I'm all bent out of shape trying to figure out what Bent's rule means. I have several formulations of it, and the most common formulation is also the hardest to understand.

Atomic s character concentrates in orbitals directed toward electropositive substituents

Why would this be true? Consider $\ce{H3CF}$.

Both the carbon and the fluorine are roughly $\ce{sp^3}$ hybridized. Given that carbon is more electropositive than fluorine, am I supposed to make the conclusion that because carbon is more electropositive than fluorine, there is a great deal of s-character in the $\ce{C-F}$ bond and most of this s-character is around the carbon?

Or is this a misunderstanding of "orbitals directed toward electropositive substituents"? The fluorine is $\ce{sp^3}$ hybridized and these orbitals are "directed" toward the carbon in that the big lobe of the hybrid orbital is pointing toward the carbon. So does electron density concentrate near the fluorine? Because that would make more sense.

And this s-character concentrated toward the fluorine has the effect of what on the bond angle? I understand that the more s-character a bond has, the bigger the bond angle - consider $\ce{sp}$ vs $\ce{sp^2}$. But since the $\ce{C-F}$ bond now has less s-character around the carbon, the $\ce{H-C-F}$ bond angle can shrink, correct?

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    $\begingroup$ Fluorine is definitely not sp3 hybridised. Usually none of the terminal atoms uses this hybridisation, they are usually always sp hybridised. $\endgroup$ – Martin - マーチン Sep 2 '14 at 6:57
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    $\begingroup$ To elaborate on Martin's point: Geometry drives hybridization. Hybridization does not drive geometry. $\endgroup$ – Lighthart Apr 5 '16 at 4:00

That's a good, concise statement of Bent's rule. Of course we could have just as correctly said that p character tends to concentrate in orbitals directed at electronegative elements. We'll use this latter phrasing when we examine methyl fluoride below. But first, let's expand on the definition a bit so that it is clear to all.

Bent's rule speaks to the hybridization of the central atom ($\ce{A}$) in the molecule $\ce{X-A-Y}$.

$\ce{A}$ provides hybridized atomic orbitals that form $\ce{A}$'s part of its bond to $\ce{X}$ and to $\ce{Y}$. Bent's rule says that as we change the electronegativity of $\ce{X}$ and \ or $\ce{Y}$, $\ce{A}$ will tend to rehybridize its orbitals such that more s character will placed in those orbitals directed towards the more electropositive substituent.

Let's examine how Bent's rule might be applied to your example of methyl fluoride. In the $\ce{C-F}$ bond, the carbon hybrid orbital is directed towards the electronegative fluorine. Bent's rule suggests that this carbon hybrid orbital will be richer in p character than we might otherwise have suspected. Instead of the carbon hybrid orbital used in this bond being $\ce{sp^3}$ hybridized it will tend to have more p character and therefore move towards $\ce{sp^4}$ hybridization.

Why is this? s orbitals are lower in energy than p orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more s character. The two electrons in the $\ce{C-F}$ bond will spend more time around the electronegative fluorine and less time around carbon. If that's the case (and it is), why "waste" precious, low-energy, s orbital character in a carbon hybrid orbital that doesn't have much electron density to stabilize. Instead, save that s character for use in carbon hybrid orbitals that do have more electron density around carbon (like the $\ce{C-H}$ bonds). So as a consequence of Bent's rule, we would expect more p character in the carbon hybrid orbital used to form the $\ce{C-F}$ bond, and more s-character in the carbon hybrid orbitals used to form the $\ce{C-H}$ bonds.

The physically observable result of all this is that we would expect an $\ce{H-C-H}$ angle larger than the tetrahedral angle of 109.5° (reflective of more s character) and an $\ce{H-C-F}$ angle slightly smaller than 109.5° (reflective of more p character). In terms of bond lengths, we would expect a shortening of the $\ce{C-H}$ bond (more s character) and a lengthening of the $\ce{C-F}$ bond (more p character).

