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Topic: Oxidation states and ionic charges/formal charge  (Read 7209 times)

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Offline Needaask

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Oxidation states and ionic charges/formal charge
« on: March 12, 2013, 10:14:40 AM »
In COO-, the oxidation states of carbon and oxygen is +4 and -2 respectively even though the ionic charge/formal charge of  one of the oxygens is -1. So what would be a guideline to find the oxidation state? Since in Cl- the formal charge/ionic charge is -1 and so is the oxidation state. So I'm puzzled over the 3 factors.

Thanks for the help

Offline Corribus

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Re: Oxidation states and ionic charges/formal charge
« Reply #1 on: March 12, 2013, 10:24:35 AM »
Oxidation states assume that all bonds are completely ionic; when a bond is formed between two atoms, all electrons in the bond are assumed to located completely on the atom with higher electronegativity (no matter the scale of the difference).

Formal charges assume that all bonds are completely covalent; when a bond is formed between two atoms, all electrons in the bond are assumed to be shared equally between the two atoms - that is, one electron for each.

Neither extreme offers a very good picture of what is actually going on - they are really just formalisms to keep track of electrons.  Typically oxidation states work better in compounds where the electronegativity differences between atoms are large and the ionic limit is more satisfactory; formal charges work better in compounds where the electronegativity differences between atoms are small and the covalen limit is more satisfactory.

Consider CO2.  Oxygen is more electronegative than carbon, but not by a whole lot.  It doesn't really make very much sense to describe the oxidation state of the oxygens as -2 and that of carbon to be +4, because electrons just aren't distributed that way.  The formal charge on each of these would be zero, which isn't completely true either, but in my view it's a better approximation of what's going on.
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Offline Needaask

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Re: Oxidation states and ionic charges/formal charge
« Reply #2 on: March 12, 2013, 10:56:51 AM »
Hi thanks for the reply but what ionic charges? Because in that COO- case one of the oxygen has an ionic charge of -1 while its oxidation state is considered to be -2 so what are the general rules of oxidation states? because for some anions like chlorine the ionic charge is equal to their oxidation state.

Thanks again

Offline Corribus

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Re: Oxidation states and ionic charges/formal charge
« Reply #3 on: March 12, 2013, 11:13:01 AM »
What is COO-?  Are you referring to the conjugate base of a carboyxlic acid, RCOO-? 
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Needaask

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Re: Oxidation states and ionic charges/formal charge
« Reply #4 on: March 13, 2013, 12:18:20 AM »
Yup sorry I made a mistake here. In RCOO- if we don't put resonance into the picture, the ionic/formal charge of one oxygen is -1 while the oxidation state is -2 so I'm a little puzzled on how to find oxidation states.

Offline Corribus

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Re: Oxidation states and ionic charges/formal charge
« Reply #5 on: March 13, 2013, 09:46:44 AM »
Finding oxidation state and formal charge is pretty easy. First, draw a lewis structure.  Then:

Formal charge (F):  F = V - N - B where N = # of valence electrons in lone pairs (nonbonding electrons), V = # of valence electrons and B = # of bonds to other atoms.  (sometimes this equation is written as B/2, with B being the number of electrons shared in bonds, and 2 meaning that they are divided between the two atoms; these two ways of expressing F are functionally equilvalent).

Oxidation state (O): O = V - N - 2B - X, where V and N are as before and now B = # of bonds to other atoms that have lower electronegativity and X = # of bonds to other atoms that have the same electronegativity.

If you think about these equations, you will see that formal charge treats all bonds as being totally covalent (because all bonding electrons are shared equally between atoms) and oxidation state treats all bonds as being totally ionic (because all bonding electrons are awarded completely to the atom with higher electronegativity), except those between atoms of the same element, which are treated as covalent.

So let's look at formate, C(=O)[O-], the simplest carboxylate.

If you draw a Lewis structure, you'll see that the H and C have no non-bonding electrons, the carboxyl oxygen has two bonds (double bond) and two lone pairs and the single-bonded oxygen has three lone pairs and one bond.

This means that the formal charges in formic acid for H, C, =O and -O are 0, 0, 0 and -1, respectively, and the oxidation states are +1, +2, -2, -2.  Note in both cases the sum of all formal charges and oxidations states are equivalent.  In a neutral molecule this sum is zero; in a charged molecule, like formate, the sum equals the overal molecular charge. 

