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 atom E –lone pair

integers

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

Types of repulsion:

• Lone pair – lone pair

• Bond pair – lone pair

• Bond pair- bond pair

Increasing strength: bond pair:bond pair<bond pair:lone pair<lone pair:lone pair

Figure 10.2 the five basic molecular shapes

lineartrigonal planar

tetrahedral

trigonal bipyramidal

octahedral

Figure 10.3 The single molecular shape of the linear electron-group arrangement.

Examples:

CS2, HCN, BeF2

Figure 10.4 The two molecular shapes of the trigonal planar electron-group arrangement.

Class

Shape

Examples:

SO3, BF3, NO3-,

CO32-

Examples:

SO2, O3, PbCl2, SnBr2

Figure 10.5 The three molecular shapes of the tetrahedral electron-group arrangement.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

Figure 10.6 Lewis structures and molecular shapes.

Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

Figure 10.8 The three molecular shapes of the octahedral electron-group arrangement.

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

Figure 10.9 A summary of common molecular shapes with two to six electron groups.

Figure 10.10 The steps in determining a molecular shape.

Molecular formula

Lewis structure

Electron domain

geometry Bond angles

Molecular shape

(AXmEn)

Electron domain geometry =No. of atoms attached + no. of lone pairs

2- linear 3- trigonalpyramidal 4- tetrahedral 5- trigonal bipyramidal 6- octahedral

Step 1

Step 2

Step 3

Step 4

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

FElectron domain geometry : tetrahedral

bond angle: 109.50

Molecular geometry: trigonal pyramidal.

PF F

F

The type of shape is AX3E

SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groupscontinued

(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

Electron domain geometry: trigonal planar

Type : AX3

Molecular geometry: trigonal planar

CCl

O

Cl

bond angle= 1200

CCl

O

Cl

124.50

1110

SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Five or Six Electron Groups

PROBLEM: (a) SbF5 and (b) BrF5.

SOLUTION: • SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs;

F

SbF

F F

FF Sb

F

F

F

F

Electron domain geometry : trigonal bypyramidalMolecular geometry shape is AX5 - trigonal bipyramidalBond angles : 120 and 90

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

BrF

F F

F

F

Electron domain geometry : octahedralMolecular geometry shape is AX5E, square pyramidal.Bond angles : 90

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

Electric field OFF

Electric field ON

Nonpolar molecules

Molecules with the following molecular geometry:

(applicable if all the bonded atoms are the same)

Linear

Trigonal planar

Tetrahedral

Square planar

Trigonal bipyramidal

Octahedral

SAMPLE PROBLEM 10.9 Predicting the Polarity of Molecules

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

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

PROBLEM: 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, and combine the concepts to determine the polarity.

SOLUTION: (a) NH3

NH

HH

NH

HH

NH

HH

bond dipoles

molecular dipole

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

SAMPLE PROBLEM 10.10 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.

S C O

C and S have the same EN (2.0) but Oxygen is more electronegative and the molecule is quite polar(DEN) so the molecule is polar overall.

SAMPLE PROBLEM Combined Concepts

PROBLEM: (a) SnCl2

SOLUTION: 18 valence e-;

Electron domain geometry : trigonal planarMolecular geometry : AX2E - bentBond angles : 120

Sn

Cl Cl

Lewis Structure Formal charge of the central atom: -1

Hybridization of the central atom: sp2

Polarity : polar

SAMPLE PROBLEM Combined Concepts

PROBLEM: (b) C2H2

SOLUTION: 10 valence e-; all electrons around central atom will be in bonding pairs;

Electron domain geometry : linearMolecular geometry : AX2 - linearBond angles : 180

Lewis Structure Formal charge of the C: 0

Hybridization of the C: sp

Polarity : nonpolar

The Central Themes of MO Theory

A molecule is viewed on a quantum mechanical level as a collection of nuclei surrounded by delocalized molecular orbitals.

Atomic wave functions are summed to obtain molecular wave functions.

If wave functions reinforce each other, a bonding MO is formed (region of high electron density exists between the nuclei).

If wave functions cancel each other, an antibonding MO is formed (a node of zero electron density occurs between the nuclei).

Properties of Molecular orbitals

*holds maximum of 2 electrons of opposite spins*have definite/discrete energy states*electron density distribution can be represented by contour

diagrams (similar to AO’s – s, p, d, f)*MO is associated with the entire molecule*MO’s are formed by the linear combination of atomic

orbitals (AO’s) and the overlap results in the formation of sigma () and pi () bonds

Figure 11.14 Contours and energies of the bonding and antibonding molecular orbitals (MOs) in H2.

The bonding MO is lower in energy and the antibonding MO is higher in energy than the AOs that combined to form them.

Bond Order = ½ (bonding electrons – nonbonding electrons)

Figure 11.15 The MO diagram for H2.

En

erg

y

MO of H2

*1s

1s

AO of H

1s

AO of H

1s

H2 bond order = 1/2(2-0) = 1

Filling molecular orbitals with electrons follows the same concept as filling atomic orbitals.

Figure 11.16 MO diagram for He2+ and He2.

