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What Do Molecules Look Like?. The Lewis Dot Structure approach provides some insight into molecular structure in terms of bonding, but what about 3D geometry?. Recall that we have two types of electron pairs: bonding and lone.

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what do molecules look like
What Do Molecules Look Like?

The Lewis Dot Structure approach provides some insight into molecular structure in terms of bonding, but what about 3D geometry?

Recall that we have two types of electron pairs: bonding and lone.

Valence-Shell Electron-Pair Repulsion (VSEPR). 3D structure is determined by minimizing repulsion of electron pairs.

slide2

Electron pairs (both bonding and lone) are distributed around a central atom such that electron-electron repulsions are minimized.

slide3

Electron pairs (both bonding and lone) are distributed around a central atom such that electron-electron repulsions are minimized.

2 electron pairs

3 electron pairs

4 electron pairs

Period 1, 2

5 electron pairs

6 electron pairs

Period 3 & beyond

arranging electron pairs

H

H

C

H

H

Arranging Electron Pairs

• Must consider both bonding and lone pairs when

minimizing repulsion.

  • Example: CH4 (bonding pairs only)

Lewis Structure

VSEPR Structure

arranging electron pairs cont

H

H

N

H

Arranging Electron Pairs (cont.)

Example: NH3 (both bonding and lone pairs).

Lewis Structure

VSEPR Structure

Note:“electron pair geometry” vs.

“molecular shape”

vsepr structure guidelines
VSEPR Structure Guidelines
  • The previous examples illustrate the strategy for applying VSEPR to predict molecular structure:
  • Construct the Lewis Dot Structure
  • Arrange bonding/lone electron pairs in space such that repulsions are minimized (electron pair geometry).
  • Name the molecular shape from the position of the atoms.

VSEPR Shorthand:

1. Refer to central atom as “A”

2. Attached atoms are referred to as “X”

3. Lone pair are referred to as “E”

Examples:

CH4: AX4

NH3: AX3E

H2O: AX2E2

BF3: AX3

vsepr 2 electron pairs

Be

F

F

Be

F

F

VSEPR: 2 electron pairs

Experiments show that molecules with multiple bonds can also be linear.

Linear (AX2): angle between bonds is 180°

Example: BeF2

Multiple bonds are treated as a single effective electron group.

F

F

Be

180°

More than one central atom? Determine shape around each.

vsepr 3 electron pairs
VSEPR: 3 electron pairs

Trigonal Planar (AX3): angle between bonds is 120°

Multiple bond is treated as a single effective electron group.

Example: BF3

F

F

120°

B

F

F

F

B

F

vsepr 4 electron pairs cont

H

H

C

H

H

VSEPR: 4 electron pairs (cont.)

Tetrahedral (AX4): angle between bonds is ~109.5°

Example: CH4

109.5°

tetrahedral e- pair geometry AND tetrahedral molecular shape

bonding vs lone pairs
Bonding vs. Lone pairs

Bond angle in a tetrahedral arrangement of electron pairs may vary from 109.5° due to size differences between bonding and lone pair electron densities.

bonding pair is constrained by two nuclear potentials; more localized in space.

lone pair is constrained by only one nuclear potential; less localized (needs more room).

vsepr 4 electron pairs

H

H

N

H

VSEPR: 4 electron pairs

Trigonal pyramidal (AX3E): Bond angles are <109.5°, and structure is nonplanar due to repulsion of lone pair.

Example: NH3

107°

tetrahedral e- pair geometry; trigonal pyramidal molecular shape

vsepr 4 electron pairs cont1
VSEPR: 4 electron pairs (cont.)

Classic example of tetrahedral angle shift from 109.5° is water (AX2E2):

104.5o

“bent”

tetrahedral e- pair geometry; bent molecular shape

vsepr 4 electron pairs cont2
VSEPR: 4 electron pairs (cont.)

