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### Topic: a paradox on finding an Oxidation number  (Read 4136 times)

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#### amirziv24

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##### a paradox on finding an Oxidation number
« on: April 16, 2024, 02:15:29 PM »
hi everyone.
I apologize in advance because my first language isn't English so you may find it a little not easy to read
So I tried to find the oxidation number of N in the N2O5

we know that for a neutral compound, the sum of the oxidation numbers of all constituent atoms must be equal to zero
the oxidation number of oxygen is usually equal to -2
therefore N+ (5*(-2)) = 0  ==> N = +5

but here's the thing,
there's another way of finding an oxidation number which is based on scoring bonds based on Electronegativity. which goes like this:
in a bond, the more negative element gets -1 and the other one get's +1
then we'll sum them

for example :

CH4:

we have four C-H bonds. C has a more negative number therefore we'll have C = -1 and H = +1
so C is 4 *(-1) = -4
in OF2 there are two O-F bonds
F is more negative so: O = +1 and F = -1
O = 2 * (-1) = -2

now I've tried the previous solution for N in N2O5 but didn't work
can someone please explain to me why?

#### Hunter2

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##### Re: a paradox on finding an Oxidation number
« Reply #1 on: April 16, 2024, 02:30:35 PM »
Why you cannot find for nitrogen 5 *(-2) / 2 = = -5 so nitrogen has +5.

#### Babcock_Hall

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##### Re: a paradox on finding an Oxidation number
« Reply #2 on: April 16, 2024, 02:30:43 PM »
I do not have time for a complete answer, but in compounds such as peroxides, the oxidation number of oxygen is -1, not -2.  I do not think that this exception applies in your case.

#### Hunter2

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##### Re: a paradox on finding an Oxidation number
« Reply #3 on: April 16, 2024, 02:33:39 PM »
Here we have nitrogen pentoxide, the anhydride of nitric acid, no peroxide group.

#### Babcock_Hall

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##### Re: a paradox on finding an Oxidation number
« Reply #4 on: April 16, 2024, 02:48:25 PM »
Here we have nitrogen pentoxide, the anhydride of nitric acid, no peroxide group.
I agree.  On the other hand, I think that the strategy that the more electronegative atom essentially owns the electrons works well here.

#### amirziv24

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##### Re: a paradox on finding an Oxidation number
« Reply #5 on: April 16, 2024, 03:19:35 PM »
Why you cannot find for nitrogen 5 *(-2) / 2 = = -5 so nitrogen has +5.
every N in N2O5 has 4 bonds. so 4 *(+1) = +4
however the correct answer is +5
so , I guess no it can't be done in this case but I don't know why

#### Hunter2

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##### Re: a paradox on finding an Oxidation number
« Reply #6 on: April 16, 2024, 03:41:54 PM »

#### Aldebaran

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##### Re: a paradox on finding an Oxidation number
« Reply #7 on: April 16, 2024, 03:53:06 PM »
@OP
I’m struggling a bit to follow your thinking process. In N2O5 oxygen has its usual 2- oxidation number which gives 5x2- = - 10 to be shared by two nitrogens giving +5 as the oxidation number for each nitrogen…..as already explained by previous responses. In oxygen difluoride the oxidation number for fluorine is -1. There’s two of them so the balancing oxygen has oxidation number of +2 not -2 as seems to be indicated in your original post. I feel I’m missing something in what you are searching for. Can you elaborate on your reasoning and where you are seeking clarification perhaps?

#### Borek

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##### Re: a paradox on finding an Oxidation number
« Reply #8 on: April 16, 2024, 04:46:20 PM »
every N in N2O5 has 4 bonds

That's a dangerous approximation, no wonder it produces strange results.

Note that ON are just an accounting device that helps in balancing redox reactions, in general there is no real world physical property that could be measured and called ON.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### amirziv24

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##### Re: a paradox on finding an Oxidation number
« Reply #9 on: April 16, 2024, 05:01:29 PM »
Each N  has 4 bonds  and a positive charge, 2 oxygen have a negative charge, what gives mesomer 5 bonds.

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSkkeqGM18NZS8fsURSyPtdVxqhmUrtJF2iWXcQsw&s=0

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTx_Ean9Hgk0RchTx0r42Gh7a4s4lqRutmU1w9sRA&s=0

interesting.
haven't studied it yet.
thanks
It's going to be scary. I have an exam soon and now I'm paralyzed with not knowing if there's an extra bond for a - or + charge -_-

#### amirziv24

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##### Re: a paradox on finding an Oxidation number
« Reply #10 on: April 16, 2024, 05:05:26 PM »
@OP
I’m struggling a bit to follow your thinking process. In N2O5 oxygen has its usual 2- oxidation number which gives 5x2- = - 10 to be shared by two nitrogens giving +5 as the oxidation number for each nitrogen…..as already explained by previous responses. In oxygen difluoride the oxidation number for fluorine is -1. There’s two of them so the balancing oxygen has oxidation number of +2 not -2 as seems to be indicated in your original post. I feel I’m missing something in what you are searching for. Can you elaborate on your reasoning and where you are seeking clarification perhaps?
well because I didn't know where I went wrong. and I'm studying for a UEE and there are some kind of questions that give you a very very huge lewis structure and ask how many atoms can you find with a specific oxidation number.

