Oxidation State of Iodine in Triiodide Ion (for Frost diagram)

[Farooq W]

A student came up with a problem on constructing a Frost diagram from a
given Latimer diagram for Iodine.

Let us assume for simplicity the the diagram is such

H5IO6 ---------> HIO3 ---------> HOI --------> I3(-) ------------> I(-)

Frost diagram requires that oxidation state of the element be plotted
against against the potential E for the half reaction X^n / X^o. Now
what is the oxidation state of triiodide ion-with the problem there are
two different types of iodine in it or conventionally,
I3(-) = I2.I(-)

I am not clear as to what do in such a case for Frost diagram. Should
we take the average as -1/3 ?

Thanks

[beav]

Post by Farooq W
A student came up with a problem on constructing a Frost diagram from a
given Latimer diagram for Iodine.
Let us assume for simplicity the the diagram is such
H5IO6 ---------> HIO3 ---------> HOI --------> I3(-) ------------> I(-)
Frost diagram requires that oxidation state of the element be plotted
against against the potential E for the half reaction X^n / X^o. Now
what is the oxidation state of triiodide ion-with the problem there are
two different types of iodine in it or conventionally,
I3(-) = I2.I(-)
I am not clear as to what do in such a case for Frost diagram. Should
we take the average as -1/3 ?
Thanks

you also need to place HI and I2 in their appropriate location on your
simple Latimer Diagram.

what's the half cell for I2 compared to the half cell for I3(-)? i'm
rusty as heck, but that should pin down the Frost diagram location.

an average -1/3 would seem unlikely....

[Lloyd Parker]

Post by beav

Post by Farooq W
A student came up with a problem on constructing a Frost diagram from a
given Latimer diagram for Iodine.
Let us assume for simplicity the the diagram is such
H5IO6 ---------> HIO3 ---------> HOI --------> I3(-) ------------> I(-)
Frost diagram requires that oxidation state of the element be plotted
against against the potential E for the half reaction X^n / X^o. Now
what is the oxidation state of triiodide ion-with the problem there are
two different types of iodine in it or conventionally,
I3(-) = I2.I(-)
I am not clear as to what do in such a case for Frost diagram. Should
we take the average as -1/3 ?
Thanks

you also need to place HI and I2 in their appropriate location on your
simple Latimer Diagram.
what's the half cell for I2 compared to the half cell for I3(-)? i'm
rusty as heck, but that should pin down the Frost diagram location.
an average -1/3 would seem unlikely....

If you draw the Lewis structure and assign ox. no., you find the middle I is
-1 and the 2 outer ones are 0.

[t***@gmail.com]

Lewis structures allow you to assign formal charges, not oxidation
states.

[Farooq W]

Post by t***@gmail.com
Lewis structures allow you to assign formal charges, not oxidation
states.

See Gary Wulfsberg's "Inorganic Chemistry", there is a detailed method
of extracting oxidation states of elements from the Lewis structures
e.g. determining the oxidation state of S atoms in S4O6 (2-).

[Lloyd Parker]

Post by t***@gmail.com
Lewis structures allow you to assign formal charges, not oxidation
states.

They do both. For ox. no., you assign shared electrons to the more
electronegative element; if both are = in EN, you split the pair.

Check out the diff. ox. no. for the carbons in, say, ethanol, and how we get
them.

[Oscar Lanzi III]

Are we REALLY sure that the Lewis striucture will give oxidaitons states
of 0, -1, 0 fot the iodine atoms?

With three main-group atoms and 22 valence electrons, triiodide ion
cannot be drawn with pairwise bonds and still satisfy the octet rule.
22 valence electrons is one pair too many. So we are faced with a
choice:

1) assume that "outer d" orbitals are available as necessary to expand
the octet. Now we can draw the two bonds, putting 10 electrons around
the central atom, and when we count up oxidation stsates of the
individual atoms we get 0, -1, 0 as claimed.

2) assume that there can be only one bond and it's shared by two atom
pairs. We implement this by supposing two contributing structures --
I(-) I-I, and I-I I(-). This correlates with a MO model where we find
a shared bond involvoing the s and p orbitals. Now the oxidaityon
stsates are -1/2, 0, -1/2.

Which is correct? I would be surprised if either one were completely
correct. The true structure is surely a resonance hybrid of all three
possible valence-bond structures, with the "expanded octet" contribution
depending on the efficacy of the outer d orbitals. A true description
requires a MO model that includes s, p, and outer d orbitals (the one
referred to in possibility (2) above ignores the outer d orbitals).

Oxidation state is somewhat of an artifice already, and the case of
triiodide illustrates what can happen when we rely too literally on it.
The only thing we KNOW about this system, without a lot of sophistated
computer calculations, is that the oxidation states somehow average to
1/3. Fortunately, as far as atom anbd charge balances are concerned in
triiodide reactions, that's all the informaiton we need.

--OL

[Lloyd Parker]

Post by Oscar Lanzi III
Are we REALLY sure that the Lewis striucture will give oxidaitons states
of 0, -1, 0 fot the iodine atoms?
With three main-group atoms and 22 valence electrons, triiodide ion
cannot be drawn with pairwise bonds and still satisfy the octet rule.

Being a third-row or below element, I does not have to satisfy the octet rule,
any more than, say, S in SF4 does.

