# Cyanide's Lewis structure - electronegativity or octet rule?

The formula for the cyanide ion is $\ce{CN-}$. That gives us a total of ten valence electrons to work with. There are two obvious ways to build the Lewis structure.

$\ce{:^-C#N:}$

$\ce{:C=N^-::}$

(I hope that was clear. I'm not sure how to make ASCII Lewis structures. In that second example, the nitrogen has two lone pairs.)

In the first example, each atom has a complete octet, but the carbon has a formal charge of -1. In the second example, the carbon atom does not have a complete octet, and the nitrogen (the more electronegative atom) has a formal charge of -1.

My understanding is that when formal charge cannot be avoided, negative formal charge should reside on the most electronegative atom.

Which rule takes precedence over the other? Using cyanide as an example, should I:

1. Follow the octet rule and go with the first example?
2. Follow the electronegativity rule and go with the second example?
3. Treat them as resonance structures?
• Octet rule should always take priority. I do not know of any counterexamples to this. (In fact, they are resonance structures. The same rules - octet, electronegativity - allow you to judge which is the greater resonance contributor. In this case the first resonance form contributes more.) – orthocresol Oct 18 '16 at 23:27
• in the first example ther is no charge on carbon. its octet is complete – Vedant Oct 19 '16 at 2:38
• Related and related. – Jan Oct 19 '16 at 12:45
• Thanks so much, @orthocresol and Jan, that was incredibly helpful! I had a feeling it was going to be a pretty straightforward answer... – Andrew LaPrise Oct 19 '16 at 15:48
• @Vedant Complete octet does not mean that there is no formal charge. Take for example the chloride ion as a trivial counterexample. Full octet, but negatively charged. – orthocresol Oct 19 '16 at 15:55