Re: NH3 and PH3 molecule shape



In article <41a2eea5.0506150344.46b0c60d@xxxxxxxxxxxxxxxxxx>, josef muller <josef.muller@xxxxxxxxx> wrote:

: i have read that the difference in the shape of molecule of NH3 and
: PH3 stems from different hybridization. can anybody explain me this?
: electron configurations of N and P are rather similar (2s2 2p3 and 3s2
: 3p3). why they are different in shape? is VSEPRT helpful in this case?

While you have so far received at least one wrong answer, I haven't seen
any explanations so far (except for the one that I had almost finished
when there was a power failure). So here's something of an answer to
your question.

One thing that you have to realize is that most people who learn introductory
chemistry are taught hybridization more or less exactly backwards. The
concept of hybridization is used to explain an already known structure, not
to predict the structure of a molecule given its formula.

Now let's consider the compounds of the general formula AH3 (A = a Group 15
(aka Group VA) atom). Since the ground-state configuration for each atom
A is s2p3, we might think a priori that the atom can form three covalent
bonds contributing one electron in each of the three p orbitals. The two
remaining electrons would still be in the (full) s orbital. According to
this model, the H-A-H angle should be 90 degrees, and indeed, for
A = P, As, Sb, that is more or less what we do see. For A = N (NH3), however,
the H-N-H angle is more like 107 degrees. How can we explain that deviation
from the expected geometry?

One model that can explain this is *hybridization*. In "hybridization,"
_n_ atomic orbitals combine to form _n_ "hybrid" orbitals, which in the
most commonly used scheme are linear combinations (sums and differences)
of the atomic orbitals. It is possible to show mathematically that the
mathematical combination of one s and one p yields two "sp" orbitals that
are identical except for their direction (they face 180 degrees apart); the
combinations of one s and two p yield three "sp2" orbitals, pointing 120
degrees apart; and the combinations of one s and three p orbitals yield
four "sp3" orbitals, which point at the corners of a tetrahedron (H-A-H
angle of about 109.5 degrees). From the point of view of this model, the
orbitals of the central N atom seem to be most accurately described as
being "sp3" hybrids. Since the H-N-H angle is a little smaller than would
"sp3 hybrids." Since the actual H-N-H angle is a bit smaller than pure
sp3 hybridization would predict, the orbitals that form the bonds are said
to have slightly more "p character" than pure sp3 orbitals would; that is,
instead of a 1:1:1:1 combination of s and three p orbitals, the s component
is slightly less than 1, and the p components larger by an equivalent amount.
According to this model, the two remaining electrons from the N atom will
be in the fourth sp3 orbital, which points toward the fourth corner of
the tetrahedron.

So far, all that we have shown is that hybridization can account for the
structure of the molecule. The next question is whether or not this
scheme is sensible from the point of view of the energy. Since the p orbitals
are higher in energy than the s orbitals, in order to create these
hybrids, we have to add energy to the atom. On the other hand, we get back
energy for two reasons: first of all, the "overlap" between the N orbital
and the H (1s) orbital will be greater for the hybrid orbital than for
the p orbital alone, which stabilizes the bond, lowering the overall
energy. In addition, if the H atoms are at the corners of a tetrahedron,
they will be further apart, which will lessen the electron-electron repulsions
between the bonding pairs. If the lowering of energy from these two
sources overcomes the energy needed to create the hybrids, then the
tetrahedral structure will be more stable than the structure using p orbitals
alone. In the case of N, the effects do compensate, and the N atom
does hybridize.

In the case of the other atoms, however, the gap in energy between the
s and p orbitals is larger than it is for N, which means that more energy
has to be added to the atom to create the hybrids. Not only that, the
atoms themselves are larger, which puts the bonding pairs further apart, so
there is less bonding pair-bonding pair repulsion to begin with. That means
that forming the tetrahedral structure costs more energy but returns less,
and so the unhybridized structure is the more stable one. In this model,
the actual H-A-H angle of 91-93 degrees indicates a very small amount of
mixing in of p character.

One piece of evidence that this model provides a reasonable explanation of
the structures is the relative basicities of the compounds: NH3 is orders
of magnitude more basic than any of the others. Since the lone pair in
NH3 is in an sp3 orbital directed away from the bonding orbitals, making
it a good proton acceptor to form NH4+. On the other hand, for the remaining
AH3 compounds, the lone pair is in an s orbital, which is spherically
symmetric and hence will only bind poorly to a proton, and we find that these
compounds are essentially *not* basic.

As for your question about VSEPR, you have to understand that the VSEPR
model is basically a fairy tale invented to account for the known structures
of compounds. In the case of AH3, it only "works" if you know in advance
that for every compound except for NH3, the lone pair remains in an s
orbital. Within the VSEPR model, if we exclude participation of the valence
s orbital, then the most reasonable structure is the one using the three
p orbitals. (A "T-shaped" structure with a "3-center, 2-electron" bond
is presumably of higher energy because of the bonding pair-lone pair
repulsions.)

I hope that this helps to answer your question.

-----
Richard Schultz schultr@xxxxxxxxxxxxxx
Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel
Opinions expressed are mine alone, and not those of Bar-Ilan University
-----
"Contrariwise," continued Tweedledee, "if it was so, it might be, and
if it were so, it would be; but as it isn't, it ain't. That's logic."
.