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 $\ce{Cl}$.