VSEPR Theory

VSEPR THEORY

Valence Shell Electron Pair Repulsion

VSEPR THEORY:AT THE CONCLUSION OF OUR TIME TOGETHER, YOU SHOULD BE ABLE TO:

1. Use VSEPR to predict molecular shape2. Name the 6 basic shapes that have no

unshared pairs of electrons

VSEPR Theory

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MOLECULAR SHAPES

Lewis structures show which atoms are connected where, and by how many bonds, but they don't properly show 3-D shapes of molecules.

To find the actual shape of a molecule, first draw the Lewis structure, and then use VSEPR Theory.

VALENCE SHELL ELECTRON-PAIR REPULSION THEORY OR VSEPR

► Molecular Shape is determined by the repulsions of electron pairsElectron pairs around the central atom stay

as far apart as possible.► Electron Pair Geometry - based on number of

regions of electron densityConsider non-bonding (lone pairs) as well

as bonding electrons.Electron pairs in single, double and triple

bonds are treated as single electron clouds.► Molecular Geometry - based on the electron

pair geometry, this is the shape of the molecule

Electron-group Repulsions And The Five Basic Molecular Shapes.

The Single Molecular Shape Of The Linear Electron-group Arrangement.

Examples:

CS2, HCN, BeF2

The Two Molecular Shapes Of The Trigonal Planar Electron-group Arrangement.

Class

Shape

Examples:

SO3, BF3, NO3-,

CO32-

Examples:

SO2, O3, PbCl2, SnBr2

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

Hideal1200

1200

larger EN

greater electron density

C O

H

H

1220

1160

real

Effect of Double Bonds

Factors Affecting Actual Bond Angles

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

Cl Cl

950

Effect of Nonbonding Pairs

The Three Molecular Shapes Of The Tetrahedral Electron-group Arrangement.

Examples:

CH4, SiCl4, SO4

2-, ClO4-

NH3

PF3

ClO3

H3O+

H2O

OF2

SCl2

The Four Molecular Shapes Of The Trigonal Bipyramidal Electron-group Arrangement.

SF4

XeO2F2

IF4+

IO2F2-

ClF3

BrF3

XeF2

I3-

IF2-

PF5

AsF5

SOF4

The Three Molecular Shapes Of The Octahedral Electron-group Arrangement.

SF6

IOF5

BrF5

TeF5-

XeOF4

XeF4

ICl4-

LET’S REVIEW VSEPR THEORY

Predicts the molecular shape of a bonded molecule

Electrons around the central atom arrange themselves as far apart from each other as possible

Unshared pairs of electrons (lone pairs) on the central atom repel the most

So only look at what is connected to the central atom

REDNECK INNOVATIONS: WHEN YOU GOTTA GO…..

LET’S CONSIDER A FEW MORE EXAMPLES:6 TYPES OF MOLECULES WITH NO UNSHARED PAIRS OF ELECTRONS

LINEAR

2 atoms attached to central atom 0 unshared pairs (lone pairs)

Bond angle = 180o

Type: AB2

Ex. : BeF2

3 EXCEPTIONS TO THE OCTET RULE

Molecules with an odd number of electrons Molecules with atoms near the boundary

between metals and nonmetals will tend to have less than an octet on the central atom. (i.e. B, Be, Al, Ga)

Molecules with a central atom with electrons in the 3rd period and beyond will sometimes have more than an octet on the central atom, up to 12, called an extended or expanded octet.

LINEAR

Carbon dioxide

CO2

TRIGONAL PLANAR

3 atoms attached to central atom 0 lone pairs

Bond angle = 120o

Type: AB3

Ex. : AlF3

TRIGONAL PLANAR

Boron Trifluoride BF3

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TETRAHEDRAL


Bond angle = 109.5o

Type: AB4

Ex. : CH4

TETRAHEDRAL

Carbon tetrachloride CCl4

TRIGONAL BIPYRAMIDAL 5 atoms attached to central atom 0 lone pairs

Bond angle = equatorial -> 120o

axial -> 90o

Type: AB5

Ex. : PF5

TRIGONAL BIPYRAMIDAL

Antimony Pentafluoride

SbF5

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OCTAHEDRAL


Bond angle = 90o

Type: AB6

Ex. : SF6

OCTAHEDRAL

Sulfur hexafluoride

SF6

EXAMPLES OF MOLECULES WITH BOTH PAIRED AND UNSHARED (LONE) PAIRS OF ELECTRONS AROUND THE CENTRAL ATOM.

