Reading Assignments:
Chapter 7 in
R. Chang, Chemistry, 9th Ed., McGraw-Hill,
2006. or previous editions.
Quantum Theory and the
Electronic Structure of Atoms
Or related topics in other textbooks.
Consultation outside lecture room:
Office Hours:
Tuesday & Thursday
10 AM -12 noon,
Wednesday
1-4 PM
Chapter 7
@Room 313-3 or by appointment
Chemistry for Engineers, SCS126
Chemistry for Engineers, SCS126
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 =
Chemistry for Engineers, SCS126
7.1
Planck’s constant (h)
h = 6.63 x 10-34 J•s
Chemistry for Engineers, SCS126
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?
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.
λ
λxν=c
E=hxν
ν
Chemistry for Engineers, SCS126
7.1
Chemistry for Engineers, SCS126
7.2
Chemistry for Engineers, SCS126
7.3
E = hν
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
E = hν
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
Chemistry for Engineers, SCS126
7.3
ni = 3
Ephoton = ∆E = Ef - Ei
ni = 3
Ef = -RH (
ni = 2
Ei = -RH (
nf = 2
1
n2f
1
n2i
)
)
nnf f==11
Chemistry for Engineers, SCS126
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 = ∆E = RH(
Ephoton = 2.18 x
1
n2i
10-18
1
n2f
Chemistry for Engineers, SCS126
7.3
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Ψ) describes:
)
1.
J x (1/25 - 1/9)
2.
Ephoton = ∆E = -1.55 x 10-19 J
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
Ephoton = h x c / λ
λ = h x c / Ephoton
λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
Chemistry for Engineers, SCS126
7.3
Chemistry for Engineers, SCS126
7.5
Schrodinger Wave Equation
Schrodinger Wave Equation
Ψ = fn(n, l, ml, ms)
Ψ = fn(n, l, ml, ms)
principal quantum number
angular momentum quantum number
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, 2, 3, 4, ….
n=1
n=2
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
n=3
Chemistry for Engineers, SCS126
l=0
l=1
l=2
l=3
7.6
l = 0 (s orbitals)
s orbital
p orbital
d orbital
f orbital
Chemistry for Engineers, SCS126
7.6
Chemistry for Engineers, SCS126
7.6
l = 2 (d orbitals)
l = 1 (p orbitals)
Chemistry for Engineers, SCS126
7.6
Schrodinger Wave Equation
Ψ = fn(n, l, ml, ms)
magnetic quantum number
ml = -1
for a given value of l
ml = -l, …., 0, …. +l
ml = 0
ml = 1
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
ml = -2
Chemistry for Engineers, SCS126
ml = -1
ml = 0
ml = 1
7.6
Schrodinger Wave Equation
ml = 2
Chemistry for Engineers, SCS126
7.6
Schrodinger Wave Equation
Ψ = fn(n, l, ml, ms)
Ψ = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Ψ.
spin quantum number
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
ms = +½ or -½
ms = +½
ms = -½
Chemistry for Engineers, SCS126
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
7.6
Chemistry for Engineers, SCS126
7.6
Schrodinger Wave Equation
Ψ = fn(n, l, ml, ms)
– electrons with the same value of n
– electrons with the same values of n and l
– 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)
Chemistry for Engineers, SCS126
7.6
Chemistry for Engineers, SCS126
Energy of orbitals in a
How many 2p orbitals are there in an atom?
7.6
electron atom
Energy only depends on principal quantum number n
n=2
If l = 1, then ml = -1, 0, or +1
2p
n=3
l=1
n=2
How many electrons can be placed in the 3d
subshell?
n=3
If l = 2, then ml = -2, -1, 0, +1, or +2
3d
l=2
n=1
Chemistry for Engineers, SCS126
7.6
Chemistry for Engineers, SCS126
7.7
Energy of orbitals in a
-electron atom
“Fill up” electrons in lowest energy orbitals (Aufbau principle)
Energy depends on n and l
n=3 l = 2
n=3 l = 0
n=2 l = 0
n=3 l = 1
??
Be
Li
B5
C
3
64electrons
electrons
22s
222s
22p
12 1
BBe
2s
Li1s1s
1s
n=2 l = 1
H
He12electron
electrons
n=1 l = 0
He
1s12
H 1s
Chemistry for Engineers, SCS126
7.7
Chemistry for Engineers, SCS126
7.9
Order of orbitals (filling) in multi-electron atom
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
C
N
Ne
O
F 1s
1s222s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
Chemistry for Engineers, SCS126
7.7
Chemistry for Engineers, SCS126
7.7
What is the electron configuration of Mg?
Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
What are the possible quantum numbers for the
last (outermost) electron in Cl?
