chm 4444, advanced inorganic chemistry. chapters 1, atomic structure 2, molecular structure &...
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CHM 4444, Advanced Inorganic
Chemistry
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapters
1, Atomic structure2, Molecular structure & bonding3, Structures of simple solids6, Molecular symmetry7, Introduction to coordination
compounds20, d-Metal complexes: electronic
structure and spectra
Chapter 1
Atomic Structure
Chapter 1: Atomic Structure
1.1 The nucleosynthesis of light elements
1.2 The nucleosynthesis of heavy elements
1.3 Spectroscopic information1.4 Some principles of quantum
mechanics1.5 Atomic orbitals1.6 Penetration and shielding1.7 The building-up principle1.8 The classification of the elements1.9 Atomic parameters
Chapter 1: Atomic Structure
1.1 The nucleosynthesis of light elements
1.2 The nucleosynthesis of heavy elements
1.3 Spectroscopic information1.4 Some principles of quantum
mechanics1.5 Atomic orbitals1.6 Penetration and shielding1.7 The building-up principle1.8 The classification of the elements1.9 Atomic parameters
Section 1.0 ObjectivesKnow what a nucleon is.
Wikipedia
nucleon: a collective name for two particles: the neutron and the proton.
Section 1.0 ObjectivesDescribe the origin of H and He in
terms of the big bang, electromagnetic force, and the strong force.
“Gödel, Escher, Bach” by D. HofstadterGödel’s Incompleteness Theorem (1st part)Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true, but not provable in the theory.From
Wikipedia
Section 1.0 ObjectiveWhat is the most abundant element in
the universe? What is the second most abundant element?
Estimated Current composition of the universe:89% H11% He
Section 1.0 ObjectivesUse the nucleon number to describe isotopes and nuclides.
Wikipedia
nuclide: an atomic species characterized by the specific constitution of its nucleus:
number of protons, Z, number of neutrons, N, and nuclear energy state.
Isotopes: have same number of protons
nucleon number: sum of Z and NP-31, 12C
Section 1.0 ObjectivesGiven a fundamental particle, state
what its properties are, and vice versa.
Subatomic particles relevant to chemistrymass
particlesymbol mass/u*numbercharge/e†spinproton p 1.0073 1 +1 ½neutron n 1.0087 1 0 ½electron e– 0.0005486 0 –1 ½positron e+ 0.0005486 0 +1 ½α particleα nucleus of He-4 4 +2 0ß particleß e– ejected from nucleus 0 –1
½photon γ 0 0 0 1γ photon γ photon from nucleus 0 0
1neutrino ν very small 0 0 ½
*Masses are given in atomic mass units; mu = 1.6605 × 10–27 kg†The elementary charge, e = 1.602 × 10–19 C
Section 1.1 ObjectivesUse “nuclear math” to complete a
nuclear reaction.
42
126C + γ 16
8O+ α ΔE = -7.2 MeV
Mass is conserved
Charge is conserved
Section 1.1 ObjectivesGiven the conversion factor, convert
between eV and kJ/mol.
1 eV = 1.602 10-19 J, or 96.48 kJ/mol
42
126C + γ 16
8O+ α ΔE = -7.2 MeV
= -1.15 × 10–12 J
= -6.94 × 108 kJ/mol
( 6.022×1023
1mol )ΔE = -7.2 MeV( 1×10
6 eV1MeV )( 1×10
− 3 kJ1 J )( 1.602×10
−19 J1eV )
Section 1.1 ObjectivesCalculate the nuclear binding energy
per nucleon of a nuclide.
Wikipedia
Nuclear binding energy: Energy required to split the nucleus of an atom into its component parts.
atom + binding energy nucleons + electrons
Nuclear binding energyCalculate nuclear binding energy using the equivalence of mass and energy: E = mc2.
atom + binding energy nucleons + electrons
matom + menergy = mnucleons + melectrons
menergy = mnucleons + melectrons – matom
E = mc2 = (mnucleons + melectrons – matom)c2
Nuclear binding energyCalculate nuclear binding energy using the equivalence of mass and energy: E = mc2.
