Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
Properties of Waves
Wavelength (λ) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
7.1
Properties of Waves
Frequency (ν) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = λ ν
7.1
Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Electromagnetic
radiation is the emission
and transmission of energy
in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
 =c
7.1
What will an electron do?
• An electron does go though both, and
makes an interference pattern.
• It behaves like a wave.
• Other matter has wavelengths too short to
notice.
Image
7.1
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
l
λν = c
n
λ = c/ν
λ = 3.00 x 108 m/s / 6.0 x 104 Hz
λ = 5.0 x 103 m
λ = 5.0 x 1012 nm
Radio wave
7.1
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy (light) is emitted or
absorbed in discrete units
(quantum).
E=h ν
Planck’s constant (h)
h = 6.63 x 10-34 J•s
7.1
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
Light has both:
1. wave nature
2. particle nature
hν
KE e-
Photon is a “particle” of light
h ν= KE + W
KE = hν - W
KE: kinetic energy of
the ejected electron
W:is the work function
7.2
When copper is bombarded with high-energy electrons,
X rays are emitted. Calculate the energy (in joules)
associated with the photons if the wavelength of the X
rays is 0.154 nm.
E=h ν
E=h c/λ
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10 -15 J
7.2
Line Emission Spectrum of Hydrogen Atoms
7.3
7.3
Bohr’s Model of
the Atom (1913)
THIS CALCULATION
HAS BEEN REMOVED
1. e- can only have specific
(quantized) energy
values No orange or
yellow light in H2
2. light is emitted as emoves from one energy
level to a lower energy
level
1
En = -RH (
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
7.3
E = hν
E = hν
7.3
THIS CALCULATION
HAS BEEN REMOVED
ni = 3
ni = 3
ni = 2
nf = 2
Balmer
Ephoton = E = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
ΔE = RH ( 2
ni
)
)
1
n2f
)
nnf f==11
Lymann
7.3
Calculate the wavelength (in nm) of a photon
emitted by a hydrogen atom when its electron
drops from the n = 5 state to the n = 3 state.
Ephoton = ΔE = RH(
1
n2i
1
n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = ΔE = -1.55 x 10-19 J
Ephoton = h x c / 
+ E only
indicates
wether or not
energy has
been added or
lost from the
system
λ = h x c / Ephoton
λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
λ = 1280 nm
7.3
Why is e- energy quantized?
De Broglie (1924) reasoned
that e- is both particle and
wave.
2pr = νλ
λ = h/mu
u = velocity of em = mass of e7.4
• Wave interference
Animation
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
λ = h/mu
h in J•s m in kg u in (m/s)
λ = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
λ = 1.7 x 10-32 m = 1.7 x 10-23 nm
7.4
Chemistry in Action: Element from the Sun
In 1868, Pierre Janssen detected a new dark line in the solar
emission spectrum that did not match known emission lines
Mystery element was named Helium
In 1895, William Ramsey discovered helium in a mineral of
uranium (from alpha decay).
Chemistry in Action: Laser – The Splendid Light
Light Amplification by Stimulated Emission of Radiation
Laser light is (1) intense, (2) monoenergetic, and (3) coherent
Chemistry in Action: Electron Microscopy
le = 0.004 nm
Scanning tunneling microscope
STM image of iron atoms
on copper surface
Heisenberg Uncertainty Principal:
It is impossible to know simultaneously both the momentum
p (defined as mass time velocity) and the position of a
particle with certainty.
Δx Δp ≥ h/4π
Δx position, Δp momentum, ≥ crude exp > rules, the
probability is never less than h/4π
This proves that Bohr was not completely right so that
brings us to Erwin Schrodinger.