TRIGONAL PLANAR

Boron Trifluoride BF3

BENT Trigonal Planar variation #1 2 atoms attached to central atom 1 lone pair

Bond angle = <120

Type: AB2E Ex. : SO2

TETRAHEDRAL

Carbon tetrachloride CCl4

TRIGONAL PYRAMIDAL Tetrahedral variation #1 3 atoms attached to central atom 1 lone pair

Bond angle = 107o

Type: AB3E Ex. : NH3

TRIGONAL PYRAMIDAL

Nitrogen trifluoride NF3

BENT Tetrahedral variation #2 2 atoms attached to central atom 2 lone pairs

Bond angle = 104.5o

Type: AB2E2

Ex. : H2O

BENT

Chlorine difluoride ion

ClF2+

TRIGONAL BIPYRAMIDAL

Antimony Pentafluoride

SbF5

SEE SAW

Trigonal Bipyrimid Variation #1

Sulfur tetrafluoride

SF4

T-SHAPED

Trigonal Bipyramid Variation #2

Chlorine tribromide

LINEAR

Trigonal Bipyramid Variation #3

Xenon difluoride XeF2

OCTAHEDRAL

Sulfur hexafluoride

SF6

SQUARE PYRAMIDAL

Octahedral Variation #1Chlorine pentafluoride ClF5

SQUARE PLANAR

Octahedral Variation #2Xenon tetrafluoride XeF4

OCTAHEDRAL

Do not need to know:T-shapeLinear

BENT

Nitrogen dioxide NO2

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Sample Problems

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

Step 1: Count the number of valence electrons. For a neutral molecule this is equal to the number of valence electrons of the constituent atoms.

Example (CH3NO2):Each hydrogen contributes 1 valence electron. Each carbon contributes 4, nitrogen 5, and each oxygen 6 for a total of 24.

REVIEW OF LEWIS STRUCTURESREVIEW OF LEWIS STRUCTURESREVIEW OF LEWIS STRUCTURESREVIEW OF LEWIS STRUCTURES

Step 2: Connect the atoms by a covalent bond represented by a dash.

Example:Example:Methyl nitrite has the partial Methyl nitrite has the partial structure: structure:


CC OO NN OOHH

HH

HH

Step 3: Subtract the number of electrons in bonds from the total number of valence electrons.

Example:Example:24 valence electrons – 12 electrons 24 valence electrons – 12 electrons in bonds. Therefore, 12 more in bonds. Therefore, 12 more electrons to assign.electrons to assign.


Step 4: Add electrons in pairs so that as many atoms as possible have 8 electrons. Start with the most electronegative atom.

Example: The remaining 12 electrons Example: The remaining 12 electrons in methyl nitrite are added as 6 pairs. in methyl nitrite are added as 6 pairs.


....CC OO NN OOHH

HH

HH

........ ::.... ....

Step 5: If an atom lacks an octet, use electron pairs on an adjacent atom to form a double or triple bond.

Example: There are 2 ways this can Example: There are 2 ways this can be be done.done.


........

CC OO NN OOHH

HH

HH

....::........

....CC OO NN OOHH

HH

HH

.... ::....

Step 6: Calculate formal charges.

Example: The left structure has Example: The left structure has formal charges that are greater formal charges that are greater than 0. Therefore it is a less stable than 0. Therefore it is a less stable Lewis structure.Lewis structure.


........

CC OO NN OOHH

HH

HH

.... ::....++ ––

........

CC OO NN OOHH

HH

HH

....::....

SAMPLE PROBLEM:

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

(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.

CCl

O

Cl

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

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

SAMPLE PROBLEM:

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

VSEPR Theory

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