Understand Lewis Dot Structures: Cl F & Triple Bonds

In summary, Lewis dot structures are a way to represent the valence electrons of atoms in a molecule. When drawing the Lewis dot structure for Cl F, it is important to remember that Cl has seven valence electrons and F has seven as well. This results in a single bond between Cl and F, with the remaining electron being unpaired on Cl. Additionally, triple bonds can also be represented in Lewis dot structures, where three pairs of electrons are shared between two atoms. This type of bond is stronger than a single or double bond and is commonly found in molecules with high electronegativity differences between the atoms. Overall, understanding Lewis dot structures is crucial in visualizing the arrangement of valence electrons in molecules and predicting their chemical properties
  • #1
Shay10825
338
0
What would the Lewis structure look like for Cl F? Would it look like this:
http://img53.exs.cx/my.php?loc=img53&image=dot8pn.png

How do you know when there is a triple bond?

~Thanks
 
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  • #2
they shared one electron only...
 
  • #3
How do you know how many electrons they share?
 
  • #4
In the case of halogens,it's the octet (8 electrons in the last shell) rule.

Daniel.
 
  • #5
So since there is one electron left that one bonds? If there were 2 left then those 2 would bond?
 
  • #6
So would H O Cl look like:

http://img60.exs.cx/my.php?loc=img60&image=dot21ni.png
 
  • #7
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.
 
  • #8
Shay10825 said:
So would H O Cl look like:

http://img60.exs.cx/my.php?loc=img60&image=dot21ni.png

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.
 
  • #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...
 
  • #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?
 
  • #11
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.
 
  • #12
In C2 H2 why can't it look like this?:
http://img143.exs.cx/my.php?loc=img143&image=dot34jd.png

Can electrons stand alone like that (the ones in red) or do they always have to be in pairs?
 
  • #13
dextercioby said:
PS.These things are really much more complicated,but,since it is a K-12 forum,no QM allowed.

What's QM stand for?
 
  • #14
Shay10825 said:
In C2 H2 why can't it look like this?:
http://img143.exs.cx/my.php?loc=img143&image=dot34jd.png

Can electrons stand alone like that (the ones in red) or do they always have to be in pairs?


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.
 
  • #15
Shay10825 said:


What's QM stand for?


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

Daniel.
 
  • #16
Ok I think I understand. So if they can make another bond then they should?
 
  • #17
Shay10825 said:
Ok I think I understand. So if they can make another bond then they should?

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


Daniel.
 
  • #18
For example:

C2 H2 should not look like this:
http://img38.exs.cx/my.php?loc=img38&image=dot39es.png
because they can form another bond since there are 2 electrons (the ones in red) left over.
 
  • #19
Shay10825 said:
For example:

C2 H2 should not look like this:
http://img38.exs.cx/my.php?loc=img38&image=dot39es.png
because they can form another bond since there are 2 electrons (the ones in red) left over.


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.
 
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  • #20
dextercioby said:
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).

Ok i understand.

Thanks :smile:
 
  • #21
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:
    + +                * *                     + +      * *
+         +                  *              +        +        *
     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.
 
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  • #22
ok I have another question. If it says NO2+ would it look like this:

http://img43.exs.cx/my.php?loc=img43&image=dot40hz.png
 
  • #23
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.
 
  • #24
No it's [itex] NO^{2+} [/itex]
 
  • #25
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.
 
  • #26
dextercioby said:
Well,then the ion u indicated does not exist.

The problem says:
Draw Lewis structures for the following polyatomic ions:
[itex] NO^{2+} [/itex] is one of the ones listed
 
  • #27
Shay, according to your picture, N has a valency of 3 or O has a valency of 4.

Moreover, it's likely [itex]NO_2^+ [/itex]. I've never heard of [itex]NO^{2+} [/itex], and neither has dexter.
 
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  • #28
Gokul43201 said:
Shay, according to your picture, N has a valency of 3 and O has a valency of 4.

I'm sorry. Could I add the other 2 electrons anywhere?
 
  • #29
I'm sorry.I've got it.I didn't think of it to be possible,though.

[tex] \cdot\overline{N}=\overline{O}\cdot [/tex]

Daniel.

PS.The unpaired electrons can be bonded with a halogen.
 
  • #30
Shay, even if it is [itex]NO^{2+} [/itex], your approach is incorrect. Read my first post with an explanation of how to tackle such problems. Do it one step at a time. You could start by adding the necessary number of electrons to make a neutral molecule, and them take them away in the end.

Dexter, you've drawn [itex]NO^+[/itex]
 
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  • #31
so would it be
[tex]\overline{N}=\overline{O}\cdot [/tex]
 
  • #32
If you're sure it's [itex]NO^{2+} [/tex], then, yes, that is one possibility.

However, it's not the best one.

Explanation : NO^(2+) has 5+ 6 - 2 = 9 valence electrons, which is an odd number. You can not make two octets out of an odd number of electrons. So, one of the two atoms must necessarily settle for a non-octet configuration.

Here's another possibility. Tell me if you think it's better or worse, and why.

[tex](+)~~\cdot N \equiv \overline {O}~~(+) [/tex]
 
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  • #33
oh well. I give up. I'll finish working on this stuff tomorrow. Thanks everyone for your help :smile: . It's extremely late over here and I need to go to bed. :zzz:
 
  • #34
That would be better because the bond is stronger? A triple bond is better than a double. So for some of them there are many answers?
 
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  • #35
Okay,Gokul,you're right:In this formula [tex] NOF_{2} [/tex] with the Lewis structure:

[tex]\cdot [/tex]​
[tex] |\overline{\underline{F}}-N=\overline{O}|[/tex]​
|​
[tex]|\underline{F}|[/tex]​
The total number of electrons is as it should be:25 and the oxydation number of nytrogen is +4,which means that the "NO" combination has the ON "+2".

I believe they are called "mesomeric" structures and have been discovered by Linus Pauling in the early 1930's.

Daniel.
 

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