0
$\begingroup$

My textbook asks this as an objective question. What I had in mind was that the lone pair on phosporous was more distributed as compared to nitrogen(since nitrogen is more electronegative and smaller in size ) and hence the statement. But the option given in the textbook says the answer is lone pair of phosphorous resides in almost pure s orbital. How do you know if the lone pair is placed in pure s orbital or pure p orbital

$\endgroup$
0
$\begingroup$

The bond angle in $\ce{PH3}$ is about $92-93^\circ$, and by the percentage $s$ character formula i.e $\cos(\theta) = s/(s-1)$ you can calculate the $s$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $p_x$ ,$p_y$, $p_z$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.

| improve this answer | |
$\endgroup$
-4
$\begingroup$

We think that both P and N contain lone pair of electrons ,but this is not true if it comes to you donating that lone pair to hydrogen ion AKA H+

Here comes back bonding into action it simply means that if an atom has got lone pair ,it would distribute it to a group to which it is bonded if that group contains (or is) an atom which has its orbital unfilled or empty or has an incomplete octet

First two reasons have got almost same meaning.

P's lone pair get distributed over its own 3D orbital hence availability of lone pair decreases in its pure orbital (lone pair can only exist in pure orbitals)so bond to H + is well ..... prohibited.

Check out this image ... it'll help..

This explanation has been interpreted from Concise Inorganic Chemistry by J.D. Lee

| improve this answer | |
$\endgroup$
  • $\begingroup$ No, phosphorus’ high-energy unpopulated d orbitals have no contribution whatsoever and there is no relevant backbonding. Not to mention that there is certainly no hybridisation between phosphorus’ s lone pair and any of the 3d orbitals. I hope Lee doesn’t state that in their book, because if they do it would be better described as highly outdated. $\endgroup$ – Jan Jul 3 at 10:23
  • 2
    $\begingroup$ @Jan After doing a keyword search and going over the relevant parts of the book here I can confirm that Lee makes no direct claims of such lone pair "distribution" as stated in this answer. This is entirely the answerer's version of interpretation of said book $\endgroup$ – Yusuf Hasan Jul 3 at 11:47

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