Electronic Structure of Atoms Chapter 4 Electronic Structure of Atoms.
Quantum Theory and the Electronic Structure of Atoms
Transcript of Quantum Theory and the Electronic Structure of Atoms
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Quantum Theory and the
Electronic Structure of Atoms
Chapter 7
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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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
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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 n7.1
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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
l x n = c7.1
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7.1
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l x n = c
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
Radio wave
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 = 5.0 x 1012 nm
l
n
7.1
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Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy (light) is emitted or
absorbed in discrete units
(quantum).
E = h x n
Planck’s constant (h)
h = 6.63 x 10-34 J•s
7.1
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Light has both:
1. wave nature
2. particle nature
hn = KE + BE
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
Photon is a “particle” of light
KE = hn - BE
hn
KE e-
7.2
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E = h x n
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
E = h x c / l
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.
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7.3
Line Emission Spectrum of Hydrogen Atoms
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7.3
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1. e- can only have specific
(quantized) energy
values
2. light is emitted as e-
moves from one energy
level to a lower energy
level
Bohr’s Model of
the Atom (1913)
En = -RH ( )1
n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
7.3
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E = hn
E = hn
7.3
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Ephoton = DE = Ef - Ei
Ef = -RH ( )1
n2f
Ei = -RH ( )1
n2i
i f
DE = RH( )1
n2
1
n2
nf = 1
ni = 2
nf = 1
ni = 3
nf = 2
ni = 3
7.3
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Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = DE = -1.55 x 10-19 J
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
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 = h x c / l
l = h x c / Ephoton
i f
DE = RH( )1
n2
1
n2Ephoton =
7.3
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De Broglie (1924) reasoned
that e- is both particle and
wave.
Why is e- energy quantized?
7.4
u = velocity of e-
m = mass of e-
2pr = nl l = h
mu
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l = h/mu
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
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
m in kgh in J•s u in (m/s)
7.4
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Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent
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Chemistry in Action: Electron Microscopy
STM image of iron atoms
on copper surface
le = 0.004 nm
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Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the e-
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 multi-electron systems.
7.5
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Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
n=1 n=2 n=3
7.6
distance of e- from the nucleus
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Where 90% of the
e- density is found
for the 1s orbital
7.6
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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
Shape of the “volume” of space that the e- occupies
l = 0 s orbital
l = 1 p orbital
l = 2 d orbital
l = 3 f orbital
Schrodinger Wave Equation
7.6
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l = 0 (s orbitals)
l = 1 (p orbitals)
7.6
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l = 2 (d orbitals)
7.6
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Y = fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
orientation of the orbital in space
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
Schrodinger Wave Equation
7.6
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ml = -1 ml = 0 ml = 1
ml = -2 ml = -1 ml = 0 ml = 1 ml = 27.6
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Y = fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
Schrodinger Wave Equation
ms = -½ms = +½
7.6
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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.
Schrodinger Wave Equation
Y = fn(n, l, ml, ms)
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
7.6
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7.6
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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
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How many 2p orbitals are there in an atom?
2p
n=2
l = 1
If l = 1, then ml = -1, 0, or +1
3 orbitals
How many electrons can be placed in the 3d
subshell?
3d
n=3
l = 2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e-
7.6
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Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
En = -RH ( )1
n2
n=1
n=2
n=3
7.7
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Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=1 l = 0
n=2 l = 0n=2 l = 1
n=3 l = 0n=3 l = 1
n=3 l = 2
7.7
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
H 1 electron
H 1s1
He 2 electrons
He 1s2
Li 3 electrons
Li 1s22s1
Be 4 electrons
Be 1s22s2
B 5 electrons
B 1s22s22p1
C 6 electrons
? ?
7.9
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C 6 electrons
The most stable arrangement of electrons
in subshells is the one with the greatest
number of parallel spins (Hund’s rule).
C 1s22s22p2
N 7 electrons
N 1s22s22p3
O 8 electrons
O 1s22s22p4
F 9 electrons
F 1s22s22p5
Ne 10 electrons
Ne 1s22s22p6
7.7
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Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s7.7
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Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
1s1
principal quantum
number n
angular momentum
quantum number l
number of electrons
in the orbital or subshell
Orbital diagram
H
1s1
7.8
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What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
7.8
Abbreviated as [Ne]3s2 [Ne] 1s22s22p6
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5 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
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Outermost subshell being filled with electrons
7.8
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7.8
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Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p7.8
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Chemistry Mystery: Discovery of Helium
In 1868, Pierre Janssen detected a new dark line in the solar
emission spectrum that did not match known emission lines
In 1895, William Ramsey discovered helium in a mineral of
uranium (from alpha decay).
Mystery element was named Helium