Orbital diagram
Cl 17 electrons
1s < 2s < 2p < 3s < 3p < 4s
H
1s1
Chemistry for Engineers, SCS126
7.8
n=3
l=1
ms =for ½
or -½
ml = -1, 0, or +1 Chemistry
Engineers, SCS126
Nuclear Chemistry
unpaired electrons
2p
Chapter 23
all electrons paired
2p
Chemistry for Engineers, SCS126
7.8
Chemistry for Engineers, SCS126
7.8
Atomic number (Z) = number of protons in nucleus
Reading Assignments:
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Chapter 23 in
R. Chang, Chemistry, 9th Ed., McGraw-Hill,
2006. or previous editions.
Mass Number
Atomic Number
A
ZX
Element Symbol
Or related topics in other textbooks.
Consultation outside lecture room:
Office Hours:
Tuesday & Thursday
10 AM -12 noon,
Wednesday
1-4 PM
proton
1H
1p
1 or 1
neutron
1n
0
electron
0
0
-1e or -1β
positron
0
0
+1e or +1 β
α particle
4He
or 42α
2
A
1
1
0
0
4
Z
1
0
-1
+1
2
@Room 313-3 or by appointment
Chemistry for Engineers, SCS126 23.1
Chemistry for Engineers, SCS126
Balancing Nuclear Equations
212Po
decays by alpha emission. Write the balanced
nuclear equation for the decay of 212Po.
1. Conserve mass number (A).
The sum of protons plus neutrons in the products must equal
the sum of protons plus neutrons in the reactants.
235
92 U
+ 10n
138
55 Cs
+
96
37 Rb
alpha particle - 42He or 42α
212Po
84
4He
2
+ AZX
+ 2 10n
235 + 1 = 138 + 96 + 2x1
2. Conserve atomic number (Z) or nuclear charge.
212 = 4 + A
A = 208
84 = 2 + Z
Z = 82
The sum of nuclear charges in the products must equal the
sum of nuclear charges in the reactants.
235
92 U
+ 10n
138
55 Cs
+
96
37 Rb
+ 2 10n
92 + 0 = 55 + 37 + 2x0Chemistry for Engineers, SCS126
23.1
Chemistry for Engineers, SCS126 23.1
Nuclear Stability and Radioactive Decay
Beta decay
+-10β + ν
14C
6
14N
7
40
19K
40Ca
20
Decrease # of neutrons by 1
+ -10β + ν
1n
0
Increase # of protons by 1
1p
1
+ -10β + ν
Positron decay
11C
6
11B
5
38
19K
38Ar
18
++10β + ν
Increase # of neutrons by 1
++10β + ν
Decrease # of protons by 1
1p
1
+ν
37
18 Ar
+ -10e
37Cl
17
55Fe
26
+ -10e
55Mn
25
1p
1
Increase # of neutrons by 1
+ν
Decrease # of protons by 1
+ -10e
1n
0
+ 01n
14C
6
14C
6
14N
7
+ 11H
+ -10β + ν
t½ = 5730 years
Uranium-238 Dating
238U
92
+ν
23.2
Radiocarbon Dating
14N
7
Electron capture decay
++10β + ν
ν and ν have A = 0 and ZChemistry
= 0 for Engineers, SCS126
Chemistry for Engineers, SCS126 23.1
Nuclear Stability and Radioactive Decay
1n
0
206Pb
82
+ 8 24α + 6-10β
t½ = 4.51 x 109 years
Alpha decay
212Po
84
4He
2
+
208Pb
82
Decrease # of neutrons by 2
Decrease # of protons by 2
Spontaneous fission
252Cf
98
1
2125
49 In + 2 0n
Chemistry for Engineers, SCS126 23.2
Chemistry for Engineers, SCS126 23.3
Nuclear Fission
Nuclear Fission
Nuclear chain reaction is a self-sustaining sequence of
nuclear fission reactions.
The minimum mass of fissionable material required to
generate a self-sustaining nuclear chain reaction is the
critical mass.
235U
92
+ 01n
90Sr
38
1
+ 143
54Xe + 3 0n + Energy
Non-critical
Energy = [mass 235U + mass n – (mass 90Sr + mass 143Xe + 3 x mass n )] x c2
Energy = 3.3 x 10-11J per 235U
Critical
= 2.0 x 1013 J per mole 235U
Combustion of 1 ton of coal = 5 x 107 J
Chemistry for Engineers, SCS126 23.5
Chemistry for Engineers, SCS126 23.5
Schematic Diagram of a Nuclear Reactor
Nuclear Fusion
Fusion Reaction
2
2
3
1
1 H + 1H
1 H + 1H
2
1H
+ 13H
4
2 He
+ 01n
6Li
3
+ 12H
2 42He
Energy Released
6.3 x 10-13 J
2.8 x 10-12 J
3.6 x 10-12 J
Tokamak
magnetic plasma
confinement
Chemistry for Engineers, SCS126 23.5
Chemistry for Engineers, SCS126 23.6