Comparison of CH4 (AX4), NH3 (AX3E), and H2O (AX2E2):

slide14

AX2E

AX3E

AX2E2

1. Refer to central atom as “A”

2. Attached atoms are referred to as “X”

3. Lone pair are referred to as “E”

molecular vs electron pair geometry

H

H

H

H

Molecular vs. Electron-Pair Geometry

F

N

O

C

Central Atom

Compound

Electron-Pair Geometry

Molecular Shape

Carbon, C

CH4

tetrahedral

tetrahedral

Nitrogen, N

NH3

tetrahedral

trigonal pyramidal

Oxygen, O

H2O

tetrahedral

bent

Fluorine, F

HF

tetrahedral

linear

slide16

What is the electron-pair geometry and the molecular shape for HCFS?

  • trigonal planar, bent
  • trigonal planar, trigonal planar
  • tetrahedral, trigonal planar
  • tetrahedral, tetrahedral
vsepr beyond the octet
VSEPR: Beyond the Octet

Systems with expanded valence shells will have five or six electron pairs around a central atom.

F

Cl

F

Cl

F

Cl

S

P

Cl

F

F

Cl

F

90°

90°

F

F

F

90°

S

120°

F

F

F

vsepr 5 electron pairs

F

F

F

F

F

F

F

F

F

F

F

F

VSEPR: 5 electron pairs
  • • Consider the structure of SF4 (34 e-, AX4E)
  • What is the optimum arrangement of electron pairs around S?

??

S

S

S

Compare e– pair angles

lone-pair / bond-pair:

twoat 90o, twoat 120o

threeat 90o

threeat 90o, three at 120o

fourat 90o, one at 120o

bond-pair / bond-pair:

Repulsive forces (strongest to weakest):

lone-pair/lone-pair > lone-pair/bond-pair > bond-pair/bond-pair

vsepr 5 electron pairs1
VSEPR: 5 electron pairs

The optimum structure maximizes the angular separation of the lone pairs.

I3- (AX2E3):

5 electron pair geometries

AX4E

AX3E2

AX2E3

5-electron-pair geometries

our previous example

vsepr 6 electron pairs
VSEPR: 6 electron pairs

Which of these is the more likely structure?

See-saw

Square Planar

slide22

AX5E

AX4E2

6-electron-pair geometries

our previous example

molecular dipole moments
Molecular Dipole Moments

We can use VSEPR to determine the polarity of a whole molecule.

  • Draw Lewis structures to determine 3D arrangement of atoms.

2. If one “side” of the molecule has more EN atoms than the other, the molecule has a net dipole.

Shortcut: completely symmetric molecules will not have a dipole regardless of the polarity of the bonds.

slide24

Molecular Dipoles

The C=O bonds have dipoles of equal magnitude but opposite direction, so there is no net dipole moment.

The O-H bonds have dipoles of equal magnitude that do not cancel each other, so water has a net dipole moment.

molecular dipoles cont
Molecular Dipoles (cont.)

symmetric

symmetric

asymmetric

molecular dipole example
Molecular Dipole Example
  • Write the Lewis dot and VESPR structures for CF2Cl2. Does it have a dipole moment?

F

32 e-

Cl

F

Cl

Tetrahedral

advanced vsepr application
Advanced VSEPR Application

Molecules with more than one central atom… methanol (CH3OH)

H

C

O

H

H

tetrahedral e- pairs

tetrahedral shape

tetrahedral e- pairs

bent shape

H

the vsepr table

# e- pairs

e- Geom.

Molec. Geom.

The VSEPR Table

the vsepr table1

# e- pairs

e- Geom.

Molec. Geom.

The VSEPR Table

slide30

What is the expected shape of ICl2+?

20 e-

AX2E2

A. linear

C. tetrahedral

D. square planar

B. bent

slide31

Valence Bond Theory

Basic Principle of Localized Electron Model:

A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei.

Rule 1: Maximum overlap. The bond strength depends on the attraction of nuclei to the shared electrons, so:

The greater the orbital overlap, the stronger the bond.

slide32

Valence Bond Theory

Basic Principle of Localized Electron Model:

A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei.

Rule 2: Spins pair. The two electrons in the overlap region occupy the same space and therefore must have opposite spins.

There may be no more than 2 electrons in a molecular orbital.

slide33

Valence Bond Theory

  • Basic Principle of Localized Electron Model:
  • A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons occupies the region between the two nuclei.
  • Rule 3: Hybridization.To explain experimental observations, Pauling proposed that the valence atomic orbitals in a molecule are different from those in the isolated atoms. We call this concept
  • Hybridization
slide34

What is hybridization?

