# Predicting electronic geometry, observable geometry, and hybridization for any atom in a molecule

Can someone please explain the intuition around the answers to this problem:

I'm finding it very confusing because it seems like you have to nitpick between two resonance structures to get the desired results. Do we just always look for the "lowest" possible VSEPR in any resonance structure? Why isn't the observable geometry for "X" linear instead of none?

• I'm not sure if it matters, but this really isn't a homework problem – Bob Dylan Sep 28 '13 at 2:47
• But it looks like a homework problem or a problem from a text book or study guide. Even if this is for self-study, our approach is still the same. We want to help you get to the answer and not give you answer. For more info, see: How do I ask homework questions on Chemistry Stack Exchange – Ben Norris Sep 28 '13 at 12:56
• There's something funky going on in this example, because resonance does not require translational motion of nuclei. can you provide the reference to this VSEPR vs. observable geometry? – bobthechemist Sep 28 '13 at 21:49

Do we just always look for the "lowest" possible VSEPR in any resonance structure?

Yes. Consider the two resonance contributors below for a N,N-dimethylacetamide, which has a functional group in common with your molecule. The rule of thumb I teach is: **if an atom appears to be $sp^2$-hybridized in one resonance contributor and $sp^3$ in another, then that atom is $sp^2$-hybridized always.

In the figure above, I have used the curved arrow notation to represent how the lone pair on the nitrogen atom is delocalized into the $\pi$-system. In order for this delocalization to occur, that lone pair needs to be in a $p$ orbital. In order for that nitrogen atom to have a $p$ orbital, it must be $sp^2$-hybridized (or possibly $sp$-hybridized if applicable). All three of the atoms (N, C, and O) in the delocalized system need a p orbital, and thus must be $sp^2$-hybridized.

Once you have hybridization, then you have limited your choices of geometry:

• $sp^3 \implies$ tetrahedral geometry group: tetrahedral, trigonal pyramidal, bent, "none"
• $sp^2 \implies$ trigonal planar group: trigonal planar, bent, "none"
• $sp^{\ \ \ } \implies$ linear group: linear, "none"

Why isn't the observable geometry for "X" linear instead of none?

You are confusing observable geometry at an atom with molecular geometry.

Observable geometries at an atom are a relationship describing the relative geometric positions of other atoms bonded to that central atom. The central atom is not counted.A linear arrangement means that the atom in question is bonded to two other atoms or groups and those two atoms or groups are $180^\circ$ apart. If there is only one atom or group bonded to the atom in question, we cannot define the geometric relationship between that atom or group and the other atoms or groups bonded to that atom because there are not other atoms or groups. Consider the following examples from the tetrahedral group. The last case is "none" or "undefined, because you need at least two things to have a relationship.

This is different than molecular geometry which deals with the shape of the entire molecule. Let's look at HF. The shape of the molecule is linear, because the two points define a line (segment). The geometry around F is "none" or undefined. We cannot describe the arrangement in space of the hydrogen atom relative to other atoms bonded to fluorine because there are none.

I've never seen this distinction made between "VSEPR geometry" and "observed geometry", and it does not make much sense to me. It would seem more correct that what they call "VSEPR geometry" includes the lone pairs, while their "observed geometry" is only the arrangement of atoms, ignoring the lone pairs. Thus, geometry around X is ill-defined ("none"), because it has only one neighbor (linear geometry require two neighbors, like the central C in O=C=O).