what do molecules look like? - seattle central...

What Do Molecules Look Like?

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

The Lewis Dot Structure approach provides

some insight into molecular structure in terms

of bonding, but what about 3D geometry?

Valence-Shell Electron-Pair Repulsion

(VSEPR). 3D structure is determined by

minimizing repulsion of electron pairs.

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

5 electron pairs 6 electron pairs

Period 1, 2

Period 3 & beyond

Electron pairs (both bonding and lone) are

distributed around a central atom such that

electron-electron repulsions are minimized.

Arranging Electron Pairs

• Example: CH4 (bonding pairs only)

• Must consider both bonding and lone pairs when

minimizing repulsion.

Lewis Structure VSEPR Structure

H C

H

H

H

Arranging Electron Pairs (cont.)

Example: NH3 (both bonding and lone pairs).

Lewis Structure VSEPR Structure

Note:

“electron pair geometry” vs.

“molecular shape”

H N

H

H

VSEPR Structure Guidelines The previous examples illustrate the strategy for applying

VSEPR to predict molecular structure:

1. Construct the Lewis Dot Structure

2. Arrange bonding/lone electron pairs in space such that

repulsions are minimized (electron pair geometry).

3. 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

Linear (AX2): angle

between bonds is 180°

Example: BeF2

180°

Experiments show that molecules with

multiple bonds can also be linear.

Multiple bonds are treated as a

single effective electron group. F Be F

F Be F

More than one central atom?

Determine shape around each.

Be F F


Trigonal Planar (AX3):

angle between bonds

is 120°

Example: BF3

Multiple bond is

treated as a single

effective electron

group.

F B F

F

B

F F

F

120°

VSEPR: 4 electron pairs (cont.)

Tetrahedral (AX4): angle

between bonds is ~109.5°

Example: CH4

109.5°

H C

H

H

H

tetrahedral e- pair geometry AND tetrahedral molecular shape

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 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

H N

H

H


Classic example of tetrahedral angle

shift from 109.5° is water (AX2E2):

“bent”

104.5o

tetrahedral e- pair geometry; bent molecular shape


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

(AX2E2):

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”

H H

H

H

C N O F

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

Molecular vs. Electron-Pair Geometry

What is the electron-pair geometry and the

molecular shape for HCFS?

a) trigonal planar, bent

b) trigonal planar, trigonal planar

c) tetrahedral, trigonal planar

d) tetrahedral, tetrahedral

C

S

H

F

VSEPR: Beyond the Octet Systems with expanded valence shells will have

five or six electron pairs around a central atom.

P

Cl

Cl

Cl

Cl

Cl

F

F

F

F

F F

S

90°

120° S

F

F

F

F

F

F 90°

90°


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

S

F F

F F

S

F F

F F

S

F

F F

F ??

lone-pair / bond-pair: two at 90o, two at 120o

Repulsive forces (strongest to weakest):

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

bond-pair / bond-pair:

three at 90o

four at 90o, one at 120o three at 90o, three at 120o

Compare e– pair angles

VSEPR: 5 electron pairs The optimum structure maximizes the angular

separation of the lone pairs.

I3- (AX2E3):

AX3E2

AX4E

AX2E3

5-electron-pair geometries

our

previous

example

Which of these is the more likely structure?

Square Planar


See-saw

AX5E

AX4E2

6-electron-pair geometries

our

previous

example

Molecular Dipole Moments

1. Draw Lewis structures to

determine 3D arrangement

of atoms.

Shortcut: completely

symmetric molecules will not

have a dipole regardless of

the polarity of the bonds.

2. If one “side” of the molecule has

more EN atoms than the other,

the molecule has a net dipole.

We can use VSEPR to

determine the polarity of a

whole molecule.

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.)

symmetric symmetric asymmetric

F

Cl

F Cl

Molecular Dipole Example

• Write the Lewis dot and VESPR structures for

CF2Cl2. Does it have a dipole moment?

