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Oct 25
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The atmosphere, part 1
Midterm … No lecture
The atmosphere, part 2
Light, blackbodies, Bohr
Postulates of QM, p-in-a-box
Hydrogen and multi – e atoms
Multi-electron atoms
Periodic properties
Periodic properties
Valence-bond; Lewis structures
Hybrid orbitals; VSEPR
VSEPR
MO theory
MO theory
Review for exam
Ch. 8
Ch. 8
Ch. 9
Ch. 9
Ch. 9
Ch.9,10
Ch. 10
Ch. 10
Ch. 11
Ch. 11, 12
Ch. 12
Ch. 12
Ch. 12
• We can split the hydrogen wavefunction
into two:
Y(x,y,z)  Y(r,q,j) = R(r) x Y(q,j)
Depends on r only
Depends on angular variables
• The solutions have the same features we
have seen already:
– Energy is quantized
• En = - R Z2 / n2
= - 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …]
– Wavefunctions have shapes which depend on
the quantum numbers
– There are (n-1) nodes in the wavefunctions
• Because we have 3 spatial dimensions, we end up with 3
quantum numbers:
n, l, ml
• n = 1,2,3, …; l = 0,1,2 … (n-1); ml = -l, -l+1, …0…l-1, l
• n is the principal quantum number – gives energy and
level
• l is the orbital angular momentum quantum number – it
gives the shape of the wavefunction
• ml is the magnetic quantum number – it distinguishes the
various degenerate wavefunctions with the same n and l
•En = - R Z2 / n2
= - 2.178 x 10-18 Z2 / n2 J [ n = 1,2,3 …]
… degenerate
Probability Distribution
for the 1s wavefunction:
3/2
1  1  -r/a0
Y100 =
e


p  a0 
Maximum probability at nucleus
A more interesting way to look at
things is by using the radial probability
distribution, which gives probabilities of
finding the electron within an annulus
at distance r (think of onion skins)
max. away from nucleus
90% boundary:
Inside this lies
90% of the probability
nodes
P-orbitals
Node at nucleus
The result (after a lot of math!)
Node at s = 2!!
Nodes at f, q = 0 !!
The Boundary Surface Representations of
All Three 2p Orbitals
A Comparison of the Radial Probability
Distributions of the 2s and 2p Orbitals
A Cross Section of the Electron Probability Distribution
for a 3p Orbital
Spatial nodes
and
Angular nodes!
Nodes at f, q = 0 !!
Nodes at s = 0 and 4
The Boundary Surfaces of All of the
3d Orbitals
Representation of the 4f Orbitals in Terms of
Their Boundary Surfaces
The Radial Probability
Distribution for the 3s, 3p,
and 3d Orbitals
Another quantum number!
Electrons are influenced by a magnetic field as though they were spinning charges.
They are not really, but we think of them as having “spin up” or “spin down” levels.
These are labeled by the 4th quantum number: ms, which can take 2 values.
This 2-valued electron spin can be shown in an experiment
In silver (and many other atoms) there is one more “spin up” electron than
“spin down” or vice versa. This means that an atom of silver can interact with a
magnetic field and be deflected up or down, depending on which type of spin
is in excess.
In multi-electron atoms things change, because
of the influence of electrons which are already there
E
Remember
the
energies
are <0
3s
3p
2s
2p
3d
1s
The degenerate energy levels
are changed somewhat to become
THE MULTI-ELECTRON ATOM
ENERGY LEVEL DIAGRAM
Remember
the
4d
5s
energies
4p
are < 0
3d
4s
E
3p
3s
2s
1s
2p
THE PAULI PRINCIPLE
No two electrons in the same atom can have
the same set of four quantum numbers (n, l, ml , ms).
An orbital is described by three quantum numbers,
Then each electron in a given orbital
must have a different ms
HOW MANY ELECTRONS IN AN ORBITAL?
each orbital may contain a maximum of two
electrons, and they must have opposite
spins.
ELECTRONIC
CONFIGURATIONS
THE BUILDING-UP PRINCIPLE.
GROUND STATE
lowest energy electronic configuration
assign electrons to orbitals one at a time
Electrons go into the available orbital of
lowest energy.
Electrons are placed in orbitals according
to the Pauli Principle.
A maximum of two electrons per orbital.
THE AUFBAU (BUILDING-UP) PRINCIPLE:
electrons are added to hydrogen-like atomic orbitals
in order of increasing energy
5s
4s
E
4d
4p
3d
3p
3s
2s
2p
1s
The electron configuration of any atom or ion…....
can be represented by an orbital diagram
ORBITAL DIAGRAM
Hydrogen has its one electron in the 1s orbital:
1s
2s
2p
1
1s
H:

Helium has two electrons: both occupy the 1s orbital
Pauli principle
with opposite spins:
1s
He: 1s2
2p
2s
2p

1s
He:
2s
ORBITAL DIAGRAM
Hydrogen has its one electron in the 1s orbital:
1s
2s
2p
1
1s
H:

Helium has two electrons: both occupy the 1s orbital
Pauli principle
with opposite spins:
1s
He: 1s2
2s
2p

helium ground state
Helium can also exist in an excited state such as:
1s
He: 1s12s1 
2s
2p

Now onto the next atoms
Lithium has three electrons, so it must use the 2s orbital:
1s
Li:
2s
2p
1s22s1
Beryllium has four electrons, which fill both the 1s and 2s
orbitals:
1s
2s
2p
Be: 1s2 2s2
Boron’s five electrons fill the 1s and 2s orbitals, and begin
to fill the 2p orbitals. Since all three are degenerate, the
order in which they are filled does not matter.
1s
B:
1s22s22p1
2s
2p
CARBON
Z=6
A CHOICE
1s
C:
2s
2p
2s
2p
1s22s22p2
OR
1s
C:
1s22s22p2
How can we decide?????
HUND’S RULE
FOR THE GROUND STATE
ELECTRONS OCCUPY DEGENERATE
ORBITALS SEPARATELY
THE SPINS ARE PARALLEL
SO FOR CARBON THE GROUND STATE IS
1s
C:
1s22s22p2
2s
2p
ENERGY LEVEL DIAGRAM FOR A MULTIELECTRON ATOM
BROMINE ELECTRONIC CONFIGURATION
5s
4s
E
3s 
2s
1s 
4d
4p 
3d
3p
2p
[Ar] 4s23d104p5
The idea of penetration
explains why the 3d orbitals
lie higher in energy than the
4s.
The valence electron configuration of the elements
in the periodic table repeat periodically!
H
1s1
He
1s2
Li
2s1
Be
2s2
B
2p1
C
2p2
N
2p3
O
2p4
F
2p5
Ne
2p6
Na
3s1
Mg
3s2
Al
3p1
Si
3p2
P
3p3
S
3p4
Cl
3p5
Ar
3p6
Every element in a group has
the same valence electron configuration!