[rozarria]

For the longest time, the easiest method that I used to identify hybridization was the number of bonds between atoms. Single bonds are associated with sp3 hybridization, double bonds with sp2, and triple bonds with sp. This rule worked without fail for me...

...except a few weeks ago, when I was posed with a molecule, the exact structure of which I can't remember, but which consisted of a straight 3-Carbon chain. There were double bonds between them. When our professor asked us what the hybridization was between Carbons, sp2 was the correct answer, which didn't surprise me. However, she then asked us the hybridization of the central Carbon atom, and I believe that she said that it was sp. I then became horribly confused.

Can someone explain how two different hybridizations can exist within the same set of Carbons?

Thanks,
Martin

[spirochete]

I assume you're referring to a molecule called an allene (look it up on wikipedia) with the general formula R₂CCCR₂

The central carbon in that molecule is SP hydridized, the outside two are SP2 hybridized.  You can figure this out using the standard rules for determining hybridization.  The central carbon is making two sigma bonds and two pi bonds.  It is also bonded to two "things."  Both are consistent with SP hybridization.

[archyjuice]

Rozarria, i used your method to figure out the hydridization of molecules. But from what i been doing lately, it seems kinda wrong to do it this way. theres a certain percentage of getting it wrong.

Anyone with a better and more specific way to determine the hydridization?

[Astrokel]

Spirochete has mentioned you could use the standard rules for determining hybdrization such as the follow.

Lone pair/single bond is treated as one unit
Double/tripe bonds is treated as one unit

The sum of the units around the atom is equal to the sum of the power of the superscripts of sp^x where s has a superscript of 1.

For example, to determine the oxygen hybdrization in water molecule,

H-O-H, oxygen has 2 single bonds and 2 lone pairs and thus total of 4 units.

4 units is the sum of the power of the superscripts of sp^x, thus x must be equal to 3 since s superscript is of 1. Thus it is sp³ hybdrization.

[AhBeng]

Yes, I'll add a couple of exampes to illustrate, paraphrase and elaborate on Astrokel's method (I always encourage students to understand the *meaningfulness* behind any mathematical formula). Here, students should appreciate the *purpose* of hybridization.

1) Ethyne.
To minimize electron repulsion, the VSEPR molecular geometry is linear, and the VSEPR electron geometry is also linear (since there are no lone pairs). To achieve linear VSEPR electron geometry, hybridization of C orbitals must be sp (without hybridization, the bond angles would result in significant electron repulsion and an unstable geometry). This leaves 2 p orbitals (s s p p --> s p p p --> sp sp p p) for overlap (with the other C's 2p orbitals) to form the 2 pi bonds (over the sigma bond; hence triple bond between the 2 Cs).

2) Ammonia.
The molecular VSEPR geometry is trigonal pyramidal, but the electron geometry is tetrahedral. Hence hybridization is sp3 (to achieve tetrahedral electron VSEPR geometry).


A related question to this :
Identify and explain whether PH3 is a stronger or weaker nucleophile and base, as compared to NH3.

The solution, is to recognize that P is one electron shell larger than N; consequently the electron repulsion (between electron pairs, regardless lone or bond) is less in PH3 as compared to NH3; consequently there is less need and therefore less extent of hybridization of electron orbitals; consequently the bond angles in PH3 are almost orthogonal as compared to (slightly less than, due to greater repulsion between lone pair and bond pairs) tetrahedral bond angles in NH3; which implies that the lone pair in PH3 is largely the s orbital as compared to sp3 orbital in NH3; consequently the s orbital lone pair of PH3 is less available for nucleophilic attack and/or accepting a proton, as compared to NH3.