Molecular Shape: VSEPR and VB theory
Major concepts

The shape of a molecule can be determined if the geometry around central atoms is known.

Valence shell electron pair repulsion can often be used (in simple molecules) to determine whether an atom’s electron geometry is linear, trigonal planar, or tetrahedral.

Wedged and dashed lines can be added to bond-line convention to show shapes.

The physical basis for the shapes of molecules can be described in terms of stable orbitals for valence electrons, and bonds can be described in terms of orbital overlap.

Using a qualitative model of orbitals, we can describe an atom in terms of its hybridization, which tells us about molecular geometry and stability of electrons.
Vocabulary

Primary, secondary, tertiary, quaternary carbons
 carbocations

VSEPR

Linear, trigonal planar, tetrahedral

Wedge/dash convention

Orbitals
 hybridization
Students should be able to:

Use VSEPR to determine geometry of an atom

Draw bond-line structures including wedges and dashes to show appropriate shapes

Determine approximate bond angles for compounds

Explain what is meant by “hybridization” and “orbital”

Given the hybridization of atoms, be able to draw orbital overlap pictures of simple molecules and ions
Daily Problems
1. Use VSEPR to indicate the geometry of the indicated atom as linear, trigonal planar, or tetrahedral.
Remember: implicit hydrogens and lone pairs affect the geometry!

2. According to VSEPR, what should the bond angles be for CH
3
+.
?
In VSEPR, three areas of electron density means the geometry is trigonal planar and a bond angle of
120 o . The picture in #3 is accurate. All the H-C-H have an angle of 120 o , and the “empty” orbital is perpendicular.
3. According to VSEPR, what should the bond angles be for NH
3
.
?
According to VSEPR, there are 4 areas of electron density, which means the geometry is tetrahedral with a bond angle of 109.5
o . The picture in #5 is accurate with a H-N- H angle of ~109.5 o .
4. Based on what you have learned about hybridization and geometry, I will expect you to be able to draw compounds well from now on. Here are some very common mistakes that I see. Explain why these drawings are not well drawn, then draw the correct structure. (Hint: Use models to check out the bond geometries.)
B C
A
B
Compound A has an sp 3 hybridized carbon where the dashed and wedged lines are; it is not drawn in a tetrahedral geometry.
Compound B has two sp hybridized carbons (one on each side of the triple bond), which should have a linear geometry.
Compound C has two sp 2 carbons on either side of the double bond which have a trigonal planar geometry, meaning the bonds coming from those carbons should be planar (not dashes or wedges, which allude to bonds going into and coming out of the plane of the page, respectively).
5. Fill in the implicit hydrogens in this molecule to indicate their proper bond angles. (You may or may not need to use wedges/dashes.)
O

Cumulative problems
6. The shape of products of these reactions is partially incorrect. A. Use VSEPR to determine which atom has the wrong geometry, then draw the product in the correct shape using wedges and dashes as necessary. B. Indicate whether this reaction leads to greater stability or less stability of the compound.
Less stable
Less stable
7. Draw 2,3,3trimethylpent-1-ene with wedges and dashes on any quaternary carbon atoms to designate molecular shape.
8. Explain why 3-chloro-3-methylpent-2-ene does not exist.
There would be 5 bonds to the center carbon.
9. Why is cyclopropene less stable than cyclopropane?
A
More stable
B

In A, all the carbons are sp 3 hybridized and desire a 109.5
o bond angle but are forced to be 60 o . In B, the bottom two carbons are sp 2 hybridized and desire to be 120 o but are forced to be 60 o , making a larger distortion of bond angle.