If you missed my first two videos on types of hybridization and hybrid orbitals, you might want to watch them before continuing here...
All caught up? Great, let's get started.
Today we're going to do a bunch of hybridization examples and hybrid orbital practice problems.
The sp3 Hybridization of NH3
ex: Completely describe the bonding in NH3
➞ we first need to figure out that NH3 has 8 valence electrons.
➞ this allows us to properly draw its Lewis Structure and determine its Molecular Geometry: trigonal pyramidal.
Now that we have its three-dimensional geometry, we can see there's 4 effective pairs around the central nitrogen atom.
This requires sp3-hybridization for NH3
Hydrogen keeps its original 1s atomic orbital. 👍
The sp-Hybridization of N2
ex: Completely describe the bonding in N2
➞ we first need to figure out that N2 has 10 valence electrons.
➞ this allows us to properly draw its Lewis Structure and determine its Molecular Geometry: linear.
The moleculary geometry is "linear," which has an arrangement of 2 effective pairs around each nitrogen atom.
Two effective pairs ➞ sp-hybridization for N2
As mentioned above, the N≡N triple bond contains one sigma bond (σ-bond) because there's two sp-orbitals, and two pi bonds (π-bonds) from each nitrogen's two 2p atomic orbitals.
The Hybridization of I3-
ex: Determine the hybridization of each iodine atom in the following anion: I3- ?
➞ if we can get the Lewis Structure, we can answer the question. I3- has 22 valence electrons.
➞ this allows us to properly draw its Lewis Structure and here we go:
* TIP - Just count the effective pairs around each atom like this:
" s...p1...p2...p3...d1...d2 " (sound 'em out as you count).
ex: For each of the following molecules or ions, predict the hybridization of each atom, and completely describe the molecular structure.
Hybridization of XeF2
a. XeF2 has 22 valence electrons. It's Lewis Structure is below.
The Xe has 5 effective pairs: dsp3
Each F has 4 effective pairs: sp3
Hybridization of CO
b. CO is just like N2 from earlier above.
It has 10 valence electrons, and both the carbon and the oxygen are sp-hybridized.
As we saw with N2, the C≡O triple bond contains one sigma bond (σ-bond) because there's two sp-orbitals, and two pi bonds (π-bonds) from both the carbon's and the oxygen's two 2p atomic orbitals.
Hybridization of BF4-
c. BF4- has 32 valence electrons. It's Lewis Structure, and it's 3-D Molecular Geometry (Shape) are shown below:
From above, we see that the BF4- ion has a tetrahedral molecular geometry.
Thus boron has 4 effective pairs, which requires sp3-hybridization...
In my next post covering SECTION 9 - Covalent Bonding and Molecular Orbitals, we'll get into the "Molecular Orbital Model."
See you there :-)