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I tried to google and watch YouTube videos on how to write Lewis structures properly, but I can't seem to understand how to do this correctly.

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 - (unbonded 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 diazomethane, but my answer (circled in green) does not match the book's answer (in red).

Lewis structures of diazomethane

What did I do wrong?

After further considerations, I realized that my structures didn't obey the octet rule. So then would this revised structure be also correct?

Lewis structure of diazomethane

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  • $\begingroup$ I think both images in the red box are showing the "formal charge" on their respective atoms. Formal charge is similar to oxidation state used in inorganic chemistry as a bookkeeping scheme. The charge in either system is not to be taken literally, but merely gives us a handle on the electrons to track their movement. $\endgroup$
    – David P Bligh
    Nov 20, 2021 at 1:50

1 Answer 1

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enter image description here

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.

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