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ATOMIC STRUCTURE (PART 2) Lesson by Dr.Chris UP, May 2014
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Shielding effects of electrons and consequences for atomic properties across the periodic table (atomic radii, electronegativities, ionization energy)

### Transcript of Atomic structure part 2

ATOMIC STRUCTURE (PART 2)

Lesson by Dr.Chris

UP, May 2014

WHAT WE WILL LEARN … PART 1:

REVIEW LESSON 1

Atomic nucleus

atomic mass unit amu

Isotopes / MS

Bohr model of the Hydrogen atom

absorption and emission spectra

<-> energy levels in the atom = orbits

Electron as standing wave

deBroglie: λ = h/p = h/(mv)

Schroedinger eq. for 3D el.waves

=> 3 quantum numbers n, l and m

Aufbau principle

Shielding of electrons

ATOMIC SPECTRA

Show us the energy levels in an atom/molecle (n) and also splitting (l)

MAIN ENERGY LEVELS (N)

Each line with a wavelength λ corresponds to

an electron transition with an energy

∆E = E2 – E1 = h * c / λ 1/ λ = ν

wavenumber in cm-1

Planck constant h = 6.626 x 10-34 J s

C speed of light = 300’000 km/s

Energy unit “electronVolt” eV = energy of 1 electron in a field of 1 V

1 eV => λ = 1240 nm

λ = 1240 nm / E [eV]

CHECK

Which light will an electron emit, when it

falls from energy level 4 to 2 ?

(is it shorter or longer than from 4 to 3 ?)

Which energy in eV will an electron bring from

its ground level to the first excited state ?

WHY ONLY CERTAIN ENERGY LEVELS ?

The Bohr model cannot explain why

electrons can only live in certain orbits !

When we look at electrons as WAVES, we

can understand that each orbit must be a mulitple of λ/2

CLASSICAL VS. QUANTUM MECHANICS

Quantum

mechanics

describes the

function

which

represents these waves

FROM WAVE TO ENERGY

Euler representation

DIFFERENTIATE

with respect to x => impuls p = m*v

- “ - t => energy E = ½ mv2

= p2/2m

FROM LINE SPECTRA TO WAVEFUNCTIONS (ORBITALS)

Model the electron as a standing wave in

3D, we can describe the most likely places of

an electron and its energy from the

Schroedinger Equation

If you want to know this in detail:

and:

http://www.physicsforidiots.com/quantum.html

This equation leads to 3 quantum numbers

which describe the energy and the

distribution of the electron in an atom

SCHROEDINGER EQUATION

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schapp.html#c1

It can describe the behavior of any small particle in the micro cosmos – in theory

Main spectral lines = n

Fine structure = l

with magnetic field:

Zeeman effect

magnetic quantum no. m

=> 3 quantum numbers

ELECTRONIC SHIELDING AUFBAU PRINCIPLE (PERIODIC TABLE)

SHIELDING EFFECT

Simplified demonstration

SLATER’S RULES ZEFF = Z – S ESTIMATION OF S:

EXAMPLE: K – WHERE IS THE 19TH ELECTRON ?

HOMEWORK (PRESENT NEXT LESSON)

Calculate the shielding for the valence electron(s) of:

Ca compare 4s2 <-> 3 d2

Sc compare 3d1 <-> 4 p1

Cu (1) compare 4s1 <-> 4 p1

Cu (2) compare 3d10 4s1 <-> 3d9 4s2

Mn compare 3d5 4s2 <-> 3d7

Co (1) compare 3d7 4s2 <-> 3 d9

Co (2) compare 3d7 4s2 <-> 3d8 4s1

Cr (1) compare 3d5 4s1 <-> 3d4 4s2 Cr (2) compare 3d5 4s1 <-> 3 d6

Questions: explain

1. How shielding determines the AUFBAU

principle

2. trend of atomic radius in PT (left to right)

3. -”- ionization energies -- “ --

4. - “ - electronegativities -- “ –

IONIZATION ENERGIES

ELECTRO NEGATIVITY

PERIODIC TABLE AND TRENDS

Watch clip on:

***** BREAK *****

3 RULES FOR CONFIGURATIONS

Aufbau Principle:

Electrons are filled according to their

lowest energy possible

Pauli exclusion principle:

Electrons must differ in one of 4 quantum

numbers

=> max 2 electrons in one orbital

Hund’s Rules: Electrons want to have maximum SPIN

ELECTRONIC SHIELDING AUFBAU PRINCIPLE (PERIODIC TABLE)

ARE THESE CONFIGURATIONS FOR GROUND STATE POSSIBLE ?

