What is "dot" sign in •NO? I know that it is radical nitric oxide, but I don't know if it is necessary to put the "dot". Is there any difference between •NO and NO?

  • 9
    $\begingroup$ The dot is the radical (unpaired electron). $\endgroup$ – Michael Lautman Jun 14 '19 at 12:32
  • 7
    $\begingroup$ The answer is NO. $\endgroup$ – electronpusher Jun 14 '19 at 22:47

Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron. This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron.

Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen. The unpaired electron is actually incorporated into the molecular orbital structure of the nitric oxide molecule, covering both atoms.


It is not necessary, but optional, to express explicitly the radical status.

In other cases, like alkyl radicals, the dot marking is mandatory, not to be confused e.g. with a functional group.

For curiosity, the ground state of the oxygen molecule - triplet oxygen - is a biradical, with 2 unpaired electrons. What we write as $\ce{O=O}$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.


Nitric oxide ($\ce{NO}$) is a free radical and hence why that dot is for. Explanation for the dot and the reason why its there have been well answered in other responses. Nevertheless, I decided to put the molecular orbital representation of $\ce{NO}$ as depicted below for your convenience (Ref.1):

MO of Nitric oxide

There are three electrons in antibonding orbitals and eight electrons in bonding orbitals. One single electron occupying the highest energy level is in antibonding $\pi$∗ orbital (HOMO). It is an unpaired electron, and therefore, $\ce{NO}$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $\ce{NO}$ molecule: $\ce{NO}$ can easily be oxidized to become nitrosonium ion ($\ce{NO+}$) and be reduced to be nitroxide ($\ce{NO-}$). $\ce{NO}$ is unique among the diatomic biomolecules because it can bind to both ferric and ferrous heme iron due to its electronic configuration.


  1. Byung-Kuk Yoo, “Investigation of the mechanisms of regulations, activation, and deactivation of Guanylate Cyclase, the endogenous NO-receptor, and NO-sensors,” PhD Thesis, École Polytechnique, Paris, France, 2010 (https://tel.archives-ouvertes.fr/tel-00557106).

The dot represents an unpaired electron. It's written that way as a reduced Lewis dot diagram.

The reduction works like this:

  1. Start with the typical dot-diagrams for Nitrogen and Oxygen atoms, i.e.$$ \begin{array}{ccc} \textbf{Nitrogen} & \qquad & \textbf{Oxygen} \\[-10px] {\Huge{\begin{array}{rcl} & \cdot \phantom{\cdot} \\[-50px] \raise{0.1ex}{.} &\mathrm{N} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot} \phantom{\cdot} \end{array}}} & & {\Huge{\begin{array}{rcl} & \cdot \phantom{\cdot} \\[-50px] \raise{0.1ex}{:}&\mathrm{O} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot} \phantom{\cdot} \end{array}}} \end{array} $$

  2. Combine them to create nitrogen oxide:$$ \textbf{Nitrogen Oxide} \\ {\Huge{ \begin{array}{rcccc} \raise{0.1ex}{.} &\mathrm{N} & {\rlap{\raise{0.1ex}{: :}}} ~~ &\mathrm{O} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot \cdot} & & \raise{0.25ex}{\cdot \cdot} \end{array} }} $$

  3. Condense the representation by drawing the double-bond as in:$$ \textbf{Nitrogen Oxide} \\ {\Huge{ \begin{array}{rcccc} \raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot \cdot} & & \raise{0.25ex}{\cdot \cdot} \end{array} }} $$

  4. Identify the paired electrons,$$ \textbf{Nitrogen Oxide} \\ {\Huge{ \begin{array}{rcccc} \raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O} & \raise{0.1ex}{\boxed{:}} \\[-500px] &\raise{0.25ex}{\boxed{\cdot \cdot}} & & \raise{0.25ex}{\boxed{\cdot \cdot}} \end{array} }} \rlap{ \qquad \underbrace{\boxed{\Huge{\cdot \cdot}}}_{\text{electron pair}} } $$

    then just go ahead and omit them, assuming that the reader is using typical electron-counting rules, such that they'll infer that those electrons are there without us explicitly drawing them, yielding$$ \textbf{Nitrogen Oxide} \\ {\Huge{ \begin{array}{rcc} \raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O} \end{array} }} $$

  5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$ \textbf{Nitrogen Oxide} \\ {\Huge{ \raise{0.1ex}{.}\mathrm{NO} }} $$

This gives us $`` \, \raise{0.25ex}{.}\mathrm{NO} \, " ,$ which could be further reduced to just $`` \, \mathrm{NO} \, "$ if we further assume that the reader knows about the unpaired electron.

  • 2
    $\begingroup$ I want to upvote this answer, but I can't condone the misconception stated in the last paragraph. The orbitals can't be treated like a cup with enough volume for two electrons and having only the bottom half full when there's only one; it's less incorrect to treat it like a sealed box that will hold one electron okay, and two under pressure, but blow its safety valve if there are more than that, so that the entire volume is still used but with less pressure when only one electron is present. Hence the dot should be located as if taking up all that space to itself, in the centre. $\endgroup$ – Nij Jun 15 '19 at 1:10
  • $\begingroup$ @Nij I'd be happy to correct it if that's the case, but do you have a reference or something? I mean, I'd thought that our current models were that the electrons could have one set of quantum numbers identifying a stable orbital, and while resonance might mean that it's not consistently one orbital being filled, the actual electron orbital structures aren't well-described as being the average of the two, but rather their superposition. But, you're saying that it's less wrong to see the electron's position as the average? $\endgroup$ – Nat Jun 15 '19 at 1:23
  • $\begingroup$ @Nij In other words, I'd say that it's more accurately represented as $\left[\sideset{_{\bullet}}{}{\mathrm{NO}} \rightleftharpoons \sideset{^{\bullet}}{}{\mathrm{NO}}\right] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. ${\small{\bullet}} \mathrm{NO} ,$ just seems off to me. I mean, that seems like saying that a benzyl ring doesn't have single- and double- bonds, but rather 1.5-times-bonds all around it, because that's kinda what it looks like if we average it over time. $\endgroup$ – Nat Jun 15 '19 at 1:29
  • $\begingroup$ In a sense, yes, as you don't know (unless by convention which may not be universal and hence either ambiguous or confusing, or by other specific means in a particular case representation) which "side" of the orbital the electron is (currently) occupying. Since the diagram is itself ambiguous between showing an average of the numbers {0,1} and the superposition of an occupied sub-orbital and an unoccupied one, it simply makes more sense to go for the middle. $\endgroup$ – Nij Jun 15 '19 at 1:31
  • $\begingroup$ @Nij That's inconsistent, though. We don't draw benzyl rings with one-and-a-half bonds all around; we draw them with a single time-dependent snapshot. Or we explicitly dot the lines to represent time-dependence. But the halfway-drawn dot is a misconception; it doesn't exist at any time. Superpositions can't be time-averaged. $\endgroup$ – Nat Jun 15 '19 at 1:32

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