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Why do all structures of $\ce N_2\ce O$ have Nitrogen as their central atom? Can we not draw a lewis structure with Oxygen as the central atom with 2 double bonds to the two nitrogens? This will still satisfy the octets of each atom.

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    $\begingroup$ Unhelpful answer: that's just how it is, really. And no, oxygen can't form two double bonds. $\endgroup$ – orthocresol Nov 22 '19 at 12:47
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    $\begingroup$ I guess we could draw a Lewis structure with O as the central atom that would satisfy all octets. The reason we don't do so is simply because the real-world molecule is not like that. $\endgroup$ – Ivan Neretin Nov 22 '19 at 12:55
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An all-octet Lewis structure for $\ce{NON}$ would look like this:

$$\ce{\overset{-}{N}=\overset{2+}{O}=\overset{-}{N}}$$

That’s all atoms bearing formal charges, the charges applied contrary to electronegativity (oxygen having the positive formal charge), and oxygen carrying a double formal charge. This violates practically every single rule of thumb that students get to decide whether structures are stable.

Compare to $\ce{NNO}$:

$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O}$$

This is a much more conventional structure. There is only one pair of formal charges and the positive charge never goes on the oxygen. Furthermore, you can draw resonance structures which typically provide additional stabilistion (no significant resonance structures are possible in the $\ce{NON}$ case). It can be expected that the left structure contributes slightly more to the overall picture by virtue of the oxygen being formally negatively charged.

Thus, even without knowing a thing about the real-world molecule the central oxygen atom does not lend itself well to any axioms about favourable structures.

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