Lewis Structure IBr3

[Neil W.]

Can anyone show me the Lewis Structure (including Geometry) for IBr3. I'm
trying to help my son with his chemistry, but we can't figure this one out.
Thanks.

[Bob]

On Sat, 27 Aug 2005 11:26:26 -0400, "Neil W." <ne...@netlib.com>
wrote:

>Can anyone show me the Lewis Structure (including Geometry) for IBr3. I'm
>trying to help my son with his chemistry, but we can't figure this one out.
>Thanks.
>

Hard to draw Lewis here. But more importantly, it is better help if we
guide and you do -- your son does. What is step 1 in drawing a Lewis
structure? How many valence electrons are in this molecule total? Now,
start drawing. At what step is there difficulty?

bob

[Uncle Al]

Do your VSEPR diddle. If it is monomeric is it a singlet molecule?
Planar

Br Br Br
\ / \ /
I I
/ \ / \
Br Br Br

looks not not unreasonable to me - but what does an organiker know?

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

[Sbharris[atsign]ix.netcom.com]

Iodine has 7 electrons. Br's each take one, so that leaves 2 lone pairs
to give everybody 8. Lewis done.

Geometry is tricky. You've got 3 single bonds and 2 lone pairs for a
total of 5, which gives you a structure that looks like two trigonal
pyramids stuck together at the bases. You'd think lone pairs would be
at the pyramid apexes (apices?), but in general lone pairs want all the
space they can get, and they can be 120 degrees appart in the
equatorial position, whereas the angle to the apex is 90 degrees,
right? So picture the two lone pairs out there at two of the three
equatorial positions, and that leaves the bromines hung out in a shape
like a "T." Weird, eh? But that should make the molecule polar, and
these halogen trihalides are indeed all polar molecules. So it works.

SBH

[ab]

On 2005-08-27, Uncle Al <Uncl...@hate.spam.net> wrote:
>
> Br Br Br
> \ / \ /
> I I
> / \ / \
> Br Br Br
>
> looks not not unreasonable to me - but what does an organiker know?
>

Hm...I'm not sure I don't understand you!

[Sbharris[atsign]ix.netcom.com]

COMMENT:

Pay no attention to Al on this one.

[Bob]

On 28 Aug 2005 12:06:56 -0700, "Sbharris[atsign]ix.netcom.com"
<sbha...@ix.netcom.com> wrote:

Well, that may be an overstatement. ICl3 is known to dimerize like
that, and it is reasonable that IBr3 does too. perhaps it is known.

However, if the OP is asking re an exercise for an intro chem class, I
am sure that what is appropriate is the Lewis and expected shape for
the given chemical. If someone wants to then discuss its chemical
behavior, that can come later. In fact, seeing the Lewis for the
monomer may at some point lead one to question the stability.

bob

[Sbharris[atsign]ix.netcom.com]

Bob wrote:
> On 28 Aug 2005 12:06:56 -0700, "Sbharris[atsign]ix.netcom.com"
> <sbha...@ix.netcom.com> wrote:
>
> >
> >ab wrote:
> >> On 2005-08-27, Uncle Al <Uncl...@hate.spam.net> wrote:
> >> >
> >> > Br Br Br
> >> > \ / \ /
> >> > I I
> >> > / \ / \
> >> > Br Br Br
> >> >
> >> > looks not not unreasonable to me - but what does an organiker know?
> >> >
> >> Hm...I'm not sure I don't understand you!
> >
> >
> >COMMENT:
> >
> >Pay no attention to Al on this one.
>
> Well, that may be an overstatement. ICl3 is known to dimerize like
> that, and it is reasonable that IBr3 does too. perhaps it is known.

COMMENT:

Yes, there is a flat I2Cl6 that looks like Unc's drawing. I can't find
reference to the analogus Br compound, perhaps because of steric
effects. In any case, if it dimerizes it's not IBr3, now is it? Don't
make this harder than it is. This isn't P2O5, which is really P4O10. It
really is IBr3.

SBH

[///Owen\\\]

"Neil W." <ne...@netlib.com> wrote in message
news:uE%Pe.794$OT1...@fe09.lga...

Here's a little-known, but useful, way of finding the number of bonding and
lone pairs for molecules that comprise a central atom surrounded by a number
of halogen or hydrogen atoms.......................

1. Work out the total number of valence electrons (add one for a - charge,
and subtract one for a + charge.

2. Divide by 8

3. The whole number gives the number of bonding pairs. Divide the remainder
by 2 and this gives the number of lone pairs.

So now you have the total number of electrons pairs, and using VSEPR you
have the geometry of the parent shape.

Your example gives 3 bonding and 2 lone pairs - total 5 pairs. So... the
parent shape is a trigonal bipyramid. The two lone pairs are at equatorial
positions (you just have to know that the lone pairs take up equatorial (not
axial) positions. You get a "T" shaped molecule.

And........................... if the molecule contains H atoms pretend
each has 7 valence electrons.

eg H2O.............. valence electrons = 7 X 2 plus 6 = 20

divide by 8.................. 2 remainder 4 ....... 2 bonding and 2 lone
pairs = 4 pairs in total............ parent shape is the tetragon

john