Page III-8-1 / Chapter Eight Lecture Notes MAR. Two s orbitals overlap. One s & one p. overlap. Two p orbitals. overlap MAR
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1 Bonding and Molecular Structure: Orbital ybridization and Molecular Orbitals Chapter 8 Page III-8-1 / Chapter Eight Lecture Notes Advanced Theories of Chemical Bonding Chemistry 222 Professor Michael Russell Atomic Orbitals Molecules Two Theories of Bonding MOLECULAR ORBITAL TEORY - Robert Mulliken ( ) valence electrons are delocalized valence electrons are in (called molecular ) spread over entire molecule. Two Theories of Bonding VALENCE BOND TEORY - Linus Pauling valence electrons are localized between atoms (or are lone pairs). half-filled atomic overlap to form bonds. electrons stabilized by 2 nuclei Two s overlap Sigma Bond Formation by Orbital Overlap Sigma Bond Formation by Orbital Overlap Two s overlap One s & one p overlap Two p overlap Page III-8-1 / Chapter Eight Lecture Notes
2 Page III-8-2 / Chapter Eight Lecture Notes Using VB Theory F Boron configuration B F F 1s 2p planar triangle angle = 120 o ow to account for 3 bonds 120 o apart using a spherical s orbital and p that are 90 o apart? Pauling said to modify VB approach with ORBITAL YBRIDIZATION - mix available to form a new set of - YBRID ORBITALS - that will give the maximum overlap in the correct geometry. hybridize orbs. 2p rearrange electrons The three hybrid are made from 1 s orbital and 2 p create 3 sp 2 hybrids. three sp 2 hybrid unused p orbital Now we have 3, half-filled YBRID that can be used to form planar B-F sigma bonds. hydridize orbs. 2p three sp 2 hybrid rearrange electrons unused p orbital An orbital from each F overlaps one of the sp 2 hybrids to form a B-F σ bond. F Bonding in C 4 ow do we account for 4 C- sigma bonds 109 o apart? Need to use 4 atomic - s, p x, p y, and p z - to form 4 new hybrid pointing in the correct direction. 109 o F B F Page III-8-2 / Chapter Eight Lecture Notes
3 Page III-8-3 / Chapter Eight Lecture Notes Bonding in a Tetrahedron - Formation of ybrid Atomic Orbitals 4 C atom hybridize to form four equivalent sp 3 hybrid atomic. Bonding in a Tetrahedron - Formation of ybrid Atomic Orbitals 4 C atom hybridize to form four equivalent sp 3 hybrid atomic. Bonding in C 4 Orbital ybridization Bonds EPG ybrid REMAINING p orbs? 2 linear sp 2 p 3 trigonal sp 2 1 p planar 4 tetrahedral sp 3 none 5 trigonal sp 3 d --- bipyramid 6 octahedral sp 3 d see: VSEPR Guide ybrid Examples Bonding in Glycine sp 3 start sp 3 N C O C sp 2 O sp 3 finish Page III-8-3 / Chapter Eight Lecture Notes
4 Page III-8-4 / Chapter Eight Lecture Notes Multiple Bonds Sigma Bonds in C 2 4 Consider ethylene, C C C sp C C sp 2 π Bonding in C 2 4 The unused p orbital on each C atom contains an electron and this p orbital overlaps the p orbital on the neighboring atom to form the π bond. 2p 3 sp 2 hybrid p orb. for π bond π Bonding in C 2 4 The unused p orbital on each C atom contains an electron, and this p orbital overlaps the p orbital on the neighboring atom to form the π bond. 2p 3 sp 2 hybrid p orb. for π bond Multiple Bonding in C 2 4 σ and π Bonding in C 2 2 C 2 2 has a triple bond Page III-8-4 / Chapter Eight Lecture Notes
5 Page III-8-5 / Chapter Eight Lecture Notes Consequences of Multiple Bonding Consequences of Multiple Bonding There is restricted rotation around C=C bond. Restricted rotation around C=C bond. trans-but-2-ene n-butane Molecular Orbital Theory Advantages of MO Theory Dioxygen should be electron paired (diamagnetic) by VB Theory, but dioxygen is actually paramagnetic. MO Theory accounts for paramagnetism of O2 Accounts for paramagnetism, color, bonding Atomic delocalize into molecular Bonding, Antibonding and Nonbonding Quite complicated, need computers; we will only look at diatomics (2 atom systems) from the first and second periods only Molecular Orbital Type Four Principles of MO Theory Principle #1: Number of Molecular Orbitals = Number of Atomic Orbitals When two atomic 1s combine, a bonding (σ) and antibonding (σ*) molecular orbital forms antibonding σ* Two 1s from two hydrogen atoms create two molecular in 2 Two 1s and two from two lithium atoms create four molecular in Li2 See Four Principles of MO andout Page III-8-5 / Chapter Eight Lecture Notes bonding σ
6 Page III-8-6 / Chapter Eight Lecture Notes Four Principles of MO Theory Principle #2: Bonding MO lower in energy than the parent orbital Antibonding MO higher in energy than the parent orbital MO Diagram for 2 Antibonding MO Atomic Orbitals Principle #3: Electrons of molecule assigned to successively higher MOs Use Pauli Exclusion Principle and und's Rule when assigning electrons Two 1s electrons from two atoms occupy the σ orbital in 2 Bonding MO MO Diagram for e 2 Antibonding MO Atomic Orbitals Bonding MO Bond Order in MO Theory Bond Order = 1 / 2 (# bonding e - - # antibonding e - ) Bond Order > 0, stable molecule Bond Order = 0 or < 0, unstable molecule Two 1s electrons in σ, Two 1s electrons in σ * In 2, Bond Order = 1 / 2 (2-0) = 1; stable Bond Order in MO Theory Bond Order = 1 / 2 (# bonding e - - # antibonding e - ) Bond Order > 0, stable molecule Bond Order = 0 or < 0, unstable molecule In e 2, Bond Order = 1 / 2 (2-2) = 0; unstable e 2 does not exist Page III-8-6 / Chapter Eight Lecture Notes Four Principles of MO Theory Principle #4: Atomic combine to give molecular only when the atomic are of similar energy Similar energy = better overlap 1s + 1s = good MO 1s + = poor MO + = good MO + 2p = poor MO 3s + = poor MO... etc....
