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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?

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    $\begingroup$ 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. $\endgroup$
    – orthocresol
    Jul 24 '20 at 18:31
  • $\begingroup$ I have rectified some inconsistencies and errors in my question, please check them out, sorry and thanks. $\endgroup$
    – SEO
    Jul 24 '20 at 18:57
  • $\begingroup$ @orthocresol can you please shed some light on this $\endgroup$
    – SEO
    Jul 27 '20 at 14:21
  • $\begingroup$ 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. $\endgroup$
    – orthocresol
    Jul 27 '20 at 14:31
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    $\begingroup$ 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. $\endgroup$
    – user98209
    Sep 8 '20 at 7:37
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$\ce{HCN}$ 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).

Your teacher pointed out something interesting, but the exact split ratio remains to be calculated.

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