1s

1s

2s

1s

2s

2p

1s

2s

SHAPES OF ORBITALS CHARACTERISTIC IS THE NO. OF NODE PLANES

2p orbital: 1 node

3d orbital: 2 nodes

FINE STRUCTURE OF HYDROGEN SPECTRA

When an electron is in an orbital, it can

cause different energies because of 2 forces:

• SPIN • ANGULAR MOMENTUM

3 MOMENTUMS: SPIN, ANGULAR AND SUM

Watch clip until t=6:20 mins http://www.youtube.com/watch?v=V7DcOXbVY70

ELECTRON HAS 3 “MOMENTUMS”

An electron rotates around itself (like the

earth) producing a SPIN S

This produces

a magnetic field

around the electron

ANGULAR MOMENTUM L

Because the electron circles around the

nucleus, it creates an angular momentum ( L ) – which creates also a magnetic field

Since only certain

can have only discrete values

SPIN-ORBIT COUPLING J

Both momentums combine to the “total angular momentum” J

Lower energy !

TOTAL ANGULAR MOMENTUM J

2D5/2

and 2D3/2

state

CONSEQUENCE

An electron in an s-orbital has angular

momentum of zero on average (L=0)

An electron in a p orbital can have 2

different energies:

depending if the spin momentum points

in the same direction of the angular

momentum or opposite.

A smaller J (L-S) means lower energy

than higher J (L+S)

TERM SYMBOLS (RUSELL-SAUNDER)

To describe the electron configuration in

an atom:

2S+1LJ

L: orbital of the electrons

(S =0, P =1, D =2, F =3)

S: total spin of all these electrons

J = L+S, L+S-1, L+S-2, ....|L-S|

Orbital angular momentum

multiplicity

Total angular momentum

EXAMPLES

Hydrogen ground state

1s1 2S1/2

Helium: 1s2 1S0

He(1s12s1) Excited State Configuration

Terms: 1S0 , 3S1

Boron B(1s22s22p1)

Terms: 2P1/2, 2P3/2

ROUSELL-SAUNDERS COUPLING (LS)

Watch the clip

APPLICATION ON THE NA SPECTRUM

HYDROGEN EXCITED STATES

WATCH A DEMO VIDEO TO LEARN ABOUT S AND L

POSSIBLE CONFIGURATIONS FOR H

POSSIBLE TRANSITIONS

ELECTRON TRANSITIONS IN NA ATOM

3 “NEW” QUANTUM NUMBERS

(1) Angular Momentum L

= the unfilled highest shell of the electron(s)

Add l of each electron in this shell:

l1 + l2 , …. , |l1 - l2| (min.0)

Example: 2 electrons in p shell: L = (1+1 ) 2 -> 1 -> 0

(2) MULTIPLICITY S

For 2 valence electrons S can be 1 or 3

(3) TOTAL ANGULAR MOMENTUM J

Example: Carbon Atom 1s2 2s2 2p2

(1) unfilled shell: p (l1, l2=1) => L= 2,1,0 (D,P,S)

(2) possible Spin: S = +1/2 +1/2 = 1

or S = +1/2 -1/2 = 0

=> multiplicity 3 (triplet) or 1 (singlet)

(3) total momentum J: S=0: L+S: 2,1,0

S =1: L+S: 3,2,1,0

States: 1S0 3S1

1P1 and 3P0, 3P1,

3P2 and

1D2 and 3D1 3D2

3D3

CONFIGURATIONS FOR CARBON

ENERGY ORDER

Apply Hund’s rules to find the lowest energy

ground state:

1. Term with highest S

2. Within the same S, the term with highest L

3. Within same S and L:

a) shell half-filled or less: lowest J

b) more than half-filled: highest J

(for excited states, we cannot get the lowest

energy from these rules)

EXAMPLE C ATOM GROUND STATE

2 p electrons with max. L in a

configuration with highest spin S=1

=> L=1 (“P”)

=> J = 2,1,0

Less than half-filled => lowest J:

=> ground state: S=1, L=1, J=0: 3P0

EXAMPLE C ATOM EXCITED STATE

Configuration : 2 p1 and 3 s1

Spin: +1/2 +1/2 or +1/2 -1/2 => S= 0,1

1 electron in s orbital => l1 = 0

and 1 el. in p orbital => l2 = 1

=> L= 1+0, 1-0 = 1 (“P”)

=> S=0: J=1

S=1: J = 2,1,0

Possible states: 1P1 and 3P0,

3P1, 3P2

MICROSTATES FOR CARBON C(1S22S22P2)

EXAMPLE FE ATOM

Configuration 4s2 3d6

(1) Spin: 2 (2) L: -2, -1, 0, 1, 2 the max.L is in this state: L=2 (-2 -1 +0 +1 +2 +2)

(3) J: 4,3,2,1,0

Hund’s rule: max. spin, L=2 (“D”) highest J (more than half-filled): 4

=> Ground state: 5D4

HOMEWORK (1)

1. Ground state conf. for He

2. Excited state (a) 1s1 2s1

3. Excited state (b) 1s1 2p1

4. Ground state conf. for Be 2s2

5. Excited state 2s1 2p1

SUMMARY

***** BREAK *****

Go to

“Molecular

Structure and Bonding”