I needed a faster way to determine the oxidation number

#### amirziv24

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##### Re: a paradox on finding an Oxidation number
« Reply #11 on: April 16, 2024, 05:08:02 PM »
every N in N2O5 has 4 bonds

That's a dangerous approximation, no wonder it produces strange results.

Note that ON are just an accounting device that helps in balancing redox reactions, in general there is no real world physical property that could be measured and called ON.
thank you verymuch
can you recommend some extra sources please

#### Corribus

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##### Re: a paradox on finding an Oxidation number
« Reply #12 on: April 17, 2024, 12:11:48 PM »
but here's the thing,
there's another way of finding an oxidation number which is based on scoring bonds based on Electronegativity. which goes like this:
in a bond, the more negative element gets -1 and the other one get's +1
then we'll sum them

This is a great question with some good lessons to be learned about oxidation states specifically and the hazards of making assumptions  without thinking about where they come from.

The most important point to understand, which Borek has already pointed out, is that oxidation state is a bookkeeping tool to help account for where electrons spend most of their time in a molecule and help predict where they are likely to go when two different molecules collide. Oxidation state is not something that can be measured in a quantitative sense and it should not be taken to imply that certain atoms possess a localized charge in discrete units.

In fact, there are two main ways to keep track of electrons. The other is formal charge. Both metrics simply count electrons and assign them to nuclear centers. Their names don't really help understand the distinction, so let's review the key difference:

Oxidation state assumes that every bond in a molecule is strictly ionic. Since each bond between two nuclei involves two electrons, oxidation state awards both electrons in a bond to the more electronegative nucleus. If the two nuclei are identical, the two electrons are split evenly.

Formal charge assumes that every bond in a molecule is strictly nonpolar covalent. In this case, each nucleus is awarded a single electron for each bond between two nuclei, regardless of its polarity. The formal charge for the nucleus is then just a matter of determining the total number of electrons around the nucleus (one for each bond plus two for a long pair), minus the core charge.

In essence, oxidation state and formal charge represent two extreme cases. Oxidation state will tend to overstate the charge density on atoms in a molecule, and formal charge will understate it. The truth, insofar as electrons can be thought to be localized anywhere at all, is somewhere in the middle. A truer representation would account for the degree of electron sharing between adjacent nuclei to get a more accurate representation of the average charge at a nuclear center based on the difference between the nuclear core charge and the average amount of electron density in the vicinity of the nucleus. But predicting electron density at a specific location is not trivial, and such precision is not necessary to understand whether two molecules are likely to react. Despite their simplicity in representing where electrons actually spend their time, oxidation state and formal charge are very useful toward understanding why different chemicals interact in the way that they do.

Anyway, I lay out that out because it’s important to understand what the definition of oxidation state is in order to understand proper way to determine it.. and to understand why you get some weird results in cases like N2O5.

Strictly speaking, to determine the oxidation state of atoms in a molecule, you want to look at each bond between neighboring atoms and award both electrons in the bond to the more electronegative atom. Since each atom begins as neutral, each extra electron awarded from an ionic bond drops the charge of that atom by one charge unit, and vice-versa. Effectively for each atom in a molecule this is equivalent to starting with an oxidation state of zero and adding +1 for each bond to a more electronegative atom and -1 for each bond to a less electronegative atom. Rinse and repeat. As you note, this works well for a molecule like CH4: as carbon is the more electronegative atom, and is bonded to four hydrogens, you award carbon (-1) for each hydrogen for a total oxidation state of (-4) and you award each hydrogen a (+1) for a total oxidation state for each hydrogen of (+1). A check yields an overall charge of (0) for the molecule, mission accomplished.

This is the “another way” that you mentioned in your opening post, but it is well-aligned with the formal definition of oxidation state.

A shortcut, as you’ve noted, to start with the fact that a neutral molecule must have an overall charge of zero, so the oxidation states of all atoms have to sum to zero, and then make certain assumptions about the typical values for oxidation states of certain atoms. For instance, in determining the oxidation states of N2O5, you started with the assumption that oxygen has an oxidation state of (-2). This is usually a good assumption that is borne from the definition of oxidation state: oxygen atoms usually bond twice, and they are usually more electronegative than atoms adjacent to them, so they usually get awarded a (-2). This does not always work, though: in a peroxide like H2O2, each oxygen atom is bonded twice, but it only gets a (-1) from one of its neighbors, the less electronegative hydrogens. So, the oxidation state of oxygen is (-1) in hydrogen peroxide. We must be careful with our assumptions.

