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Franklin Ow

Franklin Ow

Advanced Bonding Theories

Slide Duration:

Table of Contents

I. Basic Concepts & Measurement of Chemistry
Basic Concepts of Chemistry

16m 26s

Intro
0:00
Lesson Overview
0:07
Introduction
0:56
What is Chemistry?
0:57
What is Matter?
1:16
Solids
1:43
General Characteristics
1:44
Particulate-level Drawing of Solids
2:34
Liquids
3:39
General Characteristics of Liquids
3:40
Particulate-level Drawing of Liquids
3:55
Gases
4:23
General Characteristics of Gases
4:24
Particulate-level Drawing Gases
5:05
Classification of Matter
5:27
Classification of Matter
5:26
Pure Substances
5:54
Pure Substances
5:55
Mixtures
7:06
Definition of Mixtures
7:07
Homogeneous Mixtures
7:11
Heterogeneous Mixtures
7:52
Physical and Chemical Changes/Properties
8:18
Physical Changes Retain Chemical Composition
8:19
Chemical Changes Alter Chemical Composition
9:32
Physical and Chemical Changes/Properties, cont'd
10:55
Physical Properties
10:56
Chemical Properties
11:42
Sample Problem 1: Chemical & Physical Change
12:22
Sample Problem 2: Element, Compound, or Mixture?
13:52
Sample Problem 3: Classify Each of the Following Properties as chemical or Physical
15:03
Tools in Quantitative Chemistry

29m 22s

Intro
0:00
Lesson Overview
0:07
Units of Measurement
1:23
The International System of Units (SI): Mass, Length, and Volume
1:39
Percent Error
2:17
Percent Error
2:18
Example: Calculate the Percent Error
2:56
Standard Deviation
3:48
Standard Deviation Formula
3:49
Standard Deviation cont'd
4:42
Example: Calculate Your Standard Deviation
4:43
Precisions vs. Accuracy
6:25
Precision
6:26
Accuracy
7:01
Significant Figures and Uncertainty
7:50
Consider the Following (2) Rulers
7:51
Consider the Following Graduated Cylinder
11:30
Identifying Significant Figures
12:43
The Rules of Sig Figs Overview
12:44
The Rules for Sig Figs: All Nonzero Digits Are Significant
13:21
The Rules for Sig Figs: A Zero is Significant When It is In-Between Nonzero Digits
13:28
The Rules for Sig Figs: A Zero is Significant When at the End of a Decimal Number
14:02
The Rules for Sig Figs: A Zero is not significant When Starting a Decimal Number
14:27
Using Sig Figs in Calculations
15:03
Using Sig Figs for Multiplication and Division
15:04
Using Sig Figs for Addition and Subtraction
15:48
Using Sig Figs for Mixed Operations
16:11
Dimensional Analysis
16:20
Dimensional Analysis Overview
16:21
General Format for Dimensional Analysis
16:39
Example: How Many Miles are in 17 Laps?
17:17
Example: How Many Grams are in 1.22 Pounds?
18:40
Dimensional Analysis cont'd
19:43
Example: How Much is Spent on Diapers in One Week?
19:44
Dimensional Analysis cont'd
21:03
SI Prefixes
21:04
Dimensional Analysis cont'd
22:03
500 mg → ? kg
22:04
34.1 cm → ? um
24:03
Summary
25:11
Sample Problem 1: Dimensional Analysis
26:09
II. Atoms, Molecules, and Ions
Atoms, Molecules, and Ions

52m 18s

Intro
0:00
Lesson Overview
0:08
Introduction to Atomic Structure
1:03
Introduction to Atomic Structure
1:04
Plum Pudding Model
1:26
Introduction to Atomic Structure Cont'd
2:07
John Dalton's Atomic Theory: Number 1
2:22
John Dalton's Atomic Theory: Number 2
2:50
John Dalton's Atomic Theory: Number 3
3:07
John Dalton's Atomic Theory: Number 4
3:30
John Dalton's Atomic Theory: Number 5
3:58
Introduction to Atomic Structure Cont'd
5:21
Ernest Rutherford's Gold Foil Experiment
5:22
Introduction to Atomic Structure Cont'd
7:42
Implications of the Gold Foil Experiment
7:43
Relative Masses and Charges
8:18
Isotopes
9:02
Isotopes
9:03
Introduction to The Periodic Table
12:17
The Periodic Table of the Elements
12:18
Periodic Table, cont'd
13:56
Metals
13:57
Nonmetals
14:25
Semimetals
14:51
Periodic Table, cont'd
15:57
Group I: The Alkali Metals
15:58
Group II: The Alkali Earth Metals
16:25
Group VII: The Halogens
16:40
Group VIII: The Noble Gases
17:08
Ionic Compounds: Formulas, Names, Props.
17:35
Common Polyatomic Ions
17:36
Predicting Ionic Charge for Main Group Elements
18:52
Ionic Compounds: Formulas, Names, Props.
20:36
Naming Ionic Compounds: Rule 1
20:51
Naming Ionic Compounds: Rule 2
21:22
Naming Ionic Compounds: Rule 3
21:50
Naming Ionic Compounds: Rule 4
22:22
Ionic Compounds: Formulas, Names, Props.
22:50
Naming Ionic Compounds Example: Al₂O₃
22:51
Naming Ionic Compounds Example: FeCl₃
23:21
Naming Ionic Compounds Example: CuI₂ 3H₂O
24:00
Naming Ionic Compounds Example: Barium Phosphide
24:40
Naming Ionic Compounds Example: Ammonium Phosphate
25:55
Molecular Compounds: Formulas and Names
26:42
Molecular Compounds: Formulas and Names
26:43
The Mole
28:10
The Mole is 'A Chemist's Dozen'
28:11
It is a Central Unit, Connecting the Following Quantities
30:01
The Mole, cont'd
32:07
Atomic Masses
32:08
Example: How Many Moles are in 25.7 Grams of Sodium?
32:28
Example: How Many Atoms are in 1.2 Moles of Carbon?
33:17
The Mole, cont'd
34:25
Example: What is the Molar Mass of Carbon Dioxide?
34:26
Example: How Many Grams are in 1.2 Moles of Carbon Dioxide?
25:46
Percentage Composition
36:43
Example: How Many Grams of Carbon Contained in 65.1 Grams of Carbon Dioxide?
36:44
Empirical and Molecular Formulas
39:19
Empirical Formulas
39:20
Empirical Formula & Elemental Analysis
40:21
Empirical and Molecular Formulas, cont'd
41:24
Example: Determine Both the Empirical and Molecular Formulas - Step 1
41:25
Example: Determine Both the Empirical and Molecular Formulas - Step 2
43:18
Summary
46:22
Sample Problem 1: Determine the Empirical Formula of Lithium Fluoride
47:10
Sample Problem 2: How Many Atoms of Carbon are Present in 2.67 kg of C₆H₆?
49:21
III. Chemical Reactions
Chemical Reactions

43m 24s

Intro
0:00
Lesson Overview
0:06
The Law of Conservation of Mass and Balancing Chemical Reactions
1:49
The Law of Conservation of Mass
1:50
Balancing Chemical Reactions
2:50
Balancing Chemical Reactions Cont'd
3:40
Balance: N₂ + H₂ → NH₃
3:41
Balance: CH₄ + O₂ → CO₂ + H₂O
7:20
Balancing Chemical Reactions Cont'd
9:49
Balance: C₂H₆ + O₂ → CO₂ + H₂O
9:50
Intro to Chemical Equilibrium
15:32
When an Ionic Compound Full Dissociates
15:33
When an Ionic Compound Incompletely Dissociates
16:14
Dynamic Equilibrium
17:12
Electrolytes and Nonelectrolytes
18:03
Electrolytes
18:04
Strong Electrolytes and Weak Electrolytes
18:55
Nonelectrolytes
19:23
Predicting the Product(s) of an Aqueous Reaction
20:02
Single-replacement
20:03
Example: Li (s) + CuCl₂ (aq) → 2 LiCl (aq) + Cu (s)
21:03
Example: Cu (s) + LiCl (aq) → NR
21:23
Example: Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
22:32
Predicting the Product(s) of an Aqueous Reaction
23:37
Double-replacement
23:38
Net-ionic Equation
25:29
Predicting the Product(s) of an Aqueous Reaction
26:12
Solubility Rules for Ionic Compounds
26:13
Predicting the Product(s) of an Aqueous Reaction
28:10
Neutralization Reactions
28:11
Example: HCl (aq) + NaOH (aq) → ?
28:37
Example: H₂SO₄ (aq) + KOH (aq) → ?
29:25
Predicting the Product(s) of an Aqueous Reaction
30:20
Certain Aqueous Reactions can Produce Unstable Compounds
30:21
Example 1
30:52
Example 2
32:16
Example 3
32:54
Summary
33:54
Sample Problem 1
34:55
ZnCO₃ (aq) + H₂SO₄ (aq) → ?
35:09
NH₄Br (aq) + Pb(C₂H₃O₂)₂ (aq) → ?
36:02
KNO₃ (aq) + CuCl₂ (aq) → ?
37:07
Li₂SO₄ (aq) + AgNO₃ (aq) → ?
37:52
Sample Problem 2
39:09
Question 1
39:10
Question 2
40:36
Question 3
41:47
Chemical Reactions II

55m 40s

Intro
0:00
Lesson Overview
0:10
Arrhenius Definition
1:15
Arrhenius Acids
1:16
Arrhenius Bases
3:20
The Bronsted-Lowry Definition
4:48
Acids Dissolve In Water and Donate a Proton to Water: Example 1
4:49
Acids Dissolve In Water and Donate a Proton to Water: Example 2
6:54
Monoprotic Acids & Polyprotic Acids
7:58
Strong Acids
11:30
Bases Dissolve In Water and Accept a Proton From Water
12:41
Strong Bases
16:36
The Autoionization of Water
17:42
Amphiprotic
17:43
Water Reacts With Itself
18:24
Oxides of Metals and Nonmetals
20:08
Oxides of Metals and Nonmetals Overview
20:09
Oxides of Nonmetals: Acidic Oxides
21:23
Oxides of Metals: Basic Oxides
24:08
Oxidation-Reduction (Redox) Reactions
25:34
Redox Reaction Overview
25:35
Oxidizing and Reducing Agents
27:02
Redox Reaction: Transfer of Electrons
27:54
Oxidation-Reduction Reactions Cont'd
29:55
Oxidation Number Overview
29:56
Oxidation Number of Homonuclear Species
31:17
Oxidation Number of Monatomic Ions
32:58
Oxidation Number of Fluorine
33:27
Oxidation Number of Oxygen
34:00
Oxidation Number of Chlorine, Bromine, and Iodine
35:07
Oxidation Number of Hydrogen
35:30
Net Sum of All Oxidation Numbers In a Compound
36:21
Oxidation-Reduction Reactions Cont'd
38:19
Let's Practice Assigning Oxidation Number
38:20
Now Let's Apply This to a Chemical Reaction
41:07
Summary
44:19
Sample Problems
45:29
Sample Problem 1
45:30
Sample Problem 2: Determine the Oxidizing and Reducing Agents
48:48
Sample Problem 3: Determine the Oxidizing and Reducing Agents
50:43
IV. Stoichiometry
Stoichiometry I

