10
$\begingroup$

Larger the difference in electronegativities of bonded atoms, the larger the dipole moment.

(this statement has been extracted from chemwiki-dipole moment)

The electronegativity difference between carbon and fluorine is about $1.43$ (Pauling scale), $1.649$ (Allen scale). The electronegativity difference between carbon and chlorine is about $0.61$ (Pauling scale), $0.325$ (Allen scale). Although, the electronegativity difference of $\ce{C-F}$ bond is greater than electronegativity difference of $\ce{C-Cl}$ bond, dipole moment of $\ce{C-Cl}$ bond ($\pu{1.860 Debye}$) in $\ce{CH_3-Cl}$ is greater than dipole moment of $\ce{C-F}$ bond ($\pu{1.847 Debye}$) in $\ce{CH3-F}$, why?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
12
$\begingroup$

The bond dipole moment ($\mu$) of a chemical bond is equal to the magnitude of the separation of charge ($\delta$) times the displacement of the charge (or the distance over which the charges are spearated) $d$, or:

$$\vec{\mu}=\delta \vec{d}$$

Thus, there are two factors controlling dipole moment. The charge separation can be estimated from the electronegativity difference, but not perfectly. Disparities in size and polarizability also effect the separation of charge between two atoms. The displacement is the length of the bond. The $\ce{C-Cl}$ bond length (176 pm) is longer than the $\ce{C-F}$ bond length (134 pm) - data from here.

Thus:

For $\ce{C-F}$, the electronegativity difference on the Pauling scale is 1.43, and the bond length is 134 pm: $1.43\times 134 = 191$ (ignoring units at the moment).

For $\ce{C-Cl}$, the electronegativity different on the Pauling scale is 0.61, and the bond length is 174 pm: $0.61\times 174=106$.

So, my quick and dirty method did not work so well. However the disparity between these two bonds by my method $(106/191)=0.55)$ is smaller than by just comparing electronegativities $(0.61/1.43=0.42)$ and that is because of the other factor mentioned above: size and polarizability. Chlorine is bigger and more polarizable than fluorine (and carbon). The carbon-chlorine bond is polarized toward chlorine not just because of the electronegativity difference. Chlorine's larger size decreases the effectiveness of orbital overlap in bonding, and chlorine's increase polarizability implies that it can support more partial negative charge than suggested by its electronegativity alone.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I like the idea that chlorine may have a higher partial charge than fluorine given the former's larger size. Indeed, the electron affinities of the elements seems to provide evidence that fluorine simply cannot handle much charge. However, I do believe I've seen a graph in Huheey's book where a fluorine atom can stabilize a partial negative charge more than a chlorine atom all the way to around $-0.8e$, and only then do the relative stabilities switch. Do you think the halogen atoms in the methyl halides have that high of a partial charge? $\endgroup$
    – Nicolau Saker Neto
    Nov 4, 2013 at 23:39
  • $\begingroup$ When I was writing this post, I was looking for some reference, but did not find any. I did not look too hard however, and would love to see one. I was also planning to run some simulations on, for example, the methyl halides, but have not had time to do so yet. $\endgroup$
    – Ben Norris
    Nov 5, 2013 at 11:35
  • 1
    $\begingroup$ Can larger size imply that it can hold more partial charges? Could you support that with a reference? $\endgroup$
    – Tan Yong Boon
    Jan 1, 2018 at 7:03
  • $\begingroup$ Not to waaaay necro this, but I notice the same thing in my textbook recently. It says a C-H bond has a greater dipole than a C-N bond (0.3 D vs 0.22 D). Similar thoughts? $\endgroup$
    – J M
    May 15, 2018 at 14:25

Your Answer

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

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