  • Atoms adjust to meet the “needs” of the molecule.
  • In a molecule, electrons rearrange in an attempt to give each atom a noble gas configuration and to minimize electron repulsion.
  • Atoms in a molecule adjust their orbitals through hybridization in order for the molecule to have a structure with minimum energy.
  • The source of the valence electrons is not as important as where they are needed in the molecule to achieve a maximum stability.
slide35

Example: Methane

  • 4 equivalent C-H covalent bonds
  • VSEPR predicts a tetrahedral geometry
slide36

How do we explain formation of 4 equivalent C-H bonds?

The Valence Orbitals of a Carbon Atom

Carbon: 2s22p2

slide37

Hybridization: Mixing of Atomic Orbitals to

form New Orbitals for Bonding

+

+

+

+

+

+

+

slide38

Other Representations of Hybridization:

y1 = 1/2[(2s) + (2px) + (2py) + (2pz)]

y2 = 1/2[(2s) + (2px) - (2py) - (2pz)]

y3 = 1/2[(2s) - (2px) + (2py) - (2pz)]

y4 = 1/2[(2s) - (2px) - (2py) + (2pz)]

slide39

Hybridization is related to the number of

valence electron pairs determined from VSEPR:

Methane (CH4)

VSEPR: AB4

tetrahedral

 sp3 hybridized

109.47 º

Electron pair geometry determines hybridization, not vice versa!!

slide40

Hybridization is related to the number of

valence electron pairs determined from VSEPR:

Ammonia (NH3)

VSEPR: AB3E

tetrahedral

 sp3 hybridized

N

H

H

H

108.1 º

slide41

Hybridization is related to the number of

valence electron pairs determined from VSEPR:

Water (H2O)

VSEPR: AB2E2

tetrahedral

 sp3 hybridized

105.6 º

slide42

sbonding and pbonding

  • Two modes of bonding are important for
  • 1st and 2nd row elements: s bonding and p bonding
  • These two differ in their relationship to the internuclear axis:
    • s bonds have electron density ALONG the axis
  • p bonds have electron density ABOVE AND BELOW the axis
slide43

Problem: Describe the hybridization and bonding of the carbon orbitals in ethylene (C2H4)

VSEPR: AB3

trigonal planar

 sp2 hybridized orbitals for s bonding

sp2 hybridized orbitals used for s bonding

remaining p orbital used for p bonding

slide45

Problem: Describe the hybridization and bonding of the carbon orbitals in Carbon Dioxide (CO2)

  • VSEPR: AB2
  • linear
  • sp hybridized orbitals for s bonding
slide47

H

H

C2

C1

N

:

N

CH3

C

H

p

p

sp

sp

sp

sp3

Atoms of the same kind can have different hybridizations

Acetonitrile (important solvent and industrial chemical)

Bonds

s

C2: AB4

C1: AB2

2s2 2px2py

s

p

p

N: ABE

p

sp

p

sp

2s2 2px2py2pz

lone pair

slide48

What have we learned so far?

  • Molecular orbitals are combinations of atomic orbitals
  • Atomic orbitals are “hybridized” to satisfy bonding in molecules
  • Hybridizationfollows simple rules that can be deduced from the number of chemical bonds in the molecule and the VSEPR model for electron pair geometry
hybridization
Hybridization
  • sp3 Hybridization (CH4)
    • This is the sum of one s and three p orbitals on the carbon atom
    • We use just the valence orbitals to make bonds
    • sp3 hybridization gives rise to the tetrahedral nature of the carbon atom
hybridization1
Hybridization
  • sp2 Hybridization (H2C=CH2)
    • This is the sum of one s and two p orbitals on the carbon atom
    • Leaves one p orbital uninvolved – this is free to form a p bond (the second bond in a double bond)
hybridization2
Hybridization
  • sp Hybridization (O=C=O)
    • This is the sum of one s and one p orbital on the carbon atom
    • Leaves two p orbitals free to bond with other atoms (such a O in CO2), or with each other as in HC≡CH
general notes
General Notes
  • This is a model and only goes so far, but it is especially helpful in understanding geometry and expanding Lewis dot structures.
  • Orbitals are waves. Hybridized orbitals are just the sums of waves – constructive and destructive interference.
what is important to know about hybridization
What is important to know about hybridization?
  • You should be able to give the hybridization of an atom in a molecule based on the formula given.
  • Example: CH3-CH2-CHO
  • Step 1: Draw the Lewis Dot Structure
what is important to know about hybridization1
What is important to know about hybridization?
  • Step 2: What is the electron pair geometry and molecular shape?

