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:
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.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
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 -- “ –

PERIODIC TABLE AND TRENDS
Watch clip on:
http://www.youtube.com/watch?v=zG9TsVeHbNA#t=451

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

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

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
http://www.youtube.com/watch?v=j7VMZk1sISU

APPLICATION ON THE NA SPECTRUM
http://www.youtube.com/watch?v=j4Z8zMeVYW4

WATCH A DEMO VIDEO
TO LEARN ABOUT S AND L
http://www.youtube.com/watch?v=A6DiVspoZ1E

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

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

***** BREAK *****
Go to
“Molecular
Structure and
Bonding”

Atomic structure part 2

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Atomic structure part 2