x0.85-2=2.2 E=2.22/32 (E1sH) = -7.31 eV

For Cl: Z*=17-6x0.35-8x0.85-2=6.1 E=6.12/32 (E1sH) = -56.2 eV

Electronegativity decreases from top to bottomCompare Na with K:

for Na: Z*=11-8x0.85-2=2.2 E=2.22/32 (E1sH) = -7.31 eV

For K: Z*=19-8x0.85-10=2.2 E=2.22/3.72 (E1sH) = -4.81 eV

Energy decrease with n*

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IP from Slater

1 2 3 4 5 6 7 80

10

20

30

nombre d'électron de valence

Potentiel d'ionisation (eV) Valeurs expérimentales

n=2

n=3

Ionization potentials

Number of valence electronsCompletely filled column

Hunds rule

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By ordering the elements according to increasing atomic weight in vertical rows so that the horizontal rows contain analogous elements, still ordered by increasing atomic weight, one obtains the following arrangement, from which a few general conclusions may be derived.1. The elements, if arranged according to their atomic weights, exhibit a

periodicity of properties.

2. Chemically analogous elements have either similar atomic weights (Pt. Ir, Os), or weights which increase by equal increments (K, Rb, Cs).

3. The arrangement according to atomic weight corresponds to the valence of the element and to a certain extent the difference in chemical behavior, for example Li, Be, B, C, N, O, F.

D. Mendelejeff, Zeitschrift für Chemie 12, 405-6 (1869)

The opposite orientation of now

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4. The elements distributed most widely in nature have small atomic weights, and all such elements are marked by the distinctness of their behavior. They are, therefore, the representative elements; and so the lightest element H is rightly chosen as the most representative.

5. The magnitude of the atomic weight determines the properties of the element. Therefore, in the study of compounds, not only the quantities and properties of the elements and their reciprocal behavior is to be taken into consideration, but also the atomic weight of the elements. Thus the compounds of S and Tl [Te was intended], Cl and J, display not only many analogies, but also striking differences.

6. One can predict the discovery of many new elements, for example analogues of Si and Al with atomic weights of 65-75.

7. A few atomic weights will probably require correction; for example Te cannot have the atomic weight 128, but rather 123-126.

8. From the above table, some new analogies between elements are revealed. Thus Bo [Ur was intended] appears as an analogue of Bo and Al, as is well known to have been long established experimentally.

38

Dmitri Mendeleev Russian (Siberia – St Petersbourg)(1834 - 1907)

? Br

S

Te

As

An unknown compound predicted

? M=78.2 V=19 density=4.6

Se M=78.8 V=17.2 density=4.6

Better than just a classification, a periodic table!

39

Dmitri Mendeleev Russian (Siberia – St Petersbourg)(1834 - 1907)

52Te 53I

27Co 28Ni

127 126

59 58

Switch between atom classification because of chemistry (valency)

40

Antoine Laurent Lavoisier

(1743-1794) Better, than a list a

classification

41

Antoine Lavoisier (1743-1794) Better, than a list a classification

He classified the known elements into four groups: Elastic fluids

Lavoisier included light, heat, oxygen, nitrogen, and hydrogen in this group.

Nonmetals This group includes "oxidizable and acidifiable nonmetallic elements".

Lavoisier lists sulfur, phosphorus, carbon, hydrochloric acid, hydrofluoric acid, and boric acid.

Metals These elements are "metallic, oxidizable, and capable of neutralizing an acid to form a salt." They include antimony and arsenic (which are not

considered metals today), silver, bismuth, cobalt, copper, tin, iron, manganese, mercury, molybdenum, nickel, gold, platinum, lead,

tungsten, and zinc. Earths

Lavoisier's salt-forming earthy solid "elements" included lime, magnesia (magnesium oxide), baryta (barium oxides), alumina (aluminum oxide),

and silica (silicon dioxide).

42

Forming ions is endothermicexcept if environment is stabilizing

1 électron à -7.31 eV

7 électrons à -56.23 eV

8 électrons à -49.96 eV

Na

Cl

Cl -Energy loss 1.23 EV (Slater)

1.53 eV exp.

In a dipole (d=2.56 a0) the stabilizing energy is 10.63 eV

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Atomic orbital levelsE = - IP

Tjalling C. Koopmans DutchNobel Prize in Economic Sciences in 1975

H He Li Be B C N O F

1s 13.6 24.25 58 115 192 268 406 538 654

2s 5.4 9.3 12.9 16.6 20.3 28.5 37.9

2p 8.3 11.3 14.5 13.6 18.4

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Values from Extended Hückel(-eV)

H He1s 13.6 24.25

Li Be B C N O F2s 5.4 10 15.2 21.4 26 32.3 40

2p 3.5 6 8.5 11.4 13.4 14.8 18.1

Na Mg Al Si P S Cl

3s 5.1 9 12.3 17.3 18.6 20 30

3p 3. 4.5 6.5 9.2 14 13.3 15

Roald Hoffmann

45

Is abundance of atoms related to their stability?

Why?What are the most abundant atoms

on the earth surface?In the human body?

46

Relative abundance of elementson earth crust

47

OAbundance

ppb by weight ppb by atoms

Universe 10000000 800000

Sun 9000000 700000

Meteorite (carbonaceous) 410000000 480000000

Crustal rocks 460000000 600000000

Sea water 857000000 331000000

Stream 880000000 55000000

Human 610000000 240000000

Si

Abundanceppb by weight

ppb by atoms

Universe 700000 30000

Sun 900000 40000

Meteorite (carbonaceous)

140000000 100000000

Crustal rocks 270000000 200000000

Sea water 1000 220

Stream 5000 180

Human 260000 58000

CAbundance

ppb by weight

ppb by atoms

Universe 5000000 500000

Sun 3000000 300000

Meteorite (carbonaceous)

15000000 18000000

Crustal rocks 1800000 3100000

Sea water 28000 14400

Stream 1200 100

Human 230000000 120000000

49

Atom formation

• It is not related to the stability (otherwise rare gas atoms would be abundant)

• It results from the formation of the nuclei

The big bang story

50

4 steps after the big bangabout 13.7 billion years ago.

• 1) 0-100 seconds after– T is cooling and particles form. There are

many particles in a confined space. They meet to form nuclei (later on after expansion it will not be possible; after 3 minutes space is too dilute for nuclear reactions)

– Universe contains a lot of hthat destroy nuclei. (109 for 1 H+); only H and He nuclei are formed.

• 2) 300 000 years after– T is still cooling. Radiations become

ineffective. H and He atoms are formed. They still represent 98% of the mass of the universe (1 He for 12 H everywhere)

George Gamow in 1948 Russian-American1904-1968

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4 steps after the big bang

• 3) 30 000 000 000 years– Stars are formed because of gravitation– They are atom foundries– Atoms are unstable because of shocks. Nuclei meet

again and form larger ones– 3 He make a C (occurrence is weak but time is long in

a confined space: combustion of H forming He takes 9000 millions of years; that of He to C, 300 millions of years

– Fusion are exothermic up to Z=26 (Fe) thus elements (Z =1-100) are made. Beyond Z=26, the star uses its own energy. After a while, stars die

52

4 steps after the big bang

• 4) Stars explose in supernova.– Pieces of Stars feed the universe. Heavy

atoms represent 2% of the total– Universe is cold and atoms are stable; they

gather to make molecules and condense into new stars and planets.

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