AXE2

Trigonal Planar

AX3

AX4

Trigonal Planar

AX4

Tetrahedral

Tetrahedral

what is important to know about hybridization2
What is important to know about hybridization?
  • Step 3: Use the molecular shape to determine the hybridization.

AXE2

Trigonal Planar

sp2

AX3

AX4

Trigonal Planar

AX4

Tetrahedral

sp3

Tetrahedral

sp2

sp3

slide56

Major limitations of the LE model:

..

..

..

..

Example: O2

Lewis dot structure O=O

All electrons are paired Contradicts experiment!

The Localized Electron Model is very powerful

for explaining geometries and basic features of

bonding in molecules, but it is just a model.

  • Assumes electrons are highly localized between the nuclei (sometimes requires resonance structures)
  • Doesn’t easily deal with unpaired electrons (incorrectly predicts physical properties in some cases)
  • Doesn’t provide direct information about bond energies
slide57

The Molecular Orbital Model

Basic premise: When atomic orbitals interact to form a bond, the result is the formation of newmolecular orbitals

HY= EY

Important features of molecular orbitals:

1. Atomic Orbitals are solutions of the Schrödinger equation for atoms.

Molecular orbitals are the solutions of the same Schrödinger equation applied to the molecule.

molecular orbital theory
Molecular Orbital Theory

2.Atomic Orbitals can hold 2 electrons with opposite spins.

Molecular Orbitals can hold 2 electrons with opposite spins.

3. The electron probability for the Atomic Orbital is given by Y2.

The electron probability for the Molecular Orbital is given by Y2.

4. Orbitals are conserved - in bringing together 2 atomic orbitals, we have to end up with 2 molecular orbitals!

How does this work?

slide59

Molecular Orbitals are simply

Linear Combinations of Atomic Orbitals

Example: H2

santi-bonding (s*)

-

+

Molecular Orbitals have phases (+ or -)

+

sbonding

Next Question:Why does this work?

constructive and destructive interference
Constructive and Destructive Interference

Destructive interference between two orbitals of opposite sign leads to an anti-bonding orbital.

Constructive interference between two overlapping orbitals leads to a bonding orbital.

slide61

= = nucleus

+

Bonding is driven by stabilization of electrons

  • Electrons are negatively charged
  • Nuclei are positively charged

The bonding combination puts electron density between the two nuclei - stabilization

The anti-bonding combination moves electron density away from region between the nuclei - destabilization

mo diagrams
MO Diagrams
  • We can depict the relative energies of molecular orbitals with a molecular orbital diagram:

The new molecular orbital is lower in energy than the atomic orbitals

slide63

s* M.O. is raised in energy

sM.O. is lowered in energy

H atom: (1s)1 electron configuration

H2 molecule: (s1s)2 electron configuration

slide64

s*

s

Same as previous description of bonding

review of orbital filling
Review of Orbital Filling
  • Pauli Exclusion Principle:
    • No more than 2 e- in an orbital, spins must be paired (↑↓)
  • Aufbau Principle (a.k.a. “Building-Up”):
    • Fill the lowest energy levels with electrons first
      • 1s 2s 2p 3s 3p 4s 3d 4p …
  • Hund’s Rule:
    • When more than one orbital has the same energy, electrons occupy separate orbitals with parallel spins:

No

No

Yes

slide66

H2

Filling Molecular Orbitals with Electrons

1) Orbitals are filled in order of increasing Energy (Aufbau principle)

slide67

He2

Filling Molecular Orbitals with Electrons

2) An orbital has a maximum capacity of two electrons with

opposite spins (Pauli exclusion principle)

filling molecular orbitals with electrons
Filling Molecular Orbitals with Electrons

3) Orbitals of equal energy (degenerate orbitals) are half filled,

with spins parallel, before any is filled completely (Hund’srule)

slide69

# bonding #anti-bonding

electrons electrons

Bond Order =

2

The bond order is an indication of bond strength:

Greater bond order Greater bond strength

(Shorter bond length)

Bond Order

slide70

H2

He2

Bond Order: Examples

Bond order = (2-0)/2 = 1

Single bond

Stable molecule (436 kJ/mol bond)

Bond order = (2-2)/2 = 0

No bond!

