Ask A Scientist Chemistry Archive

Question: 
When predicting molecular geometries (especially in trigonal
bipyramidal arrangements), how can you predict which "bonds"
are lone pairs, and which are actual bonds?
 
For example, I3- (subscript 3, superscript -), has a linear geometry
because the two Iodines are bonded to the polar bonds, but they could
just as well have been bonded to two of the angular bonds, or to a
polar and one to an angular (for a seesaw shape)...
 
How do you know?
 jon j shotola
 
Answer 1:
Ultimately, this kind of question can only be resolved
by experiments designed to measure the geometry. For
example, ozone (O3) has a perfectly valid Lewis Dot structure
corresponding to an equilateral triangle...yet, the actual
structure is that of an isoceles triangle. Ultimately, the
Lewis dot structure rules are a "game" which in many cases
can be used to guess structures and, ultimately, wavefunctions
of molecules. But there are plenty of instances where it
fails (diatomic oxygen is the classic example...O2 is paramagnetic,
whereas Lewis dot rules predict that it is diamagnetic).
For such cases, we must go to more sophisticated "games" for
structure-guessing. Such games include molecular orbital theory
and valence bond theory... in the case of ozone, MO theory
correctly predicts that O3 is bent, not an equilateral triangle.
Such ambiguities are common in small homonuclear molecules,
and I guess that something like this is happening with I3-.
 
What's more, iodine is a large-numbered atom and therefore d orbitals
are probably involved in the bonding (they've gotta be actually;
the octet rule isn't satisfied by I3-). Basically there are no simple
rules which can be successfully used for elements that are
so far down the periodic table.
 
-topper
 
Answer 2:
I'd like to add a little more to the proceeding...
 
I3- has 22 valence electrons, right? So if you make a linear
molecule out of three iodides, well, you've got an "extra" lone
pair. If you arbitrarily try to put that lone pair on the central
I, well, that I has 5 "structural" pairs (3 lone pairs and 2
bond pairs). Using the VSEPR model, you'd predict that I3- would
therefore be bent, with an I-I-I angle of about 120 degrees
(with the structural pairs forming a trigonal bipyramid).
But nature says no: that I2 forms a "charge-transfer complex" with
I- to form a linear I3- molecule. Actually, I3- can probably be
best decribed as (I2)I-, i.e., an iodide ion stuck to an iodine
molecule.  Such complexes really can't be well-described by
the usual rules, which neglect electrostatic interactions between
non-covalently bound species.
 
-topper