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?