En

erg

y

MO of He+

*1s

1s

AO of He+

1s

MO of He2

AO of He

1s

AO of He

1s

*1s

1s

En

erg

y

He2+ bond order = 1/2 He2 bond order = 0

AO of He

1s

SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams

SOLUTION:

PROBLEM: Use MO diagrams to predict whether H2+ and H2

- exist. Determine their bond orders and electron configurations.

PLAN: Use H2 as a model and accommodate the number of electrons in bonding and antibonding orbitals. Find the bond order.

1s1s

AO of HAO of H

1s1s

AO of HAO of H++

MO of HMO of H22++

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

HH22++ does exist does exist

MO of HMO of H22--

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

H2- does exist

1s1s 1s1s

AO of HAO of H AO of HAO of H--

configuration is (1s)1configuration is (1s)2(2s)1

*2s

2s

2s2s

Figure 11.18

2s 2s

*2s

2s

Li2 bond order = 1 Be2 bond order = 0

Bonding in s-block homonuclear diatomic molecules.E

ner

gy

Li2Be2

Figure 11.19Contours and energies of s and p MOs through

combinations of 2p atomic orbitals.

Figure 11.20 Relative MO energy levels for Period 2 homonuclear diatomic molecules.

MO energy levels for O2, F2, and Ne2

MO energy levels for B2, C2, and N2

without 2s-2p mixing

with 2s-2p mixing

Figure 11.21

MO occupancy and molecular properties for B2 through Ne2

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

SOLUTION:

PROBLEM: As the following data show, removing an electron from N2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O2 has a stronger, shorter bond:

PLAN: Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results.

Explain these facts with diagrams that show the sequence and occupancy of MOs.Explain these facts with diagrams that show the sequence and occupancy of MOs.

Bond energy (kJ/mol)Bond energy (kJ/mol)

Bond length (pm)Bond length (pm)

NN22 NN22++ OO22 OO22

++

945945

110110

498498841841 623623

112112121121112112

N2 has 10 valence electrons, so N2+ has 9.

O2 has 12 valence electrons, so O2+ has 11.

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

continued

2s

2s

2p

2p

2p

2p

N2 N2+ O2 O2

+

bond orders

1/2(8-2)=3 1/2(7-2)=2.5 1/2(8-4)=2 1/2(8-3)=2.5

2s

2s

2p

2p

2p

2p

bonding e- lost

antibonding e- lost

En

erg

y

MO of HF

AO of H

1s

2px 2py

AO of F

2p

Figure 11.23 The MO diagram for HF

En

erg

yFigure 11.24 The MO diagram for NO

MO of NO

2s

AO of N

2p

*1s

1s

2sAO of O

2p

2p

2p

*2p

*2s

N O

0 0

N O

-1 +1

possible Lewis structures

Problem:(1)Construct the molecular orbital

diagram(2)give the electronic configuration

(using MOs) (3)give the bond order of the following

molecules: a. O2

2-

b. N2

a. O22-

1. Molecular orbital diagram:

2s2 * 2s

2 2p4 2p

2 * 2p4

2.Electronic configuration using MO:

3.Bond order

= ½(8-6) = 1

b. N2

1. Molecular orbital diagram:

2s2 * 2s

2 2p4 2p

2

2.Electronic configuration using MO:

3.Bond order

= ½(8-2) = 3

Summary

Figure 10.1

The steps in converting a molecular formula into a Lewis structure.

Molecular formula

Atom placement

Sum of valence e-

Remaining valence e-

Lewis structure

Place atom with lowest

EN in center

Add A-group numbers

Draw single bonds. Subtract 2e- for each bond.

Give each atom 8e-

(2e- for H)

Step 1

Step 2

Step 3

Step 4

Figure 10.12The 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

See Figure 10.1

Metallic Bonding

Metallic Bond – bonding in which bonding electrons are relatively free to move throughout the three-dimensional structure.

Physical Properties of Metals

*Metal surfaces has a characteristic luster*Metals have high electrical conductivity*Metals have high thermal/heat conductivity*Most metals are malleable and ductile

Electron Sea Model for Metallic Bonding*in this model, the metal is pictured as an array of metal cations in a “sea” of electrons.*electrons are confined to the metal by electrostatic attractions to the cations, and they are uniformly distributed throughout the structure.*above physical properties of metal can be explained by this model

Molecular-Orbital Model for Metals or Band Theory

*in a metal the number of atomic orbitals that interact or overlap is very large thus, the number of molecular orbitals is also very large.*it has an energy band – numerous and continuous tiny energy separation of between metal orbitals.*electrons available for metallic bonding do not completely fill the available molecular orbitals. It is partially filled.*electrons at the top require a little input to be promoted to still higher energy orbitals.

Metallic conductors have partially filled energy bands, as shown in (a). Insulators have filled and empty energy bands, as in (b).Notes: According to the MO theory metals are able to conduct electricity because there are more molecular orbitals in the band than are necessary to accommodate the bonding electrons. In metals an excited electron may easily move to a nearby higher orbital. The bonding and antibonding molecular orbitals of insulators such as diamond are separated by a large energy gap so there are no nearby orbitals for the electrons to move to, making diamond a poor conductor of electricity.