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Homework Help: Lewis Dot Structures

  1. Jan 13, 2005 #1
  2. jcsd
  3. Jan 13, 2005 #2
    they shared one electron only.....
     
  4. Jan 13, 2005 #3
    How do you know how many electrons they share?
     
  5. Jan 13, 2005 #4

    dextercioby

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    In the case of halogens,it's the octet (8 electrons in the last shell) rule.

    Daniel.
     
  6. Jan 13, 2005 #5
    So since there is one electron left that one bonds? If there were 2 left then those 2 would bond?
     
  7. Jan 13, 2005 #6
  8. Jan 13, 2005 #7

    dextercioby

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    No,no,remember that in one covalent bond,there must be two electrons,one coming from each atom.On the last shell halogens have 7 electrons and they want to make 8.To do that through one covalent bond,they must share one electron.So they have only one electron AVAILABLE.

    Daniel.

    PS.For oxygen,sulphur and selenium,the situation is practically the same,just that 1->2 in every sentence because they have 7->6 electrons in the last shell.
     
  9. Jan 13, 2005 #8

    dextercioby

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    Nope.Let's put one pair of electrons by one "bar",either vertical/horizontal.The hypochlorous acid has the Lewis structure:

    [tex] H-\underline{\overline{O}}-\underline{\overline{Cl}}| [/tex]

    Daniel.
     
  10. Jan 13, 2005 #9
    just memorize... in the 7th column of the perioder table, all elements has one electron to shared ... in the 6th column... all elements has two electron to shared...
     
  11. Jan 13, 2005 #10
    So there are all single bonds there. How do you know when there is a double or triple bond? Like for C2 H2 there is a double bond and N2 has a triple bond. How do you know this?
     
  12. Jan 13, 2005 #11

    dextercioby

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    The octet rule in the case of electronegative elements is essential.
    For the nitrogen molecule,the triple bond gives the only viable Lewis structure.As for ethane,the same thing would apply to the Carbon atoms.They must have 8 electrons in their last shell,since they can share only one with one H atom,they must have a double bond.

    Daniel.

    PS.These things are really much more complicated,but,since it is a K-12 forum,no QM allowed.
     
  13. Jan 13, 2005 #12
  14. Jan 13, 2005 #13
    What's QM stand for?
     
  15. Jan 13, 2005 #14

    dextercioby

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    It's better if u always try to put them in pairs.Your structure is correctly written
    [tex] H-C\equiv C-H [/tex]

    Daniel.
     
  16. Jan 13, 2005 #15

    dextercioby

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    Quantum Mechanics,a very interesting and difficult discipline of Theoretical Physics.

    Daniel.
     
  17. Jan 13, 2005 #16
    Ok I think I understand. So if they can make another bond then they should?
     
  18. Jan 13, 2005 #17

    dextercioby

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    What does the last phrase mean??They "should"... :confused: Give an example.


    Daniel.
     
  19. Jan 13, 2005 #18
  20. Jan 14, 2005 #19

    dextercioby

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    Since we're speaking about the acetylene (ethyne) molecule,there are no electrons "left over".They're all hooked up in covalent bonds.2 simple (between C and H) and one triple (between the 2 C elements).

    Daniel.
     
    Last edited: Jan 14, 2005
  21. Jan 14, 2005 #20
    Ok i understand.

    Thanks :smile:
     
  22. Jan 14, 2005 #21

    Gokul43201

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    Shay, you're missing the trick here :

    1) First draw the dot-structure for each atom individually. Make sure you give them all the right number of valence electrons : Cl = 7, F = 7, H = 1, O = 6, C = 4, etc. The total number of valence electrons from all the relevant atoms must be fixed (ie : during bonding, no valence electrons are destroyed or created). Also, if you use different symbols to represent the electrons of the different atoms, it might be easier for you.

    2) Bring the atoms together and make them "share" electrons so that, if you include the shared electrons, each atom has 8 valence electrons (except for H, which needs only 2).

    Examples :
    Code (Text):

        + +                * *                     + +      * *
    +         +                  *              +        +        *
         Cl                  F         ----->       Cl       F
    +                   *        *              +        *        *
        + +                * *                     + +      * *

    7 electrons          7 electrons              octets for each    


                                                       +
                                                       *
      ~        +         *                        ^    +    *
      H     +  C  +   *  C  *     H    ----->   H    C *  C   H
               +         *        ^               +    +    ^
                                                       *
    1 elec    4 elec    4 elec  1 elec        duet octet octet duet
     
    After you start feeling comfortable with the method you can use just dots.
     
    Last edited: Jan 14, 2005
  23. Jan 14, 2005 #22
  24. Jan 14, 2005 #23

    dextercioby

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    Nope,u've given a link to a page where you had an incorect description for the NO,not [itex] NO_{2}^{+} [/itex].

    Think about it again.

    Daniel.
     
  25. Jan 14, 2005 #24
    No it's [itex] NO^{2+} [/itex]
     
  26. Jan 14, 2005 #25

    dextercioby

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    Well,then the ion u indicated does not exist.This ion does,however:[itex] NO^{+} [/tex].The ion must have a total of 11 electrons.The oxigen mut complete its octet,which means that the nitrogen cannot.
    The only possible Lewis structures for the [itex] NO^{+} [/itex] are
    [tex] \cdot \overline{N}=\overline{O}| [/tex]

    [tex] |\overline{N}-\overline{\underline{O}}|[/tex]

    The first one seems to be the "orthodox" one,but the second one is correct as well.

    Daniel.
     
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