I have been trying to figure out the Lewis structures of carbon nitride C2N2
and can't figure out whether the carbons are triple bonded to each other or
whether the carbons are triple bonded to the nitrogens to give the proper
Lewis representation.
Calculating the formal charges on both of the molecules gives the same
result.
Does the correct structure have 4 lone pairs or 2 lone pairs? And how are
the electrons delocalized over the system?
Thanks in advance
A. Giusti
Hint: it is possible to draw a structure with no formal charges, and it
has two lone pairs. Don't take the name "C2N2" to have any structural
or connectivity implications.
Now does it make more sense?
Eric Lucas
C2N2 as such is NC-CN, cyanogen, cyanide chemistry being analogous to
the halogens.
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In fact C2N2 could also be CN-CN (isocyanogen) or even CN-NC
(diisocyanogen), both of which have been identified and isolated.
there is a comprehensive discussion of the bonding in these compounds in a
paper by Baerends in J.Phys.Chem. 1992.
Crispin Roche
Antoine Giusti <A.J.G...@maartens.nl> wrote in article
<77dkqn$hun$1...@info.service.rug.nl>...
> Can someone please help me,
>
> I have been trying to figure out the Lewis structures of carbon nitride
C2N2
> and can't figure out whether the carbons are triple bonded to each other
or
> whether the carbons are triple bonded to the nitrogens to give the proper
> Lewis representation.
>
> Calculating the formal charges on both of the molecules gives the same
> result.
>
> Does the correct structure have 4 lone pairs or 2 lone pairs? And how are
> the electrons delocalized over the system?
>
> Thanks in advance
>
> A. Giusti
>
>
It would be the dinitrile. N likes electrons just a little more than C. I
would expect a more or less ovular bag of electrons around the whole thing,
like a sausage. SP hybridization is mandated by symmetry, there's no
reason for it to twist. So, formally- :N:::C:C:::N: As you can see,
there are lots of different ways to arrange the electrons in the NCCN line
so there has to be resonance throughout but having N holding two lone pairs
would require SP at terminal N which would require a change in the
hybridization.