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Molecular GeometriesUnit 3: Chemical Bonding

V S

E P

R

+Lesson Essential Question:

How can Lewis Structures be used to determine molecular geometry?

EQ: How can Lewis structures be used to predict molecular geometries in covalently bonded molecules?

The VSEPR theory is used. VSEPR stands for: ________________________________________________

______________________________which basically means that when a molecule shares electrons it tries to

minimize repulsions between them in the arrangement formed.

HOW can we figure this out? Follow a step wise procedure

1.____________________________________________________________________________________

2. ___________________________________________________________________________________

3. ___________________________________________________________________________________

4.____________________________________________________________________________________

Remember the most stable electron arrangement is when all atoms can achieve an octet.

Continue with remaining steps as needed . . . #5., #6. #7. etc from separte handout

then explain the geometry of the shape formed.

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How can Lewis Structures be used to determine molecular geometry? How can Lewis Structures be used to determine molecular geometry?

Molecular Geometry

# of Bonded Atoms

Number of Unshared Electron Pairs

ABE notation(AXE)

Bond Angles

PredictedMoleculePolarity

Linear

Bent

Trigonal Planar

Trigonal Pyramidal

Tetrahedral

+VSEPR stands for….

ValenceShell

ElectronPairRepulsion

The structure around a given atom is determined principally by minimizing electron pair repulsions.

+How to predict a VSEPR Structure?

follow steps on the handout . .

1. Count the total number of valence electrons available from all atoms involved in the molecule. (if the molecule acts as an ion, add or subtract electrons here)2. Attach each atom to the center molecule with a single bond (2 electrons (dots) for each line)3. Complete the octet for each of the attached

atoms by distributing remaining electrons in pairs

4. Complete the octet around the center atom.

+Electron Pair Count and Geometry

Linear – two pairs on the central atom.

Trigonal – three pairs on the central atom.

Tetrahedral – four pairs on the central atom.

+How do I figure out the geometry?

Shared Electron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

2 Linear,180o

2 BP Linear

BeCl2

Be– 2 valence e-2 Cl – 14 valence e-Total = 16 valence e- C

lBe

Cl

Two Bonding Pair (BP) of electrons and no unbonded pair of electrons fits the Linear criteria.

AB2E0

+Shared Electron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

2Bent

104.5o2 BP1 LP

BentSO2

S – 6 valence e-2 O– 12 valence e-Total = 18 valence e-

O S OThis structure has resonance so the bonds between the S and O represent 2 Bonding Pairs (BP) of electrons.

One Lone Pair (LP) of electrons.

AB2E1

+Electron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

2 Bent104.5o

2 BP2 LP

BentH2O

2 H – 2 valence e-O– 6 valence e-Total = 8 valence e- H O H

3 Bonding Pairs of electrons and Two Lone Pair of electrons has this compound as a Trigonal Pyramidal.

AB2E2

+Electron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

3Trigonal Planar,120o

3 BP Trigonal Planar

BF3

B– 3 valence e-3 F – 21 valence e-Total = 24 valence e-

F B FF

Three Bonding Pair (BP) of electrons and no unbonded pair of electrons fits the Trigonal Planar criteria.

AB3E0

+SharedElectron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

3 Trigonal Pyramidal

107o

3 BP1 LP

Trigonal Pyramidal

NH3

N – 5 valence e-3 H– 3 valence e-Total = 8 valence e-

H N HH

3 Bonding Pairs of electrons and one Lone Pair of electrons has this compound as a Trigonal Pyramidal.

AX3E1

+Electron Pair #

GeometricalArrangement

Types of e- Pairs

Molecular Shapes

Ex’s Ball and Stick Model

4 Tetrahedral

109.5o

4 BPTetrahedr

al CH4

C – 4 valence e-4 H– 4 valence e-Total = 8 valence e-

H CH

HH

4 Bonding Pairs of electrons and no Lone Pair of electrons has this compound as a Tetrahedral.

AB4E0

+Dotted Line and Wedge Model

Linear AX2E0

Trigonal Planar AX3E0

Bent AX2E1

+Dotted Line and Wedge Model

Tetrahedral AX4E0

Trigonal Pyramidal

AX3E1

Bent AX2E2