VSEPR T HEORY 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 shape 2. Name the 6 basic shapes that have no unshared pairs of electrons 3. Name a few variations off of these basic shapes

VSEPR Theory

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M OLECULAR S HAPES 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.

V ALENCE S HELL E LECTRON -P AIR R EPULSION T HEORY OR VSEPR ► Molecular Shape is determined by the repulsions of electron pairs Electron pairs around the central atom stay as far apart as possible. ► Electron Pair Geometry - based on number of regions of electron density Consider non-bonding (lone pairs) as well as bonding electrons. Unshared repel the most. 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.

LET’S CONSIDER THESE BASIC SHAPES AND SOME VARIATIONS OF THEM…

L INEAR 2 atoms attached to central atom 0 unshared pairs (lone pairs) Bond angle = 180 o Type: AX 2 Ex. : BeF 2

L INEAR Carbon dioxide CO 2

The Single Molecular Shape Of The Linear Electron-group Arrangement. Examples: CO 2, BeF 2

T RIGONAL P LANAR Boron Trifluoride BF 3

T RIGONAL P LANAR 3 atoms attached to central atom 0 lone pairs Bond angle = 120 o Type: AX 3 Ex. : AlF 3

The Two Molecular Shapes Of The Trigonal Planar Electron-group Arrangement. Class Shape Examples: H 2 CO, BCl 3, NO 3 -, CO 3 2-

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. Some exceptions follow: ideal larger EN greater electron density real Effect of Double Bonds

Factors Affecting Actual Bond Angles Lone pairs repel bonding pairs more strongly than bonding pairs repel each other 95 0 Effect of Nonbonding Pairs

T RIGONAL P LANAR V ARIATION

The Second Molecular Shape Of The Trigonal Planar Electron-group Arrangement. Examples: SO 2, O 3

B ENT Trigonal Planar variation #1 2 atoms attached to central atom 1 lone pair Bond angle = <120 Type: AX 2 E Ex. : SO 2

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T ETRAHEDRAL 4 atoms attached to central atom 0 lone pairs Bond angle = o Type: AX 4 Ex. : CH 4

T ETRAHEDRAL Carbon tetrachloride CCl 4

2 T ETRAHEDRAL V ARIATIONS

The Three Molecular Shapes Of The Tetrahedral Electron-group Arrangement. Examples: CH 4, SO 4 2- NH 3 PF 3 H 2 O OF 2

T RIGONAL P YRAMIDAL Tetrahedral variation #1 3 atoms attached to central atom 1 lone pair Bond angle = 107 o Type: AX 3 E Ex. : NH 3

T RIGONAL P YRAMIDAL Nitrogen trifluoride NF 3

B ENT Tetrahedral variation #2 2 atoms attached to central atom 2 lone pairs Bond angle = o Type: AX 2 E 2 Ex. : H 2 O

B ENT Chlorine difluoride ion ClF 2 +

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R EMEMBER THE 3 EXCEPTIONS TO THE OCTET RULE ? 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 3 rd period and beyond will sometimes have more than an octet on the central atom, up to 12, called an extended or expanded octet. Molecules with an odd number of electrons

5 B OND S ITES ON THE C ENTRAL A TOM

T RIGONAL B IPYRAMIDAL 5 atoms attached to central atom 0 lone pairs Bond angle = equatorial -> 120 o axial -> 90 o Type: AX 5 Ex. : PF 5

T RIGONAL B IPYRAMIDAL Antimony Pentafluoride SbF 5

3 B IPYRAMIDAL V ARIATIONS

The Four Molecular Shapes Of The Trigonal Bipyramidal Electron-group Arrangement. SF 4 XeO 2 F 2 ClF 3 BrF 3 XeF 2 I 3 - PF 5 AsF 5

S EE S AW Trigonal Bipyrimid Variation #1 Sulfur tetrafluoride SF 4

T-S HAPED Trigonal Bipyramid Variation #2 Chlorine tribromide

L INEAR Trigonal Bipyramid Variation #3 Xenon difluoride XeF 2

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6 B OND S ITES ON THE C ENTRAL A TOM

O CTAHEDRAL 6 atoms attached to central atom 0 lone pairs Bond angle = 90 o Type: AX 6 Ex. : SF 6

O CTAHEDRAL Sulfur hexafluoride SF 6

2 O CTAHEDRAL V ARIATIONS

The Three Molecular Shapes Of The Octahedral Electron-group Arrangement. SF 6 I OF 5 IF 5 XeOF 4 XeF 4 (BrF 4 ) -

S QUARE P YRAMIDAL Octahedral Variation #1 Chlorine pentafluoride ClF 5

S QUARE P LANAR Octahedral Variation #2 Xenon tetrafluoride XeF 4

O CTAHEDRAL Do not need to know: T-shape Linear

L ET ’ S R EVIEW VSEPR T HEORY 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

R EMEMBER THE 3 EXCEPTIONS TO THE OCTET RULE ? 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 3 rd period and beyond will sometimes have more than an octet on the central atom, up to 12, called an extended or expanded octet. Molecules with an odd number of electrons

B ENT Nitrogen dioxide NO 2

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VSEPR THEORY: LET’S SEE IF YOU CAN: 1. Use VSEPR to predict molecular shape 2. Name the 6 basic shapes that have no unshared pairs of electrons 3. Name a few variations off of these basic shapes

Sample Problems

The Steps In Determining A Molecular Shape. Molecular formula Lewis structure Electron-group arrangement Bond angles Molecular shape (AX m E n ) 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 (CH 3 NO 2 ): Each hydrogen contributes 1 valence electron. Each carbon contributes 4, nitrogen 5, and each oxygen 6 for a total of 24. R EVIEW OF L EWIS S TRUCTURES

Step 2: Connect the atoms by a covalent bond represented by a dash. Example: Methyl nitrite has the partial structure: R EVIEW OF L EWIS S TRUCTURES CONOHHH

Step 3: Subtract the number of electrons in bonds from the total number of valence electrons. Example: 24 valence electrons – 12 electrons in bonds. Therefore, 12 more electrons to assign. R EVIEW OF L EWIS S TRUCTURES

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 in methyl nitrite are added as 6 pairs. R EVIEW OF L EWIS S TRUCTURES.. CONOHHH.... :....

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 be done. R EVIEW OF L EWIS S TRUCTURES.... CONOHHH.. : CONOHHH.. :..

Step 6: Calculate formal charges. Example: The left structure has formal charges that are greater than 0. Therefore it is a less stable Lewis structure. R EVIEW OF L EWIS S TRUCTURES.... CONOHHH.. :.. + –.... CONOHHH.. :..

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) PF 3 and (b) COCl 2.

SOLUTION:(a) For PF 3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be < due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. < The type of shape is AX 3 E

(b) For COCl 2, C has the lowest EN and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. 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. The Cl-C-Cl bond angle will be less than due to the electron density of the C=O Type AX 3

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) SbF 5 and (b) BrF 5.

SOLUTION: (a) SbF valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5 - trigonal bipyramidal.

(b) BrF valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal.

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