Unit 3: Chemical Bonding V S E P Molecular Geometries R Unit 3: Chemical Bonding

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. Print slides 3&4 to give to the students as guided notes to complete during PPT

Molecular Geometry # of Bonded Atoms ABE notation (AXE) Bond Angles Number of Unshared Electron Pairs ABE notation (AXE) Bond Angles Predicted Molecule Polarity Linear Bent Trigonal Planar Trigonal Pyramidal Tetrahedral How can Lewis Structures be used to determine molecular geometry? How can Lewis Structures be used to determine molecular geometry?

VSEPR stands for…. Valence Shell Electron Pair Repulsion The structure around a given atom is determined principally by minimizing electron pair repulsions. Pair Repulsion

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 # Geometrical Arrangement 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- Two Bonding Pair (BP) of electrons and no unbonded pair of electrons fits the Linear criteria. Cl Be AB2E0

2 O S AB2E1 2 BP 1 LP Bent Bent 104.5o SO2 Shared Electron Pair # Geometrical Arrangement Types of e- Pairs Molecular Shapes Ex’s Ball and Stick Model 2 Bent 104.5o Bent 2 BP 1 LP SO2 S – 6 valence e- 2 O– 12 valence e- Total = 18 valence e- One Lone Pair (LP) of electrons. O S This structure has resonance so the bonds between the S and O represent 2 Bonding Pairs (BP) of electrons. AB2E1

2 O H AB2E2 2 BP 2 LP Bent Bent 104.5o H2O Electron Pair # Geometrical Arrangement Types of e- Pairs Molecular Shapes Ex’s Ball and Stick Model 2 Bent Bent 104.5o 2 BP 2 LP H2O 2 H – 2 valence e- O– 6 valence e- Total = 8 valence e- H O 3 Bonding Pairs of electrons and Two Lone Pair of electrons has this compound as a Trigonal Pyramidal. AB2E2

3 F B AB3E0 3 BP Trigonal Planar, Trigonal Planar 120o BF3 Electron Pair # Geometrical Arrangement Types of e- Pairs Molecular Shapes Ex’s Ball and Stick Model 3 Trigonal Planar, 120o Trigonal Planar 3 BP BF3 F B B– 3 valence e- 3 F – 21 valence e- Total = 24 valence e- Three Bonding Pair (BP) of electrons and no unbonded pair of electrons fits the Trigonal Planar criteria. AB3E0

3 N H AX3E1 3 BP 1 LP Trigonal Pyramidal 107o NH3 Shared Electron Pair # Geometrical Arrangement Types of e- Pairs Molecular Shapes Ex’s Ball and Stick Model 3 Trigonal Pyramidal Trigonal Pyramidal 107o 3 BP 1 LP NH3 N – 5 valence e- 3 H– 3 valence e- Total = 8 valence e- 3 Bonding Pairs of electrons and one Lone Pair of electrons has this compound as a Trigonal Pyramidal. H N AX3E1

4 C H AB4E0 4 BP Tetrahedral 109.5o Tetrahedral CH4 Electron Pair # Geometrical Arrangement Types of e- Pairs Molecular Shapes Ex’s Ball and Stick Model 4 Tetrahedral Tetrahedral 109.5o 4 BP CH4 C – 4 valence e- 4 H– 4 valence e- Total = 8 valence e- H C 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