# Sulfur oxidation number

Oxidation Numbers

It is often useful to follow chemical reactions by looking at changes in the oxidation numbers of the atoms in each compound during the reaction. Oxidation numbers also play an important role in the systematic nomenclature of chemical compounds. By definition, the oxidation number of an atom is the charge that atom would have if the compound was composed of ions.

1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8, and aluminum metal all have an oxidation number of 0.

2. The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of sodium in the Na+ ion is +1, for example, and the oxidation number of chlorine in the Cl- ion is -1.

3. The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH4, NH3, H2O, and HCl.

4. The oxidation number of hydrogen is -1 when it is combined with a metal as in. LiH, NaH, CaH2, and LiAlH4.

5. The metals in Group IA form compounds (such as Li3N and Na2S) in which the metal atom has an oxidation number of +1.

6. The elements in Group IIA form compounds (such as Mg3N2 and CaCO3) in which the metal atom has a +2 oxidation number.

7. Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O2, O3, H2O2, and the O22- ion.

8. The elements in Group VIIA often form compounds (such as AlF3, HCl, and ZnBr2) in which the nonmetal has a -1 oxidation number.

9. The sum of the oxidation numbers in a neutral compound is zero.

H2O: 2(+1) + (-2) = 0

10. The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of the sulfur atom in the SO42- ion must be +6, for example, because the sum of the oxidation numbers of the atoms in this ion must equal -2.

SO42-: (+6) + 4(-2) = -2

11. Elements toward the bottom left corner of the periodic table are more likely to have positive oxidation numbers than those toward the upper right corner of the table. Sulfur has a positive oxidation number in SO2, for example, because it is below oxygen in the periodic table.

SO2: (+4) + 2(-2) = 0

Sours: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch2/oxnumb.html

## Oxidation States (Oxidation Numbers)

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Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts:

• oxidation and reduction in terms of electron transfer
• electron-half-equations

To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., $$\ce{V^{2+}}$$ and $$\ce{V^{3+}}$$). The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons:

$\ce{V \rightarrow V^{2+} + 2e^{-}} \label{1}$

The vanadium in the $$\ce{V^{2+}}$$ ion has an oxidation state of +2. Removal of another electron gives the $$\ce{V^{3+}}$$ ion:

$\ce{V^{2+} \rightarrow V^{3+} + e^{-}} \label{2}$

The vanadium in the $$\ce{V^{3+} }$$ ion has an oxidation state of +3. Removal of another electron forms the ion $$\ce{VO2+}$$:

$\ce{V^{3+} + H_2O \rightarrow VO^{2+} + 2H^{+} + e^{-}} \label{3}$

The vanadium in the $$\ce{VO^{2+}}$$ is now in an oxidation state of +4.

Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}).

The positive oxidation state is the total number of electrons removed from the elemental state. It is possible to remove a fifth electron to form another the $$\ce{VO_2^{+}}$$ ion with the vanadium in a +5 oxidation state.

$\ce{VO^{2+} + H_2O \rightarrow VO_2^{+} + 2H^{+} + e^{-}}$

Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. If the process is reversed, or electrons are added, the oxidation state decreases. The ion could be reduced back to elemental vanadium, with an oxidation state of zero.

If electrons are added to an elemental species, its oxidation number becomes negative. This is impossible for vanadium, but is common for nonmetals such as sulfur:

$\ce{S + 2e^- \rightarrow S^{2-}}$

Here the sulfur has an oxidation state of -2.

### Summary

The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state.

• Oxidation involves an increase in oxidation state
• Reduction involves a decrease in oxidation state

Recognizing this simple pattern is the key to understanding the concept of oxidation states. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations.

### Determining oxidation states

Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states. These rules provide a simpler method.

Rules to determine oxidation states

• The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8, and large structures of carbon or silicon each have an oxidation state of zero.
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
• Some elements almost always have the same oxidation states in their compounds:
ElementUsual oxidation stateExceptions
Group 1 metalsAlways +1
Group 2 metalsAlways +2
OxygenUsually -2Peroxides and F2O (see below)
HydrogenUsually +1Metal hydrides (-1) (see below)
FluorineAlways -1
Chlorineusually -1Compounds with O or F (see below)

The reasons for the exceptions

Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.

Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).

Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.

Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.

Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.

Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference. An example of this situation is given below.

Example $$\PageIndex{1}$$: Chromium

What is the oxidation state of chromium in Cr2+?

Solution

For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)

What is the oxidation state of chromium in CrCl3?

This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:

n + 3(-1) = 0

n = +3

The oxidation state of chromium is +3.

Example $$\PageIndex{2}$$: Chromium

What is the oxidation state of chromium in Cr(H2O)63+?

Solution

This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.

The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.

What is the oxidation state of chromium in the dichromate ion, Cr2O72-?

The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.