42m 10s

Intro
0:00
Lesson Overview
0:23
Mole to Mole Ratios
1:32
Example 1: In 1 Mole of H₂O, How Many Moles Are There of Each Element?
1:53
Example 2: In 2.6 Moles of Water, How Many Moles Are There of Each Element?
2:24
Mole to Mole Ratios Cont'd
5:13
Balanced Chemical Reaction
5:14
Mole to Mole Ratios Cont'd
7:25
Example 3: How Many Moles of Ammonia Can Form If you Have 3.1 Moles of H₂?
7:26
Example 4: How Many Moles of Hydrogen Gas Are Required to React With 6.4 Moles of Nitrogen Gas?
9:08
Mass to mass Conversion
11:06
Mass to mass Conversion
11:07
Example 5: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂?
12:37
Example 6: How Many Grams of Hydrogen Gas Are Required to React With 6.4 Grams of Nitrogen Gas?
15:34
Example 7: How Man Milligrams of Ammonia Can Form If You Have 1.2 kg of H₂?
17:29
Limiting Reactants, Percent Yields
20:42
Limiting Reactants, Percent Yields
20:43
Example 8: How Many Grams of Ammonia Can Form If You Have 3.1 Grams of H₂ and 3.1 Grams of N₂
22:25
Percent Yield
25:30
Example 9: How Many Grams of The Excess Reactant Remains?
26:37
Summary
29:34
Sample Problem 1: How Many Grams of Carbon Are In 2.2 Kilograms of Carbon Dioxide?
30:47
Sample Problem 2: How Many Milligrams of Carbon Dioxide Can Form From 23.1 Kg of CH₄(g)?
33:06
Sample Problem 3: Part 1
36:10
Sample Problem 3: Part 2 - What Amount Of The Excess Reactant Will Remain?
40:53
Stoichiometry II

42m 38s

Intro
0:00
Lesson Overview
0:10
Molarity
1:14
Solute and Solvent
1:15
Molarity
2:01
Molarity Cont'd
2:59
Example 1: How Many Grams of KBr are Needed to Make 350 mL of a 0.67 M KBr Solution?
3:00
Example 2: How Many Moles of KBr are in 350 mL of a 0.67 M KBr Solution?
5:44
Example 3: What Volume of a 0.67 M KBr Solution Contains 250 mg of KBr?
7:46
Dilutions
10:01
Dilution: M₁V₂=M₁V₂
10:02
Example 5: Explain How to Make 250 mL of a 0.67 M KBr Solution Starting From a 1.2M Stock Solution
12:04
Stoichiometry and Double-Displacement Precipitation Reactions
14:41
Example 6: How Many grams of PbCl₂ Can Form From 250 mL of 0.32 M NaCl?
15:38
Stoichiometry and Double-Displacement Precipitation Reactions
18:05
Example 7: How Many grams of PbCl₂ Can Form When 250 mL of 0.32 M NaCl and 150 mL of 0.45 Pb(NO₃)₂ Mix?
18:06
Stoichiometry and Neutralization Reactions
21:01
Example 8: How Many Grams of NaOh are Required to Neutralize 4.5 Grams of HCl?
21:02
Stoichiometry and Neutralization Reactions
23:03
Example 9: How Many mL of 0.45 M NaOH are Required to Neutralize 250 mL of 0.89 M HCl?
23:04
Stoichiometry and Acid-Base Standardization
25:28
Introduction to Titration & Standardization
25:30
Acid-Base Titration
26:12
The Analyte & Titrant
26:24
The Experimental Setup
26:49
The Experimental Setup
26:50
Stoichiometry and Acid-Base Standardization
28:38
Example 9: Determine the Concentration of the Analyte
28:39
Summary
32:46
Sample Problem 1: Stoichiometry & Neutralization
35:24
Sample Problem 2: Stoichiometry
37:50
V. Thermochemistry
Energy & Chemical Reactions

55m 28s

Intro
0:00
Lesson Overview
0:14
Introduction
1:22
Recall: Chemistry
1:23
Energy Can Be Expressed In Different Units
1:57
The First Law of Thermodynamics
2:43
Internal Energy
2:44
The First Law of Thermodynamics Cont'd
6:14
Ways to Transfer Internal Energy
6:15
Work Energy
8:13
Heat Energy
8:34
∆U = q + w
8:44
Calculating ∆U, Q, and W
8:58
Changes In Both Volume and Temperature of a System
8:59
Calculating ∆U, Q, and W Cont'd
11:01
The Work Equation
11:02
Example 1: Calculate ∆U For The Burning Fuel
11:45
Calculating ∆U, Q, and W Cont'd
14:09
The Heat Equation
14:10
Calculating ∆U, Q, and W Cont'd
16:03
Example 2: Calculate The Final Temperature
16:04
Constant-Volume Calorimetry
18:05
Bomb Calorimeter
18:06
The Effect of Constant Volume On The Equation For Internal Energy
22:11
Example 3: Calculate ∆U
23:12
Constant-Pressure Conditions
26:05
Constant-Pressure Conditions
26:06
Calculating Enthalpy: Phase Changes
27:29
Melting, Vaporization, and Sublimation
27:30
Freezing, Condensation and Deposition
28:25
Enthalpy Values For Phase Changes
28:40
Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
29:40
Calculating Enthalpy: Heats of Reaction
31:22
Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
31:23
Using Standard Enthalpies of Formation
33:53
Standard Enthalpies of Formation
33:54
Using Standard Enthalpies of Formation
36:12
Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
36:13
Enthalpy From a Series of Reactions
39:58
Hess's Law
39:59
Coffee-Cup Calorimetry
42:43
Coffee-Cup Calorimetry
42:44
Example 7: Calculate ∆H° of Reaction
45:10
Summary
47:12
Sample Problem 1
48:58
Sample Problem 2
51:24
VI. Quantum Theory of Atoms
Structure of Atoms

42m 33s

Intro
0:00
Lesson Overview
0:07
Introduction
1:01
Rutherford's Gold Foil Experiment
1:02
Electromagnetic Radiation
2:31
Radiation
2:32
Three Parameters: Energy, Frequency, and Wavelength
2:52
Electromagnetic Radiation
5:18
The Electromagnetic Spectrum
5:19
Atomic Spectroscopy and The Bohr Model
7:46
Wavelengths of Light
7:47
Atomic Spectroscopy Cont'd
9:45
The Bohr Model
9:46
Atomic Spectroscopy Cont'd
12:21
The Balmer Series
12:22
Rydberg Equation For Predicting The Wavelengths of Light
13:04
The Wave Nature of Matter
15:11
The Wave Nature of Matter
15:12
The Wave Nature of Matter
19:10
New School of Thought
19:11
Einstein: Energy
19:49
Hertz and Planck: Photoelectric Effect
20:16
de Broglie: Wavelength of a Moving Particle
21:14
Quantum Mechanics and The Atom
22:15
Heisenberg: Uncertainty Principle
22:16
Schrodinger: Wavefunctions
23:08
Quantum Mechanics and The Atom
24:02
Principle Quantum Number
24:03
Angular Momentum Quantum Number
25:06
Magnetic Quantum Number
26:27
Spin Quantum Number
28:42
The Shapes of Atomic Orbitals
29:15
Radial Wave Function
29:16
Probability Distribution Function
32:08
The Shapes of Atomic Orbitals
34:02
3-Dimensional Space of Wavefunctions
34:03
Summary
35:57
Sample Problem 1
37:07
Sample Problem 2
40:23
VII. Electron Configurations and Periodicity
Periodic Trends

38m 50s

Intro
0:00
Lesson Overview
0:09
Introduction
0:36
Electron Configuration of Atoms
1:33
Electron Configuration & Atom's Electrons
1:34
Electron Configuration Format
1:56
Electron Configuration of Atoms Cont'd
3:01
Aufbau Principle
3:02
Electron Configuration of Atoms Cont'd
6:53
Electron Configuration Format 1: Li, O, and Cl
6:56
Electron Configuration Format 2: Li, O, and Cl
9:11
Electron Configuration of Atoms Cont'd
12:48
Orbital Box Diagrams
12:49
Pauli Exclusion Principle
13:11
Hund's Rule
13:36
Electron Configuration of Atoms Cont'd
17:35
Exceptions to The Aufbau Principle: Cr
17:36
Exceptions to The Aufbau Principle: Cu
18:15
Electron Configuration of Atoms Cont'd
20:22
Electron Configuration of Monatomic Ions: Al
20:23
Electron Configuration of Monatomic Ions: Al³⁺
20:46
Electron Configuration of Monatomic Ions: Cl
21:57
Electron Configuration of Monatomic Ions: Cl¹⁻
22:09
Electron Configuration Cont'd
24:31
Paramagnetism
24:32
Diamagnetism
25:00
Atomic Radii
26:08
Atomic Radii
26:09
In a Column of the Periodic Table
26:25
In a Row of the Periodic Table
26:46
Ionic Radii
27:30
Ionic Radii
27:31
Anions
27:42
Cations
27:57
Isoelectronic Species
28:12
Ionization Energy
29:00
Ionization Energy
29:01
Electron Affinity
31:37
Electron Affinity
31:37
Summary
33:43
Sample Problem 1: Ground State Configuration and Orbital Box Diagram
34:21
Fe
34:48
P
35:32
Sample Problem 2
36:38
Which Has The Larger Ionization Energy: Na or Li?
36:39
Which Has The Larger Atomic Size: O or N ?
37:23
Which Has The Larger Atomic Size: O²⁻ or N³⁻ ?
38:00
VIII. Molecular Geometry & Bonding Theory
Bonding & Molecular Structure

52m 39s

Intro
0:00
Lesson Overview
0:08
Introduction
1:10
Types of Chemical Bonds
1:53
Ionic Bond
1:54
Molecular Bond
2:42
Electronegativity and Bond Polarity
3:26
Electronegativity (EN)
3:27
Periodic Trend
4:36
Electronegativity and Bond Polarity Cont'd
6:04
Bond Polarity: Polar Covalent Bond
6:05
Bond Polarity: Nonpolar Covalent Bond
8:53
Lewis Electron Dot Structure of Atoms
9:48
Lewis Electron Dot Structure of Atoms
9:49
Lewis Structures of Polyatomic Species
12:51
Single Bonds
12:52
Double Bonds
13:28
Nonbonding Electrons
13:59
Lewis Structures of Polyatomic Species Cont'd
14:45
Drawing Lewis Structures: Step 1
14:48
Drawing Lewis Structures: Step 2
15:16
Drawing Lewis Structures: Step 3
15:52
Drawing Lewis Structures: Step 4
17:31
Drawing Lewis Structures: Step 5
19:08
Drawing Lewis Structure Example: Carbonate
19:33
Resonance and Formal Charges (FC)
24:06
Resonance Structures
24:07
Formal Charge
25:20
Resonance and Formal Charges Cont'd
27:46
More On Formal Charge
27:47
Resonance and Formal Charges Cont'd
28:21
Good Resonance Structures
28:22
VSEPR Theory
31:08
VSEPR Theory Continue
31:09
VSEPR Theory Cont'd
32:53
VSEPR Geometries
32:54
Steric Number
33:04
Basic Geometry
33:50
Molecular Geometry
35:50
Molecular Polarity
37:51
Steps In Determining Molecular Polarity
37:52
Example 1: Polar
38:47
Example 2: Nonpolar
39:10
Example 3: Polar
39:36
Example 4: Polar
40:08
Bond Properties: Order, Length, and Energy
40:38
Bond Order
40:39
Bond Length
41:21
Bond Energy
41:55
Summary
43:09
Sample Problem 1
43:42
XeO₃
44:03
I₃⁻
47:02
SF₅
49:16
Advanced Bonding Theories

