Formal charge question

[756034]

Hey guys,

I was reviewing formal charges and came across what I originally thought was a straightforward problem (thinking otherwise now). The question is asking me to determine the formal charge on nitrogen in the following molecule:

CH3-CH=NH2

I have been using the equation [# of valence electrons - # of lone pair electrons - (1/2* # of bonded electrons), so I figured I would double check that there were no lone electrons on that nitrogen atom before I calculated anything. I drew out the Lewis structures, using the following valence electron counts:

Carbon: 2*4 = 8
Hydrogen: 6*1=6
Nitrogen: 5
TOTAL: 19 electrons

When I drew out the molecule, I only used 18 of the electrons. So I figured there was one lone electron. But...nitrogen already has a full octet at this point, so adding a lone electron to that atom does not make sense to me. Same goes for carbon, but that is not the more electronegative atom anyway. Where does that extra electron go?! Even if I assume for whatever reason it is on nitrogen, my formal charge answer is not correct (5 valence shell electrons - 1 lone pair electron - 1/2*8 bonded electrons =0...but the answer to this problem is apparently 1). I need another pair of eyes (and maybe mind). Am I miscalculating something? Overlooking something? I can't believe how much time I have spent on this one question, but it is bothering me that I cannot figure out what happened to that final electron.

Thanks!

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[betterfuture]

I think you should recognize by now that Nitrogen is fairly happy with forming 3 bonds whereas when it forms 4 bonds, Nitrogen bears a (+) charge. Think of ammonia vs ammonium and the number of bonds present. It all has to do with the octet rule.

 

[theonlytycrane]

As ^ said, the nitrogen atom holds a positive charge (4 bonds) so we should only place 18 electrons.

 

[756034]

Thanks to you both for responding. And what you said makes sense - I realize nitrogen will have a +1 charge when bonded to four other atoms (and will have a 0 charge with three bonds). But the part that is still bothering me is that unaccounted for final electron in this molecule