Transcript No Slide Title

PowerPoint Lecture Presentation
by
J. David Robertson
University of Missouri
Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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
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
lxn=c
7.1
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? or Is this AM or FM frequency?
l
lxn=c
l = c/n
l = 3.00x108 m/s/6.0x104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
n
Radio wave
7.1
Mystery #1, “Black Body Problem”:
Frequency dependence of radiation from a heated body
Classical Physics couldn’t explain it : UV Catastrophe.
Solved by Planck witth a Quantum Theory (1900)
Energy (light) is emitted or absorbed
in discrete units or a packet (quantum).
“Energy of an atom is quantized.”
E = nhn
h = 6.63 x 10-34 J•s
Planck’s constant
n: 1,2,3,4,5 etc.
7.1
What’s nature of light?
According to classical physics, it is a wave
because of the following properties.
• Reflection
• Refraction
• Diffraction: a result of interference
However, the wave nature of light couldn’t
explain the photoelectric effect.
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
hn
Electrical Current generated by light:
Presence of the Threshold frequency
KE e-
Absence of Time Lag
Light has both:
1. wave nature
2. particle nature
E = hn
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
Alkali metals works the best.
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=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/photon
= (1.29 x 10 -15 J/photon)(6.022 x 1023photon/mol)
= 7.77x108 J/mol = 7.77x105 kJ/mol = 777 MJ/mol
Cf. Energy of a UV light with 200 nm:
598kJ/mol
Energy of a red light with 671 nm (Li): 180kJ/mol
Chemical Bond Energy: 500~1,000 kJ/mol
7.2
Why medicines are stored in a brown glass bottles?
• Brown glass blocks high energy portion (green and blue) of
the visible light by absorbing G & B. It also absorb red
portion substantially to impart brown color. Many
chemicals can be destroyed by EMW with higher energy
(Blue ,Violet, UV and others).
• Brown (Dark Red) light has much less energy (<200kJ/mol),
thus cannot harm many substances. Normally 200 kJ/mol
or more of energy is needed to break many chemical
bonds.
• Red
Lower
Green
Blue
Higher Energy
Mystery #3: Line Emission Spectrum of H Atoms
HOW are they generated?
(396)
.
410
434
496
Johann Balmer (1825-1989) solved the jigsaw puzzle
656
7.3
7.3
The Jigsaw puzzle of the H emission spectra was solved
by Balmer (1885) without knowing its implacations
1/λ = 1.097x107 (1/22-1/n2) m-1
= 0.01097 (0.25-1/n2) nm-1
where n = 3,
4,
5,
6,
λ = 656 nm
λ = 486
λ = 434
λ = 410
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
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
Is this IR or UV?
7.3
The Dual Nature of Matter:
Wave-Particle Duality: “Wavicle”
• Einstein:
(1905)
Light wave is Light Particle
(“Photon”)
E = hν = hc/ λ
• De Broglie: Electron (Particle) is Wave
(1923)
λ = h/mu
Why is e- energy quantized?
De Broglie (1924) reasoned
that e- is both particle and
wave.
2pr = nl
l = h/mu
u = velocity of em = mass of e7.4
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/mu
h in J•s m in kg
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
u in (m/s)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
What is the de Broglie wavelength (in nm)
associated with an electron moving at a
speed of 2x108 m/s?
l = 6.63 x 10-34 / (9.11x10-35 x 2x108) = 0.004 nm
7.4
Chemistry in Action: Electron Microscopy
le = 0.004 nm
STM image of iron atoms
on copper surface
7.6
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).
H:
He: 668
656
486
588
502
434 410
447
Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent
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
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
7.5
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
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
l=
0
1
2
3
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)
??
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
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
7.8
Paramagnetic
Diamagnetic
unpaired electrons
all electrons paired
2p
O, S
2p
Ne, Ar
7.8