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CHRIS25
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Oxidation state and Charges
Over the last few weeks I have watched many tutorials, read many excerpts from chemistry sites. Although I am not aware of any particular definition
being given I have subconsciously inherited a misconception and not of my own doing. That oxidation number and charge number were interchangeable
terms meaning well the same thing. A sudden realisation during a conversation with aga about KMnO4 and MnO4 ripped a hole 20
metres wide through something I thought that I had grasped. The Oxidation number and charge number can not possibly mean the same thing, so I googled
the question and came up with three sites each with a different slant, in one site you can see how amateurs like myself can be easily mislead when the
words 'oxidation' and 'charge' "appear" to be thrown about quite loosely. I would very much like to close this issue by asking knowledgeable members
on this forum the correct perspective. You can read below how it can be a little confusing as to how to correctly grasp the meaning of charge number
and oxidation number
https://uk.answers.yahoo.com/question/index?qid=201103051527...
http://www.thenakedscientists.com/forum/index.php?topic=2842...
http://uachemistry13.wordpress.com/2013/07/19/charge-v-oxida...
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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DraconicAcid
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The oxidation state of an atom is what its charge would be *if* the compound its in was entirely ionic. For KMnO4, there is a K+ cation
and an MnO4- anion. The charge on the potassium ion is the same as its oxidation state (+1). The manganese itself doesn't have a charge,
because it's part of a polyatomic ion with an overall charge of -1. The manganese does have an oxidation state, though- it's +7, because if the
polyatomic ion was entirely ionic, it would be Mn7+ and four O2- ions.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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CHRIS25
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Quote: Originally posted by DraconicAcid  | | The oxidation state of an atom is what its charge would be *if* the compound its in was entirely ionic. For KMnO4, there is a K+ cation
and an MnO4- anion. The charge on the potassium ion is the same as its oxidation state (+1). The manganese itself doesn't have a charge,
because it's part of a polyatomic ion with an overall charge of -1. The manganese does have an oxidation state, though- it's +7, because if the
polyatomic ion was entirely ionic, it would be Mn7+ and four O2- ions. |
So this is where I was getting confused, I said the charge on K was +1 Mn +7 and O2 total -8 making it neutral; but it is in fact neutral but the Mn
has no charge.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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DraconicAcid
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| Quote: Originally posted by CHRIS25 | | So this is where I was getting confused, I said the charge on K was +1 Mn +7 and O2 total -8 making it neutral; but it is in fact neutral but the Mn
has no charge. |
It doesn't have a charge because it's covalently bonded to the oxygens, and the whole permanganate ion has a charge. I wouldn't call it neutral,
though, more like "not applicable" (although some people will be able to calculate the actual charge density on the manganese, it's not particularly
useful). What you say about the charges applies to the oxidation states.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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aga
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Kindly check if i got this right, as i am struggling on this one.
KMnO4
The molecule has ionic and covalent components, with the covalent bit behaving as an ion.
MnO4 being the covalent bit, and *also* the -1 charge anion with the K+1 charge cation in the ionic bond.
Treating each part as if it were ionic, the K +1 charge = it's oxidation state, as there's nothing else going on.
MnO4 however, considered as if it were ionic bonded, would mean the Mn giving up 7 of it's electrons, becoming 7+ charged.
(I am guessing that i can not go 8+ as the valence is 7 not 8, but Why is not clear)
The four O would take 2 electrons each = -8 charge, with the Missing electron coming off the K.
So K +1, MnO4 -1 CHARGE STATES
K +1, Mn +7, (O -2) x 4 = -8 OXIDATION NUMBERS ?
[Edited on 9-5-2014 by aga]
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blogfast25
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No. The oxidation state of EACH oxygen atom is -2. There's no point in defining an oxidation number for two atoms, even if they're are the same.
The oxidation states (have they started to use the term 'oxidation numbers' now?) for a neutral species must add up to 0.
K: +1
Mn: +7
O: -2
Add it up: 1 x (+1) + 1 x (+7) + 4 x (-2) = 0, so in that sense you are entirely correct.
For a polyatomic ion, the sum of the oxidation states must be equal to the charge of the ion. Example: the sulphate ion:
SO<sub>4</sub><sup>2-</sup>.
We know that the oxidation number for O in this ion is -2. Find the oxidation number of S, say X, in this ion.
A: -2 = 1 x X + 4 x (-2), ergo X = +6
Other example: H<sub>2</sub>O<sub>2</sub>. The oxidation number of H in this molecule is +1, find the oxidation number of O,
say X, in this molecule.
A: 0 = 2 x (+1) + 2 x X ergo X = -1
Homework: find the oxidation state of N in N<sub>2</sub>H<sub>4</sub> (hydrazine) knowing the oxidation state of H is +1.
