Chapter 8 - Molecular Geometry and Polarity

http://www.scl.ameslab.gov/MacMolPlt/Surface.JPG

It’s All in the Shape…

• So what’s going on?• Think back to the lab…• What is the primary reason

molecules form the geometry we find?

• Electron Pair Repulsion

VSEPR Theory

• Electron groups around the central atom will be most stable when they are as far apart as possible – we call this valence shell electron pair repulsion theory– because electrons are negatively

charged, they should be most stable when they are separated as much as possible

• The resulting geometric arrangement will allow us to predict the shapes and bond angles in the molecule

Electron-group repulsions: the five basic electron geometries.

linear trigonal planar tetrahedral

trigonal bipyramidal octahedral

Examples:

CS2, HCN, BeF2

Two electron pairs on central atom

Electron vs Molecular Geometry

• The geometry of electron pairs around a central atom is called the electron geometry.

• The arrangement of bonded nuclei around a central atom forms the molecular geometry.

• Lone pair electrons on a central atom will repel other pairs but will not be visible in the molecular geometry (no nuclei)

• If there are lone pairs on the central atom the electron geometry and the molecular geometry will differ.

Three electronpairs on central atom

Examples:

SO3, BF3, NO3-, CO3

2-

Examples:

SO2, O3, PbCl2, SnBr2

Four electron pairs on central atom

Examples:

CH4, SiCl4, SO4

2-, ClO4-

Examples: NH3, PF3, ClO3. H3O+

Examples: H2O, OF2, SCl2

Five electron pairs on central atom

Six electron pairs on central atom

Representing 3-Dimensional Shapes on a 2-Dimensional

Surface• One of the problems with drawing molecules is trying to show their dimensionality

• By convention, the central atom is put in the plane of the paper

• Put as many other atoms as possible in the same plane and indicate with a straight line

• For atoms in front of the plane, use a solid wedge

• For atoms behind the plane, use a hashed wedge

The steps in determining a molecular shape

Molecular formula

Lewis structure

Electron-group arrangement

(electron geometry) Bond

angles

Molecular geometry

Count all e- pairs around central atom

Note lone pairs and double bonds

Consider bonding e- pairs only

Step 1

Step 2

Step 3

Step 4

Factors Affecting Actual Bond Angles

Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order.

C O

H

Hideal

1200

1200

larger EN

greater electron density

C O

H

H

1220

1160

real

Lone pairs repel bonding pairs more strongly than bonding pairs repel each other.

Sn

Cl Cl

950

Effect of Double Bonds

Effect of Nonbonding(Lone) Pairs

Predicting Molecular Shapes with Two, Three, or Four Electron Groups

PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b)

COCl2.SOLUTION: (a) For PF3 - there are 26 valence electrons, 1

nonbonding pair

PF F

F

The shape is based upon the tetrahedral arrangement.

The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair.

The final shape is trigonal pyramidal.

PF F

F

<109.50

Predicting Molecular Shapes with Two, Three, or Four Electron Groups

(b) For COCl2, C has the lowest EN and will be the center atom.

There are 24 valence e-, 3 atoms attached to the center atom.

CCl O

Cl

C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond.

The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar.C

Cl

O

Cl The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

CCl

O

Cl

124.50

1110

Predicting Molecular Shapes with Five or Six Electron Groups

PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is trigonal bipyramidal.

F

SbF

F F

FF Sb

F

F

F

F

(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is square pyramidal.

BrF

F F

F

F

Predicting Molecular Shapes with More Than One Central Atom

SOLUTION:

PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O.

Find the shape of one atom at a time after writing the Lewis structure.

C C C

OH

H

H

HH

H

tetrahedral tetrahedral

trigonal planar

C

O

HC

HHH

CH

H>1200

<1200

Molecular Polarity

• Just like bonds can be polar because of even electron distribution, molecules can be polar because of net electrical imbalances.

• These imbalances are not the same as ion formation.

• How do we know when a molecule is polar?

The orientation of polar molecules in an electric field.

Electric field OFF Electric field ON

Polarity of Molecules• For a molecule to be polar it must

1. have polar bonds

electronegativity difference - theory bond dipole moments - measured

2. have an unsymmetrical shape

vector addition

• Nonbonding pairs affect molecular polarity, strong pull in their direction

Molecule Polarity

The H─Cl bond is polar. The bonding electrons are pulled toward the Cl end of the molecule. The net result is a polar molecule.

27

Molecule Polarity

The O─C bond is polar. The bonding electrons are pulled equally toward both O ends of the molecule. The net result is a nonpolar molecule.

28

Molecule Polarity

The H─O bond is polar. Both sets of bonding electrons are pulled toward the O end of the molecule. The net result is a polar molecule.

29

Predicting the Polarity of Molecules

(a) Ammonia, NH3 (b) Boron trifluoride, BF3

(c) Carbonyl sulfide, COS (atom sequence SCO)

PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable:

Draw the shape, find the EN values and combine the concepts to determine the polarity.

SOLUTION: (a) NH3

NH

HH

ENN = 3.0

ENH = 2.1N

HH

HN

HH

H

bond dipoles

molecular dipole

The dipoles reinforce each other, so the overall molecule is definitely polar.

Predicting the Polarity of Molecules

(b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar.

F

B

F

F

F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar.

1200

(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(EN) so the molecule is polar overall.

S C O

More Molecular Polarity…

• http://academic.pgcc.edu/~ssinex/polarity/polarity.htm