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This is the molecular orbital diagram for nitric oxide:

enter image description here

Now, we know that nitric oxide can dimerize. However, why does nitric oxide dimerize as to form an $\ce{N-N}$ bond? Why not an $\ce{O-O}$ bond? Wouldn't the lone electron be more likely "possessed" by the oxygen since oxygen is more electronegative (if we were to step away from molecular orbital theory for a second).

Is this a Coulombic effect? Oxygen, by nature of being more electronegative, attracts more electron density to itself. This in turn would weaken any hypothetical $\ce{O-O}$ bond due to electron-electron repulsions.

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    $\begingroup$ >This is the molecular orbital diagram for nitric oxide || It shows oxygen having orbitals with higher energy, which is obviously wrong. Please, check. As for the question, the $\ce{ONN}$ angle in N2O2 is ~100, so I would say the nature of the bond is debatable. While antibonding orbital obviously is shifted towards atom with higher initial orbitals, the angles suggests partial O-O or O-N interactions as well. $\endgroup$ – permeakra Aug 15 '14 at 21:59
  • $\begingroup$ What does bond angle suggest about the nature of the bond, exactly? How can you conclude that there are O-O interactions based on angles? $\endgroup$ – Dissenter Aug 15 '14 at 22:02
  • $\begingroup$ Oh wait nvm I see it after I drew out the diagram; the 100 degree bond angles deviate from the usual sp3 hybridized bond angle; this suggests that the oxygens may be interacting with each other and closing up the bond angle from 109.5 degrees. Are there really available orbitals on the oxygens to do this/ $\endgroup$ – Dissenter Aug 15 '14 at 22:03
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Something is not correct with your MO diagram. Here is a correct MO diagram for NO

enter image description here

Note that in this diagram the oxygen atomic orbitals are lower in energy than the nitrogen AOs. This makes sense since the AOs of the more electronegative element should always be at lower energy than the AOs of the more electropositive element (in your diagram the relative energies of the oxygen and nitrogen AOs are somehow reversed). The energetic ordering of the starting AOs is important as we'll see in a minute.

The closer in energy a MO is to an AO from which it was formed, the more character of that AO it shows. So let's examine the nature of the single electron in the $\ce{\pi_{2p}^{\ast}}$ orbital since this electron will be involved in forming the new bond when $\ce{NO}$ dimerizes. This MO is formed from both the nitrogen and oxygen p atomic orbitals, but it is closer in energy to the nitrogen p AO. Therefore, we would expect this electron to have more nitrogen and less oxygen character (in other words it will reside more around nitrogen than oxygen). Because this electron has greater density on nitrogen, dimerization to form an N-N bond is more likely than dimerization to form an O-O bond.

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    $\begingroup$ I think it would be "always", but I'm reluctant to use "always" so let me say "generally" the more EN atom will be at lower energy. $\endgroup$ – ron Aug 15 '14 at 22:11
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    $\begingroup$ "more likely" addresses the kinetics of the situation. There is a larger coefficient on N than O in the $\ce{\pi^{\ast}}$ orbital. So initial bond formation involving N is more likely, but certainly (assuming there is some coefficient on O) oxygen bond formation could occur too, just to a lesser extent. Now enter thermodynamics - no matter what predominates initially (kinetically), since the reaction is reversible, eventually we will wind up with a product equilibrium distribution matching the free energy difference between the isomers. $\endgroup$ – ron Aug 15 '14 at 22:15
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    $\begingroup$ Using standard bond energies calculate the heats of formation of the 2 isomers. If the difference is 3 kcal/m or more, then the equilibrium mix will be > 99:1 $\endgroup$ – ron Aug 15 '14 at 22:19
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    $\begingroup$ For a back of the envelope calculation - yes. Further, in this case, I would expect the two reactions to be very similar entropically. $\endgroup$ – ron Aug 15 '14 at 22:27
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    $\begingroup$ That would make another good question. $\endgroup$ – ron Aug 15 '14 at 22:39

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