Unstable molecule (0 kJ/mol bond)

slide71

He2+

Bond order = (2-1)/2 = 1/2

Half of a single bond

Can be made, but its not very stable (250 kJ/mol bond)

Fractional bond orders are okay!

H2+

Bond order = (1-0)/2 = 1/2

Half of a single bond

Can be made, but its not very stable (255 kJ/mol bond)

forming bonds
Forming Bonds
  • A s bond can be formed a number of ways:
    • s, s overlap
    • s, p overlap
    • p, p overlap

Only orbitals of the same phase (+, +) can form bonds

anti bonding orbitals
Anti-bonding Orbitals
  • For every bonding orbital we form, we also form an anti-bonding orbital:
mo theory in bonding
MO Theory in Bonding
  • Homonuclear atoms (H2, O2, F2, N2)

H2

(Only 1s orbitals available for bonding)

slide75

Covalent Bonding in Homonuclear Diatomics

  • Atomic orbitals must overlap in space in order to participate in molecular orbitals
  • Covalent bonding is dominated by the valence orbitals (only valence orbitals are shown in the MO diagrams)
slide76

Covalent Bonding in Homonuclear Diatomics

Region of shared e- density

+

+

slide77

Valence configurations of the 2nd row atoms:

Li Be B C N O F

2s1 2s2 2s22p1 2s22p2 2s22p3 2s22p4 2s22p5

So far we have focused on bonding involving the s orbitals.

What happens when we have to consider the p orbitals?

slide78

For diatomic molecules containing atoms with valence electrons in the p orbitals, we must consider three possible bonding interactions:

= nucleus

p-type

p-type

s-type

slide79

(–)

destructive

mixing

(+)

constructive

mixing

slide80

Major limitations of the LE model:

..

..

..

..

Example: O2

- Lewis dot structure O=O

- All electrons are paired Contradicts experiment!

2) Doesn’t easily deal with unpaired electrons

(incorrectly predicts physical properties in some cases)

Experiments show O2 is paramagnetic

slide81

A quick note on magnetism…

Paramagnetic

The molecule contains unpairedelectrons and is attracted to (has a positive susceptibility to) an applied magnetic field

Diamagnetic

The molecule contains only pairedelectrons and is not attracted to (has a negative susceptibility to) an applied magnetic field

slide82

Example: the O2 Diatomic

Both have degenerate orbitals

Oxygen atom has a 2s22p4 valence configuration

M.O.

O2

O atom

O atom

Bond Order = (8-4)/2 = 2

O2 is stable

(498 kJ/molbond strength)

(s2s)2(s2s*)2(s2p)2(p2p)4(p2p*)2

slide83

A prediction from the M.O. diagram of O2

..

..

..

..

O=O

The Lewis dot structure predicts O2 should be diamagnetic-all electrons are paired.

The unpaired electrons predicted by the M.O. diagram should behave as small magnets-

O2 should be magnetic!

slide84

N2 Video

O2 Video

slide85

What have we learned so far?

1. Molecular orbitals (MO) are linear combinations of atomic orbitals

2. Both s and p atomic orbitals can be mixed to form MOs

3. Molecular orbitals are bonding and anti-bonding

4. Bonding and anti-bonding MOs lead to the definition of the bond order

5. Bond orderis related to the bond strength (bond dissociation energy)

mo diagram for h 2 vs n 2

H2

MO Diagram for H2 vs. N2

N2

p*

2p*

2p

2p

2s*

Atomic orbital overlap sometimes forms both  and  bonds. Examples: N2, O2, F2

2s

slide87

M.O. Diagram for N2

s*(2p)

p*

p*

-1,155

s(2p)

-1,240

-1,240

Electron energy (kJ mol-1)

p

p

-1,479

s*(2s)

Valence

Valence

-2,965

Core

Core

s(2s)

-37,871

-37,875

1s(N) + 1s(N)

1s(N) –1s(N)

slide88

A Complication…

M.O. Diagram for O2

(similar for F2 and Xe2)

M.O. Diagram for B2

(similar for C2 and N2)

O

O2

O

slide89

A Complication…

M.O. Diagram for O2

(similar for F2 and Ne2)

M.O. Diagram for B2

(similar for C2 and N2)

No s-p mixing

s-p mixing

slide90

Why does s-p mixing occur?

