When I took high school chemistry many years ago, considerable effort was spent on teaching us to compute oxidation states of atoms in various compounds, following a set of rules that looked somewhat arbitrary to me at the time. As far as I remember, we were never told what benefits (other than passing tests) knowledge of oxidation states would give us.

They were used as part of a convoluted procedure for "balancing redox reactions", but I never saw an example of that where it wouldn't give the same results simply to require that the number of nuclei of each element, as well as the total charge, must be the same on both sides of the reaction, and solve the resulting linear Diophantine equations.

Nevertheless, the concept must be useful other than for setting homework exercises -- I see encyclopedias and other sources use much space on classifying compounds based on the oxidation state of this atom or that. Still I don't recall seeing any case where the oxidation states are used for anything (other than computing other oxidation states).

What is this concept actually used for? Just a basic example or two where knowing the oxidation states helps produce a meaningful prediction.

It doesn't seem that "oxidation state" actually encodes any particular configuration change inside the atom that tends to be preserved across reactions. Or does it somehow require effort (energy?) to change an atom from one oxidation state to another? In which case, what actually changes?

Sorry if this is too basic a question. It may be something "everybody knows". I've tried looking e.g. at the Wikipedia article, but that too seems to be entirely focused on how to determine oxidation states and doesn't explain why one would desire to know them. And all the questions here seems similarly to be about determining the oxidation states, and it is left unsaid why the asker wants to know ...

Edit: I'm looking for applications of oxidation numbers, by which I mean rules, laws, tendencies, rules-of-thumb or the like of the general form:

When you have (condition that involves oxidation numbers) then (prediction that can be verified without knowing what oxidation numbers are).

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    $\begingroup$ I'm an organic chemist, and we rarely have to balance redox equations, thank God! We rarely have to balance ANY kind of equations, really, other than in a qualitative sense, because most organic reactions are quite messy and produce a host of byproducts and impurities which are only quantified when it's vital to know everything about the reaction and its mechanism and conditions. We do perform numerous oxidations and reductions, but we'll usually just use a slight stoichiometric excess of the redox agent and let the devil see to the details. $\endgroup$ Oct 16, 2016 at 16:13
  • $\begingroup$ I will offer you this bit of mnemonic, though, of the sort we pot-boilers are so fond of, and which my students all seem to find useful: "On the chemical jungle's floor, LEO the lion GERs his awful roar." It's a way of remembering that the loss of electrons is an oxidation (LEO), and the gain of electrons is a reduction (GER). Yes. Sometimes it just comes down to that. $\endgroup$ Oct 16, 2016 at 16:13
  • $\begingroup$ When I took Chemistry at UW-Milwaukee, they skipped the redox part entirely. Years later, when I was a teaching assistant at Princeton U, I was expected to teach redox numbers and states, and I was totally flummoxed. I had to learn it from scratch basically. What I am saying is that the emphasis on redox depends strongly on who is teaching the class and at what university. My opinion now is that redox is primarily useful in analytical and electrochemistry. In a general chemistry class, it should certainly be covered, but no more than one or two lectures. If you are a student and have a hard ti $\endgroup$ Aug 12, 2022 at 5:06

9 Answers 9


Of course they are useful. Perhaps you have not come across them yet, but being able to determine the oxidation state of an atom allows us to understand the properties of chemicals and how redox reactions work.

Let's just give two very simple examples. There are tons more. (I don't even want to go into organometallic chemistry, where being able to determine the oxidation state is incredibly important.)

Potassium permanganate, $\ce{KMnO4}$, has manganese in a +7 oxidation state. This means that it would have an electronic configuration identical to the noble gas argon: $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$, and therefore its $\mathrm{3d}$ orbitals are empty.

Now, most transition metal compounds are said to be coloured because of so-called "d-d* transitions". However, this can only occur if the d orbitals are partially filled - if there are no electrons in the d orbitals, like in $\ce{KMnO4}$, then there are no d-d* transitions available.

As such, the intense purple colour of $\ce{KMnO4}$ has to be explained via a different mechanism. In this case, it is explained by ligand-metal charge transfer.

Moving to organic chemistry. Let's say you have an acyl chloride, $\ce{RCOCl}$, and you want to convert it to an aldehyde, $\ce{RCHO}$. If you calculate the respective oxidation states of the carbonyl carbons, you get +3 and +1 respectively.

