# Are more than one correct Lewis structures possible (not counting resonance)?

I am currently learning chemical bonding, specifically Lewis structures. And these nitrogen oxides, nitrogen-oxygen ions and other atoms showing variable oxidation states are confusing me. As many examples I have understood with bonding atoms having no $d$ or $f$ orbitals, the variable oxidation states are a result of dative bonds.

So when drawing the Lewis structure for $\ce{NO3-}$, I drew:

My book however has only one dative bond, and has this structure:

To me, both of them look correct. The octets seem to be satisfied in both cases.

So my questions are:

1. Are both of them correct? If yes, how?
2. If my structure is incorrect, what rule am I violating?
• nitrogen can make 3 covalent bond and has one lone pair so 1 lone pair will be distributed to 1 oxygen and then nitrogen will form 1 double bond with another oxygen now nitrogen can form only one single bond which will be possible with oxygen having negative sign just like the second structure in your question Jun 10, 2015 at 6:34
• @shashanksharma You are explaining the correctness of the structure in the book, which I know is correct. I am asking what's wrong with my structure? It too satisfies the octet. Jun 10, 2015 at 7:34
• @M.A.Ramezani I drew the first structure myself without looking anywhere. I am asking what is wrong WITh assigning the negative charge on N and forming dative bonds with 2 oxygen atoms, """INSTEAD""" of the structure drawn in my book. Jun 10, 2015 at 8:22
• @ShashankSharma What question? $\ce{NO3-}$ means negative charge on the whole ion. Writing $\ce{NO3-}$ does not mean that only oxygen can be assigned a negative sign. The net charge on the ion must be -1. That's all. There is NO strict condition for negative charge to be present only on oxygen atom. Jun 10, 2015 at 8:41
• Aha! Then you think a little about the electronegativity differences. What does electronegativity mean? Which of the the two $\ce{N}$ and $\ce{O}$ is more electronegative? Jun 10, 2015 at 16:38

As $\ce{O}$ is more electronegative than $\ce{N}$ the $\ce{1-}$ charge in the $\ce{O-N}$ will be on the oxygen. Even if it were to be a dative bond, its donation in an electron pair makes it very easy for oxygen to take. Oxygen is, after all, an oxidising agent.

• Although I now have an answer to my question (from my teacher), your answer seems to be wrong, based on the fact in $\ce{CN-}$, N is more electronegative than C, yet the -1 charge is still on C. Jun 11, 2015 at 12:09
• @Swapnil For $\ce{CN^-}$ it is a different story, note that the -1 on the C allows for both bonding atoms to have a full octet. In the previous example both the O and the N had a full octet already so the extra -1 goes the the electronegative one. Jun 11, 2015 at 22:54