[Burner]

Hi everyone,

Why O is not the central atom for the molecule N2O? Why the structure of it must be N(triple bond)N-O?

[Corribus]

Why don't you work out the formal charges on each atom for the two possibilities. This may give you some clue.

[Burner]

I know that N atoms tend to gain 3 electrons and O tends to gain 2 electrons. However, for N(triple bond)N-O, the central atom N have 9 electrons(gained 4 electrons) in the outermost electron shell and the O has 7 electrons(gained 1 electrons), which doesn't obey the octet rule either.

If O is the central atom for N2O, the structure will be N(triple bond)O(triple bond)N, the two N atoms have 8 electrons, while the O has 12 electrons. Though it is weird for O to have 12 electrons, why only the structure N(triple bond)N-O is possible(since both structures don't obey octet rule)?

[mikasaur]

If O is the central atom for N2O, the structure will be N(triple bond)O(triple bond)N, the two N atoms have 8 electrons, while the O has 12 electrons. Though it is weird for O to have 12 electrons, why only the structure N(triple bond)N-O is possible(since both structures don't obey octet rule)?

The molecule [itex]\ce{N#O#N}[/itex] is not possible because period 2 elements must follow the octet rule. There is no d-orbital expansion like what can happen for Period 3+ elements, e.g. SO₃.

You say for [itex]\ce{N#N-O}[/itex] that the central N atom has 9 electrons, but I'm unsure of how you got to 9. Corribus asked if you could figure out the formal charges of all the elements in all the proposed molecule configurations. What are the formal charges? If you have a bunch of potential formal charge arrangements for a molecule, which one is generally "correct"?

[Burner]

Oh...I think I have mixed up 'formal charge' with other terms, since I didn't learn it from school in fact. Sorry for that.

So, the formal charge of the leftmost N should be: 5-2-(6/2)=0
The formal charge of the central N atom is: 5-1-(8/2)=0
The formal charge of the rightmost O atom should be: 6-5-(2/2)=0

Thus, N≡N−O should be the structure of molecule N2O.

Please kindly correct me if I have made any mistakes. Thnaks.

[mikasaur]

You have the right structure (well one of them anyway) but your formal charges are a little off.

I imagine you're using the formula Formal Charge = Valence - Lone - 1/2 Bonded electrons.

So for N≡N-O, make sure you think about the location and number of lone pairs. Specifically, look at the central N and the O atoms. Why have you assigned them 1 and 5 lone electrons, respectively?

Once you do that, try to think if there's anything besides N≡N-O that makes sense...

[Burner]

Sorry, I have overlooked the fact that period 2 elements must follow the octet rule.

So, for the rightmost O, it must have 8 electrons in the outermost shells. It has 6 valance electrons, so to obey the octet rule it has to form a dative bond with N or double bonds. Then for N≡N-O, the formal charge or central atom N and the O atom should be +1 and -1.

If the O forms double bond with N, i.e. the structure becomes N=N=O... then I would like to know if there can be dative bond(s) in double bonds(i.e. one atom contribute 3 valance electrons and one atom contribute 1 valance electron to the double bond)? Otherwise, I don't think any other structures make sense.

[mikasaur]

I am not very familiar with dative (coordinate) bonds. That said, it looks like for :N≡N: one of those N atoms would need to donate its lone pair to the O to form a single coordinate bond (which behaves like any other covalent bond at that point). So you might be right about that. I'll leave it to someone who understands more about coordination chemistry to chime in.

Coordinate bonds can occur in double or triple bonds like in the case of carbon monoxide. If I had to guess, I'd say the O=N bonds are non-dative and one of the bonds between N=N is non-dative. So in O=N-N the central N can then donate its lone pair to form the second part of the double bond in N=N. So the N=N bond is half dative and half non-dative. Though once again someone more versed in coordination chemistry would probably explain it better than I can.

I think you're thinking a little too hard about this though. For something this simple it's not terribly important which atom provides the electrons for each bond. If the number of valence electrons is correct, you've followed the octet rule, the formal charges of the atoms make up the charge of the molecule, and you've minimized the variance in formal charge then you've probably identified the correct arrangement.

That said, how do you think N=N=O compares to N≡N-O? Might you predict that they'd be equally likely?

[Burner]

Assuming that N=N bonds in the structure N=N=O is half dative and half non-dative, we have the formal charges of the two N atoms to be -1 and +1 respectively. So, this structure seems to be equally likely to exist with N≡N-O.

[sjb]

Consider differences between N and O.

[Burner]

I am sorry, but I feel that it is too much for me to understand now, as I know little about formal charges, Lewis structures, orbitals and other related concepts. I can already convince myself to 'know' that the molecule N2O is N=N=O or N≡N−O but not O being the central atom now. Thanks for guiding me to think of this problem.

[mikasaur]

Consider differences between N and O.

Well now I'm curious. Is it that O is more electronegative than N, and so N≡N-O is more likely because of the negative formal charge on O?

[Corribus]

I think the question has become over complicated.

As a basic rule of thumb, N likes to bond three times and O likes to bond two times. In these cases the formal charge is 0. If N or O is bonded more or less than these optimal times, the formal charge is usually equal to the amount of bonds more or less than this ideal amount. Meaning that if O is only bonded once, the formal charge is -1. If O has three bonds, the formal charge is +1. We could go through and prove that but let's just take it as axiomatic right now.

Now lets consider N₂O. There are two ways the atoms could be arranged: N-N-O and N-O-N. Why is the former preferred?

Under the assumption the molecule has to be neutral, there are two reasonable resonance structures for N-N-O: N=N=O and N≡N-O (you could also have N-N≡O, but it will shortly become apparent why this is not typically included). The respective formal charges for each of these structures are {-1, +1, 0} and {0, +1, -1}. Assuming these structures are equally likely, there is a net 1/2 unit of negative charge on each terminal atom and a full unit of positive charge on the central atom.

For N-O-N, the reasonable resonance structures are N=O=N and N≡O-N (or equivalently N-O≡N). Here the formal charges are {-1, +2, -1} and {0, +2, -2}. On average, the two terminal atoms have a formal charge of -1 and the central atom a formal charge of +2.

In essence, to stabilize N-O-N requires a much greater concentration of formal charge on the various nuclear positions than N-N-O. This makes N-O-N unfavorable compared to N-N-O, because concentrating charge in once place is usually energetically unfavorable. (Add to this that oxygen is the more electronegative element, it is especially unfavorable for it to accommodate a +2 formal charge.)