1h 11m 41s

Intro
0:00
Lesson Overview
0:09
Introduction
0:38
Valence Bond Theory
3:07
Valence Bond Theory
3:08
spᶟ Hybridized Carbon Atom
4:19
Valence Bond Theory Cont'd
6:24
spᶟ Hybridized
6:25
Hybrid Orbitals For Water
7:26
Valence Bond Theory Cont'd (spᶟ)
11:53
Example 1: NH₃
11:54
Valence Bond Theory Cont'd (sp²)
14:48
sp² Hybridization
14:49
Example 2: BF₃
16:44
Valence Bond Theory Cont'd (sp)
22:44
sp Hybridization
22:46
Example 3: HCN
23:38
Valence Bond Theory Cont'd (sp³d and sp³d²)
27:36
Valence Bond Theory: sp³d and sp³d²
27:37
Molecular Orbital Theory
29:10
Valence Bond Theory Doesn't Always Account For a Molecule's Magnetic Behavior
29:11
Molecular Orbital Theory Cont'd
30:37
Molecular Orbital Theory
30:38
Wavefunctions
31:04
How s-orbitals Can Interact
32:23
Bonding Nature of p-orbitals: Head-on
35:34
Bonding Nature of p-orbitals: Parallel
39:04
Interaction Between s and p-orbital
40:45
Molecular Orbital Diagram For Homonuclear Diatomics: H₂
42:21
Molecular Orbital Diagram For Homonuclear Diatomics: He₂
45:23
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂
46:39
Molecular Orbital Diagram For Homonuclear Diatomic: Li₂⁺
47:42
Molecular Orbital Diagram For Homonuclear Diatomic: B₂
48:57
Molecular Orbital Diagram For Homonuclear Diatomic: N₂
54:04
Molecular Orbital Diagram: Molecular Oxygen
55:57
Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
1:02:16
Sample Problem 1: Determine the Atomic Hybridization
1:07:20
XeO₃
1:07:21
SF₆
1:07:49
I₃⁻
1:08:20
Sample Problem 2
1:09:04
IX. Gases, Solids, & Liquids
Gases

35m 6s

Intro
0:00
Lesson Overview
0:07
The Kinetic Molecular Theory of Gases
1:23
The Kinetic Molecular Theory of Gases
1:24
Parameters To Characterize Gases
3:35
Parameters To Characterize Gases: Pressure
3:37
Interpreting Pressure On a Particulate Level
4:43
Parameters Cont'd
6:08
Units For Expressing Pressure: Psi, Pascal
6:19
Units For Expressing Pressure: mm Hg
6:42
Units For Expressing Pressure: atm
6:58
Units For Expressing Pressure: torr
7:24
Parameters Cont'd
8:09
Parameters To Characterize Gases: Volume
8:10
Common Units of Volume
9:00
Parameters Cont'd
9:11
Parameters To Characterize Gases: Temperature
9:12
Particulate Level
9:36
Parameters To Characterize Gases: Moles
10:24
The Simple Gas Laws
10:43
Gas Laws Are Only Valid For…
10:44
Charles' Law
11:24
The Simple Gas Laws
13:13
Boyle's Law
13:14
The Simple Gas Laws
15:28
Gay-Lussac's Law
15:29
The Simple Gas Laws
17:11
Avogadro's Law
17:12
The Ideal Gas Law
18:43
The Ideal Gas Law: PV = nRT
18:44
Applications of the Ideal Gas Law
20:12
Standard Temperature and Pressure for Gases
20:13
Applications of the Ideal Gas Law
21:43
Ideal Gas Law & Gas Density
21:44
Gas Pressures and Partial Pressures
23:18
Dalton's Law of Partial Pressures
23:19
Gas Stoichiometry
24:15
Stoichiometry Problems Involving Gases
24:16
Using The Ideal Gas Law to Get to Moles
25:16
Using Molar Volume to Get to Moles
25:39
Gas Stoichiometry Cont'd
26:03
Example 1: How Many Liters of O₂ at STP are Needed to Form 10.5 g of Water Vapor?
26:04
Summary
28:33
Sample Problem 1: Calculate the Molar Mass of the Gas
29:28
Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C?
31:59
Intermolecular Forces & Liquids

33m 47s

Intro
0:00
Lesson Overview
0:10
Introduction
0:46
Intermolecular Forces (IMF)
0:47
Intermolecular Forces of Polar Molecules
1:32
Ion-dipole Forces
1:33
Example: Salt Dissolved in Water
1:50
Coulomb's Law & the Force of Attraction Between Ions and/or Dipoles
3:06
IMF of Polar Molecules cont'd
4:36
Enthalpy of Solvation or Enthalpy of Hydration
4:37
IMF of Polar Molecules cont'd
6:01
Dipole-dipole Forces
6:02
IMF of Polar Molecules cont'd
7:22
Hydrogen Bonding
7:23
Example: Hydrogen Bonding of Water
8:06
IMF of Nonpolar Molecules
9:37
Dipole-induced Dipole Attraction
9:38
IMF of Nonpolar Molecules cont'd
11:34
Induced Dipole Attraction, London Dispersion Forces, or Vand der Waals Forces
11:35
Polarizability
13:46
IMF of Nonpolar Molecules cont'd
14:26
Intermolecular Forces (IMF) and Polarizability
14:31
Properties of Liquids
16:48
Standard Molar Enthalpy of Vaporization
16:49
Trends in Boiling Points of Representative Liquids: H₂O vs. H₂S
17:43
Properties of Liquids cont'd
18:36
Aliphatic Hydrocarbons
18:37
Branched Hydrocarbons
20:52
Properties of Liquids cont'd
22:10
Vapor Pressure
22:11
The Clausius-Clapeyron Equation
24:30
Properties of Liquids cont'd
25:52
Boiling Point
25:53
Properties of Liquids cont'd
27:07
Surface Tension
27:08
Viscosity
28:06
Summary
29:04
Sample Problem 1: Determine Which of the Following Liquids Will Have the Lower Vapor Pressure
30:21
Sample Problem 2: Determine Which of the Following Liquids Will Have the Largest Standard Molar Enthalpy of Vaporization
31:37
The Chemistry of Solids

25m 13s

Intro
0:00
Lesson Overview
0:07
Introduction
0:46
General Characteristics
0:47
Particulate-level Drawing
1:09
The Basic Structure of Solids: Crystal Lattices
1:37
The Unit Cell Defined
1:38
Primitive Cubic
2:50
Crystal Lattices cont'd
3:58
Body-centered Cubic
3:59
Face-centered Cubic
5:02
Lattice Enthalpy and Trends
6:27
Introduction to Lattice Enthalpy
6:28
Equation to Calculate Lattice Enthalpy
7:21
Different Types of Crystalline Solids
9:35
Molecular Solids
9:36
Network Solids
10:25
Phase Changes Involving Solids
11:03
Melting & Thermodynamic Value
11:04
Freezing & Thermodynamic Value
11:49
Phase Changes cont'd
12:40
Sublimation & Thermodynamic Value
12:41
Depositions & Thermodynamic Value
13:13
Phase Diagrams
13:40
Introduction to Phase Diagrams
13:41
Phase Diagram of H₂O: Melting Point
14:12
Phase Diagram of H₂O: Normal Boiling Point
14:50
Phase Diagram of H₂O: Sublimation Point
15:02
Phase Diagram of H₂O: Point C ( Supercritical Point)
15:32
Phase Diagrams cont'd
16:31
Phase Diagram of Dry Ice
16:32
Summary
18:15
Sample Problem 1, Part A: Of the Group I Fluorides, Which Should Have the Highest Lattice Enthalpy?
19:01
Sample Problem 1, Part B: Of the Lithium Halides, Which Should Have the Lowest Lattice Enthalpy?
19:54
Sample Problem 2: How Many Joules of Energy is Required to Melt 546 mg of Ice at Standard Pressure?
20:55
Sample Problem 3: Phase Diagram of Helium
22:42
X. Solutions, Rates of Reaction, & Equilibrium
Solutions & Their Behavior

38m 6s

Intro
0:00
Lesson Overview
0:10
Units of Concentration
1:40
Molarity
1:41
Molality
3:30
Weight Percent
4:26
ppm
5:16
Like Dissolves Like
6:28
Like Dissolves Like
6:29
Factors Affecting Solubility
9:35
The Effect of Pressure: Henry's Law
9:36
The Effect of Temperature on Gas Solubility
12:16
The Effect of Temperature on Solid Solubility
14:28
Colligative Properties
16:48
Colligative Properties
16:49
Changes in Vapor Pressure: Raoult's Law
17:19
Colligative Properties cont'd
19:53
Boiling Point Elevation and Freezing Point Depression
19:54
Colligative Properties cont'd
26:13
Definition of Osmosis
26:14
Osmotic Pressure Example
27:11
Summary
31:11
Sample Problem 1: Calculating Vapor Pressure
32:53
Sample Problem 2: Calculating Molality
36:29
Chemical Kinetics

37m 45s

Intro
0:00
Lesson Overview
0:06
Introduction
1:09
Chemical Kinetics and the Rate of a Reaction
1:10
Factors Influencing Rate
1:19
Introduction cont'd
2:27
How a Reaction Progresses Through Time
2:28
Rate of Change Equation
6:02
Rate Laws
7:06
Definition of Rate Laws
7:07
General Form of Rate Laws
7:37
Rate Laws cont'd
11:07
Rate Orders With Respect to Reactant and Concentration
11:08
Methods of Initial Rates
13:38
Methods of Initial Rates
13:39
Integrated Rate Laws
17:57
Integrated Rate Laws
17:58
Graphically Determine the Rate Constant k
18:52
Reaction Mechanisms
21:05
Step 1: Reversible
21:18
Step 2: Rate-limiting Step
21:44
Rate Law for the Reaction
23:28
Reaction Rates and Temperatures
26:16
Reaction Rates and Temperatures
26:17
The Arrhenius Equation
29:06
Catalysis
30:31
Catalyst
30:32
Summary
32:02
Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed
32:54
Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction
35:24
Principles of Chemical Equilibrium

