What is “dot” sign in •NO?What are radicals and how do they act in radical chain reactions?What is the functional difference between a radical anion and a nucleophile?What exactly does it mean for the reaction of superoxide with non-radicals to be spin-forbidden?What is Nano Zinc Oxide?What is the geometry of an alkyl radical?What is the difference between radical initiator and radical inhibitor?Free radical vs radical cationWhat is the difference between a radical and a neutral alone atom?How long does it take for nitric oxide to turn into dinitrogen and dioxygen?What is the Major product in free radical halogenation of Alkanes. (Between 3°,2°,1°)

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What is “dot” sign in •NO?


What are radicals and how do they act in radical chain reactions?What is the functional difference between a radical anion and a nucleophile?What exactly does it mean for the reaction of superoxide with non-radicals to be spin-forbidden?What is Nano Zinc Oxide?What is the geometry of an alkyl radical?What is the difference between radical initiator and radical inhibitor?Free radical vs radical cationWhat is the difference between a radical and a neutral alone atom?How long does it take for nitric oxide to turn into dinitrogen and dioxygen?What is the Major product in free radical halogenation of Alkanes. (Between 3°,2°,1°)













2












$begingroup$


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?










share|improve this question









New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 6




    $begingroup$
    The dot is the radical (unpaired electron).
    $endgroup$
    – Michael Lautman
    16 hours ago






  • 3




    $begingroup$
    The answer is NO.
    $endgroup$
    – electronpusher
    5 hours ago















2












$begingroup$


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?










share|improve this question









New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 6




    $begingroup$
    The dot is the radical (unpaired electron).
    $endgroup$
    – Michael Lautman
    16 hours ago






  • 3




    $begingroup$
    The answer is NO.
    $endgroup$
    – electronpusher
    5 hours ago













2












2








2





$begingroup$


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?










share|improve this question









New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




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?







inorganic-chemistry nomenclature radicals






share|improve this question









New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question









New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question








edited 51 mins ago









Gaurang Tandon

5,38772969




5,38772969






New contributor



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asked 16 hours ago









user80142user80142

171




171




New contributor



user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user80142 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









  • 6




    $begingroup$
    The dot is the radical (unpaired electron).
    $endgroup$
    – Michael Lautman
    16 hours ago






  • 3




    $begingroup$
    The answer is NO.
    $endgroup$
    – electronpusher
    5 hours ago












  • 6




    $begingroup$
    The dot is the radical (unpaired electron).
    $endgroup$
    – Michael Lautman
    16 hours ago






  • 3




    $begingroup$
    The answer is NO.
    $endgroup$
    – electronpusher
    5 hours ago







6




6




$begingroup$
The dot is the radical (unpaired electron).
$endgroup$
– Michael Lautman
16 hours ago




$begingroup$
The dot is the radical (unpaired electron).
$endgroup$
– Michael Lautman
16 hours ago




3




3




$begingroup$
The answer is NO.
$endgroup$
– electronpusher
5 hours ago




$begingroup$
The answer is NO.
$endgroup$
– electronpusher
5 hours ago










4 Answers
4






active

oldest

votes


















9












$begingroup$

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 $ceNO^+$, 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.






share|improve this answer









$endgroup$




















    4












    $begingroup$

    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 $ceO=O$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.






    share|improve this answer











    $endgroup$




















      3












      $begingroup$

      Nitric oxide ($ceNO$) 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 $ceNO$ 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, $ceNO$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $ceNO$ molecule: $ceNO$ can easily be oxidized to become nitrosonium ion ($ceNO+$) and be reduced to be nitroxide ($ceNO-$). $ceNO$ 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).





      share|improve this answer











      $endgroup$




















        2












        $begingroup$

        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.$$
          beginarrayccc
          textbfNitrogen
          & qquad
          & textbfOxygen
          \[-10px]
          Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex. &mathrmN & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
          &
          & Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex:&mathrmO & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
          endarray
          $$