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    $\begingroup$ Bear in mind that Bent's rule speaks to hybrid orbitals - a part of a bond, not the entire bond. I would say, 1) the atom will re-hybridize its hybrid orbitals so as to partition s character wisely (e.g. to follow Bent's rule, use more s character (stabilize orbitals) in orbitals with more electron density as opposed to orbitals with less electron density) and 2) the s character content in all of the hybrid orbitals for a given atom must sum to 1. $\endgroup$ – ron Aug 26 '14 at 22:49
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    $\begingroup$ (1) There is no such thing as rehybridisation, as it is a mathematical concept that is a result of a certain bonding situation. (2) Bent's rule does not specifically apply to hybrid orbitals, but to the linear combination of atomic orbitals. (It is a tiny but all defining difference.) (3) It is not a rule a priori, much more a generalised observation. $\endgroup$ – Martin - マーチン Aug 27 '14 at 3:12
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    $\begingroup$ @Martin 1) I "understand", but it is a handy word. When a water molecule vibrates and the HOH angle changes, there is a change in the electron distribution around oxygen. I would say that the oxygen rehybridizes during the vibration. For me, it's a shorthand way to denote a change in electron density with a consequent change in physical properties. 2) I thought that a carbon sp3 hybrid orbital was a linear combination of the carbon 2s and 2p AOs. If I'm wrong could you elaborate a bit? 3) I agree, I meant to say that it was a tendency in my original post, but forgot. Edited and added it. $\endgroup$ – ron Aug 27 '14 at 15:20
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    $\begingroup$ @AnuragBaundwal Yes, the $\ce{C-F}$ bond length in $\ce{CH3F}$ is longer (139 pm) than that in $\ce{CF4}$ (132 pm). The $\ce{C-H}$ bond length in $\ce{CH3F}$ (108.7 pm) is slightly shorter than that in methane (109.1 pm). $\endgroup$ – ron Jun 4 '18 at 15:27
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    $\begingroup$ Why can't it be sp5......how did you decide it to be sp4? $\endgroup$ – Hydrous Caperilla Jul 10 '18 at 23:27

Have you read the Wikipedia article to Bent's rule (especially the Justification paragraph). I think it explains the things rather well. In the example of $\ce{H3CF}$ the $\ce{H}$ is more electropositive than $\ce{C}$ and the $\ce{F}$ is more electronegative than $\ce{C}$. So, using the assumption that like in $\ce{CH4}$ the $\ce{C}$ atom is $\mathrm{sp}^3$ hybridized as a starting point, Bent's rule tells us that the $\ce{C}$-orbitals that are used to form bonds between $\ce{C}$ and $\ce{H}$ will not be "pure" $\mathrm{sp}^3$ orbitals but will contain a higher $\mathrm{s}$ character whereas the $\ce{C}$-orbital that is used to form the bond between $\ce{C}$ and $\ce{F}$ will contain a higher $\mathrm{p}$ character than a "pure" $\mathrm{sp}^3$ orbital. As for bond angles: the consequences of Bent's rule for the bond angles are also explained rather well in the Wikipedia article.

  • $\begingroup$ Higher s character for the C-H bonds because the H's cannot stabilize electron density and higher p character for the C-F bond because the fluorine can better stabilize electrons? $\endgroup$ – Dissenter Aug 26 '14 at 19:11
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    $\begingroup$ @Dissenter Essentially yes. Instead of directing equivalent sp³ orbitals towards all four substituents, shifting s character towards the C-H bonds will stabilize those bonds greatly because of the increased electron density near the (more electronegative) carbon, while shifting s character away from the C-F bond will increase its energy by a lesser amount because that bond's electron density is more localized on the more electronegative F anyway and thus further from the carbon. $\endgroup$ – Philipp Aug 26 '14 at 19:14
  • $\begingroup$ How do we decide the orientation of the orbitals - i.e. where they are directed? $\endgroup$ – Dissenter Aug 26 '14 at 19:28
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    $\begingroup$ @Dissenter It is more about the bonding partners. When you have different bonding partners, say electropositve A and electronegative B, for a central atom C, then the bonds/bond orbitals from C to A will have more s character than expected and the bonds from C to B will have less s character than expected. So, the s character concentrates in bonds/orbitals directed toward electropositive substituents. That is the sense in which the word "directed" is used in the definition. $\endgroup$ – Philipp Aug 26 '14 at 19:43
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    $\begingroup$ @Dissenter Yes, because s orbitals "penetrate" closer to the nucleus such that the higher electronegativity affects them more strongly than p orbitals. $\endgroup$ – Philipp Aug 26 '14 at 19:59

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