### Polyelectronic atoms

```Many electrons systems
Polyelectronic atoms
1
Two-electron atoms
Electron
2
électron
Electron
électron 1
-e
1
r 12
2
-e
r2
r1
noy au +Ze
Nucleus
+Ze
Write the Schrödinger Equation for He or Li+
2
Two-electron atoms
Electron
2
électron
Electron
électron 1
-e
1
r 12
2
-e
r2
r1
noy au +Ze
Nucleus
+Ze
T
V
E
3
T is positive
V contains positive and negative contributions
Attraction: negative
Attraction: negative
repulsion: positive
E is negative (for bounded states)
4
T1 acts on e1 and does not on e2
T2 acts on e2 and does not on e1
V contains positive and negative contributions
Vn,e1 acts on e1 and does not on e2
Vn,e2 acts on e2 and does not on e1
Ve1,e2 acts on both
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).
5
First approach: Complete neglect of Ve1,e2
Consequence: We neglect a repulsion
overestimation of the atom stability
Atomic orbitals are solutions for one electron: Y1s
Let try the product of two orbitals: Y1s2 = Y1s(e1) Y1s(e2)
Sum of
monoelectronic
operators
Sum
6
Conclusions
Orbitalar Approximation
•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.
Y1s2s (e1,e2) = Y1s(e1) Y2s (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
7
Ionization Potential
He
He+
He+ + eHe2+ + e-
IP = 24.5 eV
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
AA + eEA = IP of the negative ion
Defined to be positive when the anion is stable
HH + e- = 0.77 eV
not equal to 13.6 eV
2-e repulsion differs
EA may be positive or negative
8
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) experimentally
The second IP is again that of the hydrogenoid
9
Third approach: The Slater Model
Each monoelectronic operator is
the hamiltonian for the hydrogenoid,
replacing Z by Z*=Z-s
= 2 (Z-s)2 E1s(H) = - 77.69 eV
IP = 23.29 eV assuming s= 0.31
10
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 f1s(1) f1s(2) f2s(3) f2s(4) f2p(5) f2p(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: f1s(1) f1s(2) - f1s(2) f1s(1)
The exchange of two electrons should give
the same expression changing sign.
A requirement for the Pauli Principle
11
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.
12
The Slater model
John Clarke Slater (1900-1976)
Generalization of the method used for 2-electrons
Requires defining the screen factors, s
Allows respecting the Pauli principle and the
principle of stability
It provides its own ordering of atomic levels.
Z* = Z - Ss
Sum over the n-1 electrons
screening the one that we
consider
13
Screening factors
Screening Electron i are classified in families: I1sI2spI3spI3dI4spI4dI…
No distinction
between s and p
i
i
j
Electron j in which we are interested
If i is inside (closer) it is screening
If i is outside, it has no influence (Gauss theorem)
s ranges from 1 to 0.
14
Screening factors
Screening Electron i
Electron j in which we are interested
15
We thus calculate atomic orbitals,
Summing electron contributions,
we calculate atomic properties
Application
IP for C
C → C+ + eIP = E(C+)-E(C)
1s22s22p1
1s22s22p2
16
We thus calculate atomic orbitals,
Summing electron contributions,
we calculate atomic properties
Application
IP for C
C → C+ + eIP = E(C+)-E(C)  E2p
1s22s22p1
1s22s22p2
The atomic functions have the same shape but differ !
17
C → C+ + eIP = 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 charge
Not the number of electrons
It never varies
18
C → C+ + eIP = 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 shell
varies
# of electron in the shell
Varies minus the one
considered - varies
19
C and C+ orbitals
Z characterizes the nucleus charge
And not the number of electrons
Never varies
20
C and C+ orbitals
# of electron in the shell
minus the one considered
- varies
21
trends
Z* increases for negatively charged species
Z* increases for orbitals of the same shell
n → n+1
Z* → Z*+(1-s)
Energy decrease with n*
Slater values originate from an ab initio energy minimization (variation principle).
Effective quantum number
n
1
2
3
4
5
6
n*
1
2
3
3.7
4
4.2
22
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)
23
The Klechkovsky rule
n
s
p
d
f
1s
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
9
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
Table of n + l values
6s
6p
24
Exceptions to the Klechkovsky rule
22s22p63s23p63d104s24p64d10
Pd
IS
1s
46
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
25
Hund(s) rule
When the filling of a shell is incomplete, ground
states are high spin states with the maximum
number of electrons with the same spin.