Post by Oscar Lanzi III
22 valence electrons is one pair too many. So we are faced with a
1) assume that "outer d" orbitals are available as necessary to expand
the octet. Now we can draw the two bonds, putting 10 electrons around
the central atom, and when we count up oxidation stsates of the
individual atoms we get 0, -1, 0 as claimed.
2) assume that there can be only one bond and it's shared by two atom
pairs. We implement this by supposing two contributing structures --
I(-) I-I, and I-I I(-). This correlates with a MO model where we find
a shared bond involvoing the s and p orbitals. Now the oxidaityon
stsates are -1/2, 0, -1/2.
Which is correct? I would be surprised if either one were completely
correct. The true structure is surely a resonance hybrid of all three
possible valence-bond structures, with the "expanded octet" contribution
depending on the efficacy of the outer d orbitals.

And what physical interpretation is there of a fractional ox. no.?

Post by Oscar Lanzi III
A true description
requires a MO model that includes s, p, and outer d orbitals (the one
referred to in possibility (2) above ignores the outer d orbitals).
Oxidation state is somewhat of an artifice already, and the case of
triiodide illustrates what can happen when we rely too literally on it.
The only thing we KNOW about this system, without a lot of sophistated
computer calculations, is that the oxidation states somehow average to
1/3.

In C2H5OH, the carbon ox. no. average out to -2 also. It doesn't mean they're
both -2 (in fact, neither is).

Post by Oscar Lanzi III
Fortunately, as far as atom anbd charge balances are concerned in
triiodide reactions, that's all the informaiton we need.
--OL

[Oscar Lanzi III]

What physical meaning is there to an oxidation no. of -1/2? Find me a
better one for oxygen in potassium superoxide, KO2. Yes it's a real
compound. While we're at it, what is the formal chgarge on each oxygen
in the acetate ion? Fractional electron counts and charges per atom
come with the territory of delocalization.

And I am certainly aware that the carbons in ethanol are not in the same
oxidation state. We can identify an osication state of -3 for one
carbon and -1 for the other. But, only because we have a definitive
bonding model for ethanol. In triiodide ion, this splitting process is
not straightforward because we may have a delocalized bond, we may have
the central atom with 10 valence electrons -- or we may have both!

--OL

[Lloyd Parker]

Post by Oscar Lanzi III
What physical meaning is there to an oxidation no. of -1/2? Find me a
better one for oxygen in potassium superoxide, KO2. Yes it's a real
compound. While we're at it, what is the formal chgarge on each oxygen
in the acetate ion?

Both oxygens have a formal charge of 0. They each have 2 lone pairs and half
of the 2 shared pairs, or 6 val. e.

Post by Oscar Lanzi III
Fractional electron counts and charges per atom
come with the territory of delocalization.
And I am certainly aware that the carbons in ethanol are not in the same
oxidation state. We can identify an osication state of -3 for one
carbon and -1 for the other. But, only because we have a definitive
bonding model for ethanol. In triiodide ion, this splitting process is
not straightforward because we may have a delocalized bond, we may have
the central atom with 10 valence electrons -- or we may have both!

Nothing wrong with 10 val. e. As I said, check out SF4.

Post by Oscar Lanzi III
--OL

[Oscar Lanzi III]

Did I say there can't be 10 valence electrons? You could have fooled
me.

--OL

[beav]

Post by Oscar Lanzi III
Did I say there can't be 10 valence electrons? You could have fooled
me.
--OL

all right - all right - break it up boys...


we're all chemists here...


nerd fights are so unattractive.


let's just leave it as the covalency of I compared to the other
halogens is well known. when's the last time we ever saw F3-, Cl3- or
Br3-?


what's the possibility of seeing At5- or Uus7- (heh heh heh)?

[Oscar Lanzi III]

Br3(-) I've actually encountered. In a college laboratory experiment
brominaiton of an activated aromatic ring was done with what was called
"pyridinium bromide perbromide," which turned out to be C5H6N(+)Br3(-).
(They said in the lab book that its MW is 319, which is a good fit.)

The lack of lighter trihalide ions reflecfts the effect of the
counterion, not any inherent lack of stability. Even for triiodide we
find that KI3 has some asymmetry in the anion; there is partial
polarization to I(-) and I2. Only a bulky cation like a
tetraalkylammonium ion gives a symmetric structure. The polarization is
bound to be worse as we go to smaller atoms and the electrostatic
effects involving the counterion are magnified.

--OL

[beav]

Post by Oscar Lanzi III
Br3(-) I've actually encountered. In a college laboratory experiment
brominaiton of an activated aromatic ring was done with what was called
"pyridinium bromide perbromide," which turned out to be C5H6N(+)Br3(-).
(They said in the lab book that its MW is 319, which is a good fit.)

huh!

you know thinking about it, NCl3, from stepwise addition of Cl2 to
amines, is not exactly unknown either.

geez. what a feeble original comment...

Post by Oscar Lanzi III
The lack of lighter trihalide ions reflecfts the effect of the
counterion, not any inherent lack of stability. Even for triiodide we
find that KI3 has some asymmetry in the anion; there is partial
polarization to I(-) and I2. Only a bulky cation like a
tetraalkylammonium ion gives a symmetric structure. The polarization is
bound to be worse as we go to smaller atoms and the electrostatic
effects involving the counterion are magnified.
--OL

noted...

[Oscar Lanzi III]

Not sure what NCl3 has to to with trihalide. NCl3 has a structure
similar to ammonia, until it blows up.

--OL

[Oscar Lanzi III]

For more on bromine ions, look up the following:

http://www.crscientific.com/atricle-bromine.html

Tribromide is mentikoned in connection with electrolytic production of
bromine. And even pentabromide is reported!

--OL