2
$\begingroup$

Somewhere along the line of researching why the $\ce{Cl-C-H}$ bond angle in methyl chloride is less than what is predicted ($\pu{109.5^\circ}$), I ran into a book which said that the electron withdrawing effects of the chlorine atom gave the $\ce{C-Cl}$ bond more "p-character".

What does this mean? Is he simply using this term to describe the elongated nature of the bond - i.e. s-orbitals are spherical and p-orbitals not spherical but rather elongated (at least as depicted by cartoon drawings in textbooks)?

Is the author saying that the electron-withdrawing effects of chlorine cause the electrons to spend more time further away from the nucleus; hence the "p-character"? I understand that in a spherical shell, electrons are basically equidistant from the nucleus. But in an elongated, "p-character" shell, electrons can go further away from the nucleus.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

S orbitals are lower in energy than P orbitals. Electrons prefer to be in as low an energy orbital as possible. Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride). The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond. So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it? Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them. It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted). The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you. Makes sense. Reduced electron density about the carbon in the carbon-chlorine bond reduces the need for s-character as s-character helps stabilize electron density. But if there isn't that much electron density immediately around the carbon there is less of a need to stabilize said electron density. $\endgroup$
    – Dissenter
    May 18, 2014 at 16:27
  • $\begingroup$ Also speaking of point-group symmetry, what resource would you recommend as a good primer on point-group symmetry? $\endgroup$
    – Dissenter
    May 18, 2014 at 16:27
  • $\begingroup$ The first book where I came across this topic was "Intoduction to Stereochemistry" by Kurt Mislow (a brilliant thinker and chemist). It is a paperback and in one of the early chapters he spends several pages on the point groups. I think it is an easy and excellent introduction to the subject. The rest of the book is a wonderful introduction to many of the other important concepts in stereochemistry. $\endgroup$
    – ron
    May 18, 2014 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.