[Edited on 9-5-2014 by blogfast25]
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Hexavalent
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| Quote: Originally posted by aga | Kindly check if i got this right, as i am struggling on this one.
KMnO4
The molecule has ionic and covalent components, with the covalent bit behaving as an ion.
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It does not behave, per se, as an ion; we evaluate it's oxidation state/oxidation number by pretending that the manganese and oxygen are ions within
the permanganate polyatomic ion, even though they are covalently bonded.
The potassium ion indeed has a charge of +1.
The formula of the compound is KMnO4. Electrically neutral compounds must have an overall charge of 0, which means that the permanganate
anion overall must have a charge of -1 to balance out the +1 of the potassium.
Because the oxidation state of oxygen is generally fixed (with the exception of some compounds such as peroxides), i.e. -2, we begin with a charge in
permanganate of (4 x -2 =)-8.
Because the overall charge on the permanganate must be -1, we can determine the charge on manganese algebraically - where charge(Mn)=x :
-1=x+(-8)
x=8+ (-1)
x=8-1
x=7
Therefore, in potassium permanganate, the "charges" on the atoms are as follows, if all of the atoms were considered to be ions;
K=+1
O=-8
Mn=+7
The charges on the manganese and the oxygen, however, are not "true". Chemists only treat them as ions to track the movement of electrons in
reactions. The negative charge on the permanganate ion is delocalised across the structure, as shown in models such as;
[Edited on 9-5-2014 by Hexavalent]
"Success is going from failure to failure without loss of enthusiasm." Winston Churchill
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CHRIS25
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Ah that was mean, I just told him that O is always -2, but forgot to add EXCEPT in H2O2
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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There are always exceptions. Peroxides are one of them: in those the oxidation number of O is always -1.
Mwwhahaha... But I see Hexavalent has outshone me straight in meanness.
Different levels of understanding: a bit like peeling an onion.
[Edited on 9-5-2014 by blogfast25]
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CHRIS25
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Maybe I could explain extra that the Mn atom has 7 valence electrons and each O atom has 2. Each of the 2 valence bond covalently with two on the Mn.
EXCEPT the last O atom which can only bond ONE of its 2 valence electrons to the Mn. So it has a spare electron wizzing around making the WHOLE
molecule negatively charged. Try drawing it, it might help.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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Töilet Plünger
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I learned about "formal charges..." basically, every pair of electrons counts as -2 to the formal charge, and every bond counts as -1. So the formal
charge of manganese in permanganate would be -1, because the manganese forms a double bond with all four surrounding oxygen atoms (correct me if I am
wrong here).
http://en.wikipedia.org/wiki/Formal_charge
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aga
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Homework ?
I am at home !
The N has an oxidation state of the square root of Popeye, which is -2
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aga
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| Quote: | | There's no point in defining an oxidation number for two atoms |
You're too fast.
I edited it.
@TP - i cannot consider Formal Charges and also Study chemistry in this country.
[Edited on 9-5-2014 by aga]
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blogfast25
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| Quote: Originally posted by CHRIS25 | | Maybe I could explain extra that the Mn atom has 7 valence electrons and each O atom has 2. Each of the 2 valence bond covalently with two on the Mn.
EXCEPT the last O atom which can only bond ONE of its 2 valence electrons to the Mn. So it has a spare electron wizzing around making the WHOLE
molecule negatively charged. Try drawing it, it might help. |
You're close. Remember: all these electrons are delocalised as per quantum mechanics. The electrons are 'smeared out' as it were, between the atoms
they bond.
Take a simple case: HCl or skeletally: H-Cl, with '-' an electon pair forming the bonding orbital. Where the electrons are exactly we don't know. But
we do know that on average they spend more time closer to the Cl than the H: that gives the Cl a partial negative charge, the
hydrogen a partial positive charge and the whole molecule is a 'permanent bipole'.
Confused yet?
Then do the work!
[Edited on 9-5-2014 by blogfast25]
[Edited on 9-5-2014 by blogfast25]
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DraconicAcid
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Bipole? Surely you mean dipole?
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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blogfast25
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Oooopsie. Dipole it is. Bugger me with a fish fork.
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aga
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Hmm. Di/Bipolar. Fish Fork. Smeared out across orbitals.
This is approaching a phase where it could be charged for as Mainstream Internet.
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CHRIS25
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@Blogfast" =====Confused yet?====" For the first time in a long time - No. The smearing I understand, it's more "electron scope"reality I
understand, but everyone teaches in a more theoretically two dimensional model and so that is how I learn it for now, but I do get it finally. Now to
do some practical....