C

F

FCl

Cl

32 e-

Tetrahedral

Advanced VSEPR Application

Molecules with more than one central atom…

methanol (CH3OH)

H C

H

O

H

H

tetrahedral e- pairs

tetrahedral shape

tetrahedral e- pairs

bent shape

The VSEPR Table

2 AX2 BeF2 linear linear

3 AX3 BF3 trigonal planar trigonal planar

AX2E O3 trigonal planar bent

4 AX4 CH4 tetrahedral tetrahedral

AX3E NH3 tetrahedral pyramidal

AX2E2 H2O tetrahedral bent

# e- pairs e- Geom. Molec. Geom.

The VSEPR Table

5 AX5 PF5 trigonal

bipyramidal

trigonal

bipyramidal

AX4E SF4 trigonal

bipyramidal

see saw

AX3E2 ClF3 trigonal

bipyramidal

T-shaped

AX2E3 I3- trigonal

bipyramidal

linear

6 AX6 SF6 octahedral octahedral

AX4E2 XeF4 octahedral square planar

# e- pairs e- Geom. Molec. Geom.

What is the expected shape of ICl2+?

A. linear

B. bent

C. tetrahedral

D. square planar

AX2E2 ICl Cl

+

20 e-

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.

Valence Bond Theory





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.

Valence Bond Theory





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

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.

Example: Methane

• 4 equivalent C-H covalent bonds

• VSEPR predicts a tetrahedral geometry

The Valence Orbitals of a Carbon Atom

Carbon: 2s22p2

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

Hybridization: Mixing of Atomic Orbitals to

form New Orbitals for Bonding

+ –

+ –

–

+

+

– + –

+ + –

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)]

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!!



N H

H

H

108.1 º

Ammonia (NH3)

VSEPR: AB3E

tetrahedral

sp3 hybridized



Water (H2O)

VSEPR: AB2E2

tetrahedral

sp3 hybridized

105.6 º

s bonding and p bonding

• 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

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

Bonding in ethylene (C2H4)

Problem: Describe the hybridization and bonding of

the carbon orbitals in Carbon Dioxide (CO2)

VSEPR: AB2

linear

sp hybridized orbitals for s bonding

Bonding in Carbon Dioxide (CO2)

Atoms of the same kind can have different hybridizations

C2: AB4

sp3

sp

C1: AB2

2s2 2px2py sp sp p p

N: ABE

2s2 2px2py2pz sp sp p p

lone pair

s

Bonds

s p p

C1 C2 N

H

H

H

Acetonitrile (important solvent and industrial chemical)

CH3 C N :

What have we learned so far?

• Molecular orbitals are combinations of

atomic orbitals

• Atomic orbitals are “hybridized” to satisfy

bonding in molecules

• Hybridization follows simple rules that can

be deduced from the number of chemical

bonds in the molecule and the VSEPR model

for electron pair geometry

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

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)

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

• 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?

• 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



• Step 2: What is the electron pair geometry and

molecular shape?

AX4

AX4

AX3

AXE2

Tetrahedral

Tetrahedral

Trigonal Planar

Trigonal Planar



• Step 3: Use the molecular shape to determine the

hybridization.

AX4

AX4

AX3

AXE2

Tetrahedral

Tetrahedral

Trigonal Planar

Trigonal Planar

sp3

sp3 sp2

sp2

The Localized Electron Model is very powerful

for explaining geometries and basic features of

bonding in molecules, but it is just a model.

Major limitations of the LE 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

Example: O2

Lewis dot structure O=O

All electrons are paired Contradicts experiment!

..

..

..

..

The Molecular Orbital Model

Basic premise: When atomic orbitals interact

to form a bond, the result is the formation of

new molecular 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.

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?

Molecular Orbital Theory

+ -

+

Molecular Orbitals are simply

Linear Combinations of Atomic Orbitals

s bonding

s anti-bonding (s*) Example: H2

Next Question: Why does this work?

Molecular Orbitals have phases (+ or -)

Constructive and Destructive

Interference

Constructive interference between

two overlapping orbitals leads to a

bonding orbital.

Destructive interference

between two orbitals of opposite

sign leads to an anti-bonding

orbital.

