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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 (λ) 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 (ν) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = λ x ν7.1
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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 radiationλ x ν = c
7.1
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7.1
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λ x ν = cλ = c/νλ = 3.00 x 108 m/s / 6.0 x 104 Hzλ = 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?
λ = 5.0 x 1012 nm
λ
ν
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 νPlanck’s constant (h)h = 6.63 x 10-34 J•s
7.1
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Light has both:1. wave nature2. particle nature
hν = KE + BE
Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905
Photon is a “particle” of light
KE = hν - BE
hν
KE e-
7.2
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E = h x ν
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 / λ
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( )1n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J7.3
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E = hν
E = hν
7.3
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Ephoton = ΔE = Ef - Ei
Ef = -RH ( )1n2
f
Ei = -RH ( )1n2
i
i fΔE = RH( )
1n2
1n2
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 = ΔE = -1.55 x 10-19 J
λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
λ = 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 / λ
λ = h x c / Ephoton
i fΔE = RH( )
1n2
1n2Ephoton =
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-
2πr = nλ λ = hmu
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λ = h/muλ = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)λ = 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 atomson copper surface
λe = 0.004 nm
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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 (Ψ) describes:
1. energy of e- with a given Ψ
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 EquationΨ = 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 thee- density is foundfor the 1s orbital
7.6
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Ψ = 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)
7.6
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l = 2 (d orbitals)
7.6
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Ψ = 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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Ψ = 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 Ψ.
Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.
Schrodinger Wave EquationΨ = 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 EquationΨ = 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
Ψ = (n, l, ml, ½) or Ψ = (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( )1n2
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 = 0 n=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 electronsLi 1s22s1
Be 4 electronsBe 1s22s2
B 5 electronsB 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 electronsN 1s22s22p3
O 8 electronsO 1s22s22p4
F 9 electronsF 1s22s22p5
Ne 10 electronsNe 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 quantumnumber n
angular momentumquantum number l
number of electronsin 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 electronsLast 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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Paramagneticunpaired electrons
2p
Diamagneticall 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