# Why chlorine doesn't show pπ-dπ bonding but fluorine shows it? [closed]

I can't understand what is backbonding and how it affects bond angle ?? According to electronegativity concept, bond angle of $\ce{PF3}$ should be less than that of $\ce{PCl3}$ but it is not. Bond angle of $\ce{PF3}$ is greater than $\ce{PCl3}$.How ??

## closed as unclear what you're asking by Mithoron, a-cyclohexane-molecule, A.K., airhuff, DrMoishe PippikAug 12 at 6:21

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• According to Wikipedia the bond angle between the chlorine atoms is 100° and between fluorine atoms it is 96,3°. So what is your reference? – HolgerFiedler Aug 11 at 5:19
• In my book,it is given that bond angle of PF3>PCl3 – Shubham Aug 13 at 8:46

In case of $\ce{PF3}$ since the electronegativity Of $\ce{F}$ is higher than p the sigma electron density of the $\ce{P-F}$ bond will be dragged towards the F atom and hence the $\ce{P}$ becomes electronically deficient. Thereafter in order to reduce the electron deficiency, $\ce{P}$ can accept back the electron density from the three $\ce{F}$ centres simultaneously in its vacant 3d orbital by the delocalized $p-\pi\ d-\pi$ backbonding. Therefore the double bond character of the $\ce{P-F}$ bond will be enhanced. This accounts for increasing the % of S character of the hybrid orbital of P directed towards the $\ce{F}$ will be enhanced. According to the formula $\mathrm{cos(\theta) = S \times (S-1) = (P-1)\times P}$ where S and P are the fractional S and P character.
But in case of Of $\ce{PCl3}$ no such backbonding can be observed in a high extend like $\ce{PF3}$ since the size of both the $\ce{P}$ and $\ce{Cl}$ is large the [please correct]. So they can not come contact and backbonding will not occur.
But your information is wrong because you have said the bond angle in $\ce{PF3}$ is higher than $\ce{PCl3}$. But in reality the $\ce{F-P-F}$ bond angle in $\ce{PF3}$ is lower than $\ce{Cl-P-Cl}$ in $\ce{PCl3}$.
Because $\ce{P}$ is $sp^3$ hybridised in $\ce{PCl3}$ while $\ce{P}$ is $p^3$ hybridised in $\ce{PF3}$ because the $p_x, p_y, p_z$ orbitals of $\ce{PCl3}$ are directly involved in bond formation with the three $\ce{F}$ centres. So the $\%$ of $S$ character of the hybrid orbital of $\ce{P}$ directed towards $\ce{Cl}$ is higher than the hybrid orbital of $\ce{P}$ directed towards $\ce{F}$.
• Hello, welcome to Chemistry.SE! Please review the edits made to be able to better answer and ask questions in the future. It discouraged to indent paragraphs just use double spaces. Also use \$ce{}\$ for representing chemical species and equations and \$\mathrm{}\$ for math equations to present your content more cleanly. – A.K. Aug 11 at 13:33
• Also $p^3$ is not a hybidized orbital. – A.K. Aug 11 at 13:37