Lecture 14 - UCI Canvas

14- 1it’s true that…● MT2 problem 20 will be regraded on

all forms. The correct answer is…

14- 2

3.5

2.5

3.03.0

14- 3

3.5

2.5

3.03.0

Answer: C

14- 4it’s true that…● MT2 problem 20 will be regraded on

all forms. The correct answer is… ● HW 6 is due Sat. Nov. 24.● I will post weighted course % scores

by Wed (tomorrow).● Our final exam has been relocated to:

PSLH 100.● game is on!

14- 5

14- 6PSLH 100

14- 7

Lecture 14Tuesday, November 20, 2018

Prof. Penner

14- 8what’s going on here?

13- 9

hierarchy of bond theory

Lewis dot + VSEPR (Ch 6)worst

best

Valence bond theory (Ch 7)

molecular orbital theory (Ch 7)

13- 10


worst

best


13- 11

H2

a covalent bond is formedwhen electron clouds overlap

13- 12

F2

here, atomic orbitals are involved in bonding…

14- 13

HF

VB theory is about orbital overlap

13- 14

what’s its bond angle?

Lewis theory also gets this wrong….

13- 15

104.5o: H2S looks like water:

13- 16

wrong. its actual bond angle is 92o

why?

13- 17the S 2p orbitals forming the bonds to H are oriented 90o apart!

13- 18these two electrons are in orthogonal orbitals

…so we should expect a 90o angle.

13- 19

Pauling (p. 111): “The difference of 14.5o between the observed value of the bond angle

[for water] and the expected value of 90o is probably to be attributed in the main to the partial ionic character of the O-H bonds,

estimated in the preceding chapter to be 39%.”

but why, then, doesn’t water do the same thing?

13- 20

3.5 2.6

2.1 2.1 2.12.1

H2SH2O

92o

the 2H’s in H2O have a higher charge than in H2S

13- 21

the VB picture of bonding between atomic orbitals works great for H2, HF, F2

and H2S…

…but what about CH4?

13- 22

…but what about CH4?

13- 23

for making four tetahedral bonds

…how do you get fouridentical C-H bonds out of this?

13- 24

we predict CH2 with a 90o angle…

13- 25

we predict CH2 with a 90o angle…

…but that’s wrong.

13- 26

VBT has a solution: hybridization

• Some atoms hybridize their orbitals to maximize bonding. More bonds = more full orbitals = more stability.

13- 27



• Hybridizing is mixing different types of orbitals in the valence shell to make a new set of degenerate orbitals: sp, sp2, sp3, sp3d, sp3d2.

13- 28




what does this word mean?

13- 29




what does this word mean?orbitals that all have the same energy

13- 30




• The same type of atom can adopt different hybridizations. Example. carbon can adopt sp, sp2, sp3 hybridizations.

13- 31

atomic orbitals

example: carbon

13- 32

atomic orbitals

hybrid atomic orbitals

this sp3 orbital is tetrahedral

example: carbon

13- 33

each lobe of thehybrid tetrahedral sp3 isskewed, to enable better orbital overlap

13- 34making a sp3 orbital

13- 35

sp3 hybrid orbital

VB methane

13- 36

making a sp2 orbital

13- 37

let’s make formaldehyde, CH2O

atomic orbitals

13- 38let’s make formaldehyde, CH2O

13- 39

the C has three e- blobs…

13- 40

sp2

the C has three e- blobs…so its hybridization is sp3

13- 41

we actually need 2 sp2 orbitals…

sp2sp2

…same with the oxygen.

13- 42

atomic orbitals

make two sp2 orbitals

let’s make formaldehyde, CH2O

13- 43

sp2 orbital

note: there is one unhybridized p orbital, on both O and C

multiple bonds involve electronsin unhybridized orbitals

sp2 orbital

13- 44

http://www.chemtube3d.com/orbitalsformaldehyde.htm

2 sp2 orbitals(ea. holds 6 electrons)

let’s make the double bond…

13- 45


unhybridized p orbitals

let’s make the double bond…

13- 46


13- 47


boom.

13- 48

13- 49terminology: sigma (σ) and pi (π) bonds…

•σ bonds - are two electron bonds that are cylindrically symmetric about the internuclear axis.

•π bonds - are two electron bonds that are NOT cylindrically symmetric about the internuclear axis.

13- 50

sp2sp2

a double bond consists ofone σ bond + one π bond

13- 51let’s make acetylene, C2H2

13- 52what is the hybridization onthe carbons here?

