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 – Shashank Jun 10 '15 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. – Swapnil Rustagi Jun 10 '15 at 7:34
• your questions says to draw a lewis structure of n o 3 - but in your structure you showed negative sign in nitrogen which is not mentioned in question. – Shashank Jun 10 '15 at 7:51
• @shashanksharma what do you mean by "not mentioned in the question"? I had to draw the structure for $\ce{NO3-}$, I drew it by assigning negative charge on N, in the book it is drawn by assigning negative charge on one of the O atoms. I am simply asking if my approach is correct or not? And if not, then why? – Swapnil Rustagi Jun 10 '15 at 8:07
• 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? – M.A.R. Jun 10 '15 at 16:38

1 Answer

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. – Swapnil Rustagi Jun 11 '15 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. – Ali Caglayan Jun 11 '15 at 22:54