So, that tells you what kind of reagent you need to effect this transformation: you need a reducing agent. Hydrogen gas, $\ce{H2}$, is one such example of a reducing agent. Why? Well, that's because of oxidation states again. $\ce{H2}$ has hydrogen in an oxidation state of 0, and when hydrogen forms a bond to carbon, it has an oxidation state of +1.

Oxidation states are key to understanding why redox reactions work and why certain species are reducing agents (e.g. $\ce{LiAlH4}$: H(-1)) and why others are oxidising agents (e.g. DMSO, $\ce{(CH3)2SO}$, in the Swern oxidation).

  • $\begingroup$ Your first example seems to be one where you say that you would have been able to use oxidation states for something, except that it doesn't work in that case? $\endgroup$ Oct 16, 2016 at 11:01
  • $\begingroup$ For the second case: Of course if you want to take something that doesn't contain any H and make something that does, you will need to provide a H from somewhere. Of course, since the numbers add up to zero, if you have a reaction where one of the numbers increase, there must be other numbers that decrease. But so what? This would be equally true no matter which arbitrary rules you use to assign the numbers. You could assign them pseudo-randomly for each compound and this would still be true. $\endgroup$ Oct 16, 2016 at 11:04
  • $\begingroup$ First example: I am using oxidation states to show that something does not occur. Second example: All I can say is - if you say so. Also, please stop using all caps in your comments, it's not nice. $\endgroup$ Oct 16, 2016 at 11:26
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    $\begingroup$ No, I don't "say so" -- I'm asking what distinguishes the particular numbers you get from the oxidation-number rules in utility from other numbers that you could get from different rules of the same general kind. In order to be able to conclude something from oxidation numbers, there must be a rule in play somewhere that says "if the oxidation numbers are such-and-such, then you can expect this-and-that" -- where "this-and-that" is not a statement about the oxidation numbers themselves. I haven't seen such a rule formulated and would like to see some. $\endgroup$ Oct 16, 2016 at 11:38
  • $\begingroup$ For example, if I want to turn RCOCl into RCHO, I can either get the hydrogen by supplying H2, or by having CH4 turn into CH3Cl in the same reaction. I guess wildly that the latter reaction doesn't happen, but is there a general rule using oxidation states that will predict for me that one of these reactions will happen and the other doesn't? What does that rule look like? $\endgroup$ Oct 16, 2016 at 11:43

I would say the main relevance of oxidation states is qualitative, not quantitative. Sure enough, it is possible to give a detailed quantitative picture of redox reactions in terms of quantum chemistry without ever mentioning oxidation states, and oxidation states ultimately are an extremely simplified reflection of that picture. However, most chemists most of the time aren't at all concerned with the detailed quantum chemistry picture, because they are working at a significantly higher level of abstraction, and so the question about how an oxidation state translates to a binding energy (cf. the comments to Ivan Neretin's answer) just doesn't come up. (I see you have a Stack Overflow profile. Exaggerating the difference somewhat, I might say that such a question is akin to asking of someone writing a Java program what the program is doing with the x86 registries.)

Two accounts, already mentioned in various ways in other answers, of how oxidation states are qualitatively helpful:

  • Half-reactions (cf. Marcel's answer). They provide a clear picture of the key changes going on in a redox reaction, and they provide a basis for assigning oxidation and reduction potentials. Half-reactions, however, are an abstraction -- an abstraction which wouldn't be nowhere near as meaningful if they weren't interpreted in terms of oxidation states. For instance, in $\ce{3 e− + 2 H2O + MnO4- -> MnO2 + 4OH−}$ the locus of interest for someone thinking in terms of redox reactions is what is going on with the manganese atoms, and the fact that it is their oxidation state that changes, and not that of the hydrogen or oxygen atoms, reflects that. By the way, even though you don't have to use half-reactions to balance redox equations, it always was my preferred method, as it makes enough sense chemically that it makes it much harder to make silly arithmetical mistakes.

  • Classification of chemical species (cf. Ivan Neretin's answer). Oxidation states can give an approximate, but very useful, idea of some properties of chemical species, such as redox reactivity. For instance, -1 chlorine compounds can be nice and stable (think table salt, $\ce{Cl-}$), while +1 chlorine compounds tend to be quite strong oxidisers (think bleach, $\ce{ClO-}$). No wonder that occasionally species are named after their key oxidation state (cf. Marcel's answer).

