A snippet from a textbook:

Therefore, the hybridization model predicts that an $\mathrm{sp}$-hybridized carbon atom is more electronegative than an $\mathrm{sp}^3$-hybridized carbon atom. Evidence for this effect is that the positive end of the dipole moment in $\ce{N#C-Cl}$ is on the $\ce{Cl}$ atom. We conclude that the carbon atom in the cyanide group is more electronegative than a chlorine atom.

The paragraph suggests that the $\ce{N#C}$ bond has a different hybrid orbital compared to the $\ce{C-Cl}$ bond, doesn't it? I am a bit confused since there is only $1~\ce{C}$ atom but somehow it has $2$ hybrid orbitals of different state (namely $\mathrm{sp}^3$ and $\mathrm{sp}^2$). Is this due to different ligands with which the central atoms connect?

And point of interest is why the conclusion isn't

the nitrogen atom in the cyanide group is more electronegative than a chlorine atom

instead of carbon? Because I would imagine that the electron density of both $\ce{N#C}$ and $\ce{C-Cl}$ bonds would lie dominantly on $\ce{N}$ and $\ce{Cl}$, respectively. Hence, it is more reasonable to me to compare the electronegativity of $\ce{N}$ and $\ce{Cl}$ instead that of $\ce{C}$ and $\ce{Cl}$.

  • $\begingroup$ There are no such things as hybridised atoms, that is very sloppy language. Please add a citation to your quote, otherwise one could consider it plagiarism. $\endgroup$ Jul 12, 2019 at 12:14

1 Answer 1


The usual explanation for $sp$ orbitals being more electronegative than $sp^3$ is the fact that more $s$ character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of $\ce{CH3-CH2-Cl}$ to $\ce{CH2=CH-Cl}$ to $\ce{HC#C-Cl}$ and the increasing acidity of ethane to ethylene to acetylene.

I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:

$[\ce{^-N=C=Cl+ <-> N#C-Cl}]$

which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.

As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still $sp$ hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.

  • $\begingroup$ can you explain why in $- \mathrm { N } = \mathrm { C } = \mathrm { Cl } ^ { + }$ $N$ is partially negative and $Cl$ partially positive? $\endgroup$
    – jayjay
    Feb 11, 2019 at 9:14
  • $\begingroup$ I added the starting structure to make it more clear. In order for Cl to form the second bond to C, it has to share electrons with C and so it becomes (+) charged. The N takes electrons from the third N-C bond, so it gains electrons and becomes (-) charged. $\endgroup$
    – Andrew
    Feb 11, 2019 at 13:10

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