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

= = nucleus +

• We can depict the relative energies of molecular

orbitals with a molecular orbital diagram:

MO Diagrams

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

s* M.O. is raised in energy

s M.O. is lowered in energy

H atom: (1s)1 electron configuration

H2 molecule: (s1s)2 electron configuration

Same as previous description of bonding

s

s*

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:

Yes No No

Filling Molecular Orbitals with Electrons

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

principle)

H2

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

opposite spins (Pauli exclusion principle)

He2



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

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

Bond Order

Bond Order =

# bonding #anti-bonding

electrons electrons

2

The bond order is an indication of bond strength:

Greater bond order Greater bond strength

(Shorter bond length)

Bond Order: Examples

H2

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

Single bond

Stable molecule (436 kJ/mol

bond)

He2

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

No bond!

Unstable molecule (0 kJ/mol

bond)

He2+

H2+

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

Half of a single bond

Can be made, but its not very

stable (250 kJ/mol bond)

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

Half of a single bond

Can be made, but its not very

stable (255 kJ/mol bond)

Fractional bond orders are okay!

• A s bond can be formed a number of ways:

– s, s overlap

– s, p overlap

– p, p overlap

Forming Bonds

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

• For every bonding orbital we form, we also form an anti-

bonding orbital:

Anti-bonding Orbitals

MO Theory in Bonding

• Homonuclear atoms (H2, O2, F2, N2)

H2

(Only 1s

orbitals

available for

bonding)

• 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)

Covalent Bonding in Homonuclear

Diatomics

Covalent Bonding in Homonuclear

Diatomics

Region of

shared e-

density

+

+

–

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?

For diatomic molecules containing atoms with valence electrons in

the p orbitals, we must consider three possible bonding

interactions:

s-type p-type p-type

= nucleus

(+)

constructive

mixing

(–)

destructive

mixing

Major limitations of the LE model:

2) Doesn’t easily deal with unpaired electrons

(incorrectly predicts physical properties in some cases)

Example: O2

- Lewis dot structure O=O

- All electrons are paired Contradicts experiment!

..

..

..

..

Experiments show O2 is paramagnetic

A quick note on magnetism…

Paramagnetic

The molecule contains unpaired electrons and is

attracted to (has a positive susceptibility to) an applied

magnetic field

Diamagnetic

The molecule contains only paired electrons and is

not attracted to (has a negative susceptibility to) an

applied magnetic field

____ s2p*

___ ___ p2p*

___ ___ ___2p ___ ___ ___ 2p

___ ___ p2p

Energy ____ s2p

____ s2s*

___ 2s ___ 2s

____ s2s

Example: the O2 Diatomic

Bond Order = (8-4)/2 = 2 O2 is stable

(498 kJ/mol bond strength)

Both have degenerate orbitals

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

*)2

Oxygen atom has a 2s22p4 valence

configuration

O atom O atom

M.O.

O2

A prediction from the M.O. diagram of O2

..

..

..

.. O=O

The unpaired electrons predicted by

the M.O. diagram should behave as

small magnets-

O2 should be magnetic!

The Lewis dot structure predicts O2

should be diamagnetic-all electrons

are paired.

N2 Video

O2 Video


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 order is related to the bond strength (bond dissociation energy)

MO Diagram for H2 vs. N2

H2

sp*

p2p*

s2p

p2p

s2s*

s2s

N2

Atomic orbital overlap sometimes forms both s

and p bonds. Examples: N2, O2, F2

1s(N) + 1s(N) 1s(N) –1s(N)

-37,875 -37,871

-2,965

Ele

ctro

n e

nerg

y (

kJ

mo

l-1)

Core Core

Valence Valence

s(2s)

s*(2s)

-1,479

-1,240 -1,240

p p

s(2p)

-1,155

p* p*

s*(2p)

M.O. Diagram for N2

A Complication…

M.O. Diagram for O2

(similar for F2 and Xe2)

M.O. Diagram for B2

(similar for C2 and N2)

O O O2

A Complication…

M.O. Diagram for O2

(similar for F2 and Ne2)

M.O. Diagram for B2

(similar for C2 and N2)

s-p mixing

No s-p mixing

Electron repulsion!!