If you want a more direct interpretation in terms of a physical property: the oxidation state loosely corresponds to electron density around an atom, relative to what it would be like if that atom was neutrally charged and isolated. You should not expect an actual formula for getting electron densities out of oxidation states (given how loose the correspondence is), nor to compare absolute electron densities around different atoms using oxidation states (given that the values are relative).

To get a better idea of what kind of correspondence I am talking about, picture a neutral chlorine atom. If an electron is added to its electron shell (thus resulting in a chloride anion, $\ce{Cl-}$, which has -1 charge), an increase in electron density happens. Now, suppose that, rather than simply getting an extra electron, the chlorine atom forms a bond with a neutral hydrogen atom, forming an $\ce{HCl}$ molecule. The situation is quite different from the first one, as the pair of electrons forming the covalent bond is shared by both atoms, in a way that can be accurately characterised in quantum chemical terms (molecular orbitals, etc.). Still, the bond is not symmetrical, but skewed towards the chlorine atom, so that the electron density around it is higher than it was in the neutral, isolated state, though lower than in the chloride anion case (and vice-versa for the hydrogen atom). The major simplification involved in assigning oxidation states is ignoring this subtlety: we say chlorine is in -1 oxidation state in both $\ce{Cl-}$ and $\ce{HCl}$. Finally, it is worth noting that adding an electron to the neutral chlorine atom promotes an electronic reconfiguration (cf. the part about permanganate in orthocresol's answer) which happens to lead to a very stable state (a noble gas electronic configuration). That explains why -1 chlorine compounds are relatively stable next to those with higher oxidation states ($\ce{Cl2}$, $\ce{ClO-}$, etc.), which tend to be powerful oxidisers.

The example illustrates how oxidation states are, from the perspective I am adopting here, an approximation (pretending all bonds are either fully symmetrical or fully skewed towards the most electronegative atom) of an approximation (using charges as proxies for electron density). Still, they are useful to the extent they:

  • Offer a simple model of electron transfer in redox reactions (and note that this is quite independent from all I said about electron densities: Faraday's work on electrochemistry, for instance, predates quantum mechanics and the discovery of the electron by several decades).

  • Can capture general trends among chemical species of an element, thus serving as basis for heuristics. Given the many layers of approximation involved, these are heuristics, and not hard rules (which explains why no one has given you hard rules here). To go beyond the heuristics, you have to look at the electronic configuration in each particular case, which, other than in very simple cases (such as $\ce{Cl-}$) is merely hinted at by the oxidation state.

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    $\begingroup$ Even qualitatively, it is frustratingly difficult to get anyone to explicitly state one of the rules you all keep telling me you use. What is the "what is going on with the manganese atoms" that you learn from oxidation states? Not the oxidation states themselves, but what does those numbers tell you about the reaction? Or about the manganese atoms? Anything that doesn't boil down to "knowing the oxidation states will tell you how the oxidation states change". $\endgroup$ Oct 16, 2016 at 20:05
  • $\begingroup$ @HenningMakholm I hadn't emphasised that because, at first, I didn't see it as the focus of your question. I have added a second part to the answer, which hopefully will make things clearer. $\endgroup$
    – duplode
    Oct 16, 2016 at 21:23

Back when we were small and the computers were huge, math books for kids were full of tricks for quick mental calculation, like "to multiply x by 9, just multiply by 10 and subtract x" or "to find a square of a number that ends with 5, for example, 65, go like $6\cdot7=42$ and append 25". Well, the oxidation states are much like that. Sure, you may balance pretty much anything without them just as well, but we are used to do it our way.

Besides, knowing your oxidation states gives you a rough idea of what to expect from the compound. Will it oxidize everything, including my skin if it gets a chance? or will it be indifferent? or will it reduce everything, to the point of igniting spontaneously in air? will it react with such-and-such?

