A question on the 1996 AP Chemistry Free Response asks:

The $\ce{N-O}$ bonds in the $\ce{NO2-}$ ion are equal in length, whereas they are unequal in $\ce{HNO2}$. Explain.

Since the nitrogen dioxide ion has resonance, the $\ce{N-O}$ bonds are equal as resonance is in reality a hybrid of all of the possible structures for a certain molecule. However, the answer then assumes that $\ce{HNO2}$ has no resonance but rather one single and one double bond ($\ce{N-O}$ and $\ce{N=O}$), and therefore not all of the $\ce{N-O}$ bonds are equal.

How is this possible? Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?


3 Answers 3


Resonance is a remnant of valence bond theory, which is necessary because it is impossible to describe delocalised bonding within a localised bonding scheme. Resonance structures are not real. One Lewis structure describes one electronic configuration with fully localised electrons. For many molecules it is completely sufficient to describe the bonding situation with one such structure in a first order approximation. For many other molecules this is completely insufficient. For $\ce{NO2-}$ this is the case, hence you need a superposition of more than one structure to get closer to a qualitative description. This caveat is not necessary in molecular orbital theory, since this is designed to treat delocalisation.

Nitrite anion

One question therefore arises from the above: Are the $\ce{N-O}$ bond lengths equal because there is resonance, or is there resonance because the $\ce{N-O}$ bond lengths are equal?

This is obviously more of a philosophical question, but it shows the dilemma of the whole concept. The bonding of any given molecule is a result of many different factors; to name but a few: repulsion of the nuclei, repulsion between the electrons, attraction between electrons and nuclei.

For $\ce{NO2-}$ there is one key question, that gives us a very important clue as to why the bond lengths are equal: Why should one oxygen atom be different from the other? In fact the molecule is symmetric (C2v), therefore the bond lengths must be equal.
If you try to write a Lewis structure that respects this symmetry, you will end up with an electron sextet on nitrogen (and a formal charge of +2), which certainly can only be an insufficient approximation. If you write a Lewis structure with fewer formal charges, keeping symmetry, you will notice that nitrogen will have 10 electrons. That would mean that it would have to use d-orbitals, which simply are not available for nitrogen (more on that another time). There is another way to write a Lewis structure, however, this one has to break the symmetry, and you'll notice that there is a second one, which is equal. In order to keep the symmetry you introduce resonance, to state, that the "true" electronic structure is a superposition of (at least) those two configurations.

TL;DR: The $\ce{N-O}$ bonds in the nitrite anion are equal, because the oxygens are equal.

Nitrous acid

It is false to say that in $\ce{HONO}$ is no resonance. And it is also wrong to say that a lack of resonance makes the bonds unequal. If the solution manual claims this, it is wrong.

Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

These are valid resonance structures of nitrous acid. You have to keep in mind though, that resonance does not change the geometry of a molecule (or molecular structure). The contribution of the Lewis structure with fewer formal charges to the total wave function will be much higher than the contribution of the other structure in this case. This is not a general statement, there are molecules where the charge separated Lewis structure has a higher contribution to the wave function.

By analogy to the anion case, there is one key question, that gives us a clue about the molecular structure: If you bind a proton to one oxygen, are both oxygen really equal anymore? In this case the symmetry of the molecule is reduced to Cs. Therefore the bonds, cannot be equal.
For this molecule you will have no problem writing a single Lewis structure that satisfies every imposed criteria.
To answer the initial question: The bonds in $\ce{HONO}$ are unequal, because the proton is bound to one of the oxygens. It is a very polar, but covalent bond. The electron density of the $\ce{N-O}$ bond is therefore shifted towards the $\ce{H-O}$ bond, and simultaneously reduced for the $\ce{N-O}$ bond when compared to the anion. Therefore the $\ce{HO-N}$ bond is longer than the $\ce{N=O}$ bond (I implied the most contributing Lewis structure here).

TL;DR: The $\ce{N-O}$ bonds in the nitrous acid molecule are unequal, because the oxygens are unequal.

There frequently is the notion of a most stable resonance structure. While this is very popular, it is completely wrong. I again refer to the question What is resonance, and are resonance structures real? Resonance is used to describe the electronic structure of a molecule, there is no change in the molecular structure allowed. It is necessary to understand, that only the description in terms of all possible resonance structures is in itself correct, i.e. the theoretical limit of valence bond theory, everything else is an approximation. However, not all resonance structures will have the same contribution to the wave function. Whenever someone speaks of the most stable resonance structure, it is likely that they mean the resonance structure with the highest contribution to the wave function. But even if you know what is meant, it is still wrong to speak of a most stable resonance structure, it must be corrected.

