Molecular Geometryand

Bonding Theories

The properties of a molecule depend on its shape andand the nature of its bonds. In this unit, we willdiscuss three models.

(1) a model for the geometry of molecules -- valence-shell electron-pair repulsion (VSEPR) theory

(2) a model about WHY molecules form bonds and WHY they have the shape they do

-- valence-bond theory

(3) a model of chemical bonding that deals with the electronic structure of molecules

-- molecular orbital (MO) theory

CO2

bond angles: the angles made by the lines joining the nuclei of a molecule’s atoms

CH4 CH2Ocarbon dioxide methane formaldehyde

180o 120o109.5o

VSEPR electron domain: a region in which at least two

electrons are found -- they repel each other because… they are all (–)

bonding domain: 2-to-6 e– that are shared by two atoms; they form a… covalent bond

nonbonding domain: 2 e– that are located on a single atom; also called a… lone pair

For ammonia, there arethree bonding domains andone nonbonding domain. N–HH–

H

..

NHH

H

..

4 e – domains

NH3

Domains arrange themselves soas to minimize their repulsions.

The electron-domain geometry is oneof five basic arrangements of domains.

-- it depends only on the total # of e– domains, NOT the kind of each domain

The molecular geometry describes theorientation of the atoms in space.

-- it depends on how many of each kind of e– domain

NHH

H

..

..

Total # of Domains

Electron-Domain Geometry Possible Molecular Geometries

2

3

4

5

6

linear linear (CO2)

trigonalplanar

trigonal planar (BF3), bent (NO2)

tetrahedral tetrahedral (CH4),trigonal pyramidal (NH3), bent (H2O)

trigonalbipyramidal

trig. bipyramidal (PCl5), linear (XeF2)seesaw (SF4), T-shaped (ClF3)

octahedral octahedral (SF6), sq. pyr. (BrF5),square planar (XeF4)

“atoms – axial”

To find the electron-domain geometry (EDG) and/ormolecular geometry (MG), draw the Lewis structure.Multiple bonds count as a single domain.

Predict the EDG and MG of each of the following.

SnCl3–

26 e– [ ]Sn–ClCl–....

..

....

....

....Cl

–..EDG: tetrahedral

MG: trig. pyramidal

O3

18 e–

EDG: trig. planarMG: bent

..O–OO= .... ..

....

SeCl2

20 e–

EDG: tetrahedralMG: bent

Cl–Se–Cl....

..

...... ....

..OO–O

=.. ...... ..

CO32–

24 e– [ ]C=OO–....

..

.... ..

.... O

2–EDG: trig. planar

MG: trig. planar

SF4

34 e–

EDG: trig. bipyr.MG: seesaw

..

S–FF–....

..

....

....

....

F

F.. ..

..

IF5

42 e– I

.. .... F–F.. .

...–F.. ..

..F–......

F.. ....

.. EDG: octahedralMG: sq. pyramidal

ClF3

28 e–

Cl–FF–....

..

....

....

.... F

..EDG: trig. bipyr.

MG: T-shaped

..

(res

.)

ICl4–

36 e–

EDG: octahedralMG: sq. planar[ ]..

I–ClCl–....

..

....

....

....

Cl

Cl

.. ..

.. ..–

For molecules with more than one central atom,simply apply the VSEPR model to each part.

Predict the EDG and MG around the three interioratoms of ethanoic (acetic) acid.

H–C–C–O–H..H

..H O....

CH3COOH PORTION EDG MG

tetra. tetra.–CH3

trig. plan. trig. plan.

tetra. bent

C=O–– ..

..–OH

..

..

Nonbonding domains are attracted to only one nucleus;therefore, they are more spread out than are bondingdomains. The effect is that nonbonding domains(i.e., “lone pairs”) compress bond angles. Domains formultiple bonds have a similar effect.

e.g.,

CH4 NH3 H2O

NHH

H

..C

HH

H

H OH

..H

..

..ClO=C

Cl...... ..

..

....

the ideal bond angle for the tetrahedralEDG is 109.5o

111.4o

EDG = trig. plan.ideal = 120o

COCl2

124.3o

124.3o

109.5o107.0o 104.5o

Polarity of Molecules

A molecule’s polarity is determined by its overalldipole, which is the vector sum of the dipoles ofeach of the molecule’s bonds.

Consider CO2 v. H2S...

O=C=O....

.... S

H

..H

..bond dipoles

overall dipole =

bond dipoles

overall dipole =(NONPOLAR) (POLAR)

zero

Boron can be extracted bythe electrolysis of moltenboron trichloride. Boron isan essential nutrient forplants, and is also a primarycomponent of control rodsin nuclear reactors.

Classify as polar or nonpolar:

PCl3

BCl3

26 e–P–ClCl–

..

......

..

....

....Cl

..polar

..

B–ClCl–.... .... ..

..Cl .... nonpolar

24 e–

Valence-Bond Theory: merges Lewis structures w/the idea of atomic orbitals (2s, 3p, etc.)