34m 9s

Intro
0:00
Lesson Overview
0:08
Introduction
1:02
The Equilibrium Constant
3:08
The Equilibrium Constant
3:09
The Equilibrium Constant cont'd
5:50
The Equilibrium Concentration and Constant for Solutions
5:51
The Equilibrium Partial Pressure and Constant for Gases
7:01
Relationship of Kc and Kp
7:30
Heterogeneous Equilibria
8:23
Heterogeneous Equilibria
8:24
Manipulating K
9:57
First Way of Manipulating K
9:58
Second Way of Manipulating K
11:48
Manipulating K cont'd
12:31
Third Way of Manipulating K
12:32
The Reaction Quotient Q
14:42
The Reaction Quotient Q
14:43
Q > K
16:16
Q < K
16:30
Q = K
16:43
Le Chatlier's Principle
17:32
Restoring Equilibrium When It is Disturbed
17:33
Disturbing a Chemical System at Equilibrium
18:35
Problem-Solving with ICE Tables
19:05
Determining a Reaction's Equilibrium Constant With ICE Table
19:06
Problem-Solving with ICE Tables cont'd
21:03
Example 1: Calculate O₂(g) at Equilibrium
21:04
Problem-Solving with ICE Tables cont'd
22:53
Example 2: Calculate the Equilibrium Constant
22:54
Summary
25:24
Sample Problem 1: Calculate the Equilibrium Constant
27:59
Sample Problem 2: Calculate The Equilibrium Concentration
30:30
XI. Acids & Bases Chemistry
Acid-Base Chemistry

43m 44s

Intro
0:00
Lesson Overview
0:06
Introduction
0:55
Bronsted-Lowry Acid & Bronsted -Lowry Base
0:56
Water is an Amphiprotic Molecule
2:40
Water Reacting With Itself
2:58
Introduction cont'd
4:04
Strong Acids
4:05
Strong Bases
5:18
Introduction cont'd
6:16
Weak Acids and Bases
6:17
Quantifying Acid-Base Strength
7:35
The pH Scale
7:36
Quantifying Acid-Base Strength cont'd
9:55
The Acid-ionization Constant Ka and pKa
9:56
Quantifying Acid-Base Strength cont'd
12:13
Example: Calculate the pH of a 1.2M Solution of Acetic Acid
12:14
Quantifying Acid-Base Strength
15:06
Calculating the pH of Weak Base Solutions
15:07
Writing Out Acid-Base Equilibria
17:45
Writing Out Acid-Base Equilibria
17:46
Writing Out Acid-Base Equilibria cont'd
19:47
Consider the Following Equilibrium
19:48
Conjugate Base and Conjugate Acid
21:18
Salts Solutions
22:00
Salts That Produce Acidic Aqueous Solutions
22:01
Salts That Produce Basic Aqueous Solutions
23:15
Neutral Salt Solutions
24:05
Diprotic and Polyprotic Acids
24:44
Example: Calculate the pH of a 1.2 M Solution of H₂SO₃
24:43
Diprotic and Polyprotic Acids cont'd
27:18
Calculate the pH of a 1.2 M Solution of Na₂SO₃
27:19
Lewis Acids and Bases
29:13
Lewis Acids
29:14
Lewis Bases
30:10
Example: Lewis Acids and Bases
31:04
Molecular Structure and Acidity
32:03
The Effect of Charge
32:04
Within a Period/Row
33:07
Molecular Structure and Acidity cont'd
34:17
Within a Group/Column
34:18
Oxoacids
35:58
Molecular Structure and Acidity cont'd
37:54
Carboxylic Acids
37:55
Hydrated Metal Cations
39:23
Summary
40:39
Sample Problem 1: Calculate the pH of a 1.2 M Solution of NH₃
41:20
Sample Problem 2: Predict If The Following Slat Solutions are Acidic, Basic, or Neutral
42:37
Applications of Aqueous Equilibria

55m 26s

Intro
0:00
Lesson Overview
0:07
Calculating pH of an Acid-Base Mixture
0:53
Equilibria Involving Direct Reaction With Water
0:54
When a Bronsted-Lowry Acid and Base React
1:12
After Neutralization Occurs
2:05
Calculating pH of an Acid-Base Mixture cont'd
2:51
Example: Calculating pH of an Acid-Base Mixture, Step 1 - Neutralization
2:52
Example: Calculating pH of an Acid-Base Mixture, Step 2 - React With H₂O
5:24
Buffers
7:45
Introduction to Buffers
7:46
When Acid is Added to a Buffer
8:50
When Base is Added to a Buffer
9:54
Buffers cont'd
10:41
Calculating the pH
10:42
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer
14:03
Buffers cont'd
14:10
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 1 -Neutralization
14:11
Calculating the pH When 0.010 mol NaOH is Added to 1.0 L of the Buffer: Step 2- ICE Table
15:22
Buffer Preparation and Capacity
16:38
Example: Calculating the pH of a Buffer Solution
16:42
Effective Buffer
18:40
Acid-Base Titrations
19:33
Acid-Base Titrations: Basic Setup
19:34
Acid-Base Titrations cont'd
22:12
Example: Calculate the pH at the Equivalence Point When 0.250 L of 0.0350 M HClO is Titrated With 1.00 M KOH
22:13
Acid-Base Titrations cont'd
25:38
Titration Curve
25:39
Solubility Equilibria
33:07
Solubility of Salts
33:08
Solubility Product Constant: Ksp
34:14
Solubility Equilibria cont'd
34:58
Q < Ksp
34:59
Q > Ksp
35:34
Solubility Equilibria cont'd
36:03
Common-ion Effect
36:04
Example: Calculate the Solubility of PbCl₂ in 0.55 M NaCl
36:30
Solubility Equilibria cont'd
39:02
When a Solid Salt Contains the Conjugate of a Weak Acid
39:03
Temperature and Solubility
40:41
Complexation Equilibria
41:10
Complex Ion
41:11
Complex Ion Formation Constant: Kf
42:26
Summary
43:35
Sample Problem 1: Question
44:23
Sample Problem 1: Part a) Calculate the pH at the Beginning of the Titration
45:48
Sample Problem 1: Part b) Calculate the pH at the Midpoint or Half-way Point
48:04
Sample Problem 1: Part c) Calculate the pH at the Equivalence Point
48:32
Sample Problem 1: Part d) Calculate the pH After 27.50 mL of the Acid was Added
53:00
XII. Thermodynamics & Electrochemistry
Entropy & Free Energy

36m 13s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Introduction to Entropy
1:37
Introduction to Entropy
1:38
Entropy and Heat Flow
6:31
Recall Thermodynamics
6:32
Entropy is a State Function
6:54
∆S and Heat Flow
7:28
Entropy and Heat Flow cont'd
8:18
Entropy and Heat Flow: Equations
8:19
Endothermic Processes: ∆S > 0
8:44
The Second Law of Thermodynamics
10:04
Total ∆S = ∆S of System + ∆S of Surrounding
10:05
Nature Favors Processes Where The Amount of Entropy Increases
10:22
The Third Law of Thermodynamics
11:55
The Third Law of Thermodynamics & Zero Entropy
11:56
Problem-Solving involving Entropy
12:36
Endothermic Process and ∆S
12:37
Exothermic Process and ∆S
13:19
Problem-Solving cont'd
13:46
Change in Physical States: From Solid to Liquid to Gas
13:47
Change in Physical States: All Gases
15:02
Problem-Solving cont'd
15:56
Calculating the ∆S for the System, Surrounding, and Total
15:57
Example: Calculating the Total ∆S
16:17
Problem-Solving cont'd
18:36
Problems Involving Standard Molar Entropies of Formation
18:37
Introduction to Gibb's Free Energy
20:09
Definition of Free Energy ∆G
20:10
Spontaneous Process and ∆G
20:19
Gibb's Free Energy cont'd
22:28
Standard Molar Free Energies of Formation
22:29
The Free Energies of Formation are Zero for All Compounds in the Standard State
22:42
Gibb's Free Energy cont'd
23:31
∆G° of the System = ∆H° of the System - T∆S° of the System
23:32
Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
24:24
Gibb's Free Energy cont'd
26:32
Effect of reactant and Product Concentration on the Sign of Free Energy
26:33
∆G° of Reaction = -RT ln K
27:18
Summary
28:12
Sample Problem 1: Calculate ∆S° of Reaction
28:48
Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous
31:18
Sample Problem 3: Calculate Kp
33:47
Electrochemistry

41m 16s

Intro
0:00
Lesson Overview
0:08
Introduction
0:53
Redox Reactions
1:42
Oxidation-Reduction Reaction Overview
1:43
Redox Reactions cont'd
2:37
Which Reactant is Being Oxidized and Which is Being Reduced?
2:38
Redox Reactions cont'd
6:34
Balance Redox Reaction In Neutral Solutions
6:35
Redox Reactions cont'd
10:37
Balance Redox Reaction In Acidic and Basic Solutions: Step 1
10:38
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Each Half-Reaction
11:22
Redox Reactions cont'd
12:19
Balance Redox Reaction In Acidic and Basic Solutions: Step 2 - Balance Hydrogen
12:20
Redox Reactions cont'd
14:30
Balance Redox Reaction In Acidic and Basic Solutions: Step 3
14:34
Balance Redox Reaction In Acidic and Basic Solutions: Step 4
15:38
Voltaic Cells
17:01
Voltaic Cell or Galvanic Cell
17:02
Cell Notation
22:03
Electrochemical Potentials
25:22
Electrochemical Potentials
25:23
Electrochemical Potentials cont'd
26:07
Table of Standard Reduction Potentials
26:08
The Nernst Equation
30:41
The Nernst Equation
30:42
It Can Be Shown That At Equilibrium E =0.00
32:15
Gibb's Free Energy and Electrochemistry
32:46
Gibbs Free Energy is Relatively Small if the Potential is Relatively High
32:47
When E° is Very Large
33:39
Charge, Current and Time
33:56
A Battery Has Three Main Parameters
33:57
A Simple Equation Relates All of These Parameters
34:09
Summary
34:50
Sample Problem 1: Redox Reaction
35:26
Sample Problem 2: Battery
38:00
XIII. Transition Elements & Coordination Compounds
The Chemistry of The Transition Metals

39m 3s

Intro
0:00
Lesson Overview
0:11
Coordination Compounds
1:20
Coordination Compounds
1:21
Nomenclature of Coordination Compounds
2:48
Rule 1
3:01
Rule 2
3:12
Rule 3
4:07
Nomenclature cont'd
4:58
Rule 4
4:59
Rule 5
5:13
Rule 6
5:35
Rule 7
6:19
Rule 8
6:46
Nomenclature cont'd
7:39
Rule 9
7:40
Rule 10
7:45
Rule 11
8:00
Nomenclature of Coordination Compounds: NH₄[PtCl₃NH₃]
8:11
Nomenclature of Coordination Compounds: [Cr(NH₃)₄(OH)₂]Br
9:31
Structures of Coordination Compounds
10:54
Coordination Number or Steric Number
10:55
Commonly Observed Coordination Numbers and Geometries: 4
11:14
Commonly Observed Coordination Numbers and Geometries: 6
12:00
Isomers of Coordination Compounds
13:13
Isomers of Coordination Compounds
13:14
Geometrical Isomers of CN = 6 Include: ML₄L₂'
13:30
Geometrical Isomers of CN = 6 Include: ML₃L₃'
15:07
Isomers cont'd
17:00
Structural Isomers Overview
17:01
Structural Isomers: Ionization
18:06
Structural Isomers: Hydrate
19:25
Structural Isomers: Linkage
20:11
Structural Isomers: Coordination Isomers
21:05
Electronic Structure
22:25
Crystal Field Theory
22:26
Octahedral and Tetrahedral Field
22:54
Electronic Structure cont'd
25:43
Vanadium (II) Ion in an Octahedral Field
25:44
Chromium(III) Ion in an Octahedral Field
26:37
Electronic Structure cont'd
28:47
Strong-Field Ligands and Weak-Field Ligands
28:48
Implications of Electronic Structure
30:08
Compare the Magnetic Properties of: [Fe(OH₂)₆]²⁺ vs. [Fe(CN)₆]⁴⁻
30:09
Discussion on Color
31:57
Summary
34:41
Sample Problem 1: Name the Following Compound [Fe(OH)(OH₂)₅]Cl₂
35:08
Sample Problem 1: Name the Following Compound [Co(NH₃)₃(OH₂)₃]₂(SO₄)₃
36:24
Sample Problem 2: Change in Magnetic Properties
37:30
XIV. Nuclear Chemistry
Nuclear Chemistry

16m 39s

Intro
0:00
Lesson Overview
0:06
Introduction
0:40
Introduction to Nuclear Reactions
0:41
Types of Radioactive Decay
2:10
Alpha Decay
2:11
Beta Decay
3:27
Gamma Decay
4:40
Other Types of Particles of Varying Energy
5:40
Nuclear Equations
6:47
Nuclear Equations
6:48
Nuclear Decay
9:28
Nuclear Decay and the First-Order Kinetics
9:29
Summary
11:31
Sample Problem 1: Complete the Following Nuclear Equations
12:13
Sample Problem 2: How Old is the Rock?
14:21
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Lecture Comments (7)

0 answers

Post by Pooja Mittal on March 23, 2017

Very helpful explanations on MO theory.