Electron repulsion!!

s2s and s2p both have significant e- probability between the nuclei, so e- in s2s will repel e- in s2p

Effect will decrease as you move across the Periodic Table

 increased nuclear charge pulls the s2s e- closer, making the s2s orbital smaller and decreasing the s2s and s2pinteraction

molecular orbitals of x 2 molecules
Molecular Orbitals of X2 Molecules

sp orbital mixing (a little hybridization)

    • lowers the energy of the 2s orbitals and
    • raises the energy of the 2p orbitals.
  • As a result, E(2p) > E( 2p) for B2, C2, and N2.
  • As one moves right in Row 2, 2s and 2p get further apart in energy, decreasing s–p mixing  E(2p) < E(2p) for O2, F2, and Ne2. See text pages 680-681.
  • Note that s–p mixing does not affect bondorder or magnetism in the commondiatomics (N2, O2, and F2). Hence it is not of much practical importance.
slide92

No s-p mixing

s-p mixing

slide93

s-p mixing only occurs when the s and p atomic orbitals are close in energy ( 1/2 filled 2p orbitals)

When does s-p mixing occur?

B, C, and N all have  1/2 filled 2p orbitals

O, F, and Xe all have > 1/2 filled 2p orbitals

  • If 2 electrons are forced to be in the same orbital, their energies go up.
  • Electrons repel each other because they are negatively charged.
  • Having > 1/2 filled 2p orbitals raises the energies of these orbitals due to e- - e- repulsion
slide95

Sample Problem:

Using MO Theory to Explain Bond Properties

Problem: Consider the following data for these homonuclear diatomic species:

N2N2+ O2O2+

Bond energy (kJ/mol) 945 841 498 623

Bond length (pm) 110 112 121 112

No. of valence electrons 10 9 12 11

Removing an electron from N2decreases the bond energy of the resulting ion, whereas removing an electron from O2increasesthe bond energy of the resulting ion. Explain these facts using M.O. diagrams.

slide96

Plan: We first draw the MO energy levels for the four species, recalling that they differ for N2 and O2. Then we determine the bond orders and compare them with the data: bond order is related directly to bond energy and inversely to bond length.

Sample Problem:

Using MO Theory to Explain Bond Properties

Problem: Consider the following data for these homonuclear diatomic species:

N2N2+ O2O2+

Bond energy (kJ/mol) 945 841 498 623

Bond length (pm) 110 112 121 112

No. of valence electrons 10 9 12 11

slide97

N2+

O2

O2+

p*

2p*

2p

2p

2s*

2s

Sample Problem - Continued

Solution: The MO energy levels are:

N2

p*

2p*

2p

2p

2s*

2s

Bond Orders:

(8-2)/2 = 3

(7-2)/2 = 2.5

(8-4)/2 = 2

(8-3)/2 = 2.5

slide98

Sample Problem:

Using MO Theory to Explain Bond Properties

Problem: Consider the following data for these homonuclear diatomic species:

N2N2+ O2O2+

Bond energy (kJ/mol) 945 841 498 623

Bond length (pm) 110 112 121 112

No. of valence electrons 10 9 12 11

Bond Order 3 2.5 2 2.5

slide99

What have we learned so far?

