What is the hybridisation in HCN molecule?

According to VSEPR theory hybridisation of the central atom should be sp. However, my teacher says that there exists a near-100% s-character in the carbon orbital that bonds with hydrogen due to some other reasons. I couldn't get why that would be the case. One can think of it as the nitrogen that polarizes the carbon atom, but shouldn't it lower the s-character of the C-H bond? What am I missing?

Is it based on quantum calculations (which are beyond my scope) , as a commentator points out?

• Ok, no need for apologies. Now, s- and p-character is used to refer to orbitals, not atoms, and certainly not molecules; regardless of how the orbitals are hybridised, carbon always has one s and three p orbitals. The orbitals can have different amounts of s- and p-character. – orthocresol Jul 24 '20 at 18:31
• I have rectified some inconsistencies and errors in my question, please check them out, sorry and thanks. – SEO Jul 24 '20 at 18:57
• @orthocresol can you please shed some light on this – SEO Jul 27 '20 at 14:21
• I'm sorry, but I'm not qualified to. I will say that I am slightly skeptical of this claim of your teacher (firstly, it is completely unsubstantiated, and secondly, why should there be a difference in the H–C bonding in H–C≡N and H–C≡CH? Nobody would dare to say the latter is not sp hybridised), but I dare not say for a fact it is wrong without some evidence, and I don't really have the time to do quantum chemical calculations now. – orthocresol Jul 27 '20 at 14:31
• Firstly, the thing you say about N polarizes carbon..you certainly mean N is more electronegative...If that is the case, the bonds along C--->N will have more p character, and to compensate on s character, C-->H ,s' character increases. Performing rudimentary calculations on Computational software, it is indeed the fact that s character increased at C-H bond...BUT, the claim about near 100% is wrong...(my calculations show near 55%s) which is nearly the same old sp hybrid. – user98209 Sep 8 '20 at 7:37

$$\ce{HCN}$$ and $$\ce{HC#CH}$$ are linear, triple bonded, with a $$π$$ system consisting of two perpendicular $$π$$ bonds. They would be symmetrical in $$\ce{HC#CH}$$, and slightly distorted in $$\ce{HCN}$$, and they leave two orbitals for the sigma system.
In $$\ce{HCN}$$, we hybridize/combine the two remaining orbitals on the carbon atom to form two bonding orbitals, one to the hydrogen, another to the atom on the other side of the carbon (a $$\ce{C}$$ or an $$\ce{N}$$). The natural first combination is a $$50$$$$50$$ split to form two $$\mathrm{sp}$$ orbitals, one directed to the $$\ce{H}$$ and the other directed toward the $$\ce{N}$$.
That's enough most of the time, but if you get picky, you could point out that the electronegativities of $$\ce{H}$$ and $$\ce{N}$$ are quite different ($$\ce{H}$$ $$2.1$$, $$\ce{C}$$ $$2.5$$, $$\ce{N}$$ $$3.0$$), so the nitrogen will be pulling on its $$\mathrm{sp}$$ bond more than hydrogen pulls on its $$\ce{sp}$$ bond, so the $$50$$-$$50$$ split readjusts to maybe $$70$$-$$30$$ (did your teacher say $$~100\%$$-$$0\%$$?), meaning that the hydrogen gets less of the carbon orbital (i.e., less $$\mathrm p$$-character, more $$\mathrm s$$-character from the carbon orbitals), while the nitrogen gets more of carbon's $$\mathrm p$$-orbital.
If this is so, the $$\ce{H}$$ atom in $$\ce{HCN}$$ should be more easily removed than a H atom in $$\ce{HC#CH}$$. This appears to be correct: the $$pK_\mathrm a$$ of acetylene is $$24$$ (which is considered to be quite acidic for a hydrocarbon), whereas the $$pK_\mathrm a$$ of hydrocyanic acid is $$9.21$$, much more acidic (although a weak acid by any other measure).