13- 53

each has two e- blobs, so…

13- 54

sp

each has two e- blobs, so…hybridization is sp

sp

13- 55making an sp orbital

you have TWO unhybridized p orbitals left over.

© 2018 Pearson Education, Inc.

making an sp orbital

© 2018 Pearson Education, Inc.

making an sp orbital

2σ bonds2π bonds

13- 58

…this results in one π bond…

px

13- 59

px

py

…and this is the other π bond…

13- 60so, a triple bond consists ofone σ bond + two π bond

14- 61

ethylene ethyne(acetylene)

1 σ bond 1 σ bond1 π bond

1 σ bond2 π bonds

π bonds are formed from unhybridized p orbitals

ethane

13- 62

now, a word about bond rotation…

13- 63

13- 64

σ bonds can rotate

13- 65

13- 66

π bonds can’t rotate

b/c they would have to break, to do so.

13- 67

these are geometric isomers

Cis & Trans 1,2-dichloroethene

13- 68types of bonds•A sigma (σ) bond results when the interacting atomic orbitals point along the axis connecting the two bonding nuclei.


•A pi (π) bond results when the bonding atomic orbitals are parallel to each other and perpendicular to the axis connecting the two bonding nuclei.


•A pi (π) bond results when the bonding atomic orbitals are parallel to each other and perpendicular to the axis connecting the two bonding nuclei.

•The interaction between parallel orbitals is not as strong as between orbitals that point at each other; therefore, σ bonds are stronger than π bonds.

13- 71

so, every e- blob geometry has a hybrid orbital:

13- 72

yes, d orbitalscan be

involved

13- 73

that thing on the As is a sp3d hybrid orbital…

13- 74

its made from 1s, 3p, and 1d orbital, obviously…

13- 75that thing on the S is an sp3d2 hybrid orbital…

13- 76

it’s made from 1s, 3p, and 2d orbitals

13- 77

BrF3

name the hybrid orbital?

13- 78

BrF3

sp3d

13- 79

BrF3

13- 80

::

acetaldehyde

name the hybrid orbitals?

13- 81acetaldehydesp2

sp2

sp3

::

13- 82

sp3

sp2sp2

notice this diagram isincorrect about the hybridization

of the O

acetaldehyde

13- 83

ethylene

name the hybrid orbital?

13- 84

ethylene

sp2 sp2

13- 85ethylene

13- 86ethylene: what can we conclude about the methylene groups?

A: they have to be coplanar

13- 87

ethylene

ok, let’s thinkmore aboutmoleculargeometry…

13- 88

ethylene

unhybridized p orbitals on the carbons point out of the screen

13- 89

ethylene propadiene

one set of unhybridized p-orbitals poke out of the screen

13- 90

ethylene propadiene

the second set is orientedin the plane of the screen

13- 91

propadiene

the second set is orientedin the plane of the screen

this means thatthe CH2 groups are oriented ata 90o angle toone another.

14- 92benzene, C6H6

14- 93

AugustKekulé

14- 94

VB benzene, C6H6

14- 95

VB benzene, C6H6

13- 96


Lewis dot + VSEPR (Ch 6)worst

best


molecular orbital theory (Ch 7)

13- 97


worst

bestmolecular orbital theory (Ch 7)

14- 98

so what’s missing in VB theory?

interference

14- 99

orbitals are e- standing waves.

constructive interference

14- 100

orbitals are e- standing waves.

destructive interference

14- 101

this logic suggests that overlappingtwo atomic orbitals should

always generatetwo new “molecular orbitals”…

14-



102

here’s how that works:

14- 103



here’s how that works:

14- 104the colors are important. red is negative, blue is positive.

…the wave function is (+)here. Two (+) will add, giveconstructive interference, andenhance the electron density.

14- 105

the colors are important. red isnegative, blue is positive.

…here, a (+) wave function overlaps with a (-) wave function and destructive interference

occurs. Electron density is reduced.

14- 106think about the binding energy of the electrons

energy of two chargesat a distance, r.

vacuum

14- 107in anti-bonding MOs, the electrons are forced further away from the nuclei

energy of two chargesat a distance, r.

anti bonding

bonding

vacuumlow

high

14- 108

there is low e-density here.r is large. this is a low

binding energy situation

14- 109

there is high e-density here.r is small. this is a high

binding energy situation

14- 110

bond order = b.o. = bonding e- - anti bonding e-

2

14- 111ok, let’s make H2

14- 112MO diagram for H2

14- 113

b.o. = (2-0)/2 = 1; H2 is stable

what’s its bond order?