1. Molecular orbitals (MO) explain the properties of valenceelectrons in molecules (Example: O2)

2. s and patomic orbitals can be mixed to form s, s*, p, and p*molecular orbitals

3. Electrons in p or p* molecular orbitals can have the same energies: Degenerate orbitals

4. The ordering of s2p and p2p molecular orbitals depends on the electron occupancy: s-p mixing

slide100

Bonding in Diatomic Molecules

Covalent

Ionic

Covalent

Ionic

slide101

Heteronuclear:

HF

Homonuclear:

H2

Electronegativity

Nonpolar covalent bond

(450 kJ/mol bond)

Polar covalent bond

(565 kJ/mol bond)

slide102

Electrons are not equally shared

in heteronuclear bonds

HF

BecauseF (EN = 4.0)is more electronegative than H (EN = 2.2), the electrons move closer to F.

This gives rise to a polar bond:

Electronegativity

H F

Figure 14.45

slide103

M.O.s of a Polar Covalent Bond: HF

sAntibonding (s*)

Mostly H(1s)

HF

HF

This approach simplifies model and only considers electrons involved in bond.

s Bonding

Mostly F(2p)

slide104

MOs OF XY MOLECULES

  • Equal or unequal e sharing between 2 atoms is reflected in the composition of the MOs:
  • When 2 atoms X and Y have the same electronegativity (purely covalent bond), their overlapping AOs have the same energy, and the bonding and anti-bonding MOs are each halfX and halfY AO. All electrons spend equal time near X and Y. Examples: N2, O2, F2.
  • If EN(Y) > EN(X) (polar covalent X+Y), the Y AO has lower energy than the X AO. The bonding MO is more like the Y AO and the anti-bondingMO more like the X AO. Bonding e spend more time near Y than X; vice versa for anti-bonding e. Example: CO.
slide105

MOs OF XY MOLECULES

Electronegativity

CAtom (4e–) Cδ+Oδ– (10e–) O Atom (6e–)

  • CO Bond Order = 3.0(same as N2).
  • CO Bond Energy = 1,076 kJ/mol (N2 = 945 kJ/mol).
  • Isoelectronic to CO and N2:CN–, NO+.
  • NO has1e–in*  bond order = 2.5; this e– is more on N than O; NO  NO+easy…
bonding in no

.

..

..

..

.

..

..

..

N=O

N=O

Bonding in NO
  • Two possible Lewis dot structures for NO
  • The simplest structure minimizes formal charges and places the lone (unpaired) electron on the nitrogen.
  • The Lewis structure predicts a bond order of 2, but experimental evidence suggests a bond order between 2 and 3.
  • How does MO theory help us understand bonding in NO?

-1

+1

slide107

When the electronegativities of the 2 atoms are more similar, the bonding becomes less polar.

..

..

.

..

N=O

2p

2p

Electronegativity

EN(N) = 3.0

EN(O) = 3.4

2s

2s

N

NO

O

Bond order = 2.5, unpaired electron is in a N-like orbital

slide108

NO is easily oxidized to form NO+. Why? What changes can we predict in the bonding and magnetism of the molecule?

NO+

NO

oxidation

Bond Order = (8-2)/2 = 3

Diamagnetic

Bond Order = (8-3)/2 = 2.5

Paramagnetic

slide109

s2p

-1307

M.O. diagram for NO

p2p *

p2p * (empty)

-597

p2p

p2p

-1444

-1374

s2s*

-1835

s2s

-3320

key points of mo theory heteronuclear molecules
Key Points of MO Theory – Heteronuclear Molecules
  • The more electronegative atom has orbitals lower in energy than the more positive atom.
  • Electrons in bonding orbitals are closer to the more electronegative atom, anti-bonding electrons are closer to the more positive atom.
  • For most diatomic molecules, s-p mixing changes the orbital energy levels, but since these orbitals are almost always fully occupied, their order is less important to us.
slide111

Combining the Localized Electron and Molecular Orbital Models (into a convenient working model)

Figure 14.47

Only the p bonding changes between these resonance structures - The M.O. model describes this p bonding more effectively.

slide113

Another example: Benzene

p atomic orbitalsp molecular orbital

s bonding:

p bonding:

mo theory expectations
MO Theory Expectations
  • You should be able to:
    • predict which atomic orbitals are higher or lower in energy (based on electronegativity differences).
    • correctly fill a molecular orbital diagram.
    • correctly calculate bond order.
    • predict molecular magnetic properties based on orbital occupation.
    • understand how molecular properties change upon ionization (oxidation or reduction) of molecules.