# Bond length comparison in substituted phosphorus pentahalide

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?

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?

• 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)? – andselisk Oct 24 '19 at 6:56
• 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 – aditya prakash Oct 24 '19 at 12:53
• 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. – Manjunath A Y Oct 14 '20 at 16:45

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