0 answers

Post by Saadman Elman on December 2, 2014

You made the atomic hybridization seem very easy. On the other hand MO theory especially when you went to the p-block. It was extremely abstract. It wasn't clear.

0 answers

Post by Muhammad Chauhan on November 12, 2014

I have one more question. My book says that a bond oder of 1/2 does indicate the bond exists. For example if we have He2+ it should exist. And it indeed does in nature. So which rule should we follow? Thank you!

0 answers

Post by Muhammad Chauhan on November 12, 2014

Why is it expected that oxygen is diamagnetic?

0 answers

Post by Gaurav Kumar on October 25, 2014

Great lecture...really helped clear things up. Thanks!

1 answer

Last reply by: Professor Franklin Ow
Sat Apr 26, 2014 5:18 PM

Post by Yay Esme on April 21, 2014

why do you have 2 sigmas when carbon only has one sigma bond ?

Related Articles:

Advanced Bonding Theories

  • Valence bond theory can explain for observed bond angles via hybrid orbitals, but doesn’t always account for a compound’s magnetic behavior.
  • MO theory is a delocalized bonding model, and views electrons as being distributed throughout the entire compound.
  • An MO diagram can be used to predict magnetism and bond order for a molecule.

Advanced Bonding Theories

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lesson Overview 0:09
  • Introduction 0:38
  • Valence Bond Theory 3:07
    • Valence Bond Theory
    • spᶟ Hybridized Carbon Atom
  • Valence Bond Theory Cont'd 6:24
    • spᶟ Hybridized
    • Hybrid Orbitals For Water
  • Valence Bond Theory Cont'd (spᶟ) 11:53
    • Example 1: NH₃
  • Valence Bond Theory Cont'd (sp²) 14:48
    • sp² Hybridization
    • Example 2: BF₃
  • Valence Bond Theory Cont'd (sp) 22:44
    • sp Hybridization
    • Example 3: HCN
  • Valence Bond Theory Cont'd (sp³d and sp³d²) 27:36
    • Valence Bond Theory: sp³d and sp³d²
  • Molecular Orbital Theory 29:10
    • Valence Bond Theory Doesn't Always Account For a Molecule's Magnetic Behavior
  • Molecular Orbital Theory Cont'd 30:37
    • Molecular Orbital Theory
    • Wavefunctions
    • How s-orbitals Can Interact
    • Bonding Nature of p-orbitals: Head-on
    • Bonding Nature of p-orbitals: Parallel
    • Interaction Between s and p-orbital
    • Molecular Orbital Diagram For Homonuclear Diatomics: H₂
    • Molecular Orbital Diagram For Homonuclear Diatomics: He₂
    • Molecular Orbital Diagram For Homonuclear Diatomic: Li₂
    • Molecular Orbital Diagram For Homonuclear Diatomic: Li₂⁺
    • Molecular Orbital Diagram For Homonuclear Diatomic: B₂
    • Molecular Orbital Diagram For Homonuclear Diatomic: N₂
    • Molecular Orbital Diagram: Molecular Oxygen
    • Molecular Orbital Diagram For Heteronuclear Diatomics: Hydrochloric Acid
  • Sample Problem 1: Determine the Atomic Hybridization 1:07:20
    • XeO₃
    • SF₆
    • I₃⁻
  • Sample Problem 2 1:09:04

Transcription: Advanced Bonding Theories

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be advanced bonding theories.0003

We are going to start off with a brief introduction as to why0010

we need these advanced bonding theories from what we already know.0015

That was really Lewis structures.0019

Pretty much the advanced bonding theories that we are going to be talking about are the following.0022

One is what we call valence bond theory.0026

The second one is what we call molecular orbital theory.0030

We will wrap up the session with a brief summary followed by a pair of sample problems.0033

What you and I have already discussed is the representation of valence electrons schematically.0041

Remember that is what we called Lewis structures.0050

We went through a step by step procedure on how to formulate a valid Lewis structure.0052

Lewis structures are sufficient because using VSEPR theory we were able to0058

come up with some predicted geometries strictly off of the Lewis structure.0067

But there exists really two main problems with Lewis theory that we never talked about before.0076

The first main problem has to do with magnetism.0082

If you noticed that all of the Lewis structures that we looked at, all electrons are paired.0087

They are either in a single bond, double bond, or triple bond for electrons.0093

And they are either in a lone pair.0098

If all electrons are paired, then we would expect everything to be diamagnetic.0101

But as we all know, that is not the case.0108

There is definitely more than enough compounds that are paramagnetic,0110

meaning that they do have at least one unpaired electron.0116

Lewis theory does not account for magnetic behavior.0119

A second problem with Lewis theory is the following.0124

Consider all of the compounds that we looked at.0130

For all of the compounds really, the valence shell has been the p orbital.0133

We looked at p orbitals before.0139

They were pretty much three p orbitals per energy level--px, py, and pz.0141

We actually graphed them.0148

We saw that because of Cartesian coordinates, the px, py,0150

and pz orbitals are going to be perpendicular to each other.0154

If you have nothing but 90 degree angles, how do we get 109.5?0159

How do we get 120, etc?0166

Lewis theory doesn't really account for all of the bond angles that are expected from VSEPR theory.0175

We do need advanced additional bonding theories to make up for the flaws of Lewis theory.0189

Valence bond theory is what we call a localized bonding model.0198

What localized means is the following.0203

It is that an atom's atomic orbitals are going to be0205

centered around that specific atom when forming covalent bonds.0209

In other words, you can think of a specific atom owning0213

those electrons rather than being distributed throughout the entire molecule.0218

Again that is what we call localized bonding model.0224

In order to achieve bond angles that are other than 90 degrees,0227

like 120, like 109.5, what valence bond theory suggests is the following.0231

It is that an atom's valence orbitals begin to overlap each other.0239

When they begin to overlap, they mix.0245

They begin to form these mixtures, these combinations of atomic orbitals which we call hybrids or hybrid orbitals.0248

What we are going to do first is we are going to take a look at one type of hybrid orbital.0260

The first type is what we call an sp3 hybridized carbon atom.0264

We go ahead and look at carbon.0270

What this sp3 means is that we have one s orbital mixing with three of carbon's p orbitals.0272

We can go ahead and show that; let me do that in a different color.0286

If I have a carbon s orbital mixing with a p orbital of carbon, we are going to get a hybrid orbital.0290

A hybrid orbital is going to have both s character and p character.0302

But it is going to be mostly p character because it is one s orbital versus three p's.0307

We are going to get a small lobe representative of the s orbital.0313

We are going to get a much larger lobe that is representative of the p orbital.0319

This is what we call a hybrid orbital.0326

Now that we know what the hybrid orbitals look like, they are going to be centered around the carbon atom.0329

Remember it is what we call a localized bonding model.0335

We are going to have the s orbital right in the middle, closest to the nucleus remember.0338

There is going to be one hybrid here; one hybrid here.0343

One hybrid here; one hybrid just in the back.0347

As you can see the way we have drawn it,0352

if you connect the vertices, we do indeed get a tetrahedron.0355

The concept of hybrid formation allows us to get these bond angles that are predicted by VSEPR theory.0366

Let's go ahead and put that into practice now.0375

Now that we physically have a sense of what a hybrid orbital can quote unquote look like.0378

Sp3 first, we are going to first start paying attention to what the superscripts mean.0385

The superscript tells us actually a great deal.0391

In sp3, there is a superscript of 1 for the s orbital.0395

There is a superscript of 3 for the p orbital.0400

What that translates to is the following.0403

That one s orbital mixes with three p orbitals to form the0405

sp3 hybrids just like we have seen in the previous slide.0412

But the question is how many hybrids do we form?0416

Again we turn to the sum of the superscripts to help us answer that.0419

Basically the sum of the superscripts equals to the number of hybrid orbitals that are going to form.0426

When you look at sp3, the sum of the superscripts is four.0435

That is why we form four hybrid orbitals.0439

For example, let's go ahead and take a look at water0445

and see how hybridization can help to explain for bonding.0452

Water has the following structure--the central oxygen, H, and two lone pair.0457

Once again the number of electron groups, it is going to be the same as the sum of the superscripts.0467

Here we have four electron groups surrounding oxygen.0491

There is going to be a total of four for the sum of the superscripts.0494

That is going to be sp3.0499

We are going to over other hybridization what we call schemes.0500

Sp3 is the first one; this is sp3 hybridization.0505

We can show this in an energy level diagram; energy.0511

Here the sp3 hybridized atom is going to be the oxygen.0517

Oxygen starts off as 2s and then 2p.0522

Let's go ahead and fill in the electrons now.0527

Oxygen is going to be 2s2 and 2p4; this is just atomic oxygen by itself.0532

But as soon as oxygen hooks up with those two hydrogens to form water, its atomic orbitals are going to hybridize.0542

Again we are going to form a total of four sp3s.0550

Where are those four going to form?0554

Are they going to form higher in energy than the 2p's?0556

Are they going to be less than the energy of the 2s's?0559

Or are they going to be intermediate?0564

It turns out that we are going to compromise.0567

We are going to have an energy that is going to be intermediate for the entire system.0570

We form four sp3 hybrid orbitals now.0577

When we go ahead now and fill electrons, let's go ahead and fill it.0581

I have a total of six electrons--one, two, three, four, five, and six.0584

When we do this successfully, the overall picture should agree with the Lewis structure.0593

Guess what?--just like the Lewis structure which has two lone pairs around the oxygen,0599

we have one lone pair here and one lone pair here.0604

What do you think these two electrons are?0608

We have two hydrogens bonded to the oxygen.0612

These are going to be bonding electrons with H right here.0616

These electrons go on to form a single bond with the hydrogen; form single bond with H.0629

We now introduce a new terminology.0642

A single bond in terms of valence bond theory is a specific type of bond we call a σ bond; σ bond.0645

