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