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## PowerPoint Slideshow about 'Polyelectronic atoms' - alagan

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### Polyelectronic atoms

Many electrons systems

V contains positive and negative contributions

Attraction: negative

Attraction: negative

repulsion: positive

E is negative (for bounded states)

T1 acts on e1and does not on e2

T2 acts on e2and does not one1

V contains positive and negative contributions

Vn,e1 acts on e1and does not on e2

Vn,e2 acts on e2and 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).

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

- 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.
- 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

He He+ + e- IP = 24.5 eV

He+ 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 ion

Defined to be positive when the anion is stable

H- H + e- = 0.77 eVnot equal to 13.6 eV

2-e repulsion differs EA may be positive or negative

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

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

Orbitalar approximation : Every wave function describing a polyelectronic atom will be expressed as a product of atomic orbitals.The expressionf1s(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

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.

The Slater model

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

John Clarke Slater(1900-1976)

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

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)
- ranges from 1 to 0.

We thus calculate atomic orbitals, orbital energies and orbital radii. Summing electron contributions, we calculate atomic propertiesApplicationIP for CC → C+ + e-IP = E(C+)-E(C)

1s22s22p1

1s22s22p2

We thus calculate atomic orbitals, orbital energies and orbital radii. Summing electron contributions, we calculate atomic propertiesApplicationIP for CC → C+ + e- IP = E(C+)-E(C) E2p

1s22s22p1

1s22s22p2

The atomic functions have the same shape but differ !

C → C+ + e-IP = E(C+)-E(C)

1s22s22p1

1s22s22p2

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

The2sp in2s22p1 energy differs from the2sp one in2s22p2

There have not the same Z*

Z characterizes the nucleus charge

Not the number of electrons

It never varies

C → C+ + e-IP = E(C+)-E(C)

1s22s22p1

1s22s22p2

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

The2sp in2s22p1 energy differs from the2sp one in2s22p2

There have not the same Z*

# of electron in the shell

Varies minus the one considered - varies

# of electron in the shell

varies

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

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)

The Klechkovsky rule

Table of n + l values

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

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

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?

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

lines

Start a new period filling each new ns type orbital

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

columns

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

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

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

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<36

Row 4

Column 12

54<80<86

Row 6

f electrons

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*

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.

- 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.

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.

Better than just a classification, a periodic table!

An unknown compound predicted

? M=78.2 V=19 density=4.6

SeM=78.8 V=17.2 density=4.6

Dmitri Mendeleev

Russian (Siberia – St Petersbourg)

(1834 - 1907)

S

As

?

Br

Te

52Te

53I

126

59

58

27Co

28Ni

Dmitri Mendeleev

Russian (Siberia – St Petersbourg)

(1834 - 1907)

Switch between atom classification because of chemistry (valency)

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

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 (silicondioxide).

Forming ions is endothermicexcept if environment is stabilizing

Energy loss 1.23 EV (Slater)

1.53 eV exp.

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

Atomic orbital levelsE = - IP

Tjalling C. Koopmans

Dutch

Nobel 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

Values from Extended Hückel(-eV)

H He

1s 13.6 24.25

Li Be B C N O F

2s 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

Is abundance of atoms related to their stability?Why?What are the most abundant atoms on the earth surface?In the human body?

Abundance (masses) in human body; the human body is made of 65 to 90% of water

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

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 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 universe (1 He for 12 H everywhere)

George Gamow

in 1948

Russian-American

1904-1968

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

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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