When we go ahead and redraw water, H, H, and O, we are going to see that0657

the lone pairs, they are going to exist in sp3 hybrid orbitals of oxygen.0667

We are going to have another sp3 here and another sp3 here, each containing one electron.0676

Those sp3 orbitals are going to overlap and mix with the hydrogen 1s orbitals which also have one electron.0689

Same thing here; that is what we call a σ bond.0698

Again we get the very important term, what we call a σ bond, from valence bond theory.0706

Let's go ahead and look at one more example of an sp3 hybridized atom.0717

The example I would like to look at is ammonia, NH3.0727

Step one is always do the Lewis structure.0735

Here the Lewis structure for ammonia is going to be here just like that.0742

It is going to be three electron groups plus a lone pair0751

giving you a total of four electron groups around the nitrogen atom.0755

Remember the number of electron groups equals to the sum of the superscripts so this is also sp3.0760

Step two, we are going to now come up with our bonding scheme0771

and show the formation of these hybrids; show hybrid formation.0776

Hopefully again when we are done with this, we can correlate it with the Lewis structure.0785

Draw your energy diagram again; there is energy.0801

Nitrogen, just bare atomic nitrogen, is going to be 2s2 and 2p3.0807

But as soon as the nitrogen starts to form bonds with hydrogen to form ammonia, it is going to mix.0818

It is going to form again four sp3 hybrid orbitals.0827

Let's go ahead and put the electrons in.0833

I have a total of five electrons--one, two, three, four, and five.0835

Let's go ahead and see if we can correlate it with the Lewis structure.0842

The Lewis structure has one lone pair.0846

The lone pair is right there; that is so far so good.0849

You notice that there is three single bonds with hydrogen.0853

Remember we call the single bond a σ bond.0856

σ here, σ here, and σ here; guess what?0859

We have three electrons that are each going to form a σ bond with the hydrogen.0862

Once again they each form a σ bond with a hydrogen 1s atomic orbital.0870

Good, we have a pretty good understanding now of what we mean by sp3 hybridization.0885

The next bonding type, the next hybridization type is what we call sp2.0893

Again let's look at the sum of the superscripts; that tells us a great deal.0900

Sp2, the sum of the superscripts is a grand total of three which means0906

an atom is going to be sp2 hybridized when it has three electron groups around it.0915

In addition remember that the sum of the superscripts also tells us how many hybrids we are going to form.0923

Here we are going to form exactly three sp2 hybrids.0931

One more thing, the superscripts also tells us the number of each atomic orbital that is going to be used.0940

Here we are going to be having one s orbital mixing with exactly two p orbitals.0948

Let's think about that.0959

We know that every energy we said has a px, py, and pz, three p orbitals.0961

But if only two p orbitals are being used out of the three, that tells us the difference.0967

One of the p orbitals remains unused, unhybridized.0974

I am going to put that in; one p orbital is going to be unhybridized.0981

We need to now account for that in our energy diagram.0990

Let's go ahead and do that.0995

A nice example of something that is going to be sp2 hybridized, let's go ahead and look at BF3.0997

BF3 here; here is energy.1013

Regular boron itself is going to be 2s2 and then just 2p1.1019

When that goes ahead and mix, we are going to form exactly three sp2s intermediate in energy.1030

Remember that?--always intermediate in energy approximately.1037

This one p orbital that remains unhybridized, if it remains unhybridized, that means1042

it is going to be the same energy as the initial p orbitals.1048

I am going to put that approximately the same just like that.1056

Again we have exactly three sp2s; and we have a 2p orbital that remains unhybridized.1059

We are going to go ahead and fill the electrons now.1066

In BF3, boron just has three electrons--one, two, and three.1069

Guess what?--there is no lone pairs.1075

There is going to be exactly three σ bonds.1077

When we draw out the Lewis structure for boron trifluoride, that is exactly what we get.1080

Boron has zero lone pairs in this structure and exactly three σ bonds.1087

That looks pretty good right now.1098

But let's go ahead and look at another example.1100

Here we are going to learn now a new type of bonding.1103

An example I want to look at has the following Lewis structure.1106

This is going to be a CH, a double bond, O just like that.1110

This is what we call formaldehyde.1120

Here formaldehyde, the central carbon atom has three electron groups.1125

That is also going to be sp2 hybridized.1131

Let's go ahead and set up the energy level diagram now for the carbon here.1136

Carbon is 2s and 2p; it is going to be 2s2 2p2.1140

That is going to hybridize.1148

Again we are going to form exactly three sp2s--one, two, three.1150

We are going to have one of the 2p orbitals remaining unmixed just like that.1154

Let's go ahead and fill electrons now; carbon has four valence electrons.1159

Let's go ahead and populate the orbitals with four--one, two, and three.1163

I am going to put a big asterisk here because this is where we start to deviate.1169

Where do we put the fourth electron?1173

Going by typical Aufbau principle, we would always populate the lower energy1175

orbitals first before proceeding on to the higher atomic orbitals.1182

But in valence bond theory, we are going to go left to right regardless of energy.1187

We are going to go every orbital one time before doubling up.1193

I am going to put my fourth electron here, not in the sp2.1196

That is this special case for valence bond theory is really1202

a big deviation from what we know from Aufbau principle.1207

Please again keep that in mind.1210

When we do this see, it is going to work out because look at the Lewis structure.1214

The Lewis structure has zero lone pairs for the carbon atom.1218

As you can see in our bonding scheme, there is zero lone pairs.1222

We have the following; we have a σ bond here and a σ bond here; σ, σ.1227

And we have a double bond.1234

We have to now discuss what we mean by double bond.1237

A double bond we learned from Lewis theory has exactly four valence electrons.1241

But what type of bonds?1245

It turns out that a double bond is composed of the following.1248

It is composed of one σ bond and a new type of bond, what we call a π bond.1250

Our σ bond is just going to be the regular σ bond from the hybrid orbitals.1258

Guess what?--this electron that we have right here in the unhybridized atomic orbital,1262

that is going to overlap with oxygen to form a π bond.1268

One σ plus one π is equal to a double bond.1272

We can make a pretty big conclusion right now from these two exercises.1279

Lone pairs and σ bonds are going to come from hybrid orbitals.1288

π bonds are going to come from atomic orbitals that are unhybridized.1300

You may be wondering then what type of bond is stronger, is it σ or π?1311

The lower the energy, the more stable it is going to be.1318

The higher the energy, the more unstable it is going to be.1322

We are going to learn that very soon in our next discussion of thermodynamics.1326

σ bonds are typically going to be stronger than π bonds.1333

The reason why σ bonds are stronger is because once again1341

lower in energy but also we have more overlap of the orbitals.1345

If we have more overlap, the attraction between the nuclei is going to be much stronger.1355

That again is sp2 hybridization.1363

The next one I want to go over with you is what we call sp hybridization.1367

Let's go ahead and take a look at the sum of the superscripts.1372

Here is really s1 p1.1375

Remember if you don't see a number, 1 is always implied.1378

That means we are going to have two electron groups around the central atom.1381

Also this tells us that we are going to form a hybrid here by one s mixing with exactly one p.1392

Also we are going to form exactly two hybrid orbitals; two sp hybrids.1404

Let's go ahead and take a look; let's look at hydrogen cyanide; hydrogen cyanide.1415

Step one, it has the following Lewis structure--H, C, triple bond, N.1424

As you can see, carbon is two electron groups; that is going to be sp.1430

Also nitrogen also has two electron groups around it.1436

That too is going to be sp hybridized.1439

Let's first tackle carbon.1441

Then we will tackle nitrogen to see if it agrees with the Lewis structure.1443

Energy; carbon is again 2s 2p.1447

Put in its electrons; carbon is 2s2 2p2.1454

That is going to go ahead and mix.1461

Again we are going to form exactly two sp orbitals; two sp hybrids.1463

It tells us that to form the two sp's, it is the mixing of one s and one p.1470

But remember there is three p orbitals per energy level; px, py, and pz.1476

If I am only using one of those three, that means I have two remaining unhybridized.1481

We are also going to put that in; where do we put it in?1487

We put it in that it has exactly the same energy level as the initial two p orbitals.1491

Let's fill electrons--one, two, and don't forget this is where we deviate.1497

We go fill left to right all the orbitals first before double up regardless of energy.1502

One electron, two electrons, and now electron number three, and electron number four.1508

So far so good; we don't see any lone pairs at all.1514

The Lewis structure has no lone pairs around the carbon.1518

Let's see what HCN has around carbon; no lone pairs.1523

This bond right here is a σ bond; I am going to label that σ.1531

Now the nature of the triple bond.1535

We saw that a single bond is what we call a σ.1539

We saw that a double bond is one σ and one π.1542

A triple bond is going to be one σ but two π this time.1546

Again a triple bond is going to be one σ and two π.1551

We know that σ comes from hybrid orbitals; there is our σ.1557

We know that π bonds come from unhybridized atomic orbitals right here; π here and π there.1561

As you can see, we have two σ and two π's around carbon which is agreeing with the Lewis structure.1569

Let's see if we can get the same agreement with nitrogen this time.1577

Energy; nitrogen, 2s 2p; nitrogen is 2s2 and now 2p3.1582

Again as we already discussed, this nitrogen atom is also sp hybridized.1595

That is going to be one, two.1600

We are going to form two p's right here; exactly two of them unhybridized.1602

Let's go ahead and fill in the electrons.1606

This time it is going to be five--one, two, three, four...1607

OK, now we can double up again with the opposite spin.1611

According to the Lewis structure for nitrogen in HCN, there is going to be one lone pair.1617

There is going to be a triple bond around nitrogen which is one σ and two π.1626

Guess what?--it agrees perfectly.1632

Here is our lone pair; here is our one σ; here is the two π's.1636

Again we see that valence bond theory helps us to count for the1645

nature of a single, a double, and now a triple covalent bond.1650

There are two other atomic hybridizations that you should be familiar with.1658

So far we saw that sp3 is going to be for four electron groups.1663

That is going to be a tetrahedral electron geometry.1671

We saw that sp2 is going to be three electron groups.1679

That is going to be a trigonal planar geometry; again electron geometry.1684

We just did sp; we saw that sp was two electron groups.1692

That was a linear electron geometry.1696

Now sp 3d; sp 3d, now we start talking about the d block orbitals.1700

That is a grand total of five electron groups.1708

If you recall, five electron groups was a trigonal bipyramid electron geometry.1712

Finally sp3 d2, that is six electron groups.1722

That is going to be an octahedral electron geometry.1728

Again it depends on your instructor and your textbook.1734

But I will just present to you the five basic atomic hybridization schemes.1739

That is the formation of hybrid orbitals and valence bond theory.1746

Valence bond theory as we just saw is successful in explaining for the observed bond angles predicted by VSEPR theory.1753

However we still have that one issue about magnetism.1764

Take oxygen for example; oxygen's Lewis structure is right here.1768

If you look at this, every electron is paired; we would expect diamagnetism; expected.1776

In other words, if we take molecular oxygen, it should be repelled by a magnetic field.1787

But experimentally when you pour liquid oxygen through a magnet,1794

it actually sticks to the magnet showing that O2 is actually paramagnetic.1801

Obviously valence bond theory does not always successfully account or predict the actual magnetic behavior of a compound.1814