So basically, to find oxidation state: you need to draw a lewis structure, with all bonds and lone pairs.  Then consider each bond as independent of all the rest and assign which atom has the high electronegativity.  Then the oxidation state is determined by taking the valence electrons and subtracting all the electrons that belong to that atom: each nonbonding electron and two electrons for every bond connecting to an atom of lesser negativity.  The carbon in formate, for example, is +2 because the valence electrons equal 4, it has no nonbonding electrons and of the four bonds it has formed, only one (two the hydrogen) is to an atom with lower electronegativity (and zero to other carbons), so only the electrons in the C-H bond belong to carbon.  4 - 0 - 2 - 0 = +2.  Compare this to the formal charge on carbon, which is determined assuming that one electron from each bond belong to carbon, because all bonds are considered covalent.  Here, carbon has 4 valence electrons, no nonbonding electrons and 1 electron from each of its four bonds = 4 - 0 - 4 = 0.

This approach works generally.  Try acetic acid now, CC(=O)[O-]

For oxidation states, you'll find that all the 3H's are +1, as before, the methyl carbon is -3, the carbonyl carbon is +3, and the two oxygens are again -2.  Overall charge is -1, in agreement with what we expect, which is a nice internal check.  Why are the carbons -3 and +3?  Well, the methyl carbon is -3 because it has 4 valence electrons, three bonds to hydrogens (in which it is awarded 2 electrons per bond) and one bond to another carbon (in which it is awarded 1 electron, because this bond is treated as covalent).  So 4 - 6 - 1 = -3.  The carbonyl bond, on the other hand, is +3 because it has 4 valence elctrons, three bonds to oxygens (in which it is awarded zero electrons, because oxygen is MORE electronegative than carbon), and one bond to another carbon (as before, awarded 1 electron), so 4 - 0 - 0 - 1 = +3.  From formic acid to acetic acid, the carbonyl carbon has changed oxidation state because in formic acid, it was bonded to a hydrogen (awarded both electrons of the bond), but in acetic acid, it is bonded to a carbon (shares the two electrons equally).

I think from this you'll see that neither oxidation states or formal charges are particularly good representation of actual electron location.  While it's true that the two carbons in acetic acid probably have different charge densities, because one is closer to two very electron-withdrawing oxygens (which means the formal charge representation is no good), there's no reason to expect there to be a charge differential of SIX ELECTRONS between them!  That's absolutely crazy.  I think the formal charge representation is much better in this case, but even it isn't very good.  And that's neglecting the fact that neither model includes resonance forms.

Anyway, hope that helps answer your question.  Any other questions, just ask.
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Offline Needaask

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Re: Oxidation states and ionic charges/formal charge
« Reply #6 on: March 14, 2013, 05:16:41 AM »
Hi thanks so much for the help I didn't know there was a formula for oxidation states :)

Actually for that formula how should we apply it to ions? For example Cl- ion. Applying the formula here O=7-8=-1 and for Na+ O=1-8=-7. So I'm not sure where the N, B and X come in here.

Also, I was thinking about resonance now, so why is it that we have a RCOO- then we would say that the negative formal charge is split between the 2 oxygen atoms giving them a partial negative charge? Why don't we use oxidation states instead here?

Lastly, what is ionic charge? I learned that an ionic charge is gained from the lost or gain of an electron giving it a positive or negative charge. But I can't seem to associate it with the other 2 concepts of oxidation state and formal charges. How are they linked to each other?

And thanks so much for the reply I really appreciate your help :)
« Last Edit: March 14, 2013, 05:30:34 AM by Needaask »

Offline Corribus

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Re: Oxidation states and ionic charges/formal charge
« Reply #7 on: March 14, 2013, 09:56:51 AM »
Hi thanks so much for the help I didn't know there was a formula for oxidation states :)
Neither did I, so I came up with one myself. :D  But I do think it works, from the few examples I tested it on.

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Actually for that formula how should we apply it to ions? For example Cl- ion. Applying the formula here O=7-8=-1 and for Na+ O=1-8=-7. So I'm not sure where the N, B and X come in here.
You mean isolated ions?  It works here, too, although since you already know what these ions are, it's sort of redundant.  But just for kicks:

Chlorine atom has 7 valence electrons, and chloride has 8 nonbonding electrons and no electrons in bonds, so as you wrote, O = 1-8 = -7.