2n + 7(-2) = -2

n = +6

Example $$\PageIndex{3}$$: Copper

What is the oxidation state of copper in CuSO4?

Solution

Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states.

In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem:

• Recognize CuSO4 as an ionic compound containing a copper ion and a sulfate ion, SO42-. To form an electrically neutral compound, the copper must be present as a Cu2+ ion. The oxidation state is therefore +2.
• Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below).

### In naming compounds

You will have come across names like iron(II) sulfate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.

This can also be extended to negative ions. Iron(II) sulfate is FeSO4. The sulfate ion is SO42-. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion.

The sulfite ion is SO32-. The oxidation state of the sulfur is +4. This ion is more properly named the sulfate(IV) ion. The -ate ending indicates that the sulfur is in a negative ion.

FeSO4 is properly named iron(II) sulfate(VI), and FeSO3 is iron(II) sulfate(IV). Because of the potential for confusion in these names, the older names of sulfate and sulfite are more commonly used in introductory chemistry courses.

### Using oxidation states to identify what has been oxidized and what has been reduced

This is the most common function of oxidation states. Remember:

• Oxidation involves an increase in oxidation state
• Reduction involves a decrease in oxidation state

In each of the following examples, we have to decide whether the reaction is a redox reaction, and if so, which species have been oxidized and which have been reduced.

Example $$\PageIndex{4}$$:

This is the reaction between magnesium and hydrogen chloride:

$\ce{Mg + 2HCl -> MgCl2 +H2} \nonumber$

Solution

Assign each element its oxidation state to determine if any change states over the course of the reaction:

The oxidation state of magnesium has increased from 0 to +2; the element has been oxidized. The oxidation state of hydrogen has decreased—hydrogen has been reduced. The chlorine is in the same oxidation state on both sides of the equation—it has not been oxidized or reduced.

Example $$\PageIndex{5}$$:

The reaction between sodium hydroxide and hydrochloric acid is:

$NaOH + HCl \rightarrow NaCl + H_2O$

The oxidation states are assigned:

None of the elements are oxidized or reduced. This is not a redox reaction.

Example $$\PageIndex{6}$$:

The reaction between chlorine and cold dilute sodium hydroxide solution is given below:

$\ce{2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O} \nonumber$

It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Checking all the oxidation states verifies this:

Chlorine is the only element to have changed oxidation state. However, its transition is more complicated than previously-discussed examples: it is both oxidized and reduced. The NaCl chlorine atom is reduced to a -1 oxidation state; the NaClO chlorine atom is oxidized to a state of +1. This type of reaction, in which a single substance is both oxidized and reduced, is called a disproportionation reaction.

### Using oxidation states to determine reaction stoichiometry

Oxidation states can be useful in working out the stoichiometry for titration reactions when there is insufficient information to work out the complete ionic equation. Each time an oxidation state changes by one unit, one electron has been transferred. If the oxidation state of one substance in a reaction decreases by 2, it has gained 2 electrons.

Another species in the reaction must have lost those electrons. Any oxidation state decrease in one substance must be accompanied by an equal oxidation state increase in another.

Example $$\PageIndex{1}$$:

Ions containing cerium in the +4 oxidation state are oxidizing agents, capable of oxidizing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). Cerium is reduced to the +3 oxidation state (Ce3+) in the process. What are the reacting proportions?

Solution

The oxidation state of the molybdenum increases by 4. Therefore, the oxidation state of the cerium must decrease by 4 to compensate. However, the oxidation state of cerium only decreases from +4 to +3 for a decrease of 1. Therefore, there must be 4 cerium ions involved for each molybdenum ion; this fulfills the stoichiometric requirements of the reaction.

The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.

Here is a more common example involving iron(II) ions and manganate(VII) ions:

A solution of potassium manganate(VII), KMnO4, acidified with dilute sulfuric acid oxidizes iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.

The oxidation state of the manganese in the manganate(VII) ion is +7, as indicated by the name (but it should be fairly straightforward and useful practice to figure it out from the chemical formula)

In the process of transitioning to manganese(II) ions, the oxidation state of manganese decreases by 5. Every reactive iron(II) ion increases its oxidation state by 1. Therefore, there must be five iron(II) ions reacting for every one manganate(VII) ion.

The left-hand side of the equation is therefore written as: MnO4- + 5Fe2+ + ?

The right-hand side is written as: Mn2+ + 5Fe3+ + ?

The remaining atoms and the charges must be balanced using some intuitive guessing. In this case, it is probable that the oxygen will end up in water, which must be balanced with hydrogen. It has been specified that this reaction takes place under acidic conditions, providing plenty of hydrogen ions.

The fully balanced equation is displayed below:

$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \nonumber$

Sours: https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation_States_(Oxidation_Numbers)
Hint- Sulfuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$) also called oil of vitriol or hydrogen sulfate is dense, colourless, oily, corrosive liquid and one of the most commercially important of all chemicals.

