Why does sp2 hybridization imply adjacent sp2 hybridization? EDIT:  I misquoted my teacher; he said something more along the lines of: in conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate? 



My teacher said that if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. 

1) Why is this true?
2) Is this even true? 

I'm guessing this has to do with physical limitations. For example, consider a carbonyl oxygen. Oxygen is $\ce{sp^2}$ hybridized. The carbon the oxygen is bound to has already "used up" two of its four bonds. The carbon can only form two more sigma bonds in addition to the one sigma bond it already has in the $\ce{C=O}$ double bond. 

But what about "hypervalent" (if there are such beasts) molecules? Here we come to a self-consistency issue. My teacher still believes in hypervalent molecules - i.e. he draws sulfuric acid with six bonds. 

So let's consider sulfuric acid in its hypervalent form, with two double bonded $\ce{sp^2}$ oxygens. There are a total of six bonds and four sigma bonds, implying $\ce{sp^3}$ hybridization. 

So my revised set of questions is as listed:

1) Is my teacher being inconsistent with himself here? 

2) Does $\ce{sp^2}$ hybridization still imply $\ce{sp^2}$ hybridization for linked atoms? Note that sulfuric acid does not disprove my teacher's statement if we take sulfuric acid to have four bonds, not six (i.e. sulfuric acid is properly represented as charge separated due to the high ionic character of its $\ce{S-O}$ bonds. If I remember correctly, on a time-average basis, 1.7 electrons are in oxygen's valence and only 0.3 electrons are in sulfur's valence per bond.
If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?


Absolutely not, here are a few examples