hi-
i'm looking at a problem at of the cliff's dat book, and i'm pretty sure their answer is wrong.
the question: how many pi electrons are in CH3C=N
a. 0
b. 2
c. 3
d. 6
3. 4

i think the answer is 2, but they say it's 4. any help would be appreciated, thanks!

the question is asking about electrons not bonds.
therefore, there are 4 electrons which are involved in 2 pi bonds.

thanks! but i guess the thing that's confusing me is that doesn't a double-bond have 1 sigma bond and one pi bond, so that there's only 2 electrons in the pi bond?

the CN sould be a triple bond, if the answer is 4 electrons. so there are 1 sigma bond and two pi bonds.
yes, each pi bond has 2 electons.

Acetonitrile, or methyl cyanide, as you have drawn, is wrong. There should be a triple bond between C and N. The way you have it drawn would actually have formal charges and is the incorrect lewis structure. This is the reason for your confusion. Therefore, a triple bond with a sigma and two pi bonds, will have 4 e in it's pi bonds, with both C and N using 2 sets of extra "p"orbitals to construct the pi bonds.