SCH 4U - Hybridization, Part 1

Intro

We have used the Octet Rule and Lewis structures to talk about covalent bonding.  However, we need something better to explain how bonding occurs. 

Looking at the atomic orbitals of each atom doesn’t always explain the molecular shapes.  For instance, C has electrons in p orbitals which are oriented at 90° to each other. However, when C bonds to four Cl, the molecule is tetrahedral in shape, with bonds at approximately 110° to each other.  So, something must happen to “rearrange” the C’s atomic orbitals (AO) to allow the atom to bond tetrahedrally.

 

Valence Bond Theory (Linus Pauling)

Bonding occurs when two half-filled orbitals in the bonding atoms (atomic orbitals (AO)) overlap to a make filled molecular orbital (MO), which is an orbital that engulfs both atoms.

Hybridization

Procedure the carry out hybridization:

1. Focus on the central atom.  In part (a), we will focus on Be. First, create the ground state electron configuration (GSEC) for Be (it is shown in both spectroscopic notation(SN) and orbital box notation (OBN) below).  Since Be has 4 electrons, 2 are placed in the 1s orbital and 2 are placed in the 2s orbital.
2. Notice that Be has no half-filled orbitals, which are required to bond to H.  So, we will carry out a promotion, which involves moving an electron from a lower orbital to the next highest orbital that has the same principal quantum number. In this case, one of the 2s electrons is promoted to the 2p_x orbital.  Notice that gives Be two half-filled orbitals.
3. Unfortunately, the two half-filled orbitals must be of the same energy or Be will have unequal bonds to the two Hs.  This requires hybridization. Hybridization requires the "mixing" of the two half-filled orbitals (2s and 2p_x) to create two new equivalent hybrid orbitals, which are designated "sp."  The unaffected 2p_y and 2p_z continue to be empty.
4. To draw the orbital overlap diagram, bear in mind that the sp hybrid orbitals contain an electron each.  Since electrons repel each other, the hybrid orbitals* will arrange themselves in a linear fashion.  The H's half-filled 1s orbitals come in and overlap the Be's sp orbitals.

* sp orbitals have 50% s character and 50% p character.  To illustrate this, we draw sp orbitals with one large lobe and one small lobe (as if the s orbital has been engulf by one of the lobes of the p orbital.

In the diagrams below, I have highlighted the small lobe for each of the two sp orbitals, so they stand out a bit.  Although, it is not necessary, hopefully this makes it easier to see.

Here's a video.

 

Note that we will draw all hybrid molecular orbitals with one small and one large lobe, regardless of which atomic orbitals go into the mix.  For instance, the sp² hybrid orbitals, in the next example, have 33% s character and 67% p character and should be drawn to indicate that ratio.  But, we don't tend to bother.

 

Here's a video.

Here's a video. Please note that I originally misspeak and say that we have created 3 arms for C, when I meant to say that we have created *4* arms.  As the video goes on, I correct myself and say that C has 4 arms.  Oops!

 

The next compound continues to employ the above procedure, but there is a slight twist.  We still start with the GSEC for the central atom (N). Since N already has 3 half-filled orbitals, there is no need for a promotion, so we simply write out the GSEC again.  It may seem that there is no need for hybridization either.  However, the half-filled p orbitals do not have the 109.5° angles required for the trigonal pyramidal shape of ammonia.

To hybridize, we take the 2s and three 2p orbitals and mix them together.  Since certain VSEPR shapes always correspond to certain hybrid orbitals (sp = linear, sp² = trigonal planar, tetrathedral = sp³), the trigonal pyramidal ammonia will use sp³ hybrid orbitals (since tetrahedral, trigonal pyramidal and bent shapes are all under the same shape umbrella). 

 

As with methane, above, this gives us four sp³ orbitals around the central atom.  In the case of methane, all four orbitals were half-filled and ready to bond to the four Hs.  In ammonia's case, one of the orbitals is full (lone pair) and the other three are half-filled and will be used to bond to the three Hs.

 

 

TryIt!  Using the procedure above, take yourself through the next compound, which is water.  

Answer

 

Homework:  Show the three lines of hybridization (you only need to use SN for this - I used OBN in the examples so the placement and movement of electrons was more visual) and draw the orbital overlap diagram for PCl₅ & SF₆.