Download - Chapter 7 Powerpoint Le M
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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.
PowerPoint Lecture Presentation byJ. David RobertsonUniversity of Missouri
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Properties of Waves
Wavelength () 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 () is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = x 7.1
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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 x c
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x = c
= c/ = 3.00x108 m/s/6.0x104 Hz = 5.0 x 103 m
Radio wave
A photon has a frequency of 6.0 x 104 Hz. Convertthis frequency into wavelength (nm). Does this frequencyfall in the visible region? or Is this AM or FM frequency?
= 5.0 x 1012 nm
7.1
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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 = nh
h = 6.63 x 10-34 J•s Planck’s constant n: 1,2,3,4,5 etc.
7.1
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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.
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Light has both:1. wave nature 2. particle nature
h = KE + BE
Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905
Photon is a “particle” of light
KE = h - BE
h
KE e-
7.2
Electrical Current generated by light:
Presence of the Threshold frequency
Absence of Time Lag
E = h
Alkali metals works the best.
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E = h x
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
E = h x c /
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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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 Green Blue Lower Higher Energy
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7.3
Mystery #3: Line Emission Spectrum of H Atoms
HOW are they generated?
Johann Balmer (1825-1989) solved the jigsaw puzzle
(396) 410 434 496 656 .
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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, λ = 656 nm
4, λ = 486
5, λ = 434
6, λ = 410
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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 ( )1n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J7.3
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E = h
E = h
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Ephoton = E = Ef - Ei
Ef = -RH ( )1n2
f
Ei = -RH ( )1n2
i
i fE = RH( )
1n2
1n2
nf = 1
ni = 2
nf = 1
ni = 3
nf = 2
ni = 3
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Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = E = -1.55 x 10-19 J
= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
= 1280 nm Is this IR or UV?
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 /
= h x c / Ephoton
i fE = RH( )
1n2
1n2
Ephoton =
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The Dual Nature of Matter: Wave-Particle Duality: “Wavicle”
• Einstein: Light wave is Light Particle
(1905) (“Photon”)
E = hν = hc/ λ
• De Broglie: Electron (Particle) is Wave
(1923) λ = h/mu
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De Broglie (1924) reasoned that e- is both particle and wave.
2r = n = h/mu
u = velocity of e-
m = mass of e-
Why is e- energy quantized?
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= h/mu
= 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
= 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?
What is the de Broglie wavelength (in nm) associated with an electron moving at a speed of 2x108 m/s?
m in kgh in J•s u in (m/s)
7.4 = 6.63 x 10-34 / (9.11x10-35 x 2x108) = 0.004 nm
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Chemistry in Action: Electron Microscopy
STM image of iron atomson copper surface
e = 0.004 nm
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7.6
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Chemistry in Action: Element from the SunIn 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
H: 656 486 434 410
He: 668 588 502 447
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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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Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
Wave function () describes:
1. energy of e- with a given
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.
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Schrodinger Wave Equation
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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e- density (1s orbital) falls off rapidly as distance from nucleus increases
Where 90% of thee- density is foundfor the 1s orbital
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= 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 = 0n = 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 orbitall = 1 p orbitall = 2 d orbitall = 3 f orbital
Schrodinger Wave Equation
7.6
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l = 0 (s orbitals)
l = 1 (p orbitals)
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l = 2 (d orbitals)
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= fn(n, l, ml, ms)
magnetic quantum number ml
for a given value of lml = -l, …., 0, …. +l
orientation of the orbital in space
if l = 1 (p orbital), ml = -1, 0, or 1if 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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= fn(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
Schrodinger Wave Equation
ms = -½ms = +½
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Existence (and energy) of electron in atom is described by its unique wave function .
Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.
Schrodinger Wave Equation
= fn(n, l, ml, ms)
Each seat is uniquely identified (E, R12, S8)Each seat can hold only one individual at a time
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7.6
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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
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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 atomEnergy only depends on principal quantum number n
En = -RH ( )1n2
n=1
n=2
n=3
7.7
l = 0 1 2 3
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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
? ?
7.7
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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
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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 quantumnumber n
angular momentumquantum number l
number of electronsin the orbital or subshell
Orbital diagram
H
1s1
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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
Ne, ArO, S