(I currently don't have the time to provide a full fledged answer, so instead I was trying to give some pointers in the right direction. I might come back to amend it whenever I have more time.)

  • $\begingroup$ If the the electron density of the N−O bond is shifted towards the O-H bond, shouldn't that decrease the bond length? $\endgroup$
    – Rick
    Commented Dec 18, 2017 at 13:05
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    $\begingroup$ @Rick No, usually when you remove electron density, the bond get weaker and longer. The attraction between electrons and nuclei pulls the latter together. If you remove electron density, the relative strength of the nuclei-nuclei repulsion gets stronger. $\endgroup$ Commented Dec 18, 2017 at 13:13

In $\ce{NO2-}$ ion the resonance structures are equivalent and are stable enough (since there is negative charge on more electronegative element). Resonance in $\ce{NO2-}$

In the contrary, canonical structure (II) of $\ce{HNO2}$ is very instable as the +ve charge is present on more electronegative element O (which is more electronegative than N). Also, the positive and negative charges are fairly apart and hence the stability is greatly reduced. Resonance in $\ce{HNO2}$

Thus, it can be concluded that due to more stable state of the structure (I), the $\ce{N-O}$ bond length are unequal in $\ce{HNO2}$.

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    $\begingroup$ This is almost equally as wrong as the answer previously given. There is no such thing as the most stable resonance structure! Read my comment to the other answer for more information. Also $\ce{NO2-}$ is bent, not linear! $\endgroup$ Commented Dec 18, 2017 at 6:47
  • $\begingroup$ @Martin-マーチン I have answered this question according to my analysis and it might be wrong due to lack of any other considerations. But, what makes you say that there is no such thing as most stable resonance structure??? In many of my books I have found the concept of stability of canonical structures. Why, just google it and you can find ample instance. $\endgroup$
    – DJ Koustav
    Commented Dec 18, 2017 at 6:56
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    $\begingroup$ I know that this is a concept taught in many schools, in many textbooks, it is very predominant on the internet. Unfortunately sheer quantity does not make it correct. In fact it has been taught wrong and it is one of the most common misconceptions in chemistry. It needs to be eradicated. This question explains very well, why the very notion of a most stable resonance structure is wrong. $\endgroup$ Commented Dec 18, 2017 at 7:03
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    $\begingroup$ No. Wrong is wrong. Just because your answer is accepted in your level, it does not make it correct. Even worse, it does make it sad. If your course is not able to get such an elementary theory right, who knows what other misconceptions are taught as well. It is a tough fight to get these false theories out of the heads of the students, and it is truly awful to know that the next generations of students in India will learn complete and utter garbage. It is even worse that some of them might go on to university and have to learn everything anew. It makes me very, very sad. $\endgroup$ Commented Dec 18, 2017 at 8:34
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    $\begingroup$ @ballaneypranav yes, that would be better. It needs to be taught still, that this is an approximation and that if taught without the other contributors, it will still lead to wrong conclusions. There really is little point to teaching resonance and then using only one contributor - it kind of defeats the whole purpose. But at least starting it that way, would not be wrong. $\endgroup$ Commented Jan 2, 2018 at 23:04

The idea is that in a resonant structure, all the structures are equally stable -- for instance, in CO3: CO3

, you can see that the formal charges on the Os initially are -1 on 2 of them and 0 on the last; these charges are consistent in each structure. However, in your example, the structures are: enter image description here

The formal charges on the Os in this case are 0 and 0 on the left, as compared to 1 and -1 on the right. These charges aren't the same, so this is a case of more stable structure rather than resonant structure.

Image sources: http://www.chem.ucla.edu/~harding/tutorials/resonance/draw_res_str.html https://s3mn.mnimgs.com/img/shared/content_ck_images/images/HNO2.png

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    $\begingroup$ No nitpicking, simply pointing out how wrong the answer is. The idea is that in a resonant structure, all the structures are equally stable [...] There is no most stable resonance structure, resonance structures in itself cannot be stable as they do not exist on their own (compare here). Even if that were not an issue with the statement, the very notion that they have equal contribution is wrong, too. You only display 3 resonance structures, there are many more with minuscule contributions. [...] $\endgroup$ Commented Dec 18, 2017 at 5:30
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    $\begingroup$ [...] Using equilibrium arrows for resonance is dead wrong. The same applies to $\ce{HNO2}$, formal charges are simply that: formal. A resonance hybrid is a superposition of all possible resonance structures, and it is only necessary in the valence bond framework. Basically everything in this answer is wrong, and it does not explain why the $\ce{N-O}$ bonds are different in length. To be clear: I down-voted this. $\endgroup$ Commented Dec 18, 2017 at 5:30

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