Atomic structure part 2

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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 cm1
Planck constant h = 6.626 x 1034 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:
http://www.youtube.com/watch?v=7LBPXP09KC4
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
QUICK OVERVIEW ABOUT SCHROEDINGER EQ.
From: http://www.youtube.com/watch?v=cdaIsu4jBtA
SCHROEDINGER EQUATION
http://hyperphysics.phyastr.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
http://www.youtube.com/watch?v=y8fQKLgU70A
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  “ –
ATOMIC RADII
IONIZATION ENERGIES
ELECTRO NEGATIVITY
PERIODIC TABLE AND TRENDS
Watch clip on:
http://www.youtube.com/watch?v=zG9TsVeHbNA#t=451
***** 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
radius are possible, L
can have only discrete values
SPINORBIT 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 sorbital 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 (LS) means lower energy
than higher J (L+S)
TERM SYMBOLS (RUSELLSAUNDER)
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+S1, L+S2, ....LS
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
ROUSELLSAUNDERS COUPLING (LS)
Watch the clip
http://www.youtube.com/watch?v=j7VMZk1sISU
APPLICATION ON THE NA SPECTRUM
http://www.youtube.com/watch?v=j4Z8zMeVYW4
HYDROGEN EXCITED STATES
WATCH A DEMO VIDEO TO LEARN ABOUT S AND L
http://www.youtube.com/watch?v=A6DiVspoZ1E
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 halffilled or less: lowest J
b) more than halffilled: 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 halffilled => 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, 10 = 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 halffilled): 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”