Transcript Chapter 7

Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
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Max Planck (1858 - 1947)
=> new era of physics ~ 1900
scientists tried to understand the atom using classical
mechanics
=> How are atoms held together?
=> Why is radiation emitted at certain temperatures?
Planck suggested the revolutionary idea that atoms &
molecules are not governed by classical mechanics
=> Planck suggested that energy comes in discrete
quantities or quanta
=> Need to understand wave to look at Planck’s theory
=> Need to understand wave to look at Planck’s theory
Wave can be thought of as a vibrating disturbance
by which energy is transmitted.
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
Calculate the wavelength of a wave that has a frequency
of 87.4 Hz and a speed of 1.52 x 103 cm/s.
The speed (u) of the wave = l x n
u = 1.52 x 103 cm/s
n = 87.4 Hz
l=u/n
l = 1.52 x 103 cm/s = 17.4 cm
87.4 1/s
=> Let the units help you remember the equation.
There are many types of waves ……….
water, sound, light, ……...
Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Perpendicular
planes
Electromagnetic
radiation is the emission
and transmission of energy
in the form of
electromagnetic waves.
Two components
with the same
wavelength and
frequency.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
lxn=c
Usually in nm (10-9 m)
High
Energy
l
n
Low
Energy
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
lxn=c
n
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
7.1
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Amount of radiation emitted by an object at a
certain temperture depends on its wavelength.
Energy (light) is emitted or
absorbed in discrete units
Smallest quantity of energy
(quantum).
that can be emitted (or
absorbed) in the form of
electromagnetic radiation
E=hxn
Planck’s constant (h)
h = 6.63 x 10-34 J•s
Energy is in multiples of hn
hn, 2hn, 3hn, 4hn,
but not 1.23hn or 5.15hn
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
hn
Convert radiant energy
(light) to electrical energy.
KE e-
Light has both:
1. wave nature
2. particle nature
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
KE = Kinetic Energy
BE = Binding Energy
BE
KE
Energy of incident light
Cannot
generate an eunless hn >BE
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=hxn
E=hxc/l
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
Einstein’s work helped explain another “mystery”.
Emission spectra of atoms
Newton showed that sunlight is composed of
various color components
Emission spectrum is either a continuous or line
spectrum of radiation emitted by substances
Ex: Energize Fe sample with heat
Visible glow
Fe
Visible light
Heat
“red hot”
“white hot”
 IR light
Emission spectrum from atoms in gas phase emit
light at certain wavelengths
Line spectra
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
Model takes ideas from
astronomy – planets moving
around the sun => electrons in
orbit around the nucleus
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,…
Smallest
quantum
number is
nearest the
nucleus
RH (Rydberg constant) = 2.18 x 10-18J
Negative sign signifies that energy is lower than a “free electron”
Bohr’s Model of the Atom Has
Electrons Moving in Circular Orbits
Each orbit has a
particular energy
 Energy is quantized
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
n = 1 has the most negative value
 Ground state
 Ground level
 Lowest energy state
n=2
=3
=4

Higher energy state
Excited level
Excited state
E  = -RH(
1
2 )
0 = “free electron”
Energy can be emitted or absorbed
 Must go between energy levels
DE = Ef - Ei
E = hn
n=5
n=4
E = hn
n=3
n=2
n=1
= hn
1 1
= -RH ( 2- 2 )
n f ni
Energy
levels get
closer
together as
n increases.
Energy can be emitted or absorbed
 Must go between energy levels
E = hn
DE = E4 – E5
= hn
1
1
= -RH ( 2 - 2 )
4
5
= - 0.0225 RH
= - 4.91 x 10-20J
emit light
E = hn
DE = Ef - Ei
n=5
n=4
n=3
= hn
1 1
= -RH ( 2- 2 )
n f ni
RH = 2.18 x 10-18J
n=2
n=1
DE = E1 – E3
= hn
1
1
= -RH ( 2 - 2 )
1
3
= - 0.888 RH
= - 1.94 x 10-18J
emit light
ni = 3
ni = 3
ni = 2
nf = 2
Ephoton = DE = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
DE = RH( 2
ni
)
)
1
n2f
)
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
According to de Broglie, electrons bound to a
nucleus behave as a standing wave.
Think of a violin or guitar string.
node = point that does not move
# of nodes depends on the length (l) and the wavelength(λ)
See only half or whole waves
Why is e- energy quantized?
De Broglie (1924) reasoned
that e- is both particle and
Wave property
wave.
2pr = nl
l = h/mu
Particle property
u = velocity of em = mass of e-
Mismatch
in waves
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/mu
h in J•s m in kg u in (m/s)
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
7.4
Bohr’s model did not explain all the properties of
emission spectra
more complicated atoms
extra lines for H in a magnetic field
how do you specify the “position” of a wave?
Heisenberg Uncertainty Principle:
It is impossible to know both the momentum (p =
mass x velocity) and position of a particle
simultaneously with certainty.
(Δx)(Δp) > h / 4π
position
momentum
 If x is known precisely then
p is not known precisely
Δx 
Δp 
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Y) describes:
1. energy of e- with a given Y
2. probability of finding e- in a volume of space
Atoms are mostly space  electron density
 where do we find electrons?
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must
approximate its solution for
multi-electron systems.
Atomic orbital can be throught of as the
wavefunction of an electron in an atom.
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
A collection of orbitals with the same n value are
referred to as a shell, the different l ’s refer to subshells.
l = 0 (s orbitals)
l = 1 (p orbitals)
7.6
l = 2 (d orbitals)
7.6
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
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
Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
n=3
n=2
En = -RH (
1
n2
)
n=1
7.7
Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=3 l = 1
n=2 l = 1
n=1 l = 0
7.7
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
C 6 electrons
B 1s22s22p1 B 5 electrons
Be 1s22s2
Be 4 electrons
Li 1s22s1
Li 3 electrons
He 2 electrons
H 1 electron
He 1s2
H 1s1
7.7
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
Ne 10 electrons Ne 1s22s22p6
F 9 electrons F 1s22s22p5
O 8 electrons
N 7 electrons
C 6 electrons
O 1s22s22p4
N 1s22s22p3
C 1s22s22p2
7.7
Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
7.7
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
Outermost subshell being filled with electrons
7.8
ns2np6
ns2np5
ns2np4
ns2np3
ns2np2
ns2np1
d10
d5
d1
ns2
ns1
Ground State Electron Configurations of the Elements
4f
5f
8.2
7.8
Are there any exceptions to the filling rules
for the orbitals?
Cr, Cu, Mo, Ag and Au do not follow the rules
Since filled orbitals and half-filled orbitals are the
most stable and the energy levels are so close in
energy, electrons go into d orbitals instead of s
orbitals
Cr
1s22s22p63s23p64s13d5
Cu
1s22s22p63s23p64s13d10
Mo
1s22s22p63s23p64s23d104p65s14d5
Ag
1s22s22p63s23p64s23d104p65s14d10
Au
1s22s22p63s23p64s23d104p65s24d105p66s14f145d10
Ground State Electron Configurations of the Elements
Exceptions
8.2
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
7.8