[ajkjmspe]

I apologize as I'm sure this question has been asked before and also I'm sure I'm confusing or overlooking something, but my question is why selenium can form one double bond and one single bond as in the molecule SeO2.

My textbook gives the answer to "Write a Lewis structure" for this molecule as the following two resonance structures (with three lone pairs of electrons around the O with the single bond, one lone pair around the Se, and two lone pairs around the O with the double bond):

• O-Se=O <-> O=Se-O

However, as I understand it, selenium is in Group 6A and thus needs only two valence electrons to form an octet.  As I see it, that should limit selenium to only forming either two single bonds with two partner atoms or one double bond with a partner atom.

(I am aware of expanded octets, but this does not appear to be the case here as both resonance structures in the answers provided by the book show selenium as having an octet with no more or less electrons.)

I do not understand: if selenium can form three bonds, would that not mean the number of valence electrons is not a limiting factor on the number of bonds?  And thus any element can form however many bonds as each situation requires?

I am curious as to what makes selenium unique (and if it is unique in this fashion) and why other elements cannot form as many bonds as required.  For example, if selenium can form three bonds, can hydrogen form two?  Can carbon form more than four?

I kindly request some clarification on this.

[Borek]

what makes selenium unique

What about SO₂? O₃?

[ajkjmspe]

Drawing the Lewis structures for SO2 and O3, I get these (resonance structures not shown):

• O-S=O
• O-O=O

For both SO2 and O3, I have three lone pairs on the single bond terminal O, one lone pair on the central O, and two lone pairs on the double bond terminal O that matches the requisite total of 18.

It would appear that selenium is not unique in this regard, but I still would request some clarification as to how and why this works.

S and O (like Se) are both Group 6A with 6 valence electrons and need two additional electrons to complete their octets.

According to these examples, it would appear that the number of valence electrons to form an octet (get to the next higher noble gas configuration) is not a limiting factor in the number of bonds these elements can form?

Or is it that each "electron group" (in this case, the double bond) counts as 1 bond?

If it is either of these cases, then I would like to know if S and O can form three or four bonds like:
X -
     O - X
X -
  <or>
X -
     O = X
X -
  <or>
X -   - X
     O
X -   - X

If yes, then why is C noteworthy for being able to form four bonds?
If not, why can S/Se/O form three bonds (one more than the number of valence electrons they need to complete their octet) but not four (or more)?
I am guessing the answers to all these questions somehow involves the double bond - should the double bond be counted as one bond (at least in terms of how many bonds an element can form)?
But if the double bond is counted as one bond, then why can H not form double bonds (the 1s orbital supports two electrons)?

I know there is most definitely a really obvious simple thing I am missing, but I just cannot seem to understand why certain elements can apparently ignore the "number of needed valence electrons to form an octet" being a bond-number limiting factor while others cannot?

[Corribus]

First, it needs to be pointed out that the most stable form of SeO₂ is a solid linear polymer. Se has similar bonding chemistry to that of other chalcogens (O, S, Se, Te). However as with most groups, as you go down the periodic table, pi bonds become weaker and sigma bonds become more favorable. This is because electron repulsion effects and p-orbital overlap are both weaker as you go down the periodic table. This is why, for example, the pure form of oxygen at STP is the double bonded dimer O₂, the pure form of sulfur is clusters of single-bonded S nuclei, and the pure form of Se and Te is more metallic, 3-dimensional solid.

You may also consider the oxides of each of the chalcogens, O₃, SO₂, SeO₂, and TeO₂. Ozone is a gaseous molecule with high reactivity, SO₂ is a fairly stable gaseous molecule, SeO₂ is a linear polymeric solid that can exist in the gas phase in a molecular form, and TeO₂ is a crosslinked polymeric solid with pretty high melting point.  This trend is due to the stability of chalcogen single and double bonds as you go down the group.

Formally, you could draw the same Lewis structure for each of these oxides. But as you see, it becomes less in agreement with reality as you move down the group, illustrating that Lewis structures are only useful to a point and do not take into account the various sigma and pi bonding energies that determine what the actual bonding structure is.

[Corribus]

If yes, then why is C noteworthy for being able to form four bonds?
If not, why can S/Se/O form three bonds (one more than the number of valence electrons they need to complete their octet) but not four (or more)?
I am guessing the answers to all these questions somehow involves the double bond - should the double bond be counted as one bond (at least in terms of how many bonds an element can form)?
But if the double bond is counted as one bond, then why can H not form double bonds (the 1s orbital supports two electrons)?

You are overthinking things. You need only start with a few simple rules. The rest is counting electrons to make sure the total valence electrons in the molecule is equal to the total valence electrons coming from each atom. The rules are, for most stable molecules, hydrogen nuclei like to have 2 electrons around them, and everything else likes 8. Lone pairs are worth 2 and each bond is worth 2. That's it.

Take the simpler SeH₂. Count valence electrons: you get 8, one each from the hydrogens and six from the selenium. Since hydrogens like two electrons, they can only bond once. Therefore the only option in this molecule is H-Se-H. Four of the valence electrons go to the bonds, leaving four that get left on Se (two long pairs). This is the same formal bonding structure as water. Similar logic gives the structure of methane, which has also a total of eight valence electrons. One bond each for the four hydrogen atoms, leaving none left over for carbon because each bond takes two. The carbon has no lone pairs, just four C-H sigma bonds.

Now SeO₂. Here you have six valence electrons from each nucleus, for a total of 18. It's just a matter of arithmetic really, there's only one way you can distribute bonds and lone pairs to make the electrons add up so that each nucleus has eight total, and that's one oxygen with two bonds (double bond) and two lone pairs, one oxygen with one bond (single bond) and three lone pairs, and the selenium with three bonds (double bond to one oxygen, single bond to the other) and one lone pair. To see why this is the only possibility that works, consider an alternate possibility where each Se-O bond is a double bond. That would mean that the selenium is surrounded by four bonds, which means it doesn't have any lone pairs (since it wants 8 - that's the rule we have to work with). Each oxygen thus has two bonds and, to make eight electrons, must also have two lone pairs. That would seem to work out, but now count the total valence electrons in the molecule and you find a problem: this molecule would have only 16 total valence electrons (eight from lone pairs and eight from bonds). Ergo, that Lewis structure doesn't work for neutral SeO₂.