E = mc2 = (mnucleons + melectrons – matom)c2
neutron 1.008665 u
proton 1.007277 uelectro
n 0.000548 u
u kg1.66054 × 10-27 kg/ u
c2.99793 × 108 m/s
J eV1.60218 × 10-19 J/eV
Constants
ExampleCalculate the nuclear binding energy per nucleon in MeV for , which has a mass of 14.00307 u.
neutron
1.008665 u
proton1.007277 u
electron
0.000548 u
Constants
Mass of N-14 components:7 p7 n7 e-
7 × 1.007277 u7 × 1.008665 u7 × 0.000548 u
= 7.050939 u= 7.060655 u= 0.003836 u14.115430 u
Difference in Mass: 0.11236 u
ExampleDifference in Mass: 0.11236 u
u kg1.66054 × 10-27 kg/u
c2.99793 × 108 m/s
J eV1.60218 × 10-19 J/eV
Constants
E = mc2
= 1.67688 × 10-11 J
E = 0.11236 u( 1.66054×10−27 kg
u )( 2.99793×108 m
s )2
E = 1.67688 × 10-11 J( 1 eV
1.60218×10−19 J )( 1 MeV
1×106 eV )
ExampleDifference in Mass: 0.11236 uE = mc2
= 1.67688 × 10-11 J
E = 0.11236 u( 1.66054×10−27 kg
u )( 2.99793×108 m
s )2
E = 1.67688 × 10-11 J( 1 eV
1.60218×10−19 J )( 1 MeV
1×106 eV )= 104.663 MeV
Binding energy per nucleon = 104.663 MeV 14
Binding energy per nucleon = 7.1759 MeV
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81-9
-8
-7
-6
-5
-4
-3
-2
-1
0 1
2
3
4
5
6
7
8
9
10 26
82
83
Atomic number
–Bin
din
g e
nerg
y p
er
nucl
eon/M
eV
Nuclear Binding Energy per Nucleon vs. Atomic Number
Box 1.1 ObjectivesPredict whether a nuclide will undergo
fusion or fission.
Fusion: nuclei merge
Fission: nuclei split
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81-9
-8
-7
-6
-5
-4
-3
-2
-1
0 1
2
3
4
5
6
7
8
9
10 26
82
83
Atomic number
–Bin
din
g e
nerg
y p
er
nucl
eon/M
eV
Nuclear Binding Energy per Nucleon vs. Atomic Number
fusion fissionnuclei merge nuclei split2 Ne1020 → Ca20
40 +energyU92
236 → Xe54140 + Sr38
93 +3 n01 + energy
Box 1.1 ObjectivesGiven nuclear binding energies per nucleon, calculate the energy change for a fusion or fission reaction.
Example ProblemCalculate the energy change for the
following reaction:
Binding energies per nucleon:U-236: 7.6 MeVXe-140: 8.4 MeVSr-93: 8.7 MeV
Answer: 191.5 MeV
23692U
10n+ 3 140
54Xe+ Sr9338
Example ProblemCalculate the energy change for the
following reaction:Binding energies per nucleon:
7.6 MeV 8.4 MeV 8.7 MeV
23692U
10n+ 3 140
54Xe+ 9338Sr
(140 × –8.4 MeV) + (93 × –8.7 MeV) – (236 × –7.6
MeV) = –191.5 MeV
Section 1.1 ObjectivesGiven a plot of nuclear binding energies, explain trends in elemental abundances in the sun.
0 20 40 60 80-14
-12
-10
-8
-6
-4
-2
0 Series1
Abundance of the Elements:the Sun
Atomic number
Lo
g (a
tom
fra
cti
on
)
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71 76 81-9
-8
-7
-6
-5
-4
-3
-2
-1
0 1
2
3
4
5
6
7
8
9
10 26
82
83
Atomic number
–Bin
din
g e
nerg
y p
er
nucl
eon/M
eV
Binding Energy per Nucleon vs. Atomic Number
0 20 40 60 80-16
-14
-12
-10
-8
-6
-4
-2
0
Series1
Abundance of the Elements:the Earth's Crust
Atomic Number
Lo
g (
ato
m f
racti
on
)
Section 1.1 ObjectivesKnow that carbon catalyzes conversion of H to He.
Carbon-catalyzed synthesis of He from HProton capture
Positron decay
Proton capture
Proton capture
Positron decay
Proton capture
Sum:
126C
136C
42a
e+ n11p
+
137N g1
1p ++136C+13
7N147N++ g
147N
11p
158O+ g
158O
157N+ e+ n+
157N
11p
126C++42a
11p +4 2e+ 2n+ +3g
+
Section 1.1 ObjectivesGiven the reaction,
explain how carbon dating works.
147N
10n+ 14
6C+ 11p
Source of neutrons: cosmic rays (mainly protons) hit O-16, producing neutrons, among other things.--From Wikipedia
147N
10n+ 14
6C+ 11p
CO214CO2
O2
t½ = 5730 y
Fraction of 14CO2 in atmosphere is constant
t½ = 5730 y
Chapter 1: Atomic Structure
1.1 The nucleosynthesis of light elements
1.2 The nucleosynthesis of heavy elements
1.3 Spectroscopic information1.4 Some principles of quantum
mechanics1.5 Atomic orbitals1.6 Penetration and shielding1.7 The building-up principle1.8 The classification of the elements1.9 Atomic parameters