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.

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**The sp**^{3} Hybridization of NH_{3}

^{3}Hybridization of NH

_{3}

*ex:* Completely describe the bonding in NH

_{3}

_________

**answer:**➞ we first need to figure out that NH_{3} 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 sp^{3}-hybridization for NH_{3}

Hydrogen keeps its original 1s atomic orbital. 👍

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**The sp-Hybridization of N**_{2}

_{2}

*ex:* Completely describe the bonding in N

_{2}

_________

**answer:**➞ we first need to figure out that N_{2} 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 N_{2}

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.

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**The Hybridization of I**_{3}^{-}

_{3}

^{-}

*ex:* Determine the hybridization of each iodine atom in the following anion: I

_{3}

^{-}

_________

**answer:**➞ if we can get the Lewis Structure, we can answer the question. I_{3}^{-} 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).

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**Hybridization Examples**

*ex: * For each of the following molecules or ions, predict the hybridization of each atom, and completely describe the molecular structure.

a. XeF_{2}

b. CO

c. BF_{4}^{-}

_________**answers:**

**Hybridization of XeF**_{2}

_{2}

**a.** XeF_{2} has * 22 valence electrons*. It's Lewis Structure is below.

The Xe has __5 effective pairs__: dsp^{3}

Each F has __4 effective pairs__: sp^{3}

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**Hybridization of CO**

**b.** CO is just like N_{2} from earlier above.

It has * 10 valence electrons*, and both the

__carbon__and the

__oxygen__are sp-hybridized.

As we saw with N_{2}, 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.

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**Hybridization of BF**_{4}^{-}

_{4}

^{-}

**c.** BF_{4}^{-} 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 BF_{4}^{-} ion has a __tetrahedral__ molecular geometry.

Thus boron has **4 effective pairs**, which requires sp^{3}-hybridization...

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In my next post covering **SECTION 9 - Covalent Bonding and Molecular Orbitals**, we'll get into the "Molecular Orbital Model."

We'll discuss Bonding and Antibonding Orbitals, Bond Order, and Molecular Orbital Diagrams.

See you there :-)