Deviations from Expected Shape Molecular Geometry Deviations from Expected Shape

What we have so far: The geometry of a molecule depends on the number of pairs of electrons around the central atom in the structure. That, combined with the VSEPR enables us to predict a “shape” of the molecule.

You should already be familiar with the following: Two bonds – linear Three bonds – trigonal planar Four bonds – tetrahedral Five bonds – trigonal bipyramidal Six bonds - octahedral

Factors that lead to a shape that is different than predicted: Double bonds Lone pairs of electrons on the central atom. Lone pairs distort the expected geometry because the un-shared pair of electrons has a greater repulsion effect than do bond pairs. This will cause the bond angle to be a bit less than expected. For purposes of expected geometry, a double bond is treated as if it was a single bond. This is because the pi bond component of the double bond is essentially parallel to the sigma bond between the atoms.

Let’s start with the lone pairs Consider the molecule of ammonia which has the formula NH3 . Its dot structure appears to the right: Notice how it has three bond pairs of electrons and 1 lone pair – for a total of 4 pairs of electrons around the central atom. N H

Working through the deviation: With four total pairs of electrons around the central atom, we expect a geometry that is tetrahedral. The bond angle should be 109.5⁰ N H

Continuing: We now describe the shape as “pyramidal” The lone pair has a greater repulsion factor than the bond pairs. Therefore, it “squishes” the bond pairs closer together to each other. The bond angle ends up at a value of 107.3⁰ H N We now describe the shape as “pyramidal”

Next example: H2O If you draw out a Lewis Dot Structure for this molecule, you see that once again, we have a central atom with a total of 4 electron pairs surrounding. But…this time there are 2 bond pairs and 2 lone pairs. O H

Continuing this example: With 4 total pairs of electrons around the central atom, we start with a tetrahedral geometry. But the two lone pairs “squish” the bond angle even more than the single lone pair did in the previous example. O H

Finishing this one: The molecule is typically drawn in this format. O The bond angle is reported as 104.5⁰ The shape is described as “bent”. The name of the shape is determined by the actual location of atoms. With two bonds, we normally expect a linear shape. However, initial orientation of four total pairs of electrons combined with the fact that two of them are lone pairs, essentially “bends” the arms of the linear structure.

The Odd Examples: Consider the molecule with the formula ICl3 . The dot structure for this compound has a total of 5 pairs of electrons around the central atom. That would lead you to expect a trigonal bipyramidal geometry. I Cl

Working this example: Cl The key to success on this example is remembering that lone pairs on the central atom in the trigonal bipyramidal structure will always “go on the pizza”. The result of this is shown to the right.

The end result: Cl The lone pairs actually push on the arms of the axis and cause the bond angle to be reduced to a value < 90⁰ . (no number to memorize here!) We describe the shape as a “distorted T”. Cl Cl

Two for you to do: Draw the Lewis structure and determine the expected geometry and the actual shape of the XeF2 molecule. Draw the Lewis structure and determine the expected geometry and the actual shape of the XeF4 molecule.

Pi Bonding and Molecular Geometry Part II Pi Bonding and Molecular Geometry

Double Bond Effects On their own, double bonds (and triple bonds too) do not affect the shape of a molecule. The initial geometry is only based on the number of “sigma orbitals” around the central atom. By “sigma orbitals” , we mean the total of sigma bonds + the number of lone pairs.

Consider the SO2 molecule When you draw the molecule (use the stick and valence strategy), you get a resonant structure that appears to the right. S O •

Analyzing While it is true that there are three bonds, one of them is a pi bond. Therefore, there are only two sigma bonds and one lone pair for a total of three “sigma orbitals”. S O •

Working through this: With three sigma orbitals, what initial geometry do you expect? What bond angle do you expect? What will the presence of the lone pair do to the expected geometry? What is the generic formula of the actual molecule? What is the shape of the molecule?

Now you do these: Draw the structure for the molecule SO3 and work through its bonding, hybridization, expected geometry, and actual shape. Draw the structure for the molecule CO2 and work through its bonding, hybridization, expected geometry and actual shape.

Now try this: The benzene molecule has the formula C6H6 and is illustrated to the right. Does the molecule have a resonant form?. Determine the hybridization in the carbon atoms and explain how the molecule has the structure that it does.

Just for fun: Go to the website http://worldofmolecules.com and find the benzene molecule in its library. The site will permit you to rotate the structure to see it in 3-D format.