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I'm a little confused as to why we ignore the effect the lone pair has on the geometry of the molecule when it is participating in resonance. Wouldn't it be the case that the molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal? Besides, the resonance with trigonal planar shape has a less importance because it has formal charges and the other does not.

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    $\begingroup$ "molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal?" Pretty much that, but trigonal planar has more importance, as far as geometry is concerned. $\endgroup$
    – Mithoron
    Jun 5 at 13:50
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    $\begingroup$ chemistry.stackexchange.com/questions/4390/… $\endgroup$
    – Mithoron
    Jun 5 at 13:58
  • $\begingroup$ chemistry.stackexchange.com/questions/30943/… $\endgroup$
    – Mithoron
    Jun 5 at 13:58
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    $\begingroup$ Molecular structure is determined by the balance of electronic effects. Resonance is a model necessary to address the shortcomings of Lewis structures; what is depicted as molecular structures are hypothetical electron distributions belonging to a single nuclear arrangement, the disposition of which lead to a good approximation of the electronic structure. See also: chemistry.stackexchange.com/q/51632/4945 $\endgroup$ Jun 5 at 18:48
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Wouldn't it be the case that the molecule is actually in some hybrid state of trigonal planar and trigonal pyramidal?

"Hybrid state" refers to the distribution of electrons, not nuclei. Different resonance structures can have different electron distributions, but they can't have nuclei in different places, which means that they can't have different geometries. Either you have both of them be pyramidal, or both of them be planar.

Besides, the resonance with trigonal planar shape has a less importance because it has formal charges and the other does not.

Let's briefly digress. At its core, the question is what geometry does the amide have at nitrogen: is it pyramidal or planar?

  • Most nitrogens prefer to be pyramidal because of lone pair–bond pair repulsion, i.e. classic VSEPR theory. If the amide nitrogen adopts a pyramidal geometry, it therefore gains some stability from having the lone pair further away from the bond pair.

  • On the other hand, if it is pyramidal then the second resonance structure can't contribute at all (the bond angles at a doubly bonded nitrogen must be planar). So the resonance can only truly kick in if it is planar.

So it boils down to whether the LP–BP repulsion is more important, or the stabilisation from resonance is more important. How do you tell which one is more important, especially when neither of these can be measured quantitatively (with your current knowledge)...?

There are two possible answers to this:

  1. Come up with a quantitative way to measure the repulsion, as well as the stabilisation, so that you can actually say for sure which one is larger. This means lots of theoretical calculations and quantum mechanics.

  2. Take an amide, go into a lab, and measure its bond angles (or some other useful molecular property). You'll find that it turns out to be planar at nitrogen. Logically, that means that the resonance must be more important. How much more important we don't know; but we know it is more important.

So, it doesn't really matter what you or I "think" about the importance of the resonance: the fact is that the amide is planar. The explanation using resonance is merely a way to rationalise the experimental observation that amides are planar.

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    $\begingroup$ Thanks for the thorough response. I still am not sure of "Different resonance structures can have different electron distributions, but they can't have nuclei in different places, which means that they can't have different geometries." But the lone pairs themselves contribute to geometry because of repulsive forces. So why wouldn't different electron distributions cause different geometries? $\endgroup$
    – Ameji B.
    Jun 5 at 14:41
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    $\begingroup$ @AmejiB. Hmmm... That is true for molecules, but resonance structures aren't molecules which have come to equilibrium after figuring out all the forces involved. They are hypothetical, nonexistent things, which represent possible states of molecules, which in all likelihood don't represent equilibrium states. So, yes, the resonance structure with planar N and lone pair on N, is probably not very good. But this is not how the molecule ends up, the molecule is not the resonance structure. $\endgroup$
    – orthocresol
    Jun 5 at 15:21
  • $\begingroup$ Why can't you have nuclei in different places? $\endgroup$
    – Vikki
    Jun 5 at 22:53
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    $\begingroup$ @Vikki-formerlySean It's not really a rule, it's more about what resonance structures are. Resonance structures are not tools to explain the overall structure of a molecule incl. nuclear geometry. They are meant as tools to describe the electronic wavefunction of a molecule: the idea is something like this: resonance structure A is $\psi_A$, resonance structure B is $\psi_B$, and the real electronic wavefunction (aka the hybrid) is the superposition $c_A\psi_A + c_B\psi_B$ (where $|c_A|$ and $|c_B|$ are somewhere between 0 and 1). $\endgroup$
    – orthocresol
    Jun 5 at 23:02
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    $\begingroup$ Crucially, all of these $\psi$'s are wavefunctions of electrons only, not nuclei. The entire framework is contingent on having already chosen a geometry for the nuclei. I'm not really sure the comments are the correct place to get into this in detail, but you may want to look up the Born–Oppenheimer approximation, which puts all of this on a much firmer theoretical basis. (I was not sure whether to go into B–O in the answer: I thought it would be a bit too much.) $\endgroup$
    – orthocresol
    Jun 5 at 23:03

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