Sodium atom has 1 valence electron and zero nonbonding electrons and no electrons in bonds, so here O = 1 - 0 = +1.  Note when we speak of # of nonbonding electrons we are talking about nonbonding electrons in the valence shell for the atom.  Sodium's valence shell is the 3s-3p level.  So it starts with 1 electron in the 3s level for sodium atom and loses that electron for the sodium ion, which leaves it with ZERO in the valence 3s-3p level, not 8 as you've written.

Also note that when an atom isn't bonded to anything, the formal charge and the oxidation state are the same, as you'd expect because these two concepts differ in how they treat bonds.  If there are no bonds, then they should be the same, which is what is observed.

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Also, I was thinking about resonance now, so why is it that we have a RCOO- then we would say that the negative formal charge is split between the 2 oxygen atoms giving them a partial negative charge? Why don't we use oxidation states instead here?
We say that because the double bond rapidly cycles (in a manner of speaking) between the two oxygens, as does the extra electron.  The negative charge is delocalized over the the two oxygen atoms - it is located on neither one exclusively.  This gives the carboxyate functional group a lot of stability, and it is the primary reason that carboxylic acids are acidic at all - compare their pKas to a hydroxide functional group, where the negative charge on the oxygen (after the proton leaves) cannot be delocalized.  Big difference.

Unfortunately neither the formal charge or oxidation state concept is really equipped to do deal with resonance structures and partial charges.  The best we can say is that if the formal charge on one oxygen is 0 and on the other it is -1, but the two oxygens are in a state of equilibrium, then the average formal charge on each is -0.5, just as the average bond order for each C-O bond is 1.5.

Quote
Lastly, what is ionic charge? I learned that an ionic charge is gained from the lost or gain of an electron giving it a positive or negative charge. But I can't seem to associate it with the other 2 concepts of oxidation state and formal charges. How are they linked to each other?
I don't know that there is any rigorous 'ionic charge' formalism. 

When one atom bumps into another and an exchange electrons takes place - that is, electrons that "belong" to one atom now "belong" to another, we would say that both atoms become ions and that both have a nonzero overall charge because the total number of protons (positive charge) in the ion exceed or are exceeded by the total number of electrons (negative charge).  The two oppositely charged ions may still stick together, because their now mutual electrostatic attraction results in a lower overall energy together than if they were separated (unless there is an even better thermodynamic gain by them being separated, as in the case of dissolution in, say, water).  We would call this an ionic "bond" and we would best determine the "charge" on each ion by applying an oxidation state formalism.  This formalism works best when the difference between the electronegativity of the two starting atoms is large, and the description of the more electronegative atom completely stealing an electron from the other is most accurate.

The other situation is a covalent bond.  Here, electrons are shared between two atoms rather than residing wholly on one atom or the other.  Both types of bonding interactions occur because most atoms have unpaired electrons, which have a high interaction energy with the surrounding environment.  Bonding - both ionic and covalent - allows electrons to pair with each other.  The difference between the two is the degree to which electrons are shared, which in turn is essentially related to how strong is the pull of the positive charge deriving from the atom's nucleus.  When two atoms with unpaired electrons meet each other, and they both have a similar affinity for electrons, covalent bonds are formed. 

Look at it this way - in all bonds between two atoms (call them A and B), electrons in the bond spend some amount of time in the vicinity of atom A and some in the vicinity of atom B.  In the ionic limit, ALL of the electrons in the bond spend ALL of the time in the vicinity of the more electronegative atom/ion, leaving atom A electron deficient (compared to the isolated atom A) and atom B electron rich (compared to the isolated atom B) - or vice-versa.  In the covalent limit, all of the electrons spend exactly 50% of their time in the vicinity of atom A and 50% of their time in the vicinity of atom B.  This can only happen if the atoms are exactly the same.  But in most cases there is some difference in electronegativity, so most bonds lie somewhere between the two extremes.  Still, even in the covalent limit, there can be a 'formal charge' on one of the atoms.  In the oxygen on carboxylate, for example, the nuclear charge is +6, there are six nonbonding electrons that spend all of their time in the vicinity of the oxygen, and there are two electrons that spend 50% of their time near oxygen and 50% of their time near the adjacent carbon, which means - on average - one of these two bonding electrons is always in the vicinity of oxygen.  So - on average - oxygen has seven electrons around it all the time.  Combined with its +6 nuclear charge, this gives it a "formal charge" of -1.  We call it a formal charge because it is an average differential between the atom's core charge and number of electrons surrounding it at all times.  However keep in mind this is just a "formal" charge - in reality, electron densities are constantly fluctuating.  The "formal" charge represents an average picture.  At any instant, both bonding electrons may be near carbon (in which case the charge on oxygen would be zero) and sometimes they would be near oxygen (in which case the charge on oxygen would be -2).  -1 is just an average, and this doesn't even take into consideration the fact that the electrons probably spend a little more time near oxygen than carbon because of their electronegativity difference.  In reality, the "true" average charge of any atom in a molecule is somewhere between the oxidation state limit and the formal charge limit.