V-B theory says… covalent bonding occurs whenvalence orbitals of adjacentatoms overlap; then, two e–sof opposite spin (one fromeach atom) combine to forma bond

Lewis theory says… covalent bonding occurs whenatoms share electrons

V-B theory is like Velcro: “No overlap, no bond.”

Consider H2, Cl2, and HCl...

H

Cl [Ne]

H2 Cl2

HCl

(unpaired e– is in 1s orb.)

(unpaired e– is in 3p orb.)

= orbital overlap region (responsible for bond)

There is always an optimum distance between twobonded nuclei. At this optimum distance, attractive (+/–) and repulsive (+/+ and –/–) forcesbalance.

H–H distance

Energy

0

optimum dist.(min. PE)

H2 molecule

Hybridized Orbitals

V-B theory can’t explain some observations aboutmolecular compounds without the concept of hybridized orbitals.

i.e., covalent (NM/NM) compounds

In…

BeF2

(LS) …the central atomSHOULD be…

…but is ACTUALLYhybridized to be…

…the resultthen being…

2 hyb. sp orbs.2 unhyb. 2p orbs.2s 2p

BF3 3 hyb. sp2 orbs.1 unhyb. 2p orb.2s 2p

2psp

sp2 2p

CH4 4 hyb. sp3 orbs.2s 2p sp3

H2O 4 hyb. sp3 orbs.2s 2p sp3

PF53s 3p sp3d3d 3d

5 hyb. sp3d orbs.4 unhyb. 3d orbs.

XeF2sp3d5s 5p 5d 5d

5 hyb. sp3d orbs.4 unhyb. 5d orbs.

XeF4sp3d25s 5p 5d 5d

6 hyb. sp3d2 orbs.3 unhyb. 5d orbs.

: BeF :

::F

::

B :

::F

:F

::

:F:

:

CH HH

H

OHH

:

:

P :

::F

:F

::

:F:

:

:: :F

:

::F

: XeF :

::F

::

: ::

: XeF :

::F

::

::F: :

F::

::

KEY: EDG hybridization of central atom

linear sptrig. planar sp2

tetrahedral sp3

trig. bipyr. sp3doctahedral sp3d2

sp hybridization

sp hybridization occurs when one s and one of the three porbitals hybridize. It results in the creation of two sp hybridorbitals (orange). The other two p orbitals (purple) areunhybridized; they retain the “figure-8” or “dumbbell” shape.

Multiple Bonds

s (sigma) bonds are bonds in which the e– density isalong the internuclear axis.

-- These are the single bonds we have considered up to this point. -- e.g., s-s, s-p, or p-p overlap, and also p-sp hybrid overlap

Multiple bonds result from the overlap of two p orbitals(one from each atom) that are oriented perpendicularlyto the internuclear axis. These are p (pi) bonds.

sigma (s) = single

pi, multiple, unhybridized

NHH

H

..

The bondsin ammoniaare s bonds.

p bonds are generally weaker than s bonds becausep bonds have less overlap.

C C

H

H

H

H

For ethene (C2H4)…

C=C–––

– H

H

H

H

For each carbon atom, thereare 3 sp2 hybridizedorbitals; these form s

bonds (- - - -) with C or H.

Where unhybridizedorbitals overlap,

a p bond ( ) is formed.

Single bondsare s bonds.

Double bonds consist ofone s and one p.

Triple bonds consist ofone s and two p.

e.g., C2H2

e.g., C2H4

e.g., C2H6

Experiments indicate that allof C2H4’s atoms lie in thesame plane. This suggeststhat p bonds introducerigidity (i.e., a reluctance to rotate) into molecules.

-- p bonding does NOT occur with sp3 hybridization, only sp and sp2

-- p bonding is more prevalent with small molecules (e.g., C, N, O) (Big atoms don’t allow enough

p-orb. overlap for p bonds to form.)

resonance structures.

Delocalized p Bonding

Localized p bonds are between…

-- e.g., C2H2, C2H4, N2

Delocalized p bonds are“smeared out” andshared among…

-- These are common for molecules with…

-- The electrons involved in these bonds are delocalized electrons.

The nitrate ion (NO3–) has

delocalized electrons.

[ ]N=OO–....

....

..

..

....O

–

> two atoms.

two atoms only.

Consider benzene, C6H6. -- Each carbon atom is ___ hybridized. -- This leaves... a 2p

orbital on each Cthat is oriented to plane of m’cule.

C

C

C

C

C C

H

H

H

H

H

H

-- 6 e– shared equally by 6 C atoms

sp2

Which of the following exhibit delocalized bonding?

H3C–C–CH3

O=

SO42–

SO2

sulfate ion

sulfur dioxide18 e–

32 e–

O–S=O...... ..

....

“YEP.”

HH–C–C–C–H

O=

.. ..

––H –H

H

–

“NOPE.”

“NOPE.”[ ]S–OO–....

..

....

....

....

O

O

.. ....2–

24 e–

propanone

(res.)

Molecules respond to themany wavelengths of light.The wavelengths that areabsorbed and then re-emitteddetermine an object’s color, while the wavelengths that areNOT re-emitted raise thetemperature of the object.

