Chemistry - The Lewis structure of HNO3

Solution 1:

Nitrogen is in the second row with no $d$ orbital in the valence shell. It obeys the octet rule and cannot have more than 8 electron.

There are exceptions to the octet rule. Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.

As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule. However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.

According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts. However, their contributions are not even. some of them are more favorable than others.

For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds. Based on octet rule alone, there are 3 possible resonance structures that are favorable.

Resonance structures of HNO3

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges. As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.

It is also common to write a mixture of resonance structures as the hybrid form

Hybrid form of HNO3

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms. The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.

Solution 2:

Well! here is a short quick reply..!!

I was pretty sure that the lewis stucture of $\ce{HNO3}$ would be the one with 0 formal charge and than I found this. Though this link only shows how to calculate formal charge on $\ce{HNO3}$ but it gives a little hint.

Secondly a thought came to my mind that if you will focus on Resonance, there will be double bond character on both bonds between N & O.


Solution 3:

Historically, there was nothing wrong with structures like the one you drew on the right. Iwan Ostromisslensky had no problems drawing 4-chloro-1,2-dinitrobenzene with a pentavalent nitrogen in 1908.[1] (Yes, a nitro group is not nitrate or nitric acid, but after some time of searching I just took what I found to prove my point.)

pentavalent nitrogen

Staudinger and Meyer drew dimethylanilinoxide similarly in 1919:[2]

dimethylanilin oxide

I wasn’t able to find the point in time when preferences changed so I gave up. It must have been sometime around when the quantum chemistry of orbitals was becoming better and better understood, and it was realised that there are only four orbitals nitrogen has access to for bonding (2s and three 2p). Henceforth, people described structures with nitrogen in the centre with four bonds only, better reflecting reality.

For elements of higher periods, most notably sulfur and phosphorus, many people still choose to draw too many bonds rather than write charge-separated structures. This is often explained with ‘d-orbital participation’ — but from a practical point of view, the 3d orbital has a very similar energy to 4s yet nobody suggests 4s participation. All ‘extended octet’ structures can be drawn in a way that conforms to the octet rule, so maybe it is only a matter of time until $\ce{P=O}$ bonds in phosphate disappear.

As for rules to determine the likelihood of Lewis structures, this is a better set:

  1. Lewis structures in which all atoms have an octet (doublet for hydrogen) are preferred.

  2. If 1. cannot be fulfilled, Lewis structures which have the least number of elements with sub-octet structures are preferred.

  3. If 1. or 2. generate a set of possible structures choose one that has a minimal number of formal charges.

  4. If 3. leaves a set of possible structures, choose one where the formal charges are distributed according to electronegativity (electronegative elements having negative formal charges).

  5. If 4. leaves a set of possible structures, choose one in which the formal charges are closer together.

  6. If your final structure has an expanded octet on a main group element, start again at 1.


[1]: I. Ostromisslensky, J. Prakt. Chem. 1908, 78, 263. DOI: 10.1002/prac.19080780121.

[2]: H. Staudinger, J. Meyer, Helv. Chim. Acta 1919, 2, 608. DOI: 10.1002/hlca.19190020161.