Let’s now take a specific look at N2O5.

Here each N is bonded to two outside Os, with the inside O bridging the two Ns. The molecule is typically drawn as two NO2 groups bonded to a bridging O, I.e. each N is surrounded by one N=O bond and two N-O bonds. But the two outside NO bonds are in resonance, such that they each have a bond order of 1.5. Anyway, it makes no difference: we can call one of them a N=O bond and one an N-O bond as long as we realize that they are indistinguishable, so we should average whatever result we get.

Using the shortcut approach, each oxygen is assumed to have a (-2) charge. The overall charge must be neutral, so the nitrogen atoms must each be assigned a charge of (+5).

Using the strict formal definition of oxidation state, the two outside oxygen atoms have a total of three bonds between them to neighboring nitrogen atom (2 for one and 1 for the other, or 1.5 each if you apply resonance). Therefore, the two outside oxygens in each NO2 group combined get a -3, resulting an oxidation state of (-1.5) each. The central oxygen gets a (-1) each from its two bonds to its neighboring nitrogen atoms for a total oxidation state of (-2). This results in an “average” oxidation state for all the oxygen atoms of (-8/5). And nitrogen is assigned (+4), one for each of its four bonds (two single bonds and one formal double bond) to oxygen atoms. This sounds like a weird result with a fractional average oxidation state for O but note that it does satisfy our requirement that the overall molecule is still neutral.

So here we have two results:

Quick method: 5 Oxygen (-2), 2 Nitrogen (+5)
Formal method: 4 Oxygen (-1.5), 1 Oxygen (-2), 2 Nitrogen (+4)

What result is correct?

The first answer is the one you will find more frequently on the internet, including Wikipedia – since the shortcut method is what most people use without even thinking about whether the assumptions built into it apply. The second answer is the one that is in stricter adherence to the definition of “oxidation state”. Therefore, I would call the second answer the more correct one. (I couldn’t tell you which answer is more likely to be marked correct on a test, but if I were to bet, I’d bet on the first even though it is less correct.) On the one hand, since oxidation state is mostly a bookkeeping tool, there’s no rigorous way to determine experimentally which result is truly “correct” in any quantitative sense. And yet, we do have tools to measure and predict electron density. A search for electronic structure of N2O5 yields quantum chemical calculations that do indeed indicate that the inner oxygen atom has a considerably higher electron density around it than the oxygen atoms on the periphery (see., e.g., McNamara and Hiller, J. Phys. Chem. A 2000, 104, 22, 5307–5319). Specifically Mulliken charges on the outer O's, inner O, and N's are +0.06, +0.16, and -0.20, respectively for N2O5.* These predictions indicate opposite charge density on the nitrogen and oxygen atoms, as expected, and imply that the central oxygen should be assigned a lower oxidation state than the oxygen atoms on the periphery. This result supports the second method as superior for determining oxidation state because it predicts that the oxidation state for the two types of oxygen atoms should be different. That the more rigorous method for assessing oxidation state is more aligned with quantum chemical calculations should not be surprising since the shortcut method simply assumes that the oxygen atoms all have identical (-2) charge, despite the fact that some of the oxygen atoms are bonded more than twice.

(An additional consideration here is that some of the bonds in N2O5 are actually coordinate/dative bonds, not normal covalent bonds. It is not really clear how such bonds ought to be treated for the purposes of determining oxidation state.)

So, in conclusion, the reason you get different results for the two methods in the case of N2O5 is that the first “shortcut” method uses certain assumptions about the oxidation state of oxygen that are not justified in this molecule, whereas the second method is more aligned with the formal definition of oxidation state. There are three lessons to be learned here:

(1)   For oxidation states in particular: the structure of the molecule matter. The better method is the one that takes bonding characteristics into consideration rather than just assuming all atoms of the same type are identical. The shortcut method is useful, but cannot be generally applied, and you need to know when that is likely to be the case.

(2)  A good lesson in chemistry, perhaps more than other sciences, is to always know what assumptions you are making when you perform any calculation, and make sure you know what the formal definition is of anything you are trying to calculate so that you know whether those assumptions are justified. Chemistry is the science of exceptions. It is why chemistry is difficult. If you don't know what the exceptions are, you are screwed.

(3)   A lot of what you find on the internet is wrong.

*Note: Mulliken charges, often called Mulliken populations, are not oxidation states but reflect the extent to which molecular orbitals in the vicinity of the nucleus are predicted to be populated with electron density - so a nucleus with a more positive value is one around which electrons spend more time, on average, which would translate to a lower oxidation state. In short: more positive Mulliken population = higher electron density = more negative oxidation state.
« Last Edit: April 17, 2024, 12:56:08 PM by Corribus »
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