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    $\begingroup$ "knowing your oxidation states gives you a rough idea of what to expect from the compound" -- how? You seem to claim that these values can be used to make predictions of something that is not ITSELF about oxidation states, but the examples you give ("will it oxidize everyting") are themselves in terms of oxidation states and do not seem to lead out of a closed world where the only "facts" you get are things about "oxidation states" themselves. $\endgroup$ Oct 16, 2016 at 10:53
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    $\begingroup$ You either have the reaction, or you don't. Say, $\ce{Mn^7+}$ is a strong oxidant, hence it belongs to the first category. It will oxidize your skin. This is a real, very palpable fact. It is not about oxidation states. It is about those nice brown spots on you palms which would stay for about a week. $\endgroup$ Oct 16, 2016 at 11:13
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    $\begingroup$ What is the rule you use to predict something from knowledge of oxidation states here? You seem to be claiming that you can conclude from oxidation states that a certain reaction will happen -- what is the rule for getting from "the oxidation states are such-and-such" to "such-and-such reaction is likely to happen"? $\endgroup$ Oct 16, 2016 at 11:18
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    $\begingroup$ Well, it's not much of a rule, but... Long story short, I know my elements and what oxidation states they "prefer" to be in; if they are in unusual states, then something is likely to happen. $\endgroup$ Oct 16, 2016 at 12:11
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    $\begingroup$ I just want some conclusion that's not about the oxidation states themselves. $\endgroup$ Oct 16, 2016 at 16:51

Oxidation states are used for "electron bookkeeping". This is useful in redox reactions, where we need to know what species are oxidised and reduced. By extension, oxidation states are an important tool for determining reacting ratios.

For example:

$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) +Cu(s)}$

$\ce{Cu: +2->0}$

$\ce{Zn: 0-> +2}$

Hence we know that zinc is oxidised, and copper is reduced. While this is obvious from the reaction equation, this will not be the case in "messier" redox reactions, and oxidation states help us understand what is going on.

They also allow us to distinguish between compounds by using oxidation numbers - for example it is the oxidation number that allows us to tell the difference between iron (II) oxide and iron (III) oxide.

We should keep in mind, though, that oxidation states are a purely theoretical concept, and do not actually exist as physical quantities. When we use oxidation states, we are essentially assuminig that all bonds have a purely ionic character - this is rarely the case.

  • $\begingroup$ This is another "how do I find them?" explanation that says nothing at all about "what do I use them for once I have found them?" $\endgroup$ Oct 16, 2016 at 9:50
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    $\begingroup$ I gave two examples of uses for oxidation states: differentiating between different compounds of the same two elements, and keeping track of electron transfer in redox reactions. $\endgroup$
    – Marcel
    Oct 16, 2016 at 10:14
  • $\begingroup$ "differentiating between different compounds of the same two elements" apparently mean that you use the oxidation states to know what the oxidation states are! That's not an application as far as I can see. Your "keeping track of electron transfer in redox reactions" d0esn't look like an application either. You write down those two extra lines under the reaction, but WHAT DO YOU GET OUT OF WRITING THOSE LINES? Again, the claimed benefit of knowing the oxidation states simply seems to be that now you can write them down -- what then? $\endgroup$ Oct 16, 2016 at 10:48
  • $\begingroup$ 1) We distinguish between the compounds by using the knowledge that they have different oxidation states. 2) A change in oxidation states is the very definition of a redox reaction - hence we need oxidation states to keep track of what is happening in the reaction. $\endgroup$
    – Marcel
    Oct 16, 2016 at 11:10
  • $\begingroup$ (2), yes, so saying "redox reaction" is not an answer to how one uses oxidation states to predict something that is not itself about oxidation states. Saying that "we need oxidation states to keep track of the oxidation states" is just tautological. I'm asking for an application where oxidation states makes a prediction where determining whether the prediction came true or not does not involve knowing anything about oxidation states. $\endgroup$ Oct 16, 2016 at 11:15
  1. For simple compounds of elements with vastly different electronegativity oxidation states are useful to distinguish compound by oxidative/reductive power. Said powers may, to some extent, be used to predict how likely particular reactions to proceed. For example, halogens in higher oxidation states are all good oxidizers and carbon is a reducer. Thus, oxohalogenates are likely to react with carbon, sometimes violently.

Oxidative power manifests as electrode potential and thus objectively exists. Sometimes, however, electrode potential is really hard to obtain or can be measured only in exotic conditions.

Similarly, each element has 'preferred' oxidation states, corresponding to most stable compounds.

This use, however, is relevant only in inorganic chemistry of simple compounds of elements with highly different electronegativity. When catenation is involved or elements connected have close electronegativity, formal electronegativity loses predictive power. The situation can be somewhat salvaged by ad-hoc 'corrections', like certain atom neighbors stabilizing higher/lower/nonstandard oxidation states, but ...