@aga, I saw some photos of electrons wizzing around atoms, it really is so different than the blackboard and chalk, it's a whole new world. So
smearing would be like blurring two halves of each of two photographs together where the one of the photos has more of its width un blurred and the
other photo is drawn more into the photo leaving less of its image sharp. Or have I gone doo-la-lilly?
[Edited on 9-5-2014 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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aga
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Nope. You got it.
If you're going nuts, then you're on the right track.
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blogfast25
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Here's a good one: when thiosulphate reacts with iodine, it is reduced to tetrathionate:
S<sub>4</sub>O<sub>6</sub><sup>2-</sup>. Oxidation number of the S?
-2 = 4 x X + 6 x (-2) ergo X = +2.5 !!!
Turns out that two of these S atoms in that anion are at 0, the two others are at +5! The average is +2.5.
http://en.wikipedia.org/wiki/Tetrathionate
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CHRIS25
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And for those of you like me who hate not understanding something for more than 10 minutes in one go.....here is the solution:
http://thetriplebond.blogspot.ie/2014/03/the-unusual-oxidati...
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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kmno4
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Formal charges, oxidation numbers ... etc, are toyish tools for balancing chem. equations... etc. Sometimes they are very useful, but one must
remember than in many cases given values are arbitrary and give idiotic values for some molecules, for example Sn93- (and many
other Zintl ions).
Much more real values (but still not free from abitrariness) can be obtained from X-ray experiments, Mössbauer spectroscopy... etc.
See linked paper - not easy to read if you are not familiar with Bader's AIM theory, but in Table 5 (p.500) you get "net atomic charges" - much better
estimators of real charges than infantile oxidation numbers.
| Code: | http://www.chem.gla.ac.uk/~louis/xdworkshop/kmno4/src/manuscript.pdf |
Слава Україні !
Героям слава !
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aga
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I should suggest that as Advanced as Current Knowledge may seem, it is likewise Infantile to even start to Imagine that we Know Anything at all.
CHRIS25 is asking about a specific method, not whether the method is valid in reality.
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blogfast25
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kmno4 is correct though, aga. See e.g. the model for MnO<sub>4</sub><sup>-</sup> Hexavalent put up. Many polyatomic ions need
that treatment and then it becomes apparent charge distribution is not an easy matter. Even in simple covalent compounds it's not that easy.
We have to deal with reality too.
[Edited on 12-5-2014 by blogfast25]
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woelen
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You can use oxidation state as a bookkeeping method, but it nothing more than that. It is quite useful though in many situations.
It does tell something about the "extent to which a certain element is oxidized". The higher the oxidation state, the "more oxidized" the element is.
If you look at ions MnO4(-) and MnO4(2-), then in the former, the oxidation state of manganese is +7 and in the other it is +6. So, in the former it
is "more oxidized" than in the latter.
The bookkeeping comes with some rules (higher rules go before lower rules):
- Fluorine has always oxidation state -1, except in F2 (where it has oxidation state 0).
- Oxygen nearly always has oxidation state -2, except in the case of peroxides, peroxo complexes, superoxides and ozonides.
- Other halogens have oxidation state -1.
- Alkali metals always have oxidation state +1, except for the elements and alloys of alkali metals (in the latter cases they have oxidation state 0).
- Earth alkali metals always have oxidation state +2, except for the elements and alloys of them.
- Aluminium has oxidation state +3, except for the element.
- Metals have positive oxidation states.
Some examples:
OF2: Apply rule 1, so F has oxidation state -1, hence oxygen has oxidation state +2.
KMnO4: Oxygen has oxidation state -2, potassium has oxidation state +1, so Mn has oxidation state +7.
SO4(2-) ion: Oxygen has oxidation state -2, total charge is -2, so sulphur must have oxidation state +6. This is the highest possible oxidation state
of sulphur and hence is the element in the most oxidized state.
Sometimes oxidation states of elements are hard to tell and are ambiguous. E.g. look at the ion S2O3(2-), the thiosulfate ion. This has the structure
of a sulfate ion, with one oxygen replaced by sulphur. Some people say that the central sulphur atom has oxidation state +6 and the outer sulphur atom
has oxidation state -2. But in reality, the central sulphur atom is in a much less oxidized state as in sulphur. So, other people say that it has
oxidation state +4, while the outer sulphur atom has oxidation state 0, and there even are people who say that both sulphur atoms have equal oxidation
states +2.
What is correct? Here you see the flaw of the concept of oxidation state. It is a nice bookkeeping method and can be very useful, but when you use it,
you must keep in mind that the concept has its limitations.
One more weird oxidation state:
Oxygen in the superoxide ion: O2(-), e.g. in KO2. Here oxygen has oxidation state -1/2, a fractional value. Both oxygens are equal in the superoxide
ion, so there is no other decent answer.
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