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Is the pi bond of a $\ce{sp}$ hybridised carbon in conjugation with another π-bond capable of resonance? Examples:

  • $\ce{={C}=C-C=}$
  • $\ce{#{C}-C=}$

Is resonance possible?

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  • $\begingroup$ Where would that charge come from? In both examples the carbon atom with the plus has a full octet it equally shares with its neighbour atoms, so four electrons. That makes it neutral. $\endgroup$ – Karl Dec 9 '17 at 10:01
  • $\begingroup$ If the charge is not present, is resonance possible?? $\endgroup$ – DJ Koustav Dec 9 '17 at 10:21
  • $\begingroup$ Can you explain to me what exactly you think is "resonance"? Also please draw a complete molecule, and try it out. With the fragments in your examples, it's hard to tell. $\endgroup$ – Karl Dec 9 '17 at 10:33
  • $\begingroup$ It's delocalisation of pi electrons $\endgroup$ – DJ Koustav Dec 9 '17 at 10:33
  • $\begingroup$ Sorry that's an incomplete answer. Resonance means you can draw at least two resonance structures. $\endgroup$ – Karl Dec 9 '17 at 10:35
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Yes, principally, any pi orbital can overlap with similarly oriented neighbouring pi orbitals. However the two orbitals at an sp hybridised carbon atom are orthogonal to each other, so only one of them can overlap with the same orbital at a neighbouring atom.

And you have to differentiate conjugation(=orbital overlap) from resonance(=total delocalisation=more than one structure formula for the same compound).

In the compound a) below, the double bonds are just conjugated. Resonance is typical for aromats, but you can imagine resonance structures here as well:

From a) to b) below, all pi electron pairs flip to the right, leaving a carbocation on the left, an a carbanion on the right. That is resonance, although the second version b) is not very favourable, energetically, as it requires a charge separation. You can draw more intermediate resonance structures where only one or two or three electron pairs flip. Under normal conditions, the energy for that charge separation is not available, so there will be no resonance..

a) $\ce{R-C=C-C=C-C=C-C=C-R}$

b) $\ce{R-\overset{+}{C}-C=C-C=C-C=C-\overset{-}{C}-R}$

In structure a) itself, there is conjugation, i.e. partial pi electron delocalisation. Resonance only comes in when you take into account further formula: $[a \leftrightarrow .. \leftrightarrow b]$

Now with sp hybridised carbon atoms (e.g. the two in the middle of the structures below), the aditional pi bond orbital is orthogonal to the other pi system, and does not take part in it:

$\ce{R-C=C=C=C-R}\leftrightarrow\ce{\overset{+}{R}=C-C\equiv C-C=\overset{-}{R}}$

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