So all that said - were I forced to answer what an ionic charge is, I would say this is probably just referring to what the charge is on an isolated ion.  I've never heard this terminology used with respect to molecular bonding, where we usually refer to oxidation states in the limit of ionic bonding and formal charges in the limit of covalent bonding.
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Offline Needaask

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Re: Oxidation states and ionic charges/formal charge
« Reply #8 on: March 14, 2013, 11:17:53 PM »
Woah that's amazing  :D

Actually what if the ions are still in a crystal lattice? Like in NaCl because in this case I know they the oxidation states remain the same as before -1 and +1 but how will this affect the formula?

As for the resonance structure question, why would we say that the -1 formal charge is delocalised between the 2 oxygen? Why not say since both atoms has an oxidation state of -2 the charge is already delocalised.

So I guess ionic charge is an actual increase of charge from a neutral species and oxidation states and formal charges just add up to get the ionic charge?

Offline Corribus

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Re: Oxidation states and ionic charges/formal charge
« Reply #9 on: March 15, 2013, 11:02:53 AM »
I think you're still confusing charge and oxidation state.

Oxidation state only represents the real charge on an atom in the ionic limit.  However C and O have very similar electronegativities.  While the oxidation state is technically -2 on both oxygens in a carboxylate, as I've said it's not a really accurate representation of what's going on.  The formal charge is a better way to express it, because C and O have similar electronegativities and therefore the bond between them is more covalent. 

So what you really need to do first is make a decision of what is the better way to represent how charge is distributed in a given molecule.  Oxidation state is more useful in salts and metal complexes where electrons are less likely to be shared to a significant extent between bonded atoms.  Formal charge is a better representation in just about everything else, particularly organic molecules where C's, O's and N's are the primary bonded atoms. 

Now, in carboxylate as we've discussed, one of the O's has a formal charge of zero (because it's doubled bonded to carbon) and the other O has a formal charge of -1, because it  has a single bond and six nonbonded electrons.  That's according to the Lewis Structure.  However, because of resonance, the double bond is actually equally shared between the two C-O bonds, and that nonbonding pair of electrons is shared between the two oxygens as well.  So what you actually have is two C-O bonds with a bond order of 1.5 and two oxygens with a formal charge of -0.5.

Remember, the "charge" on an atom is really just a time average that sums the average amount of electrons around the atom (average negative charge) and the average amount of protons in the atom (the average positive charge).  The amount of protons doesn't change, so really it's just a function of the average amount of negative charge around the atom plus the nuclear charge.  In carboxylate, the "average" state for each C-O bond is equal, because of resonance.  So the find the "real" formal charge, you just average the formal charge on each oxygen.

At the risk of adding a pinch more confusion, I'll remind you that this is all in the formal charge limit, which assumes the C and O have identical electronegativies.  But we know this really isn't true.  The O is more electronegative than C, so the "real" charge on each oxygen is going to be a little more negative than -0.5, and the formal charge on the central carbon is going to be a little more positive than zero.  However to determine the real charge you're going to have to use some much more sophisticated models to predict the real degree to which electron density is biased toward oxygen (i.e., the real degree of covalency of the bond).  That requires some quantum mechanical treatments, at the very least molecular orbital theory.  It can certainly be done, but obviously there's no simple way estimate the real charge.  This is the reason we still use oxidation states and formal charges, because they at least give us a crude idea of how charge is distributed in a molecule.   
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Offline Needaask

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Re: Oxidation states and ionic charges/formal charge
« Reply #10 on: March 16, 2013, 01:05:19 AM »
Ohh so it's like oxidation states and formal charges are used to represent ionic charge?

About the last part, is it because actually due to some electronegativity difference, actually on one oxygen the charge could be -1.2 and the other -0.2 causing the delocalised charge to be greater? Actually how would this affect the bond then?

And lastly, thanks so much Corribus for taking the time to explain all this with me. You have a great way with words :)

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