Then we need yet another bonding theory.1826

This bonding theory is what we call molecular orbital theory, also known as MO theory.1829

Here is the main difference.1840

In contrast to Lewis and valence bond theory, MO theory is a delocalized bonding model; not localized but delocalized.1842

In other words, the electron density is distributed throughout the1853

entire molecule rather than being centered around any individual atom.1857

You remember our discussion about wave functions.1866

It turns out that MO theory is a highly mathematically based theory.1870

If you recall the nature of a wave function, a wave function is representative of atomic orbitals.1878

The actual function if looked at them contain cosine and/or sine trigonometric functions.1883

If we try to think back to that, your basic cosine or sine wave function is a wave.1891

From your early math classes, you learned that these waves have both1901

positive and negative what are called phases; positive and negative phases.1909

Because atomic orbitals are the visual manifestation of a wave function, we can also infer that1923

atomic orbitals also can be represented by different phases, a + phase and a ? phase.1932

For general chemistry, we are only going to limit our discussion to s and p orbitals.1945

Let's first take a look at how s orbitals can interact.1952

I can have an s orbital of this phase mixing with another s orbital of the same phase.1958

When that happens, I form what is called a σ bonding orbital or a σ bonding MO.1970

How does that look like?1980

If here the dots represent the nuclei and if you have exactly two of these s orbitals1982

mixing with each other, we get a smeared out electron density throughout both nuclei.1990

Here is the nuclei.1997

But now we are going to get something that looks like this.1999

Again this is what we call a σ bonding MO.2005

Instead of two separate s orbitals, they now mix and form one entity.2009

As you can see from this diagram, it is completely distributed among both nuclei.2014

This is what we call delocalized.2023

The other possibility can occur.2026

What happens when we have s orbitals mixing that are of opposite phases?2028

To represent opposite phases, I am going to have now shading in one of these s orbitals just like that.2034

Now when I have s orbitals of opposite phases interacting,2044

this is going to be equivalent to destructive interference.2049

Remember we can interpret orbitals as wave-like properties.2056

When we have destructive interference, we actually get the following this time.2062

We are going to get something that looks like this now.2069

This is actually a node.2075

Remember we talked about nodes where we have zero electron probability.2077

This type of MO is not a σ bonding anymore.2085

But it is called a σ anti bonding MO; a σ antibonding MO.2089

For our purposes right now, something you have to remember is that2099

σ bonding will always be lower in energy than σ antibonding.2102

Once again σ bonding is going to lower in energy to σ antibonding.2109

Remember mother nature is always going to favor lower energy.2114

If an orbital is going to result in bonding of the molecule, it is going to be lower in energy.2117

Once again we went over σ bonding and σ antibonding.2124

σ antibonding will typically be represented as a σ*.2129

We talked about s orbitals; let's now talk about p orbitals.2135

P orbitals can interact in one of two ways.2139

They can interact head on or in a parallel fashion.2143

Let's talk about head on first.2152

When I mean head on, if we have a molecule AB like that, the bonding axis is going to be z-axis.2154

I am going to represent this as x; then represent that as y.2166

Once again the z-axis is going to be colinear with the bonding axis.2169

When we have a head on overlap of p orbitals, it is going to be the pz orbitals.2174

Here is one pz; here is another pz interacting head on.2182

We can have the same phases of the pz orbitals overlapping with each other like that.2189

When that happens, we form a σ bonding orbital.2203

Again we form a σ bonding orbital.2210

The MO is going to be distributed throughout the entire molecule around both nuclei.2214

We are going to get something that looks like this now2221

where the two nuclei are here, here, here, and here.2230

Again this is what we call a σ bonding MO.2238

You may ask then what is the other possibility?2242

The other possibility is for the pz orbitals now to interact with their opposite phases like that.2244

When the opposite phases interact, it is just like we saw last time with the s orbitals.2266

That is going to be antibonding; this is going to be σ*.2270

That is going to be represented the following way where the two nuclei again are here and here.2276

Let's go ahead and draw the nodes in, shall we?2295

Here you see that there is one node here, one node here.2298

You see here for the antibonding one node, one node, and one node.2302

I want you to compare and contrast.2307

You notice that the antibonding MO for the p orbitals has three nodes2310

while the bonding MO for the p orbitals has two nodes.2316

The same pattern occurs for the s orbitals.2320

You notice that the antibonding MO for the σ has one node2324

while the bonding MO for the σ has zero nodes.2331

The antibonding MOs typically have more nodes than the bonding MOs.2338

Let's go ahead and look at the parallel interaction.2346

Parallel interaction is going to occur for the remaining two p orbitals, basically px and py.2349

If I have parallel interaction this time, I can have p orbitals occurring like that.2355

When we have parallel interaction, the overlap is not as great.2365

This is not a σ MO anymore.2370

Instead it is going to be a π MO.2372

Because it is the identical phases interacting with each other,2375

this is going to be what is called a π bonding MO.2378

We are going to get an MO that looks like this now distributed among both nuclei just like that.2381

Here we have exactly one node.2395

The other possibility is for parallel interaction but now of opposite phases like that.2401

When we have opposite phases again that is what we call an antibonding MO.2409

This is going to be π*.2413

π* is going to look like this distributed throughout the entire compound.2415

Let me go ahead and erase that.2428

As you can see once again the antibonding MO has more nodes than the corresponding bonding MO.2434

One other possibility is having an s orbital directly combined with a p orbital in a head to head fashion.2448

That is going to be an s orbital combining with the pz orbital just like that.2455

Again that is going to be σ bonding.2464

Again we can also have s and p opposite phases.2467

That is going to be σ antibonding.2472

We pretty much covered all our bases and the different possibilities of the different types2477

of MOs that can form from s and p and how they look like.2482

Now that we have the combinations down, let's put it all together.2490

Let's go ahead and construct what we call a typical MO diagram.2494

The simplest type of compounds we are going to be looking at again are going to be homonuclear diatomics.2499

Remember two things; σ bonding is going to be less than σ antibonding.2507

Similarly π bonding is going to be less than energy of π antibonding.2518

σ versus π, σ bonding versus π bonding, that is going to become a major issue pretty soon.2527

But let's just go over the small ones first.2535

We will come across this issue here.2540

Hydrogen H2, molecular H2, two hydrogen atoms are going to combine.2542

Hydrogen is very simple.2550

We just have the hydrogen 1s orbital here and a hydrogen 1s orbital here.2551

Each of them contains one electron.2555

When hydrogens come together, we get the formation of MOs.2560

When an s orbital combines with an s orbital, remember what the two possibilities were?2566

It is going to be σ bonding.2571

The other possibility at higher energy is going to be σ antibonding.2574

Again bonding is going to be lower in energy than antibonding; that applies there.2580

There is some things I want to point out.2591

The number of MOs that you form must always equal to the number of atomic orbitals that you initially start with.2593

That is another way you check your work.2601

Here I start with a total of two atomic orbitals.2603

I wind up with two MOs--one bonding, one antibonding.2606

Let's go ahead and fill electrons going by the standard Aufbau principle.2609

That is going to be one electron here and then one electron here; spin up, spin down.2615

Now so what?--what is the big deal?--the big deal is the following.2621

We can now calculate something we call bond order.2626

Remember bond order?--it was one of those characteristics of a bond.2629

The equation for bond order is going to be the following.2634

It is going to be equal to the 1/2 number of bonding electrons minus the number of antibonding electrons.2638

When we do it for H2, we get 1.2654

Remember what bond order of 1 means?2657

A bond order of 1 means we have a single bond formed, expected to form.2659

We know H2 exists, we know hydrogen exists as a diatomic gas.2668

It is one of those several elements that occur diatomically.2673

Yes, hydrogen is expected to form.2678

Again all of these experiments are done in the gas phase.2680

Yes, we know that hydrogen exists as diatomic hydrogen in the gas phase.2686

A bond order of 1 is what we expect.2690

Remember if we have a bond order of 2, we expect double bond.2693

We have a bond order of 3, we expect a triple bond.2698

What happens if we have bond order less than 1?2703

If we have a bond order of less than 1, that is less than a single bond which means2706

we don't expect the molecule to exist at all in the gas phase; should not form.2709

You see the power of MO theory already.2716

It allows us to predict if a molecule is going to be stable or not in the gas phase.2718

Let's talk about helium; helium is a noble gas.2724

We know that noble gases exist as monoatomic gases, not diatomically.2729

Expectation for He2 is it should not exist.2734

In this case, we expect a bond order of less than 1.2741

BO less than 1 expected.2746

Let's go ahead and look at the energy diagram here.2750

Once again helium here, helium here, and again 1s, 1s.2754

Each helium is 1s2; now let's go ahead and make the MOs.2760

Again anytime we have 1s combining, one possibility is going to be σ bonding.2766

The other possibility is going to be σ antibonding.2772

Now let's go ahead and fill the electrons; one, two, three, and four.2777

When we go ahead and look at the bond order equation for He2,2782

this is going to equal to 1/2, 2 minus 2, which is going to be 0.2787

Again a bond order of less than 1 is what we expected.2793

That is what we actually get.2797

We are going to keep on moving down.2800

Now we are going to look at diatomic lithium.2802

Let's see if we expect diatomic lithium to exist in the gas phase.2805

Here lithium and lithium; lithium is going to be 2s, 2s.2810

Remember we only care about valence electrons; here 2s1 2s1.2823

Once again when we have s orbitals, we can expect formation of2830

a σ bonding orbital and then of a σ antibonding orbital.2835

Let's go ahead and put the electrons in--one electron here, one electron here.2841

Lithium, Li2, this bond order is going to be 1/2 times 2 minus 0 which is going to be 1.2845

Yes, Li2 is observed experimentally in the gas phase.2853

A bond order of 1 does make sense.2858

Let's move on to the following.2864

It turns out that not only can we do MO diagrams for2865

neutral species but we can also do it for cations.2869

Now this is Li21+; Li, Li, Li21+ in the middle.2874

Again lithium is a 2s atomic orbital; they are going to mix.2882

We are going to form σ bonding and then σ antibonding.2888

Lithium is 2s1.2897

Again because this is going to be 1+, that means we lose one of the electrons.2900

I am going to leave this one blank.2904

I am going to leave this one blank to show that we remove one electron.2906

Its MO is just going to be that, just one electron in the bonding MO.2913

BO or the bonding order is going to be equal to 1/2 times 1 minus 0.2917

We get an order of 1/2.2922

This suggests that the Li2 cation should not occur in the2925

gas phase because here the bond order is less than 1.2931

Good; we now move on to B2.2938

The reason why this is now going to be slightly different is2943

because as soon as we enter boron, we enter the p block.2946

We have not have not talked about how the p orbitals are going to be expected to be looking.2953

What do you get more overlap with?2960

Is it going to be s and s?--or is it going to be p and p?2963

The s orbitals are going to be closer to the nucleus.2967

We expect them to be lower in energy; we know that from Aufbau.2974

For example, 2s is usually less in energy than 2p.2978

It turns out that the σ bonding here is going to be less than the π bonding.2982

What we expect because typically we have more overlap with the σ's than we do with π's.2993

Remember σ is more direct overlap; π is parallel interaction.3000

We expect σ bonding to be less than π bonding; let's see what happens.3005

Boron, boron; boron is going to be a 2s orbital here, 2s orbital here, and 2p's, 2p's.3016