14- 114

1. The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.

2. The more stable the bonding MO, the less stable the corresponding antibonding MO.

3. The filling of MOs with electrons proceeds from low to high energies.

4. Each MO can accommodate up to two electrons.5. Use Hund’s rule when adding electrons to MOs of the

same energy.6. The number of electrons in the MOs is equal to the

sum of all the electrons on the bonding atoms.

Molecular Orbital (MO) Diagrams

14- 115ok, let’s make He2

14- 116boom.

14- 117

b.o. = (2-2)/2 = 0; He2 is unstable

what’s its bond order?

14- 118He2+

14- 119boom.

14- 120

b.o. = (2-1)/2 = 0.5; He2+ is stable

never heard of it, does it exist?

14- 121Li2

b.o. = (2-0)/2 = 1.0; Li2 is stable

14- 122Be2

14- 123

b.o. = (2-2)/2 = 0; Be2 is unstable

Be2

14- 124p-orbitals oriented orthogonal to the bond axis overlap weakly.

14- 125

σ*2p

σ2p

14- 126these are the two p orbitals

that are aligned with the bond axis

14- 127

π*2p

π2p

14- 128p-orbitals oriented orthogonal to the bond axis overlap weakly.

14- 129ok, ready?Let the confusion

begin.

for B2,C2, and N2…

14- 130

14- 131

14- 132

14- 133

14- 134

14- 135

14- 136

Draw an MO energy diagram and determine the bond order for the

N2– ion. Do you expect the

bond in the N2– ion to be stronger or weaker

than the bond in the N2 molecule?

Is N2– diamagnetic or paramagnetic?

N2-

14- 137






N2-

14- 138





b.o. = (6-1)/2 = 5/2

N2-

14- 139


N2– ion. Do you expect the bond

in the N2– ion to be stronger or weaker


N2-

14- 140


N2– ion. Do you expect the bond

in the N2– ion to be stronger or weaker



N2-

14- 141

look!O2 is paramagnetic!

14- 142

O2

14- 143

just say NO

14- 144

just say NO

14- 145MO diagram for heteronuclear diatomic molecules

Similar MO diagrams with the following differences:

• in polar covalent bonds, atomic orbitals having greater electronegativity have lower energy, resulting in it making the largest contribution to the lowest MO's

• highest energy atomic orbitals make the largest contributions to the highest energy MO's

• when constructing MO diagrams, atomic orbitals form the more electronegativity atoms are placed lower

14- 146

E

CO

14- 147C O

1s1s

2s2s

2p2p

σ1s

σ1s*

σ2s

σ2s*π2p

π2p*

σ2p

σ2p*

E

CO

14- 148C O

1s1s

2s2s

2p2p

σ1s

σ1s*

σ2s

σ2s*π2p

π2p*

σ2p

σ2p*

E

CO

14- 149C O

E

CO

1s1s

2s2s

σ1s

σ1s*

σ2s

σ2s*

2p2p

π2p

π2p*

σ2p

σ2p*

14- 150C O

E

CO

1s1s

2s2s

σ1s

σ1s*

σ2s

σ2s*

2p2p

π2p

π2p*

σ2p

σ2p*

HOMO

LUMO

highest occupiedMO

lowest unoccupiedMO

14- 151

● b.o. = (6-0)/2 = 3.0● diamagnetic

about CO, we can say:

14- 152HF

14- 153

● b.o. = (2-0)/2 = 1.0● diamagnetic

about HF, we can say:

14- 154what if the number ofatoms gets large?

14- 155what if the number ofatoms gets large?

MO theory predicts a band of states develops

14- 156

band theory of solids

http://eng.thesaurus.rusnano.com/wiki/article853

filled states

empty states

14- 157


http://eng.thesaurus.rusnano.com/wiki/article853

filled states

empty states

metals have partially filled bands. electrons are highly mobile.

14- 158


filled states

empty states

semiconductors have disjoint valenceand conduction bands, separated byan energy of 3.5 eV or less

14- 159


filled states

empty states

insulators have disjoint valenceand conduction bands, separated by

>3.5 eV.

14- 160btw - what is the thermal energy at room temperature?

Boltzmann’s const., k = 1.381 x 10-23 J/K

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Lecture 14 - UCI Canvas

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