theoretical chemistry - Hybridization in PH3

Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?

Arguments against hybridization:

1. 
   

   $\ce{PH3}$ is less basic than $\ce{NH3}$.

   
   



2. 
   

   This jibes with the supposition that $\ce{PH3}$ keeps its lone pair in what is essentially an unhybridized $\ce{s}$-orbital.

   
   


3. 
   

   $\ce{s}$-orbitals are symmetrical and therefore do not concentrate electron density anywhere.

   
   


4. 
   

   This lack of concentrated electron density as opposed to $\ce{NH3}$ which is $\ce{sp^3}$ hybridized makes $\ce{PH3}$ a poorer base.

   
   

However, my professor writes in his book that it can be argued that $\ce{PH3}$ exhibits $\ce{sp^3}$ hybridization.

Another professor told me that the wave function "blows up" when putting an electron pair in an s-orbital. What does that mean?

Answer

Pre: My views may be clashing with some people. I express them as far as I know.

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To start with, hybridization is a hypothetical concept$^{*1}$ just to explain all facts and make our work easier, otherwise we would have to take into account every single interaction of electron-proton and also other phenomenons; many theories developed which explained all the major/important facts at a macro-level. And what I have been saying to take into account was all micro-level analysis. Attributing hybridization is purely personal choice because neither does it account all facts (for some molecules) nor it can be dropped altogether (for its $\cdots$), if then we would look for another theory$^{*2}$ to explain observed phenomenon. Parallels similarity between this and the situation of Ideal-Gas equation, in general no perfect theory has been developed, but the most easy, practical and useful are the hybridization$^{*3}$ and van-der-Waals equation.

As we move downwards in the periodic table the hybridization concept fails at many places due to overpowering of other factors. I wonder you didn't mentioned the bond angle$^{*4}$ ($93.5^\circ$) for a $\ce{sp^3}$ hybridized (ideally $109.5^\circ$).

In general atom/molecule doesn't seek to follow a theory, it does what it is best comfortable with, it is us, who mould the theory to the observations.

As a general rule, for sake of convenience, assume all molecules show hybridisation and the tendency to hybridise decreases down the group.

In conclusion, I would say it depends on the need of the situation and such dubious statements such as "Does $\ce{PH3}$ exhibit $\ce{sp^3}$ hybridization?" should be replaced with "Does hybridization explain all (actually not all) properties of $\ce{PH3}$"

Also there are weights in $\ce{sp^\alpha}$ where $\alpha$ ranges from 1 to 3(even upto 4$^{*5}$ and maybe beyond) and is in between these for $\ce{PH3}$ and some may continue to treat it has a hybridised molecule with suitable weights.

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$^{*1}$(as far as I know)
$^{*2}$(probably the MOT)
$^{*3}$(many may argue for other theories which is genuine, but in context this follows)
$^{*4}$(One may argue as we move down the group electronegativity decreases and atomic size increases. In case of $\ce{NH3}$ due to higher bond pair bond pair repulsion (since electronegativity of $\ce{N}$ atom is very high hence it attracts bonded electrons of $\ce{N-H}$ bond towards itself) bond pair moves away from each other and hence shows greater bond angle. This may be considered one of the major factors for less hybridization.)
$^{*5}$[$\ce{O}$ in $\ce{H2O}$ which means that they have 20% s character and 80% p character, but does not imply that they are formed from one s and four p orbitals]