# Where does the 9th electron go in a N=O bond?

In the first resonating structure you can see 5 unpaired electrons and 4 shared electrons on nitrogen, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving? In the second resonating structure you can see that oxygen has 5 unpaired electrons on nitrogen and 4 shared electrons, then isn't this a extended octet? If it is so, then in which orbital that 9th electron is revolving? Can anyone explain completely about the above questions? If this question is already been asked please comment me the link of that question. Thank you in advance. Please explain with boxes if you can.

Nitric oxide is a tough molecule to represent as a Lewis structure. However, you have made one common mistake in your structures. I also want to clear up a misconception about resonance that's present in your post. Finally, I will need to introduce a difference bonding picture of $\ce{NO}$ to answer your question about which orbital the unpaired electron is in.

The Lewis Structure of $\ce{NO}$

The mistake you made in drawing those Lewis Structures is that you used all of the electrons in the nitrogen and oxygen atoms and not just the valence or outer-shell electrons.

A Nitrogen atom has 7 total electrons in the following configuration: $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. Of these seven electrons, only the 5 in the second shell (energy level) are available for bonding. These are the valence electrons. The $1s$ electrons are core electrons. They are not available for bonding because their probability density function maxes at less than the atomic radius.

Similarly, an oxygen atom can only use the six electrons in its second shell for bonding: $2s^2 2p_x^2 2p_y^1 2p_z^1$.

The bonding in $\ce{NO}$ thus only has 11 electrons and not 13 (like you have). With two fewer electrons, it's easy to place the unpaired electron in a way that does not violate the octet rule.

\begin{aligned} &.. \ \ \ .. &&\ &&&\ ..\ \ \ ..\ \\ \ce{.} &\ce{N=O:} &&\ce{<->} &&&\ce{:N=O.} \end{aligned}

Resonance

The most common misconception related to resonance is unfortunately suggested by the name of the phenomenon. The two resonance structures drawn do not exist as separate structures interconverting. They are attempts to represent a compound using Lewis structures that has bonding that is not localized. The true structure of nitric oxide is some hybrid of these resonance contributors.

Where is the unpaired electron?

To answer this question, we nned molecular orbital theory. A molecular orbital diagram describes the bonding in terms of constructive and destructive overlap of atomic orbitals. A MO diagram dispenses with the need for resonance - all electrons are clearly shown in their orbitals: $\sigma$ or $\pi$, bonding or antibonding. Since this answer is running long already, I won't go into how to construct such a diagram.

Here is the MO diagram for nitric oxide:

The unpaired electron (highligted in blue)is in a $\pi$ antibonding orbital that is formed from the destructive overlap of out of phase $p$ orbitals, one from nitrogrn and one from oxygen.

• Excellent answer as always, Ben. I have a slight nitpick about the molecular orbital diagram you found though; most references seem to have the relative energies of the $\pi_{2p}$ and $\sigma_{2p}$ switched. I found this excerpt of a book, which on page 379 has an excellent discussion of the electronic structure of $\ce{NO}$ using molecular orbital theory. – Nicolau Saker Neto Feb 2 '14 at 12:47
• The creator of this image must have thought $\ce{NO}$ to be more similar to $\ce{N2}$ than to $\ce{O2}$. The increasing degree of $s-p$ interaction causes the relative position of the $\sigma_{2p}$ and $\pi_{2p}$ to switch between $\ce{N2}$ and $\ce{O2}$. – Ben Norris Feb 2 '14 at 13:32
• Don't forget to start at a lower energy for the oxygen atom orbitals, because of the higher electronegativity (whatever that is, right...). Because of that, the single electron is energetically closer to N than O. – tschoppi Feb 2 '14 at 13:41

Your Lewis structures contain too many electrons. A molecule of $\ce{NO}$ has 11 valence electrons (5 from nitrogen, 6 from oxygen). Your proposed Lewis structures have 13 electrons, so both of them wound up having one atom with 9 electrons to compensate. With simple valence bond theory, the most accurate structure has a bond order of 2.5 resulting from the combination of two canonical structures, one with a double bond and another with a triple bond, and where both structures have the odd electron on the nitrogen atom.

Edit: As Ben Norris mentions in his answer, there actually is no satisfactory Lewis structure for $\ce{NO}$, which actually has a bond order of 2.5. The closest representation is a doubly-bonded molecule, and it is not possible to write a canonical structure with triple bond without going over the octet in one of the atoms. This happens because naive valence bond theory lacks the concept of antibonding bonds.