Molecular Orbitals-- The VSEPR and valence-bond theories don’t explain the excited states of molecules, which come into play when molecules absorb and emit light.

-- This is one thing that the molecular orbital (MO) theory attempts to explain.

molecular orbitals: wave functions that describe the locations of electrons in molecules

-- these are analogous to atomic orbitals in atoms (e.g., 2s, 2p, 3s, 3d, etc.), but MOs are possible locations of electrons in molecules (not atoms)

-- MOs, like atomic orbitals, can hold a maximum of two e– with opposite spins

-- but MOs are for entire molecules

MO theory is more powerful than valence-bond theory;its main drawback is that it isn’t easy to visualize.

Hydrogen (H2) The overlap of two

atomicorbitals produces two

MOs.

1s 1s

H atomic orbitals

+

H2 molecular orbitals

s1s

(bonding MO)

(antibonding MO)s*1s

-- The lower-energy bonding molecular orbital concentrates e– density between nuclei.-- For the higher-energy antibonding molecular orbital, the e– density is concentrated outside the nuclei.-- Both of these are s molecular orbitals.

Energy-level diagram (molecular orbital diagram)

1s 1s

s1s

s*1s

H atom H2 m’cule

H atom

1s 1s

H atomic orbitals

+

H2 molecular orbitals

s1s

(bonding MO)

(antibonding MO)s*1s

Consider the energy-level diagram for thehypothetical He2 molecule…

1s 1s

s1s

s*1s

He atom He2 m’cule

He atom

2 bonding e–, 2 antibonding e–

No energy benefit to bonding.

He2 molecule won’t form.

no bond

bond order = ½ (# of bonding e– – # of antibonding e–)

-- the higher the bond order, the greater the bond stability

-- a bond order of... 0 = single bond1 =double bond2 =triple bond3 =

-- MO theory allows for fractional bond orders as well.

What is the bond order of H2+?

1 e–

total1 bonding e–,zero antibonding e– BO = ½ (1–0) = ½

James Bond7 =007 =

Second-Row Diatomic Molecules

1. # of MOs = # of combined atomic orbitals

2. Atomic orbitals combine most effectively with other atomic orbitals of similar energy.

3. As atomic orbital overlap increases, bonding MO is lowered in energy, and the antibonding MO is raised in energy.

4. Both the Pauli exclusion principle and Hund’s rule apply to MOs.

Li3

6.941 Be

4

9.012 B

5

10.811 C

6

12.011 N

7

14.007 O

8

15.999 F

9

18.998 Ne

10

20.180

Li2

Li Li s1s

s*1s

1s 1s

2s 2s

s*2s

s2s

Use MO theory topredict whether Li2and/or Be2 couldpossibly form.

BO = ½ (4–2) = 1 “YEP.”

Be2

Be Be

2s 2s

s*2s

s2s

BO = ½ (4–4) = 0 “NOPE.”

Bonding and antibonding e– cancel eachother out in core energy levels, so any

bonding is due to e– in bonding orbitals of outermost shell.

Molecular Orbitals from 2p Atomic Orbitals

The 2pz orbitals overlap in head-to-head fashion,so these bonds are...s bonds.

-- the corresponding MOs are: s2p and s*2p

The other 2p orbitals (i.e., 2px and 2py) overlap insideways fashion, so the bonds are... p bonds.

-- the corresponding MOs are: p2p (two of these) and p*2p (two of these)

z

x

y

Rule 3 above suggests that, from low energy tohigh, the 2p MOs SHOULD follow the order:

s2p < p2p < p*2p < s*2pLOW

ENERGYHIGH

ENERGY

General energy-level diagrams for MOsof second-row homonuclear diatomic molecules...

don’t fit on this slide.(And so, they’re on the next one…

…if that’s all right with you.)(And if not, you can…

…quit and go make pancakes.)

Here, the interaction betweenthe 2s of one atom and the 2pof the other is strong. Theorbital energy distribution isaltered.

Here, the interaction is weak.The energy distribution isas expected.

(1s MOs are down here)

For B2, C2, and N2... For O2, F2, and “Ne2”...

s2s

s*2s

p2p

p*2p

s2p

s*2p

s2s

s*2s

s2p

s*2p

p*2p

p2p

same forboth

“Mr. B”(or Mr. C)(or Mr. N)

paramagnetism: describes the attraction of molecules with unpaired e– to a magnetic field

diamagnetism: describes substances with no unpaired e–

-- such substances are VERY weakly (almost unnoticeably) repelled by a magnetic field

Use the energy diagramsabove to tell if diatomicspecies are paramagneticor diamagnetic.paramagnetism of liquid oxygen

(“di-” = two; diamagnetic = “dielectron”)~

Paramagnetic or diamagnetic?

B2

C2

N2

O2

F2

O2+

O22–

C22–

(10)

(12)

(14)

(16)

(18)

(15)

(18)

(14)

s2s

s*2s

p2p

p*2p

s2p

s*2p

s2s

s*2s

s2p

s*2p

p*2p

p2p

s1s

s*1s

s1s

s*1s

P

D

D

P

D

P

D

D