2
$\begingroup$

What would be the comparison between $\ce{P-F}$ and $\ce{P-Cl}$ bond length in phosphorus tetrafluoride chloride $\ce{PF4Cl}$ and phosphorus trifluoride dichloride $\ce{PF3Cl2}$ in the equitorial hybrid orbital arrangements?

1: PF4Cl; 2: PF3Cl2

While determining the bond length between the two compounds I am using Bent's rule, since in the case of $\ce{PF4Cl}$ there are two $\ce{F}$ atoms in equatorial region and thus they decrease their %s character in turn resulting in a larger bond length and since there are two $\ce{F}$ atoms here the %s would be relatively less as compared to the $\ce{P-F}$ bond in $\ce{PF3Cl2}$, so this decrease in %s character in flourine atom should result in increase of %s character in $\ce{P-Cl}$ bond and reducing the bond length and $\ce{P-F}$ bond length would be greater in $\ce{PF4Cl}$.

8. Bond lengths (pm) in substituted phosphorous pentahalides are given below:

$$ \begin{array}{lcc} & \ce{PF4Cl} & \ce{PF3Cl2} \\ \ce{P-F_{eq}} & x & y \\ \ce{P-Cl_{eq}} & a & b \end{array} $$

$$\text{(I)}~x > y \qquad \text{(II)}~y > x \qquad \text{(III)}~a > b \qquad \text{(IV)}~b > a$$

Choose correct code:

(a) Only I
(b) II and IV
(c) I and III
(d) Only IV

So if I'm assuming things right my answer about the $\ce{P-F}$ bond length is right, but here in the option for $\ce{P-Cl}$ bond length is that it's greater in $\ce{PF4Cl}$. Why? Can someone please explain?

$\endgroup$
3
  • 1
    $\begingroup$ Could you please cite the source of the problem you've posted a photo of (e.g. author(s), title, edition, publisher, year, page number(s) and ID(s), if it's a textbook)? $\endgroup$
    – andselisk
    Oct 24 '19 at 6:56
  • $\begingroup$ This question is from a book called V Joshi by cengage publications. It only mentions the correct option but no explanation for the question so I'm totally clueless what's happening here $\endgroup$ Oct 24 '19 at 12:53
  • $\begingroup$ It can be analysed based on Bent's rules. Basically P-F bond will have more p character and hence p-Cl more s character. $\endgroup$ Oct 14 '20 at 16:45
2
$\begingroup$

According to (1) and (2), the values in your table would be as follows (all values in Å):

$$\begin{array}{c c c} & \ce{PF4Cl} & \ce{PF3Cl2} \\ \ce{P-F_{eq}} & 1.541 & 1.546 \\ \ce{P-Cl_{eq}} & 1.999 & 2.004 \\ \end{array}$$

we have $x < y$ and $a < b$, which means that (b) is the correct answer


According to Bent's rule:

atomic s character concentrates in orbitals directed towards electropositive substituents

Bonds with larger s-character are shorter in nature. Keeping these two points in mind and analyzing the structures, in $\ce{PF4Cl}$, almost all the s-character is directed towards $\ce{Cl}$, which means that $\ce{P-Cl}$ bond length in $\ce{PF4Cl}$ is less than that in $\ce{PF3Cl2}$. Similarily, considering that the $\mathrm{sp^3d}$ hybridization can be considered as $\mathrm{sp^2+pd}$, in the equatorial plane, the s-character shared between the two fluorine atoms in $\ce{PF4Cl}$ would be more than the s-character given to the single fluorine atom in $\ce{PF3Cl2}$, hence the shorter bond length.

Another perspective is that the above set of arguments are largely hand-waving, as Bent's rule is at best empirical and the bond length differences are $\pu{\approx 5 pm}$, which is very small. An easier question would be to compare bond lengths between $\ce{PCl5}$, $\ce{PF3Cl2}$ and $\ce{PF5}$, but for this question, I don't have a 'solid' explanation per se, other than the lengths from the references and the elaboration of Bent's rule above.

References:

  1. Leiding, Jeff, et al. “Bonding in PF2Cl, PF3Cl, and PF4Cl: Insight into Isomerism and Apicophilicity from Ab Initio Calculations and the Recoupled Pair Bonding Model.” Theoretical Chemistry Accounts, vol. 133, no. 2, Feb. 2014, p. 1428. doi:10.1007/s00214-013-1428-7.
  2. French, Richard J., et al. “Dichlorotrifluorophosphorane (PCl2F3): Molecular Structure by Gas-Phase Electron Diffraction and Quadratic Force Field.” Inorganic Chemistry, vol. 24, no. 18, Aug. 1985, pp. 2774–77. doi:10.1021/ic00212a014
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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