        2. Combine them to create nitrogen oxide:$$
          textbfNitrogen Oxide \
          Huge
          beginarrayrcccc
          raise0.1ex. &mathrmN & rlapraise0.1ex: : ~~ &mathrmO & raise0.1ex:
          \[-500px]
          &raise0.25excdot cdot & & raise0.25excdot cdot
          endarray

          $$


        3. Condense the representation by drawing the double-bond as in:$$
          textbfNitrogen Oxide \
          Huge
          beginarrayrcccc
          raise0.1ex. &mathrmN & = &mathrmO & raise0.1ex:
          \[-500px]
          &raise0.25excdot cdot & & raise0.25excdot cdot
          endarray

          $$



        4. Identify the paired electrons,$$
          textbfNitrogen Oxide \
          Huge
          beginarrayrcccc
          raise0.1ex. &mathrmN & = &mathrmO & raise0.1exboxed:
          \[-500px]
          &raise0.25exboxedcdot cdot & & raise0.25exboxedcdot cdot
          endarray

          rlap
          qquad
          underbraceboxedHugecdot cdot_textelectron 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$$
          textbfNitrogen Oxide \
          Huge
          beginarrayrcc
          raise0.1ex. &mathrmN & = &mathrmO
          endarray

          $$



        5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
          textbfNitrogen Oxide \
          Huge
          raise0.1ex.mathrmNO

          $$


        This gives us $`` , raise0.25ex.mathrmNO , " ,$ which could be further reduced to just $`` , mathrmNO , "$ if we further assume that the reader knows about the unpaired electron.






        share|improve this answer











        $endgroup$








        • 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
          3 hours ago











        • $begingroup$
          @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
          3 hours ago











        • $begingroup$
          @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
          3 hours ago











        • $begingroup$
          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
          3 hours ago










        • $begingroup$
          @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
          3 hours ago












        Your Answer








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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9












        $begingroup$

        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 $ceNO^+$, 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.






        share|improve this answer









        $endgroup$

















          9












          $begingroup$

          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 $ceNO^+$, 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.






          share|improve this answer









          $endgroup$















            9












            9








            9





            $begingroup$

            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 $ceNO^+$, 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.






            share|improve this answer









            $endgroup$



            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 $ceNO^+$, 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.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 16 hours ago









            Oscar LanziOscar Lanzi

            17.3k22853




            17.3k22853





















                4












                $begingroup$

                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 $ceO=O$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.






                share|improve this answer











                $endgroup$

















                  4












                  $begingroup$

                  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 $ceO=O$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.






                  share|improve this answer











                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    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 $ceO=O$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.






                    share|improve this answer











                    $endgroup$



                    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 $ceO=O$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 12 hours ago









                    Ian Bush

                    1,1491715




                    1,1491715










                    answered 15 hours ago









                    PoutnikPoutnik

                    3,546620




                    3,546620





















                        3












                        $begingroup$

                        Nitric oxide ($ceNO$) 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 $ceNO$ 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, $ceNO$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $ceNO$ molecule: $ceNO$ can easily be oxidized to become nitrosonium ion ($ceNO+$) and be reduced to be nitroxide ($ceNO-$). $ceNO$ 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).





                        share|improve this answer











                        $endgroup$

















                          3












                          $begingroup$

                          Nitric oxide ($ceNO$) 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 $ceNO$ 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, $ceNO$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $ceNO$ molecule: $ceNO$ can easily be oxidized to become nitrosonium ion ($ceNO+$) and be reduced to be nitroxide ($ceNO-$). $ceNO$ 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).





                          share|improve this answer











                          $endgroup$















                            3












                            3








                            3





                            $begingroup$

                            Nitric oxide ($ceNO$) 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 $ceNO$ 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, $ceNO$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $ceNO$ molecule: $ceNO$ can easily be oxidized to become nitrosonium ion ($ceNO+$) and be reduced to be nitroxide ($ceNO-$). $ceNO$ 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).





