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These are the structures of nitric acid and nitrous acid respectively:

Nitric acid Nitrous acid

To compare acid strength, we compare the stability of the corresponding conjugate base. Here conjugate bases are nitrate and nitrite anions respectively.

Conjugate base of nitric acid

Now, my point is that in $\ce {NO2-}$, the resonance effect is pronounced, but in the case of $\ce {NO3-}$, we have a case of cross resonance. Put simply, in $\ce {NO2-}$ the negative charge de-localizes to double bonded oxygen atom. In the case of $\ce {NO3-}$, same thing happens, but we have another $\ce {O-}$ here due to which there is a cross of 2 resonance effects (If I am using the correct terminology). So, resonance effect of nitrate ion should be weaker than nitrite ion and therefore nitrite ion should be more stable.

If I am true then nitrous acid must be a stronger acid, but all my sources suggest otherwise. They insist that in $\ce {NO3-}$, the negative charge is distributed over 3 oxygen atoms, and hence its resonance is more powerful. My confusion is that there are also 2 negatives that need to be stabilized as opposed to one negative in $\ce {NO2-}$.

So what concept am I missing here?

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    $\begingroup$ Well, you seem to not get the difference between "formal charge" and real charge. Would be good if you wouldn't even denote them in the same way. $\endgroup$
    – Mithoron
    Commented Mar 12, 2018 at 18:06

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The correct terminology is "cross conjugation" and "extended conjugation" (but the latter makes more sense in the context of more than two pi bonds, as in penta-1,3,5-triene).

Now, you have correctly identified that the nitrate anion has cross conjugation while the other has simple conjugation. However, you've incorrectly concluded one of them to be more stable. The conclusion is incorrect because there is a much dominant factor at play instead.

The three canonical resonating structures of the nitrate anion are "equivalent" i.e. wherever the negative charge may delocalise, the new resonance structure is exactly the same as the original one. Therefore, all the resonance structures are "equivalent" in their potential energy. This leads to a much rapid delocalisation of the lone pair, between the three oxygen atoms, leading to a rapid charge dispersal.

Both nitrate and nitrite anions have equivalent resonating structures. But, in the former, the negative charge is dispersed over three oxygen atoms (instead of only two in nitrite). From this, we conclude that the nitrate anion is much more stable than the nitrite anion.

Therefore, the next time you apply the fact that cross conjugation imparts lesser stability than extended conjugation, do look out for this additional dominant factor of "equivalent resonance"!

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    $\begingroup$ I did not overlook that, but I dismissed that because both had equivalent structures. One thing though, you are absolutely correct that in nitrate the negative charge is dispersed over three atoms whereas in nitrite the negative charge is dispersed over 2 atoms only, but I again insist that nitrate ion also has 2 negative charges that need to be dispersed as opposed to the only negative charge in nitrite anion. So, why are we treating the presence of 2 negative charges as equivalent to presence of 1 negative charge? $\endgroup$
    – sarthak-ag
    Commented Mar 12, 2018 at 6:09
  • $\begingroup$ @Sarthak123 That is a good point, but note that the negative charge that you see on the other oxygen atom is not a real negative charge but actually a "formal" negative charge. The bond between that $\ce{O}$ and $\ce{N}$ is a coordinate covalent bond - a lone pair is donated by $\ce{N}$ to $\ce{O}$. Hence, we depict a +ve sign on $\ce{N}$ and a -ve sign on the corresponding $\ce{O}$. These are "formal" charges only. $\endgroup$
    – Gaurang Tandon
    Commented Mar 12, 2018 at 6:24
  • $\begingroup$ oh, well I think that his has to do with my poor understanding of formal charges. I never considered the +/- on nitrate ion as formal charge, my understanding was that the hypothetical charges (much like oxidation states) are formal charges whereas these actual +/- are not formal charges. I would love to be enlightened on this issue, can you elaborate please? $\endgroup$
    – sarthak-ag
    Commented Mar 12, 2018 at 6:31
  • $\begingroup$ @Sarthak123 Oxidation states are a different thing altogether. The only use I have ever figured out for formal charges is that they are just another way to depict coordinate covalent bonds. In the structure you posted for nitric acid above, instead of showing a line with an arrow-head (indicating coordinate lone pair donation), the author has simply drawn a solid line and denoted the associated formal charges on both the atoms involved. You see a similar formal charge in the ammonium ion. Notice that in ammonium ion, the formal charge of N is only +1, whereas it's O.N. is -4. Understood? $\endgroup$
    – Gaurang Tandon
    Commented Mar 12, 2018 at 6:42
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    $\begingroup$ @Sarthak123 That formal -ve sign just indicates that neither of the electrons in the bond is from oxygen itself (in a normal covalent bond, one electron in the bond comes from each atom involved); similarly, the formal +ve sign indicates that both the electrons of the bond were given by this atom; your question basically boils down to "why is the lone pair on a real negatively charged O atom only considered while that of the formal negative charged O atom ignored"? (when comparing stability of nitrate and nitrite anions) $\endgroup$
    – Gaurang Tandon
    Commented Mar 12, 2018 at 7:48

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