  1. Oxidation states are useful for balancing redox equations, though there the order of electronegativity used to calculate them does not matter.

Admittedly, it IS possible to balance equation without them, either ad-hock, or using half-reaction method.

Oxidation states are of marginal use in organic and metaloorganic chemistry and in chemistry of polyhedral clusters. On the other hand, definition 'compounds of elements with strongly different electronegativities' covers most common chemicals and everyday substances. Thus, oxidation states ARE useful.

  • $\begingroup$ For (2), can you give an example where the oxidation states gives us something that is not already a consequences of counting nuclei of different elements? $\endgroup$ Oct 16, 2016 at 10:57
  • $\begingroup$ For (1), "distinguish compound by oxidative/reductive power" is itself a statement about oxidation states. I agree that if you know the oxidation states assigned to atoms in different compounds, you can figure out whether that number increases or decreases during a reaction, but so what? Does nature care about that number increasing or decreasing? I'm looking for an application that is not about oxidation states themselves, one that predicts whether this or that will happen, where "this or that" is not just a rephrasing of a claim about oxidation states themselves. $\endgroup$ Oct 16, 2016 at 10:58

Oxidation states are based on oxidation numbers. An oxidation number is the charge, or apparent charge, of an atom or group of atoms when combining in a compound.

The oxidation number of an atom is determined by the octet rule. This states that atoms are particularly stable when they have eight electrons in their outer shell. Outer shell is somewhat of a simplification compared to the quantum mechanical model of the atom but it works very well for particular purposes. What's important is that these outer electrons are the way the atom presents itself to the world and interacts with other atoms. As you progress up the periodic table, each atom has an additional electron to electrically balance the additional proton. These electrons are added in such a way that there are never more than eight in the outer shell. One notable exception to the octet rule is the very first shell which can only hold a maximum of two electrons.

Let's begin with oxygen which has eight electrons. Two of these are in the innermost shell leaving six in the 2nd shell which in oxygen's outer shell. Byt the octet rule, oxygen needs 2 more electrons to give it an octet. Its oxidation number is -2. Why negative? If it gains these electrons to complete its octet, it will then be negatively charged.

Let's now look at sodium. 2 electrons in the 1st shell, 8 in the 2nd and 1 in the 3rd shell. So Na needs to gain 7 electrons to form an octet in the 3rd shell. That's asking for too much, its nucleus isn't up to the electrical task. But what if it loses that lone outer electron? Then the 2nd shell becomes the outer shell and, since it has an octet, the Na will be stable. It will also be charged +1 which is sodium's typical oxidation number.

So where does oxygen get the two electrons it need to form an octet? To where does sodium dump its electron to form an octet? One way is for two sodium atoms to each give up an electron to one oxygen atom. This means sodium and oxygen would be expected to combine in a 2 to 1 ratio. Indeed, Na2O is formed since the now negative oxygen is attracted to the now positive sodiums. In this compound the oxidation state of oxygen is -2, that of the sodium is +1.

It's simplistic to imagine the electrons are completely transfered except in the case where ions are formed. In water, electrons are shared between two hydrogen atoms and one oxygen atom. Each hydrogen needs one more to fill its outer shell (the 1st shell needing only two to fill). The oxygen, as usual, needs 2 nore to make an octet. So, H2O results. But the electrons are not shared equally, the oxygen has a tighter pull on them and gains a partial charge of -2, its oxidation number in water. The hydrogens get to share one of those electrons of oxygen to complete its outer shell but both that electron and the one the hydrogen had of its own are pulled away from the hydrogen toward the more powerful attraction of the oxygen. This results in a partial positive charge on the hydrogen compared to its atomic state. Oxidation number, and oxidation state, -1.

This is just the beginning of a complete discussion of this. The bottom line is that oxidation numbers help us understand, and write chemical formulas for, chemical compounds. Proper formulas help us write and balance chemical equations. Proper chemical equations help us pass Chemistry class and graduate.


It's been years since I got my BS in chemistry (which I regrettably haven't used since), but there's two contexts where oxidation states come up --

(1) as emphasized in many of the answers, this can provide a shorthand for balancing chemical equations in general. In this usage, it's not that great or that necessary at this point in time.

