Sciencemadness Discussion Board - Oxidation state and Charges

Oxidation state and Charges

CHRIS25 - 9-5-2014 at 08:33

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...

DraconicAcid - 9-5-2014 at 08:51

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 O^2- ions.

CHRIS25 - 9-5-2014 at 08:58

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 O^2- 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.

DraconicAcid - 9-5-2014 at 09:05

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.

aga - 9-5-2014 at 12:15

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]

blogfast25 - 9-5-2014 at 12:33

Quote: Originally posted by aga

K +1, Mn +7, (O -2) x 4 = -8 OXIDATION NUMBERS ?

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: SO42-.

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: H2O2. 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 N2H4 (hydrazine) knowing the oxidation state of H is +1.

[Edited on 9-5-2014 by blogfast25]

Hexavalent - 9-5-2014 at 12:39

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.

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]

CHRIS25 - 9-5-2014 at 12:40

Ah that was mean, I just told him that O is always -2, but forgot to add EXCEPT in H₂O₂

blogfast25 - 9-5-2014 at 12:45

Quote: Originally posted by CHRIS25

Ah that was mean, I just told him that O is always -2, but forgot to add EXCEPT in H₂O₂

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]

CHRIS25 - 9-5-2014 at 12:49

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.

Töilet Plünger - 9-5-2014 at 12:55

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

aga - 9-5-2014 at 13:04

Homework ?
I am at home !

The N has an oxidation state of the square root of Popeye, which is -2

aga - 9-5-2014 at 13:05

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]

blogfast25 - 9-5-2014 at 13:09

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?

Quote: Originally posted by aga

Homework ?
I am at home !

Then do the work!

[Edited on 9-5-2014 by blogfast25]

[Edited on 9-5-2014 by blogfast25]

DraconicAcid - 9-5-2014 at 13:11

Bipole? Surely you mean dipole?

blogfast25 - 9-5-2014 at 13:14

Quote: Originally posted by DraconicAcid

Bipole? Surely you mean dipole?

Oooopsie. Dipole it is. Bugger me with a fish fork.

aga - 9-5-2014 at 13:40

Hmm. Di/Bipolar. Fish Fork. Smeared out across orbitals.

This is approaching a phase where it could be charged for as Mainstream Internet.

CHRIS25 - 9-5-2014 at 13:43

@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]

aga - 9-5-2014 at 13:52

Nope. You got it.

If you're going nuts, then you're on the right track.

blogfast25 - 10-5-2014 at 05:25

Here's a good one: when thiosulphate reacts with iodine, it is reduced to tetrathionate: S4O62-. 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

CHRIS25 - 10-5-2014 at 07:04

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...

kmno4 - 11-5-2014 at 14:03

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 Sn₉^3- (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

aga - 11-5-2014 at 15:24

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.

blogfast25 - 12-5-2014 at 04:59

kmno4 is correct though, aga. See e.g. the model for MnO4- 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]

woelen - 12-5-2014 at 07:03

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.

DraconicAcid - 12-5-2014 at 08:31

Quote: Originally posted by woelen

- Metals have positive oxidation states.

Usually. There are carbonyl complexes such as K₂Fe(CO)₄ in which the metal has a negative formal oxidation state. John Ellis, in Minnesota, did quite a bit of work on trying to make compounds with the most negative possible formal oxidation states on the metals (I almost did a post-doc with him, but took a different offer first).

http://pubs.acs.org/doi/abs/10.1021/ja00448a021?journalCode=... has Mn(CO)₄^-3, which has a metal in its -3 oxidation state (formally, at least, although much of the negative charge will actually be on the oxygens of the carbonyl ligands).

And then there are aurides, in which gold is int he -1 oxidation state:
http://www.fkf.mpg.de/134290/Aurides

[Edited on 12-5-2014 by DraconicAcid]