# Lewis structure of diazomethane

I tried to google and watch youtube videos on how to do lewis structures properly, but I cant seem to understand how to do this properly.

My teacher, when class was in session, gave me a set of instructions to complete Lewis structures:

1. Sum Valence electrons

2. Add electrons to outer atoms, while obeying octet or duet rule.

3. Then add electrons to center atom, obey octet rule, unless they're third row element or below.

4. Do formal charges for each atom Valance charge - (unbounded electrons + 1/2(bounded electrons))
5. Once formal charges have been determined, minimize formal charges by using unbound electrons to make double or triple bonds
6. Redo formal charges to check if formal charges are as low as possible , for central atom.

So I attempted to do this for the above compound, but my answer (circled in green) does not match the book's answer (in red).

What did I do wrong?

EDIT:

After reading comments, I realized that my left structures didn't obey the octet rule. So then would this structure obey?

• The second nitrogen has 6 electrons – orthocresol Jun 6 '15 at 19:36
• For evaluating Lewis structures of $\ce{C,N,O }$, all atoms obey the octet > no atoms have formal charge – Ben Norris Jun 6 '15 at 19:37
• The structure you put after the edit is your book's! And I think it's the most contributing structure to the resonance also, though I might need further checking. – M.A.R. Jun 6 '15 at 20:34
• Oh and BTW, writing self-answers is highly encouraged in StackExchange! You could write an answer of your own to complete the process of learning. – M.A.R. Jun 6 '15 at 20:39
• You didn't follow octet rule and minimizing formal charges – ShankRam Feb 3 '16 at 13:35

The image above depicts three mesomeric structures of diazomethane ($\ce{CH2N2}$). When you count the valence electrons for every atom, you will find that only the left and middle structure satisfy the octet / duet (for hydrogen) rule, i.e. that every $\ce{C}$ and $\ce{N}$ atom has 8 valence electrons, and every $\ce{H}$ has two valence electrons. The structure on the right side does not obey the octet / duet rule because the positively-charged nitrogen only has an electron sextet. The same is true for the two structures you have drawn with an $\ce{N-N}$ single bond. In both cases, at least one nitrogen has only 6 valence electrons.