(2) More specifically, it's useful for characterizing the states of metal atoms in primarily ionic bonds. This is in a sense a variation on the more general answer, but here it can genuinely be useful. From what I remember, there are many occasions where you're looking to trade out the metallic part of an ionic bond (of course "ionic" vs. "covalent" is a bit of an abstraction), but metal ions behave differently depending on oxidation state. One specific different is their electromagnetic properties -- the difference between oxidation states reflects a difference in the arrangement of their electrons (to an abstraction). Thus, ionization states provide a way to eyeball that you're going to balance things out by just looking at that number. Want to precipitate out gold? Knowing The oxidation state will tell you the easiest reaction.

Particularly when dealing with salts, we often don't care (too much) about what's on the other side of the Cl- we are going to use for our reaction, but we do care what the oxidation state is to know how much X+Cl- to add.


1.use of oxidation number in nomenclature. You must have heard of Iron(III) sulphate Fe2(SO4)3,the reason of writing it as Iron(III) sulphate is because each iron atom have a oxidation state of +3 here. Further,we indicate Magnetite (Fe3O4) as iron(II,III)oxide this is because two iron atom shows an oxidation state of +3 and one iron atom shows an oxidation state of +2 in Magnetite.

2.Oxidation states of elements are helpful in balancing redox reaction.The elementry reactions such as Mg+O2 ----->MgO can be balanced by simply hit and trial but when it comes to balance a redox reaction,the oxidation states of elements involved should be known.It simply helps in determining which copound will behave like oxidising agent and which will behave like reducing agent.

I hope you got my message.


Question: How many potassium and how many calcium do I need to neutralize sulphuric acid?

  • Answer without oxidatin numbers: How the hell I can know?
  • Answer #2 without oxidation numbers: Just mix it and see what happens.
  • Answer with oxidation numbers:

Sulphuric acid contains sulphur in oxidation state $\ce{S^{VI}}$ and acids, except for halogen ones, are formed with appropriate ammount of oxygen and hydrogen. Rule of thumb taught to kids is "divide the oxidation number of the [suplhur] by two. Round it up and you have number of oxygens. then add one hydrogen. If you don not need to round, add one oxygen and two hydrogens."

Therefore sulphuric acid is $$\ce{H_2SO_4 = H_2^IS^{VI}O_4^{-II}}$$ and in water it dissolves $$\ce{H_2SO_4 + 2H_2O -> (SO_4)^{-II} + 2(H_3O)^+}$$

Now move to our question, how to neutralize this acidic solution? Write reactions: $$\ce{(SO_4)^{-II} + 2(H_3O)^+ + Ca^0 -> Ca^{II}(SO_4)^{-II} + H_2^0 + 2(H_2O)^0}$$ $$\ce{(SO_4)^{-II} + 2(H_3O)^+ + 2K^0 -> K_2^{I}(SO_4)^{-II} + H_2^0 + 2(H_2O)^0}$$

The answer is that for neutralisation of one mole of sulphuric acid one need one mole of calcium or two moles of potassium.

The oxidation numbers and other rules tells you thats $\ce{C_2O_5}$ doesn't exist and why. Same for $\ce{CaSO_4, KS, HO_4}$, etc.

  • $\begingroup$ Without oxidation numbers? If I have H2SO4 and deprotonate it fully, I'll end up with a sulfate ion SO4(2-). In order to make that into a neutral salt together with Ca(2-) ions there needs to be one calcium per sulfate. In order to make it into a neutral salt with potassium I'll need two K- per sulfate. No oxidation states there, just plain old ion charges. $\endgroup$ Oct 18, 2016 at 9:55
  • $\begingroup$ Oxidation states are where we pretend that sulfate is made from one S(6+) ion and four O(2-) ions. This pretense seems to be to be completely useless in your example; the sulfate never breaks apart during the reaction and can conveniently be treated like an unit when doing the balancing. $\endgroup$ Oct 18, 2016 at 10:00
  • $\begingroup$ @HenningMakholm Got you! You have used oxidation numbers. $SO_4(2-)$ addresses that. $\endgroup$
    – Crowley
    Oct 18, 2016 at 10:02
  • $\begingroup$ x @Crowley: No I haven't. -2 is the electric charge of the sulfate ion, and I don't dispute that charges are useful to know. $\endgroup$ Oct 18, 2016 at 10:03
  • $\begingroup$ @HenningMakholm Try to make the molecule neutral using available electronic configurations of the atoms. $\endgroup$
    – Crowley
    Oct 18, 2016 at 10:03

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