2s
2p
C
No order among the
p levels here
26
Paramagnetism - diamagnetism
Diamagnetism = 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?
27
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
28
are core electrons
lines
Start a new period filling each new ns type orbital
columns
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)d10npc12
29
Building the Mendeleev table
1s2
2s22p6
3s23p6
4s23d104p6
5s24d105p6
6s24f145d106p6
7s25f146d107p6
configuration
2
+8 =10
+8 = 18
+18 = 36
+18 = 54
+32 = 86
+32 = 118
number of e
core electrons for the next period
Total number of electrons =
Z (for atoms)
core
2-10-18-36-54-86
+
of the rare gas
valence
n° of column
30
alkaline
Alkaline
earth
Transition
metals
http://www.webelements.com/
Noble
metals
Halogens
Rare
gases
31
Where is Ga(Z=31) in the table?
Where is Hg(Z=80) in the table?
32
Where is Ga(Z=31) in the table?
Column 13
31 = 18 + 13
18<31<36
Row 4
Where is Hg(Z=80) in the table?
Column 12
80 = 54 + 14 + 12
54<80<86
Row 6
f electrons
33
Trends
deduced from the Slater model
Electronegativity increases from left to right
Z* increases for orbitals of the same shell
n → n+1
Z* → Z*+(1-s)
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 bottom
Compare 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*
34
IP from Slater
30
Pot ent iel d'ionis at ion ( eV) Valeurs expériment ales
Ionization potentials
Hunds rule
20
n=2
10
n=3
nombre d'élect ron de v alence
Number
of valence electrons
Completely
filled column
0
1
2
3
4
5
6
7
8
35
D. Mendelejeff, Zeitschrift für Chemie 12, 405-6
(1869)
The opposite orientation of now
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.
36
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.
37
Better than just a classification, a
periodic table!
An unknown compound predicted
? M=78.2 V=19 density=4.6
Se M=78.8 V=17.2 density=4.6
Dmitri Mendeleev
Russian (Siberia – St Petersbourg)
(1834 - 1907)
S
As
?
Br
Te
38
127
126
59
58
52Te
53I
27Co 28Ni
Dmitri Mendeleev
Russian (Siberia – St Petersbourg)
(1834 - 1907)
Switch between atom classification because of chemistry
(valency)
39
Antoine Laurent
Lavoisier
(1743-1794)
Better, than a list a
classification
40
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
41
(magnesium oxide), baryta (barium oxides), alumina (aluminum oxide),
and silica (silicon dioxide).
Forming ions is endothermic
except if environment is stabilizing
Na
1 électron à -7.31 eV
Energy loss 1.23 EV
(Slater)
Cl
-
1.53 eV exp.
Cl
8 électrons à -49.96 eV
In a dipole (d=2.56 a0) the
stabilizing energy is 10.63
eV
7 électrons à -56.23 eV
42
Atomic orbital levels
E = - IP
Tjalling C. Koopmans
Dutch
Nobel Prize in Economic Sciences
in 1975
1s
2s
2p
H
He
13.6
24.25
Li
Be
B
C
N
O
F
58
115
192
268
406
538
654
5.4
9.3
12.9
16.6
20.3
28.5
37.9
8.3
11.3
14.5
13.6
18.4
43
Values from Extended Hückel
(-eV)
2s
H
13.6
Li
5.4
He
24.25
Be
10
B
15.2
C
21.4
N
26
O
32.3
F
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
1s
Roald Hoffmann
44
Is abundance of atoms related to
their stability?
Why?
What are the most abundant atoms
on the earth surface?
In the human body?
45
Relative abundance of elements
on earth crust
46
O
ppb by weight
Abundance
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
C
Si
Abundance
ppb by weight
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
Abundance
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
47
Abundance (masses) in human body; the human body is made of
65 to 90% of water
element
abundance
Carbon
18%
Hydrogen
10%
Nitrogen
3%
Calcium
1.5%
Phosphorus
1.0%
Potassium
0.35%
Sulfur
0.25%
Sodium
0.15%
Magnesium
0.05%
Copper, Zinc, Selenium, Molybdenum, Fluorine, Chlorine, Iodine, Manganese, Cobalt, Iron
0.70%
trace amounts
48
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
49
4 steps after the big bang
• 1) 0-100 seconds after
George Gamow
in 1948
Russian-American
1904-1968
– 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 hn that 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
50
universe (1 He for 12 H everywhere)
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
51
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.
52
53
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