#### Transcript Quantum

```Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
Properties of Waves
Wavelength (l) 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 (n) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = l x n
7.1
Maxwell (1873), proposed that visible light
consists of electromagnetic waves.
Electromagnetic
and transmission of energy
in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
lxn=c
7.1
7.1
Relationship Between
Frequency and Wavelength

Frequency and Wavelength are inversely
related
 C = ln

Where c = speed of light, l is the
wavelength, and n is the frequency
Example 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?
C = ln
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
No, this is a radio wave
7.1
Example 2

Red light has a wavelength that ranges from 7.0 x
10-7 m to 6.5 x 10-7 m. What is the frequency of
light found in the red region?
3.0 x 108 m/s = (7.0 x 10-7 m)(n)
n=4.29 x 1014 Hz
3.0 x 108 m/s = (6.5 x 10-7 m)(n)
n=4.62 x 1014 Hz
nrange=4.29 x 1014 Hz - 4.62 x 1014 Hz
Example 3
A gamma ray has a frequency of 3.0 x 1021 Hz.
What is the wavelength, in nm, of the ray?
3.0 x 108 = l(3.0 x 1021 s-1)
l= 1.0 x 10-13 m
l= 1.0 x 10-4 nm
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy had been thought to be emitted
continuously at all wavelengths. However, energy
(light) is emitted or absorbed in discrete units called
quanta.
E=hxn
Planck’s constant (h)
h = 6.626 x 10-34 J•s
7.1
Mystery #2, “Photoelectric
Effect”
Solved by Einstein in 1905
Discovered that light
has both:
1. wave nature
2. particle nature
hn
KE e-
Photon is a “particle” of light
7.2
Relationship Between
Energy and Frequency
E = hn
h=Planck’s Constant
h=6.626 x 10-34 J s
This constant was determined experimentally
Example 1

What is the energy of a wave that has a
frequency of 1.05 x 1016 Hz?
E = hn
E = (6.626 x 10-34 J s)(1.05 x 1016 s-1)
E = 6.96 x 10-18 J
Example 2

What is the wavelength of a photon of light
that has 4.79 x 10-19 J of energy? What color
light is represented?
E = hn
4.79 x 10-19 J = (6.626 x 10-34 J s) n
n=7.23 x 1014 Hz
c = ln
3.0 x 108 m/s = l (7.23 x 1014 s-1)
l = 4.1 x 10-7 m; violet
Line Emission Spectrum of Hydrogen Atoms
7.3
7.3
Bohr’s Model of
the Atom (1913)
1. e- can only have specific
(quantized) energy
values
2. light is emitted as emoves from one energy
level to a lower energy
level
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
7.3
E = hn
E = hn
7.3
ni = 3
ni = 3
ni = 2
nf = 2
Ephoton = DE = Ef - Ei
1
Ef = -RH ( 2 )
nf
1
Ei = -RH ( 2 )
ni
1
1
DE = RH( 2
)
2
ni
nf
nnf f==11
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 = DE = RH(
1
n2i
1
n2f
)
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = DE = -1.55 x 10-19 J
Ephoton = h x c / l
l = h x c / Ephoton
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
7.3
Why is e- energy
quantized?
De Broglie (1924) reasoned
that the e- is both particle and
wave (he called this the
wave/particle duality).
l= h/mn
h=Planck’s Constant
m=mass
7.4
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
l = h/mn
h in J•s m in kg n in (m/s)
l = 6.63 x 10-34 J s/ (2.5 x 10-3 kg x 15.6 m/s)
l = 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: Electron Microscopy
le = 0.004 nm
STM image of iron atoms
on copper surface
Heisenberg’s Uncertainty
Principle

Heisenberg stated that it is impossible to
know precisely both the velocity and the
position of a particle

*If you measure the position of a particle it
will be disturbed therefore altering the
velocity
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eEY = HY
Wave function (Y) describes:
1. energy of e- with a given Y
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 multielectron systems.
7.5
Probability of Finding an
Electron

Max Born: later found it was more useful to
determine the probability of finding an
electron in a certain location

Y2
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
Y = 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
Y = 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
three p orbitals lie 90o apart
in space
 The
l = 2 (d orbitals)
7.6
f Orbitals
For l = 3,
orbitals
f sublevel with 7
Schrodinger Wave Equation
Y = 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
Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½
7.6
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Y.
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
Y = 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
Y = (n, l, ml, ½) or Y = (n, l, ml, -½)
An orbital can hold 2 electrons
7.6
How many 2p orbitals are there in an atom?
n=2
2p
If l = 1, then ml = -1, 0, or +1
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)
7.7
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
7.7
Order of orbitals (filling) in multi-electron
atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
7.7
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
Outermost subshell being filled with
electrons
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.
This process is called hydridization
Try These!
Write
for:
Cu
W
Au
the shorthand notation
[Ar] 4s1 3d10
[Xe] 6s1 4f14 5d5
[Xe] 6s1 4f14 5d10
Exceptions to the Aufbau
Principle
next most common are f1 and f8
electron goes into the next d
orbital
 Example:
 La [Xe]6s2 5d1
 Gd [Xe]6s2 4f7 5d1
 The
 The
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