                            share|improve this answer











                            $endgroup$



                            Nitric oxide ($ceNO$) 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 $ceNO$ 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, $ceNO$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $ceNO$ molecule: $ceNO$ can easily be oxidized to become nitrosonium ion ($ceNO+$) and be reduced to be nitroxide ($ceNO-$). $ceNO$ 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).






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 6 hours ago









                            owjburnham

                            44337




                            44337










                            answered 9 hours ago









                            Mathew MahindaratneMathew Mahindaratne

                            9,0991131




                            9,0991131





















                                2












                                $begingroup$

                                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.$$
                                  beginarrayccc
                                  textbfNitrogen
                                  & qquad
                                  & textbfOxygen
                                  \[-10px]
                                  Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex. &mathrmN & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  &
                                  & Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex:&mathrmO & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  endarray
                                  $$


                                2. Combine them to create nitrogen oxide:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & rlapraise0.1ex: : ~~ &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$


                                3. Condense the representation by drawing the double-bond as in:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$



                                4. Identify the paired electrons,$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1exboxed:
                                  \[-500px]
                                  &raise0.25exboxedcdot cdot & & raise0.25exboxedcdot cdot
                                  endarray

                                  rlap
                                  qquad
                                  underbraceboxedHugecdot cdot_textelectron 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$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcc
                                  raise0.1ex. &mathrmN & = &mathrmO
                                  endarray

                                  $$



                                5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  raise0.1ex.mathrmNO

                                  $$


                                This gives us $`` , raise0.25ex.mathrmNO , " ,$ which could be further reduced to just $`` , mathrmNO , "$ if we further assume that the reader knows about the unpaired electron.






                                share|improve this answer











                                $endgroup$








                                • 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
                                  3 hours ago











                                • $begingroup$
                                  @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
                                  3 hours ago











                                • $begingroup$
                                  @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                  3 hours ago











                                • $begingroup$
                                  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
                                  3 hours ago










                                • $begingroup$
                                  @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
                                  3 hours ago
















                                2












                                $begingroup$

                                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.$$
                                  beginarrayccc
                                  textbfNitrogen
                                  & qquad
                                  & textbfOxygen
                                  \[-10px]
                                  Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex. &mathrmN & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  &
                                  & Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex:&mathrmO & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  endarray
                                  $$


                                2. Combine them to create nitrogen oxide:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & rlapraise0.1ex: : ~~ &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$


                                3. Condense the representation by drawing the double-bond as in:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$



                                4. Identify the paired electrons,$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1exboxed:
                                  \[-500px]
                                  &raise0.25exboxedcdot cdot & & raise0.25exboxedcdot cdot
                                  endarray

                                  rlap
                                  qquad
                                  underbraceboxedHugecdot cdot_textelectron 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$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcc
                                  raise0.1ex. &mathrmN & = &mathrmO
                                  endarray

                                  $$



                                5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  raise0.1ex.mathrmNO

                                  $$


                                This gives us $`` , raise0.25ex.mathrmNO , " ,$ which could be further reduced to just $`` , mathrmNO , "$ if we further assume that the reader knows about the unpaired electron.






                                share|improve this answer











                                $endgroup$








                                • 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
                                  3 hours ago











                                • $begingroup$
                                  @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
                                  3 hours ago











                                • $begingroup$
                                  @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                  3 hours ago











                                • $begingroup$
                                  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
                                  3 hours ago










                                • $begingroup$
                                  @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
                                  3 hours ago














                                2












                                2








                                2





                                $begingroup$

                                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.$$
                                  beginarrayccc
                                  textbfNitrogen
                                  & qquad
                                  & textbfOxygen
                                  \[-10px]
                                  Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex. &mathrmN & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  &
                                  & Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex:&mathrmO & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  endarray
                                  $$


                                2. Combine them to create nitrogen oxide:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & rlapraise0.1ex: : ~~ &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$


