Mdthbs

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?

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):

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.

Reference:

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.