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Ψ)psi describes:
1. energy of e- with a given Ψ
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
It requires advanced calculus to solve
7.5
Schrodingers Equation
It is important to know that the equation
incorporates both particle behavior
(mass) and wave behavior Ψ
QUANTUM NUMBERS
The shape, size, and energy of each orbital is a function
of 3 quantum numbers which describe the location of
an electron within an atom or ion
n (principal)
---> energy level
l (orbital) ---> shape of orbital
ml (magnetic) ---> designates a particular
suborbital
The fourth quantum number is not derived from the
wave function
s (spin)
---> spin of the electron
(clockwise or counterclockwise: ½ or – ½)
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3
7.6
Where 90% of the
e- density is found
for the 1s orbital
e- density (1s orbital) falls off rapidly
as distance from nucleus increases
7.6
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … (n-1)
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
7.6
Types of Orbitals (l)
s orbital
p orbital
d orbital
l = 0 (s orbitals)
l = 1 (p orbitals)
7.6
p Orbitals
this is a p sublevel
with 3 orbitals
These are called x, y, and z
3py
orbital
There is a PLANAR
NODE thru the
nucleus, which is
an area of zero
probability of
finding an electron
p Orbitals
• The three p orbitals lie 90o apart in
space
l = 2 (d orbitals)
7.6
f Orbitals
For l = 3,
orbitals
f sublevel with 7
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
7.6
ml = -1
ml = -2
ml = 0
ml = -1
ml = 0
ml = 1
ml = 1
ml = 2
7.6
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½
7.6
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Ψ.
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
7.6
7.6
Schrodinger Wave Equation
 = fn(n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
 = (n, l, ml, ½) Or  = (n, l, ml, -½)
An orbital can hold 2 electrons
7.6
How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d
subshell?
n=3
3d
l=2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e7.6
Orbital Diagrams
Energy of orbitals in a multi-electron atom
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=3 l = 1
n=2 l = 1
Energy
depends
on n and l
n=1 l = 0
7.7
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
Be
Li
B5
C
3
64electrons
electrons
22s
222s
22p
12 1
BBe
Li1s1s
1s
2s
H
He12electron
electrons
He
H 1s
1s12
7.7
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
Ne97
C
N
O
F
6
810
electrons
electrons
electrons
22s
222p
22p
5
246
3
Ne
C
N
O
F 1s
1s222s
7.7
Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
7.7
Outermost subshell being filled with electrons
7.8
Why are d and f orbitals always
in lower energy levels?
• d and f orbitals require LARGE amounts of
energy
• It’s better (lower in energy) to skip a
sublevel that requires a large amount of
energy (d and f orbtials) for one in a higher
level but lower energy
This is the reason for the diagonal rule! BE
SURE TO FOLLOW THE ARROWS IN
ORDER!
Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
H
1s1
7.8
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons
1s22s22p63s23p5
1s < 2s < 2p < 3s < 3p < 4s
2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n=3
l=1
ml = -1, 0, or +1
ms = ½ or -½
7.8
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
7.8
7.8
Exceptions to the Aufbau Principle
• Remember d and f orbitals require LARGE
amounts of energy
• If we can’t fill these sublevels, then the next best
thing is to be HALF full (one electron in each
orbital in the sublevel)
• There are many exceptions, but the most
common ones are
d4 and d9
For the purposes of this class, we are going to
assume that ALL atoms (or ions) that end in d4
or d9 are exceptions to the rule. This may or
may not be true, it just depends on the atom.
Exceptions to the Aufbau Principle
d4 is one electron short of being HALF full
In order to become more stable (require less
energy), one of the closest s electrons will
actually go into the d, making it d5 instead of d4.
For example: Cr would be [Ar] 4s2 3d4, but since
this ends exactly with a d4 it is an exception to
the rule. Thus, Cr should be [Ar] 4s1 3d5.
Procedure: Find the closest s orbital. Steal one
electron from it, and add it to the d.
Try These!
• Write the shorthand notation
for:
Cu
[Ar] 4s1 3d10
1 4f14 5d5
[Xe]
6s
W
1 4f14 5d10
[Xe]
6s
Au
Exceptions to the Aufbau Principle
• The next most common are f1 and f8
• The electron goes into the next d orbital
• Example:
– La [Xe]6s2 5d1
– Gd [Xe]6s2 4f7 5d1
Keep an Eye On Those Ions!
• Electrons are lost or gained like they
always are with ions… negative ions
have gained electrons, positive ions
have lost electrons
• The electrons that are lost or gained
should be added/removed from the
highest energy level (not the highest
orbital in energy!)
Keep an Eye On Those Ions!
• Tin
Atom: [Kr] 5s2 4d10 5p2
Sn+4 ion: [Kr] 4d10
Sn+2 ion: [Kr] 5s2 4d10
Note that the electrons came out of
the highest energy level, not the
highest energy orbital!