How do we know ##BF_3## has Lewis structure but ##Al_2S_3## does not?

In summary, the first thing we notice is that we have a metal and a nonmetal, so we would think about an ionic model. These elements form the ions ##Al^{3+}## and ##S^{2-}##. These ions form a lattice structure and we get an ionic crystal. But apparently no bond is perfectly ionic or covalent. The electronegativities of aluminium and sulfur are 1.6 and 2.6, respectively. The difference is 1, which is a relatively low value, indicating a high covalent character to the bond formed. However, I am not sure if this difference in electronegativity is relevant when we are analyzing an ionic compound.
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Homework Statement
My question is about concepts of Lewis structures and VSEPR model. I will go through my reasoning about the following problem and questions will come up along the way.
Relevant Equations
Consider ##Al## and ##S## atoms and reason about the types of bonds formed between such atoms.
The first thing we notice is that we have a metal and a nonmetal, so we would think about an ionic model.

These elements form the ions ##Al^{3+}## and ##S^{2-}##.

These ions form a lattice structure and we get an ionic crystal.

But apparently no bond is perfectly ionic or covalent.

The electronegativities of aluminium and sulfur are 1.6 and 2.6, respectively. The difference is 1, which is a relatively low value, indicating a high covalent character to the bond formed. However, I am not sure if this difference in electronegativity is relevant when we are analyzing an ionic compound. I learned it in the context of ascertaining the ionic character of a covalent bond.

However, if we think in terms of polarization and polarizing power, then we have an ion with a relatively small radius and high charge (##Al^{3+}##) and an anion (##S^{2-}##). We should expect some distortion in the electron cloud of the anion and this occurrence means the bond has relatively high covalent character. This covalent character would be even larger if the anion were a larger, heavier atom. Such atoms have lower effective nuclear charge and more shells and their electron clouds are more easily distorted.

Question 1: Does any of this mean we can form a Lewis structure for a compound like ##Al_2S_3##?

After all, there is a Lewis structure for ##BF_3##, and ##B## forms ##B^{3+}## ions.

Instead of being an ionic compound it is covalent with a central boron that has an incomplete octet. It does, however, have high ionic character since the difference in electronegativity is ##\chi_{F}-\chi_{B}=2##.

Question 2: How do we know we can't have the same thing for ##Al_2S_3##?

Now, I'm asking this from the point of view of Lewis Structures and VSEPR.

It has been frustrating to study these topics because many things seem ad-hoc.

Here is another example

##N_2O##

One rule of thumb is to assume the central atom is of the element with the lowest ionization energy. That would be oxygen.

Apparently (by way of googling this molecule), another rule of thumb is to choose the element with the lowest electronegativity. That would be nitrogen.

So, how in the heck do we ever know, given a formula, what Lewis structure to build?

My guess is that these are all experimental results, and actually there is nothing that tells you whether oxygen or nitrogen is the central atom other than listing out exceptions upon exceptions to basic rules of thumb.

EDIT: just after writing this I realized that ##O## can't be the central atom because of another rule of sorts. If ##O## were the central atom it would have expanded valence. But only p-block atoms of elements in period 3 or later can have expanded valence because these elements have empty d-orbitals.

I just want to make sure that this is really the way to study this stuff. I'm self-studying Chemistry. It seems like a bunch of cases and some rules.

Question 3: Is there a light when one learns about quantum mechanics?

I do have one more question though. When we do write out the Lewis structure of ##N_2O##, we get

1690093077804.png


Question 4: Now, ##N## has five valence electrons. It appears only four are being shared in the structures above, and the fifth one went over to the oxygen. Either that or three electrons are being shared, and the fourth bond uses both electrons from ##N##.
 
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Basically you are discovering that simplified models fail when analyzed too deeply.

At the same time you are using intuitions you have built thanks to these simplified models, so they served their purpose.

And yes, in the end QM should give the final answer. To some extent it does - but you may not like the way it does. With exception of the simplest cases solving Schroedingers equation can be done in two ways - exact (which produces numbers, but no insight) and approximate - which produces insight and generalizations, but in the end they are always limited in their applicability, just like models you are using now.
 
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1. How do we determine the Lewis structure of a molecule?

The Lewis structure of a molecule is determined by following the octet rule, which states that atoms tend to gain, lose, or share electrons in order to achieve a stable outer electron shell with eight electrons.

2. Why does ##BF_3## have a Lewis structure while ##Al_2S_3## does not?

##BF_3## has a Lewis structure because it follows the octet rule, with each atom having a full outer electron shell. However, ##Al_2S_3## does not have a Lewis structure because it does not follow the octet rule. The aluminum atoms only have six electrons in their outer shell, while the sulfur atoms have 12 electrons, making it difficult for the atoms to achieve a stable electron configuration.

3. Can molecules with incomplete octets have Lewis structures?

Yes, molecules with incomplete octets can have Lewis structures. These molecules, also known as hypervalent molecules, have atoms with more than eight electrons in their outer shell. Examples include molecules with central atoms from the third row of the periodic table and beyond, such as sulfur hexafluoride (##SF_6##).

4. How can we predict the Lewis structure of a molecule?

The Lewis structure of a molecule can be predicted by looking at the number of valence electrons in each atom and determining how many bonds and lone pairs are needed for each atom to achieve a stable outer electron shell. The structure can also be confirmed by using experimental techniques such as X-ray crystallography.

5. Are there exceptions to the octet rule?

Yes, there are some exceptions to the octet rule. These include molecules with an odd number of valence electrons, molecules with expanded octets (more than eight electrons in the outer shell), and molecules with incomplete octets as mentioned earlier. Additionally, some molecules may have resonance structures, where the electrons are delocalized and the octet rule is not strictly followed.

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