7 Page III-8-7 / Chapter Eight Lecture Notes Example: Dilithium, Li 2 Note: no overlap between 1s and Bond Order = 1 / 2 (4-2) = 1 p and π bonds Three possible p on each atom - six total p MO Two p create 2 σ MO bonds Four remaining p create 4 π MO bonds Would you expect Be 2 to exist? Why? Stable molecule Four p atomic create four π molecular, π = bonding (2) π * = antibonding (2)... but there's a catch! p and π bonds For B, C and N, π lower energy than σ orbital π * lower energy than σ * orbital p and π bonds For O, F and Ne, σ orbital lower energy than π π * lower energy than σ * orbital σ* 2pz π 2px π 2py 2p x 2p y 2p z π 2px π 2py 2p x 2p y 2p z Example: B 2 σ 2pz Bond Order = 1 σ σ See MO Diagram (B 2 - N 2 ) andout 1s σ 1s σ 1s 1s p and π bonds For O, F and Ne, σ orbital lower energy than π π * lower energy than σ * orbital Paramagnetism Paramagnetism exists when unpaired electrons in MO diagram Example: O 2 Bond Order = 2 O 2 is paramagnetic; unpaired electrons in two π * See MO Diagram (O 2 - Ne 2 ) andout Page III-8-7 / Chapter Eight Lecture Notes
8 Page III-8-8 / Chapter Eight Lecture Notes Paramagnetism Paramagnetism exists when unpaired electrons in MO diagram Molecular Orbital Notation Used to abbreviate the MO diagrams Ignore core electrons Write in order of increasing energy N 2 is diamagnetic; all electrons paired For N 2 : [core electrons](σ ) 2 (σ * ) 2 (π 2p ) 4 (σ 2p ) 2 Sigma and Pi Bonds Determine sigma and pi bonds using: # σ bonds = 1 / 2 (# σ bonding e - - # σ antibonding e - ) # π bonds = 1 / 2 (# π bonding e - - # π antibonding e - ) and # σ bonds + # π bonds = bond order MO Diagram for Diatomics For N 2 : [core electrons](σ ) 2 (σ * ) 2 (π 2p ) 4 (σ 2p ) 2 # σ bonds = 1 / 2 (4-2) = 1 σ bond # π bonds = 1 / 2 (4-0) = 2 π bonds bond order = 1 / 2 (8-2) = 3 = 1 σ + 2 π bonds Changes in MO diagrams due to s-p mixing and/or electron repulsion Ionic Diatomic Molecules Ionic Diatomic Molecules Predicting Ionic Diatomic MO diagrams simple Use und and Pauli Predicting Ionic Diatomic MO diagrams simple Use und and Pauli Example: O 2 + Remove electron from π * 2p orbital Check bond order, paramagnetism Example: O 2 - Using the O 2 diagram on the right, where would you place the extra electron? Is O 2 - more or less stable than O 2? Why? Page III-8-8 / Chapter Eight Lecture Notes
9 Page III-8-9 / Chapter Eight Lecture Notes Application: Vision Molecular Orbital Theory helps to describe the process of vision - photochemistry Application: Band Theory In metallic bonding, electrons delocalized over metallic lattice - a sea of electrons MO energies identical, excellent overlap elps explain conductivity, malleability, more End of Chapter 8 MO Diagram for Mo(CO)5=C2 See: Chapter Eight Study Guide Chapter Eight Concept Guide Page III-8-9 / Chapter Eight Lecture Notes
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