[AhBeng]

Why is PH3 a polar molecule when P and H have the same electronegativity? Is it because of the lone pair on P, making electron density greater on P? The dipole moment is 0.58 D. Is there slightly greater electron density towards the P or the Hs? Are the covalent bonds herein polar or non-polar?


There is also the matter (which may or may not be relevant) that the bond angels are close to 90 (93.5) degrees, and that "The low dipole moment and almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals. The high positive chemical shift of the P atom in31P NMR spectrum accords with the conclusion that the lone pair electrons occupy the 3s orbital and so are close to the P atom. This electronic structure leads to a lack of nucleophilicity and an inability to form hydrogen bonds."

Please advise. Thanks!

[AhBeng]

So after thinking about it a bit, and from the Wikipedia info, the following are my thoughts. If anyone has anything further to add, please do so.

As the electron geometry is tetrahedral (the molecular geometry is trigonal pyramidal), you might expect the P bonding orbitals to be sp3 hybridized, in accordance to the bond angles.

However, note that because of the following (from Wikipedia) :

>>> PH3 is a trigonal pyramidal molecule with C3v molecular symmetry. The length of the P-H bond 1.42 Å, the H-P-H bond angles are 93.5°. The dipole moment is 0.58 D, which increases with substitution of methyl groups in the series: CH3PH2, 1.10 D; (CH3)2PH, 1.23 D; (CH3)3P, 1.19 D. In contrast, the dipole moments of amines decrease with substitution, starting with ammonia, which has a dipole moment of 1.47 D. The low dipole moment and almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals. The high positive chemical shift of the P atom in31P NMR spectrum accords with the conclusion that the lone pair electrons occupy the 3s orbital and so are close to the P atom. This electronic structure leads to a lack of nucleophilicity and an inability to form hydrogen bonds. <<<
 
(edited : For those who might be confused, I will give a simplified explanation of the statement "the almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals". Recall that p orbitals are px, py and pz, 90 deg angles. Whilst sp3 hybridized orbitals have 109.5 deg angles. Since in PH3 the angles are much closer to 90 deg than 109.5 deg, you would expect the P-H bonds to be mainly sigma bonds from the overlap of p electrons (from P) and s electrons (from H). Notice then, that the s orbitals are largely not involved in bonding, ie. the lone pair occupies the s orbital. Recall that the electrons in the s orbital, on average, are closer (ie. spend more time nearer) to the nucleus, compared to the electrons in the p orbitals. In conclusion, the lone pair is close to the P nucleus, and hence P has slightly increased electron density, which has an effect on polarity of the molecule.)

Hence, due to the particular nature of the PH3 molecular orbitals and the trigonal pyramidal molecular geometry, specifically that the lone pair occupying the 3s orbital and thus being relatively closer to the P atom than the bond pairs between P and H, there is consequently a slightly greater electron density and a slightly partial negative charge on the P atom (even as P and H have the same electronegativity), and hence a slight dipole moment, making PH3 a slightly polar molecule.

Therefore, you would expect the permanent dipole-dipole interactions to be weak, unlike the stronger hydrogen bonding in NH3. Induced dipole-dipole van der Waals interactions, that are present between all molecules (whether polar or non-polar), would play a significant role in the intermolecular attractive forces for PH3.

If anyone has anything further to add, please do so. Thanks in advance!

[tamim83]

I just have some problems with "lone pairs".  There really is no such thing as lone pairs.  Electrons delocalize throughout the molecule, they don't just stay in one place.  In this case, they don't even spend most of their time on one atom because there is no electronegativity difference. 

I ran a calculation on PH3 using Gaussian, just to look at the MOs.  Here are some results using RMP2/6-31g(d)

Dipole Moment: 0.9907D (Way too large)
H-P-H angle: 94.598
P-H bond length: 1.41456 Angstroms

The MOs look a great deal like ammonia's MOs except the 2s and 2p orbitals do not participate in bonding at all.  The lowest valence orbital is all bonding with the 3s orbital of P and the 1s orbitals of H.  The others are two degenerate orbitals with one node (the 2 3p orbitals of P bonding with the antibonding and nonbonding combination of the 3 1s orbitals of H).  The HOMO is the interesting one, it is the 3pz orbital of P bonding with the all bonding combo of the 1s H orbitals.  I think that this orbital is responsible for the change in bond angle from the predicted 109.5.  If the H-P-H angle contracts closer to 90 degrees, it makes that bonding interaction much better and lowers the energy considerably.  Also, the 1s orbitals of H and the 3p orbitals of P are probably pretty far away in energy so this bond angle contraction helps the energetics a great deal. 

As for lone pairs, one could say that the lone pair sits in this HOMO orbital, but I don't think it sits in the 3s orbital at all since the 3s orbital does not really exist anymore.  I think it sits in this HOMO orbital.  I also do not believe the lone pair is responsible for the slight dipole moment since that implies that it just sits on the P atom most of the time. 
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[parranoya]

Almost every molecule will dissolve in water to some extent. Thanks to London dispersion forces. You can get complicated and start looking at molecular orbitals and electron density clouds. This works but chemists use different methods to determine solubilities. One way we use in high school is to look at their molar solubilities at 25ºC. Wikipedia has this chemical listed as having a solubility of 0.70 g/100mL which is about 0.864 mol/L. This means it will dissolve in water but only to a small extent. This is under the limit of 0.100mol/L that we use to describe mostly nonpolar/nonsoluble compounds.

There are only a few totally nonpolar covalent bonds and the rest, any other atom bonded to another, are all slightly polar. There are also many ways of measuring electronegativities. No 2 atoms can have the same electronegativity as another based on the quantum mechanical theory of the atom. Hydrogen has an electronegativity of 2.20 and phosphorus has a value of 2.9. So they do not share the electron equally in a covalent bond. Not enough to attract water though.

You can also look at waters ability to be attracted to - and + dipoles in a molecule. If it can bond to both this inceased its solubility in water. C-H bonds are polar but in hydrocarbonswater cannot make it into the - dipole on the carbon atom. All water can see are the + hydrogens on the outside of the atom so cannot dissolve well in water. They only attract  slightly. They do not repel as you can hold a drop of water on the bottom of a candle.

Hope this helps....this is a pretty hard topic to explain as there is no one real reason that molecules are polar. Almost all areas of  chemistry are grey areas that can go both ways. Yeah for chem!