Molecular Geometry and Polarity

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

Bond Angles in Carbon Compoundselectron configuration = 1s22s22p2

If they can, the bond angles should be 90o.

But…the bond angles are 109.5o!

2p orbitals with oneelectron in each.

Orbitals with oneelectron in each willoverlap to form singlebonds.

Can p orbitals with oneelectron in each findthe place where the3rd p orbital shouldbe?

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 - Valence Shell Electron Pair Repulsion Theory

Each group of valence electrons around a central atom is located as faraway as possible from the others in order to minimize repulsions.

These repulsions maximize the space that each object attached to thecentral atom occupies.

The result is five electron-group arrangements of minimum energy seenin a large majority of molecules and polyatomic ions.

The electron-groups are defining the object arrangement, but themolecular shape is defined by the relative positions of the atomic nuclei.

Because valence electrons can be bonding or nonbonding, the sameelectron-group arrangement can give rise to different molecular shapes.

AXmEn

A - central atom X -surrounding atomE -nonbonding valence electron-group

integers

Silberberg, Principles of Chemistry

Figure 10.3

Electron-group repulsions and the five basic molecular shapes.

linear trigonal planar tetrahedral

trigonal bipyramidal octahedral

Silberberg, Principles of Chemistry

Figure 10.4 The single molecular shape of the linear electron-group

arrangement.

Examples:

CS2, HCN, BeF2

Silberberg, Principles of Chemistry

Figure 10.5 The two molecular shapes of the trigonal planar electron-group arrangement.

Class

Shape

Examples:

SO3, BF3, NO3-, CO3

2-

Examples:

SO2, O3, PbCl2, SnBr2

Silberberg, Principles of Chemistry

Figure 10.6The three molecular shapes of the tetrahedral electron-

group arrangement.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

Silberberg, Principles of Chemistry

Figure 10.8

The four molecular shapes of the trigonal bipyramidal electron-group arrangement.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

Silberberg, Principles of Chemistry

Figure 10.9

The three molecular shapes of the octahedral electron-group arrangement.

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

Silberberg, Principles of Chemistry

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

Silberberg, Principles of Chemistry

Figure 10.10 The steps in determining a molecular shape.

Molecular formula

Lewis structure

Electron-group arrangement

Bond angles

Molecular shape

(AXmEn)

Count all e- groups around central atom (A)

Note lone pairs and double bonds

Count bonding and nonbonding

e- groups separately.

Step 1

Step 2

Step 3

Step 4

Silberberg, Principles of Chemistry

SAMPLE PROBLEM 10.6

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

The type of shape is

AX3E

Silberberg, Principles of Chemistry

SAMPLE PROBLEM 10.6

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

continued

(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

Type AX3

Silberberg, Principles of Chemistry

SAMPLE PROBLEM 10.7

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 AX5 - 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 AX5E, square pyramidal.

BrF

F F

F

F

Silberberg, Principles of Chemistry

SAMPLE PROBLEM 10.8

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.

PLAN: 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

CHH

>1200

<1200Silberberg, Principles of Chemistry

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?

Figure 10.12 The orientation of polar molecules in an electric field.

Electric field OFF

Electric field ON

SAMPLE PROBLEM 10.9

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:

PLAN: 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

HHH

NH

HH

bond dipoles

molecular dipole

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

SAMPLE PROBLEM 10.9

Predicting the Polarity of Molecules

continued

(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

Silberberg, Principles of Chemistry

More Molecular Polarity…

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

• Work through the site listed above.