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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?

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    $\begingroup$ Short answer: Not really. I am sure there are many comments about this molecule on the site. There are various reasons why there is only little to none hybridisation. $\endgroup$
    – Martin - マーチン ♦
    Jul 31, 2014 at 18:02
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    $\begingroup$ I'm quite sure that I read in "Organic Chemistry" by Clayden et al. that $\ce{PH3}$ is unhybridized. $\endgroup$
    – Philipp
    Aug 1, 2014 at 1:20
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    $\begingroup$ See also here: chemistry.stackexchange.com/q/14087/4945 I might be able to boil it down for you later, but it is written there between the lines. $\endgroup$
    – Martin - マーチン ♦
    Aug 1, 2014 at 3:01

2 Answers 2

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

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.

$^{*1}$(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]

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  • $\begingroup$ How would you rationalize the 93.5 degree bond angle? $\endgroup$
    – Dissenter
    Aug 13, 2014 at 16:12
  • $\begingroup$ @Dissenter my point $*4$ might answer $\endgroup$
    – RE60K
    Aug 13, 2014 at 16:14
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    $\begingroup$ @Dissenter - Hybridization is a fine model for the smaller atoms, but breaks down in the third period. See my answer to this question: chemistry.stackexchange.com/questions/33818/… $\endgroup$
    – Ben Norris
    Jul 25, 2015 at 13:24
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I think in NH3 N-H bonds are very close to each other so if lone pair does not participate in the hybridisation then repulsion will take place so it participate in the hybridisation to decrease the repulsion but not completely because hybridisation increases the energy of the s orbital because P orbitals give it direction and it is also far than s orbital so bond angle is not 109.5 it is approx 107 but in the case of PH3 the P-H bonds are far from each other and H is also a small atom and its electronegativity is also near the P atom so hybridisation almost does not take place and bond angle is near 90 degree.but in PCl3 Cl is a big atom and its electronegativity is also very high than P because of this it decreases the size of P and hybridisation take place.

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    $\begingroup$ The gist of this is correct, it is just badly phrased and not well presented. $\endgroup$
    – Jan
    Jan 16, 2017 at 22:26

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