Why does s-p mixing occur?

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 s2p interaction

Molecular Orbitals of X2

Molecules

sp orbital mixing (a little hybridization)

• lowers the energy of the s2s orbitals and

• raises the energy of the s2p orbitals.

• As a result, E(s2p) > E(p 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(s2p) <

E(p2p) for O2, F2, and Ne2. See text pages 680-681.

• Note that s–p mixing does not affect bond order or

magnetism in the common diatomics (N2, O2, and

F2). Hence it is not of much practical importance.

s-p mixing No s-p mixing

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

When does s-p mixing occur?

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

s-p mixing only occurs when the s and p atomic orbitals are close in energy (

1/2 filled 2p orbitals)

Relating the M.O. Diagrams to Physical

Properties

Sample Problem: Using MO Theory to Explain Bond Properties

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

N2 N2+ O2 O2

+

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 N2 decreases the bond energy of the resulting ion,

whereas removing an electron from O2 increases the bond energy of the resulting

ion. Explain these facts using M.O. diagrams.



N2 N2+ O2 O2

+


Bond length (pm) 110 112 121 112


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 - Continued

Solution: The MO energy levels are:

N2

sp*

p2p*

N2+

s2p

p2p

s2s*

s2s

O2 O2+

sp*

s2s

s2s*

p2p

p2p*

s2p

Bond Orders:

(8-2)/2 = 3 (7-2)/2 = 2.5 (8-4)/2 = 2 (8-3)/2 = 2.5



N2 N2+ O2 O2

+


Bond length (pm) 110 112 121 112


Bond Order 3 2.5 2 2.5


1. Molecular orbitals (MO) explain the properties of valence electrons in molecules

(Example: O2)

2. s and p atomic 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

Bonding in Diatomic Molecules

Covalent

Ionic

Ionic

Covalent

Homonuclear:

H2

Nonpolar covalent bond

(450 kJ/mol bond)

Heteronuclear:

HF

Polar covalent bond

(565 kJ/mol bond)

Electronegativity

Ele

ctro

neg

ativ

ity

Because F (EN = 4.0) is more

electronegative than H (EN = 2.2), the

electrons move closer to F.

This gives rise to a polar bond:

H F

Figure 14.45

Electrons are not equally shared

in heteronuclear bonds

HF

s Antibonding (s*)

Mostly H(1s)

s Bonding

Mostly F(2p)

H F

H F

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

This approach simplifies model and only

considers electrons involved in bond.

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 half X and half Y 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-bonding MO

more like the X AO. Bonding e spend more time near Y than X; vice versa for

anti-bonding e. Example: CO.

MOs OF XY MOLECULES ____ s*

___ ___ p*

___ ___ ___2p ___ ___ ___ 2p

↑ ____ s

Energy ___ ___ p

____ s*

___ 2s ___ 2s

____ s

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 has 1e– in p* bond order = 2.5; this e– is more on N than O; NO

NO+ easy…

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

___ ___ ___ 2p

___ 2s

Ele

ctron

eg

ativ

ity

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?

.

..

..

.. N=O

.

..

..

.. N=O

+1 -1

When the electronegativities of the 2 atoms are

more similar, the bonding becomes less polar.

EN(N) = 3.0

EN(O) = 3.4

.

..

..

.. N=O

Ele

ctron

eg

ativ

ity

2s 2s

2p 2p

NO N O

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

NO+

oxidation

NO

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

and magnetism of the molecule?

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

Paramagnetic

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

Diamagnetic

s2s

s2s*

p2p p2p

p2p *

-3320

-1835

-1444 -1374

s2p -1307

-597 p2p * (empty)

M.O. diagram for

NO

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.

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.

Figure 14.51

Atomic Orbitals Molecular Orbitals

Another example:

Benzene

p bonding:

s bonding:

p atomic orbitals p molecular orbital

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.

what do molecules look like? - seattle central...

Documents