Resonance structures for nitric oxide

[Qube]

Homework Statement

Is N triple bond O with a radical on oxygen a valid LS for the nitric oxide molecule?

Homework Equations

Max number of valence electrons in a shell - 2n².

Oxygen is in second row of periodic table. Max valence electrons: 2(2)² = 8.

The Attempt at a Solution

I remember once I got reprimanded by my teacher for drawing some nitrogen-based molecule with five bonds afforded to nitrogen in a vain attempt to make nitrogen's formal charge 0. Unfortunately, nitrogen doesn't form five bonds regularly. Nitrogen isn't a hypervalent molecule. Nitrogen is in the second row of the periodic table and accessing the extra orbitals needed to form more than 4 bonds doesn't happen on a regular basis.

However, my teacher recently drew this depiction of the nitric oxide molecule and says it's a valid LS.

Uh, what?

Your thoughts, please. I understand that oxygen does have empty d-orbitals, and for that matter, empty f-orbitals, but really? Oxygen forming a triple bond on top of having a radical and a lone pair? How legit is the latter LS below?

http://i.minus.com/jHCK0jK0k7uVu.png

 

[AGNuke]

First things first, Oxygen doesn't have d-orbitals, or f-orbitals for that matters.

Secondly, the second LS is perfectly fine, it is just difficult to explain and imagine. You may say that it is quite contradictory to the Lewis Dot Structure conventions. Just talk about Oxygen molecule. You may show it as having a sigma and a pi bond, but the fact is that it isn't present that way. Each Oxygen molecule shares one sigma electron and one whole pi electron pair, such that each oxygen have one lone electron on them, enabling them to exhibit paramagnetic behavior, which can't be explained on the basis of your conventional Lewis Dot Structure. NO is also similar to that case.

Try learning about MO structure of NO to get a better insight at this.

 

[Qube]

First things first, Oxygen doesn't have d-orbitals, or f-orbitals for that matters.

Well it has empty d-orbitals and f-orbitals. I recall my teacher mentioning this.

Secondly, the second LS is perfectly fine, it is just difficult to explain and imagine. You may say that it is quite contradictory to the Lewis Dot Structure conventions. Just talk about Oxygen molecule. You may show it as having a sigma and a pi bond, but the fact is that it isn't present that way. Each Oxygen molecule shares one sigma electron and one whole pi electron pair, such that each oxygen have one lone electron on them, enabling them to exhibit paramagnetic behavior, which can't be explained on the basis of your conventional Lewis Dot Structure. NO is also similar to that case.

Try learning about MO structure of NO to get a better insight at this.

Can you elaborate? I'm not seeing it.

EDIT: I tried drawing the MO diagram for NO. I got four populated orbitals in the 2p level. What does this tell me? Again how is oxygen forming five bonds here?

EDIT 2: Okay, I realize now. MO diagram supplements the LS because MO diagrams show both empty orbitals as well as populated and partially populated orbitals. Oxygen isn't actually forming 5 bonds here because the picture my teacher drew is a resonance form. Just because he placed the unpaired electron on oxygen doesn't mean it exists there; it just shows that there is resonance and the bond order implied by the LS is lower than the actual bond order of 2.5. The four filled orbitals in the MO show that we are using four atomic orbitals on oxygen and nitrogen.

Nitric oxide (Figure 1b) is electronically equivalent to dinitrogen (N2) plus an electron, and as a consequence it is paramagnetic with one unpaired electron. The location of the unpaired electron in the π* orbital (Figure 6a) results in a bond order of 2.5 rather than the triple bond observed for N2 (Figure 6b).

 

[AGNuke]

Well, it may have any amount of orbitals, but the size of atom is such that the energy difference between 2p and 3s is so overwhelming that Oxygen can't even access it, leave alone 3d and 4f. So it is as good as non-existent.

 

[Qube]

Well, it may have any amount of orbitals, but the size of atom is such that the energy difference between 2p and 3s is so overwhelming that Oxygen can't even access it, leave alone 3d and 4f. So it is as good as non-existent.

Thanks for reiterating something I mentioned in my OP.