Pi Orbitals

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Pyridinium.png


3
 
yep, just count the double bonds. Each double bond is one pi and each triple bond is 2 pi.
 
i thought, two pie orbitals make up one pie bond.

therefore, the answer is 6.

am i wrong?
 
Tricky question depends on how it is worded. Both of you are correct. However i think that i saw that question somewhere and the answer was 6 if i remember correctly.
 
The answer is actually 7. I originally thought it was 6. Does anyone know why it's 7???
 
I stand corrected, it is 6. There are 3 pi bonds, and 6 pi orbitals. 7's a bit of a stretch.
 
Wouldn't there be a lone electron on the nitrogen that could contribute to pi bonding?

Normally the N would have a pair, but since it is + and makes 4 bonds, there would be 1 free electron
 
I would say 6; Nitrogen is sp2 ( so 3 sp2 orbitals.) and one P orbital which participates in the pi system of the ring(the double bond with the next carbon in the ring). The lone pair on nitrogen is in one of its 3 sp2 orbitals and not delocalized into the aromatic pi-system. In acidic solution pyridine is protonated and forms this positively charged cation. ( so bond between nitrogen and H+ is co-ordinate bond)...right?

if the answer is 7, then I have no idea
 
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