Let's fill the electrons in for boron; each boron is 2s2 2p1, 2s2 2p1.3028

Let's go ahead and tackle what we know so far already.3035

We know that the 2s orbitals can mix.3038

They are going to form our typical σ bonding and σ antibonding.3042

But what about the p orbitals?--remember the p orbitals.3048

Remember that when pz interacts with pz, we can get σ bonding and σ antibonding.3053

Remember that px and py are the same.3062

Those guys are going to give us π bonding and π antibonding.3071

Notice that there is two sets.3080

It turns outs that when we talk about diatomic boron, it turns out that the π orbitals are3086

going to be lower in energy than the σ orbitals just like that and3098

that the π*s are going to be lower in energy than the σ*.3116

That is not what is expected.3125

Remember we expect σ bonding to be less than π bonding.3126

What happens is the following; the reason why σ is above π.3132

It is due to the phenomenon that we call sp mixing.3139

What happens is the following.3145

That a boron 2s orbital is actually going to mix or interact with the boron 2p orbital.3147

What that does is it results in a lowering of the σ bonding MO from 2s3162

but a raising in energy of the σ bonding orbitals from the 2p's.3176

Essentially one goes down lower in energy; one goes up.3186

It makes sense that the 2s goes down because it is more stabilized due to the mixing.3191

Remember 2s's are going to be closer to the nucleus.3195

They get affected more than 2p's due to this interaction.3198

For B2, again the big take-home message is that the π bonding MOs are going to be3203

lower in energy than the σ bonding MOs of the 2p's due to sp mixing.3209

That is our MO diagram; let's go ahead and fill it up now.3216

I have a total of six electrons--one, two, three, four, five, and six.3220

When we go ahead and calculate now the bond order, the bond order here is going to be equal to 1/2...3226

number of bonding electrons is 4, two antibonding electrons.3232

That is going to be a bond order of 1.3236

Yes, we do expect diatomic boron to exist in the gas phase.3238

Let's look at another example, N2.3246

N2 is something we have looked at already in terms of Lewis structure.3249

Look at that; that is our Lewis structure of N2.3253

We expect a bond order of 3; let's see if we get that.3255

Nitrogen, nitrogen, this is 2s, 2s; then here is nitrogen's 2p orbitals just like that.3266

N2 is right in the middle.3276

Once again the 2s orbitals are going to mix to form σ and σ*.3279

Again the π orbitals from the 2p's are going to be lower in energy than the σ orbitals from the 2p's.3288

Then followed by π* and σ*.3299

Let's go ahead and populate the orbitals with the electrons.3307

Nitrogen has a configuration of 2s2 2p3, 2s2 2p3.3311

Now let's go ahead and fill; that is a total of ten electrons.3321

One, two, three, four, five, six, seven, eight, nine, and ten.3324

When we go ahead and calculate the bond order, 1/2 number of bonding electron which is going to be 8.3335

Number of antibonding electrons is 2.3342

Look at that; we get a bond order of 3 which is what we expect.3345

The accepted Lewis structure for N2 is a triple bond.3351

The next compound is going to be molecular oxygen.3359

Molecular oxygen, that is the one where we get paramagnetism from experiment.3363

Paramagnetism expected which means the MO diagram should have at least one unpaired electron.3373

Let's go ahead and look at the MO diagram; oxygen here--2s, 2s, and 2p.3397

O2 right in the middle; let's go ahead and form our MOs.3413

Again the 2s's can form σ bonding and σ antibonding and 2p's.3420

Let's just for now go by what we have already done3433

for the previous compounds where the π's were less than σ.3436

This is π; this is σ; and then this is π* and then σ*.3445

Let's put the electrons in--2s2 2p4, 2s2 2p4.3455

That is a grand total of twelve electrons that we have to use.3467

Let's fill them--one, two, three, four, five, six, seven, eight, nine, ten, and eleven and twelve.3470

Let's go ahead now and see do we have lone pair expected?3482

Yes, we have lone pairs right there; paramagnetism is good to go.3487

But let's go ahead and look at the bond order.3490

The bond order is the following.3492

1/2 times the number of bonding electrons minus the number of antibonding electrons.3496

That is going to be number of bonding electrons is 83501

minus number of antibonding electrons is going to be 4.3505

That is going to be 2; yes, double bond expected.3510

So far this sample diagram does work.3514

However experimentally it turns out that this is not the correct MO diagram for O2.3517

It turns out that in the actual MO diagram, these orbitals here,3525

the one we had that little controversy with is flipped.3537

It turns out that in the actual MO diagram, the σ is going to be less than π.3542

Let's go ahead and draw that.3547

I will explain why in a second why that makes sense.3549

2s2 and then 2p4; let's go ahead and fill everything.3552

Here we have our σ bonding, σ antibonding.3575

Here is the actual order; σ on the bottom, π on the bottom, then π* here, then σ* here.3585

Let me go ahead and label that; σ, π, π*, and σ*.3595

Let's go ahead and fill electrons right now.3601

One, two, three, four, five, six, seven, eight, nine, and ten, and then eleven and twelve.3604

It turns out that this is going to be the accepted one where σ is less than π.3618

The reason is because of the following--the sp mixing that we discussed earlier stops.3624

The reason why it stops is because it turns out when we reach oxygen3635

in the periodic table, it is a relatively heavier element than say nitrogen.3638

What that does is the following.3645

It is that the s orbital is now too low in energy to mix with the p orbital.3647

As the atom gets heavier and heavier, most of the interaction is going to come with the s orbital.3662

That is going to if you will benefit first.3674

That is going to be much lower in energy.3677

Because of that, the s orbital of oxygen is just too low in energy compared to the p orbital of oxygen.3682

Sp mixing stops.3689

We return to what is expected where the σ bonding MO orbital from the p orbitals is3692

going to be less than the energy of the π bonding MO orbitals of the 2p's.3698

Again this is always going to be the case; please make a note.3710

This is always the case for O2 and higher; F2 counts, Cl2 counts, etc.3714

Really everything up to and including N2 is one situation.3727

Then O2 and after is going to be this situation.3732

So far we have looked at only homonuclear diatomics.3738

We can do MO diagrams also for heteronuclear diatomics too.3741

Let's consider hydrochloric acid; hydrochloric acid has the following Lewis structure.3746

Here is hydrogen; here is chlorine; hydrogen is 1s.3756

Because the chlorine is more electronegative than hydrogen, its 2s orbital is going to be lower in energy than the 1s.3765

Then chlorine's 2p is going to be relatively higher in energy there.3774

Hydrogen is 1s; chlorine is 3s 3p.3780

Chlorine is going to be 3s2 and then 3p5 just like that.3792

In the Lewis structure for hydrochloric acid, we have a σ bond right there.3806

But chlorine has three lone pairs here which means...3816

These three lone pairs means that the six electrons are not involved in bonding, are nonbonding.3823

In terms of MO purposes, nonbonding electrons will occur not in a3832

bonding MO, not in an antibonding MO, but in a nonbonding MO.3841

That is a third type of molecular orbital that we now introduce.3849

For our purposes in GChem, nonbonding MOs are going to be at3855

the same energy as their atomic orbitals; same energy as atomic orbitals.3858

Let's go ahead and see what happens then.3871

It turns out that the 3s orbital of fluorine is just too low in energy to interact with anything.3879

Here is our first nonbonding MO.3894

I am going to go ahead and put those two electrons there just like that.3898

The next item is now the 1s orbital interacting with the 3p orbitals.3907

When this happens, it turns out we are going to get the following experimentally.3916

The 3p and the 1s are going to mix.3922

They are going to form a σ.3925

Of course if they form σ, that means they must also form a σ*.3927

Remember we have to conserve orbitals; we have a total of five atomic orbitals.3936

We have only used three so far in the MOs.3943

That means I have two remaining.3946

Those two remaining are going to be additional nonbonding orbitals just like that.3948

Let's go ahead and fill everything up.3955

We have a grand total of eight electrons; two is already accounted for.3957

One, two, three, four, five, six, seven, and eight.3964

Guess what?--our MO diagram agrees with our Lewis structure.3969

Here is one lone pair nonbonding; here is another lone pair nonbonding.3974

Here is another lone pair nonbonding; there we have our σ bond.3980

When you calculate the bond order, the bond order is equal to 1/2 number of bonding electrons which is 2.3986

You notice that there is no electrons in antibonding MOs.3993

You do not count nonbonding electrons in this equation.3996

There we go--a bond order of 1 which agrees with the Lewis structure of3999

hydrochloric acid where we just get a single covalent bond.4004

That is how we tackle heteronuclear diatomics.4008

To summarize, valence bond theory was the first bonding theory we talked about today.4013

It explains for observed bond angles but doesn't always account for magnetic behavior.4018

MO theory is pretty much the one that is going to supersede everything else.4022

That is the delocalized bonding model which views electrons as being distributed throughout4028

the entire molecule rather than just being centered around any individual atom.4033

Let's go ahead and look at a pair of sample problems.4041

In this one, determine the atomic hybridization around the central atom in the following molecules.4044

I picked these because these were Lewis structures that we have previously seen in the last lecture.4050

Here is the Lewis structure for XeO3, for xenon trioxide.4057

It is going to look like that.4061

Here we have a total of four electron groups around the xenon.4063

That is going to be sp3 hybridization.4068

SF6, SF6 is going to look like this.4071

There are lone pairs around each of the fluorines.4080

Remember sulfur is one of those elements that can have more than an octet.4086

This is completely valid Lewis structure.4092

Here we have six electron groups around the sulfur.4094

That is going to be a sp3 d2 hybridization.4097

Now I31-, I31-, its Lewis structure is going to be this.4102

When we go ahead and look at this, this is a total of five bonding groups around the iodine.4124

That is going to be sp3 d.4133

That is atomic hybridization prediction for a central atom from a Lewis structure.4137

Now moving on to the final sample problem, sample problem two.4145

Draw a qualitative MO diagram for F21- and comment on its expected stability.4149

Let's draw it--energy; fluorine, fluorine, and then F21-.4156

Fluorine has the 2s orbital and then its 2p's; let's go ahead and populate.4165

Fluorine is 2s2 2p5 and again 2s2 2p5.4177

Don't forget we have a 1- charge.4190

I am just going to go ahead and add that to here.4192

It doesn't matter which one; they are the same.4195

2s orbitals are going to mix.4198

They are going to form a σ MO and a σ* MO.4202

Here fluorine is after nitrogen; that is when the sp mixing does not occur anymore.4208

In this case, from the 2p orbitals, we are going to get σ first4218

followed by the π's and then π* and now σ*.4224

σ, π, π*, σ*; let's go ahead and fill the electrons in.4233

I am going to have a grand total of fifteen electrons to use.4239

One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, and fifteen.4244

Let's go ahead and calculate its bond order in order to comment on the stability.4258

The bond order is equal to 1/2 number of bonding electrons--that is going to be 8.4262

Minus number of antibonding electrons--that is going to be 7.4270

In this case, we get a bond order of 1/2 which is less than 1.4278

Therefore we do not expect F21- to occur in the gas phase.4282

That is another qualitative MO problem.4292

Thank you all for your attention; I will see you all next time on Educator.com.4297

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