When drawing for example the Lewis structure of nitrate ion (NO3)^-1 whould it be wrong to draw nitrogen and oxygen separately and then try to figure out the structure of the ion? In that case does the nitrogen give one of its electrons to the oxygen? cause a nitrogen atom does have a lone pair of electons but in the nitrate ion they are gone.enter image description here


1 Answer 1


Each Oxygen atom has 8 electrons surrounding it and so does Nitrogen. 2 electrons form a bond pair and hence a bond, thus they are no more a ‘lone pair available for donation.

In each N-O bond 2 electrons are used. The point is that nitrogen uses all its 8 surrounding electrons (which constitute an octet) in forming a total of 4 bonds, 3 single bonds and 1 double bond with Oxygen atoms.

So it has no electrons availed freely to become lone pairs, it thus has a formal positive charge on it and no lone pairs available for donation. Essentially what you did in a way, is absolutely correct, just the last reasoning you missed.

Note: N, when forms 3 bonds usually has 1 lone pair for donation, when 4 bonds are formed, it has none and +1 charge (generally).

Hope it helps.

  • $\begingroup$ Yes nitrogen doesn't have any lone pairs in the nitrate ion and i am asking why, since it has a lone pair in its atomic form. Is this because it gave one electron, of its lone pair to the oxygen beside and made a bond using the other electron of that lone pair. Thank you for trying to explain this $\endgroup$ Commented Jun 5, 2022 at 17:36
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    $\begingroup$ Nitrogen atom has a lone pair and three free electrons around it. If you surround the nitrogen atom with three oxygen atoms, three electrons of the nitrogen atom can make a bond with an electron belonging to the oxygen atoms. This would give to the nitrogen atom a structure respecting the octet rule. But not the oxygen atoms. They do not respect this octet rule, even with adding a supplementary electron to make $\ce{NO3^-}$. The only way to give oxygen an octet is to break the lone pair, as you did previously.. $\endgroup$
    – Maurice
    Commented Jun 5, 2022 at 19:33

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