Ok, there is more to this matter, but here is a brief
summary. 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.