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Unlike ammonia, the lone pair of phosphorous does not add to the dipole moment of phosphine. My textbook says this is due to the presence of lone pair of phosphorous in $\ce{PH3}$ in an s-orbital. I couldn't understand my book's explanation. How can the orbital of lone pair determine its contribution to the dipole moment?

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  • $\begingroup$ I elaborate on that in the second half of my answer here. Also, user ron already seems to have answered that in another question you made, so you might want to reread it. $\endgroup$ – Nicolau Saker Neto Jun 29 '14 at 11:04
  • $\begingroup$ Ron has beautifully answered that ; i also understood the 1st part but couldn't decipher how the orbital of lone pair is important in determining the dipole moment. I only want to know why it happens so? $\endgroup$ – user5764 Jun 29 '14 at 11:13
  • $\begingroup$ @Nikolau Saker :How does the asymmetry of an orbital give rise to dipole moment? $\endgroup$ – user5764 Jun 29 '14 at 11:19
  • $\begingroup$ @user36790 Do you understand why the lone pair in ammonia contributes to the dipole moment? $\endgroup$ – ron Jun 29 '14 at 13:25
  • $\begingroup$ @ron : Sir,yes but no! Sir, I could not understand how an asymmetric hybridised orbital can cause polarity (in case of ammonia) and why does a symmetric one cannot add to polarity (in case of phosphine). $\endgroup$ – user5764 Jun 29 '14 at 14:10
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Let's see if we can use the following picture to help clarify things.

enter image description here

Let's look at the molecule $\ce{H3X-Z}$. Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$. Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment. Are you with me so far?

If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased. If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$. These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture. The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be. The orbital and the electrons in it still have directionality. They still contribute to a dipole moment. Can you see this?

Let's continue assuming you can. Let's repeat all that we've done above with one change. Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized. Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital. An $\ce{s}$ orbital is spherically symmetric. It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric. So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.

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  • $\begingroup$ @ ron:sir,if instead of taking $s$ ,if I take other pure atomic orbitals,then also will I get the same result? $\endgroup$ – user5764 Jun 30 '14 at 2:51
  • $\begingroup$ @user36790 No, all other orbitals, p, d and f have directionality; the distribution of electrons in them is not spherically symmetric. Only the electrons in an s orbital are distributed in a spherically symmetric manner. $\endgroup$ – ron Jun 30 '14 at 2:57
  • $\begingroup$ @ ron:Thanks for such an excellent answer. It is clear the orbital which has direction can be responsible for polarity. $\endgroup$ – user5764 Jun 30 '14 at 4:46
  • $\begingroup$ Great, I think you've got it now! $\endgroup$ – ron Jun 30 '14 at 13:13

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