                                3. Condense the representation by drawing the double-bond as in:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$



                                4. Identify the paired electrons,$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1exboxed:
                                  \[-500px]
                                  &raise0.25exboxedcdot cdot & & raise0.25exboxedcdot cdot
                                  endarray

                                  rlap
                                  qquad
                                  underbraceboxedHugecdot cdot_textelectron 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$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcc
                                  raise0.1ex. &mathrmN & = &mathrmO
                                  endarray

                                  $$



                                5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  raise0.1ex.mathrmNO

                                  $$


                                This gives us $`` , raise0.25ex.mathrmNO , " ,$ which could be further reduced to just $`` , mathrmNO , "$ if we further assume that the reader knows about the unpaired electron.






                                share|improve this answer











                                $endgroup$



                                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.$$
                                  beginarrayccc
                                  textbfNitrogen
                                  & qquad
                                  & textbfOxygen
                                  \[-10px]
                                  Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex. &mathrmN & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  &
                                  & Hugebeginarrayrcl & cdot phantomcdot \[-50px] raise0.1ex:&mathrmO & raise0.1ex: \[-500px] &raise0.25excdot phantomcdot endarray
                                  endarray
                                  $$


                                2. Combine them to create nitrogen oxide:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & rlapraise0.1ex: : ~~ &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$


                                3. Condense the representation by drawing the double-bond as in:$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1ex:
                                  \[-500px]
                                  &raise0.25excdot cdot & & raise0.25excdot cdot
                                  endarray

                                  $$



                                4. Identify the paired electrons,$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcccc
                                  raise0.1ex. &mathrmN & = &mathrmO & raise0.1exboxed:
                                  \[-500px]
                                  &raise0.25exboxedcdot cdot & & raise0.25exboxedcdot cdot
                                  endarray

                                  rlap
                                  qquad
                                  underbraceboxedHugecdot cdot_textelectron 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$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  beginarrayrcc
                                  raise0.1ex. &mathrmN & = &mathrmO
                                  endarray

                                  $$



                                5. Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
                                  textbfNitrogen Oxide \
                                  Huge
                                  raise0.1ex.mathrmNO

                                  $$


                                This gives us $`` , raise0.25ex.mathrmNO , " ,$ which could be further reduced to just $`` , mathrmNO , "$ if we further assume that the reader knows about the unpaired electron.







                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited 2 hours ago

























                                answered 4 hours ago









                                NatNat

                                165129




                                165129







                                • 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
                                  3 hours ago











                                • $begingroup$
                                  @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
                                  3 hours ago











                                • $begingroup$
                                  @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                  3 hours ago











                                • $begingroup$
                                  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
                                  3 hours ago










                                • $begingroup$
                                  @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
                                  3 hours ago













                                • 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
                                  3 hours ago











                                • $begingroup$
                                  @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
                                  3 hours ago











                                • $begingroup$
                                  @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                  3 hours ago











                                • $begingroup$
                                  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
                                  3 hours ago










                                • $begingroup$
                                  @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
                                  3 hours ago








                                2




                                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
                                3 hours ago





                                $begingroup$
                                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
                                3 hours ago













                                $begingroup$
                                @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
                                3 hours ago





                                $begingroup$
                                @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
                                3 hours ago













                                $begingroup$
                                @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                3 hours ago





                                $begingroup$
                                @Nij In other words, I'd say that it's more accurately represented as $left[sideset_bulletmathrmNO rightleftharpoons sideset^bulletmathrmNOright] ,$ if we're talking about resonance. But to say that a resonance structure is its average, e.g. $smallbullet mathrmNO ,$ 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
                                3 hours ago













                                $begingroup$
                                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
                                3 hours ago




                                $begingroup$
                                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
                                3 hours ago












                                $begingroup$
                                @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
                                3 hours ago





                                $begingroup$
                                @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
                                3 hours ago











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