It is often useful to follow chemical reactions by looking at changes in the oxidation numbers of the atoms in each compound during the reaction. Oxidation numbers also play an important role in the systematic nomenclature of chemical compounds. By definition, the oxidation number of an atom is the charge that an atom would have if the compound was composed of ions.
As we know that the charge on hydrogen atom is usually +1
Oxidation number of hydrogen (H) = +1
Also, the charge on oxygen atom is usually -2
Oxidation number of oxygen (O) = -2
Since, we know that the charge on any neutral compound is always equal to zero. Here, sulfuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$) is a neutral compound i.e., overall charge on ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is zero
As, the total charge on any compound is the sum of the charges on individual constituent atoms present in the compound.
i.e., 2(Oxidation number of H) + 1(Oxidation number of S) + 4(Oxidation number of O) = 0
$\Rightarrow$ 2(+1) + 1(Oxidation number of S) + 4(-2) = 0
$\Rightarrow$ 2 + Oxidation number of S – 8 = 0
$\Rightarrow$ Oxidation number of S = 8 – 2 = 6
Therefore, the oxidation number of sulfur (S) in ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is +6.

Hence, option D is correct.

Note- The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in ${{\text{O}}_2}$, ${{\text{O}}_3}$, ${{\text{S}}_8}$, etc all have an oxidation number of zero. The oxidation number of simple ions is equal to the charge on the ion. The sum of the oxidation numbers in a neutral compound is always zero.

Using oxidation states to identify what's been oxidised and what's been reduced

This is easily the most common use of oxidation states.

Remember:

Oxidation involves an increase in oxidation state

Reduction involves a decrease in oxidation state

In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced.

Example 1:

This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:

Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:.

The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced.

Example 2:

The reaction between sodium hydroxide and hydrochloric acid is:

Checking all the oxidation states:

Nothing has changed. This isn't a redox reaction.

Example 3:

This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is:

Obviously the chlorine has changed oxidation state because it has ended up in compounds starting from the original element. Checking all the oxidation states shows:

The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? Yes! Both! One atom has been reduced because its oxidation state has fallen. The other has been oxidised.

This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced.

Using oxidation states to identify the oxidising and reducing agent

This is just a minor addition to the last section. If you know what has been oxidised and what has been reduced, then you can easily work out what the oxidising agent and reducing agent are.

Example 1

This is the reaction between chromium(III) ions and zinc metal:

The chromium has gone from the +3 to the +2 oxidation state, and so has been reduced. The zinc has gone from the zero oxidation state in the element to +2. It has been oxidised.

So what is doing the reducing? It is the zinc - the zinc is giving electrons to the chromium (III) ions. So zinc is the reducing agent.

Similarly, you can work out that the oxidising agent has to be the chromium(III) ions, because they are taking electrons from the zinc.

Example 2

This is the equation for the reaction between manganate(VII) ions and iron(II) ions under acidic conditions. This is worked out further down the page.

Looking at it quickly, it is obvious that the iron(II) ions have been oxidised to iron(III) ions. They have each lost an electron, and their oxidation state has increased from +2 to +3.

The hydrogen is still in its +1 oxidation state before and after the reaction, but the manganate(VII) ions have clearly changed. If you work out the oxidation state of the manganese, it has fallen from +7 to +2 - a reduction.

So the iron(II) ions have been oxidised, and the manganate(VII) ions reduced.

What has reduced the manganate(VII) ions - clearly it is the iron(II) ions. Iron is the only other thing that has a changed oxidation state. So the iron(II) ions are the reducing agent.

Similarly, the manganate(VII) ions must be the oxidising agent.

Using oxidation states to work out reacting proportions

This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation.

Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons.

Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else.

This example is based on information in an old AQA A' level question.

Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions?

The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate.

But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion.

The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.

Or to take a more common example involving iron(II) ions and manganate(VII) ions . . .

A solution of potassium manganate(VII), KMnO4, acidified with dilute sulphuric acid oxidises iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.

The oxidation state of the manganese in the manganate(VII) ion is +7. The name tells you that, but work it out again just for the practice!

In going to manganese(II) ions, the oxidation state of manganese has fallen by 5. Every iron(II) ion that reacts, increases its oxidation state by 1. That means that there must be five iron(II) ions reacting for every one manganate(VII) ion.

The left-hand side of the equation will therefore be: MnO4- + 5Fe2+ + ?

The right-hand side will be: Mn2+ + 5Fe3+ + ?

After that you will have to make guesses as to how to balance the remaining atoms and the charges. In this case, for example, it is quite likely that the oxygen will end up in water. That means that you need some hydrogen from somewhere.

That isn't a problem because you have the reaction in acid solution, so the hydrogens could well come from hydrogen ions.

Eventually, you will end up with this:

Personally, I would much rather work out these equations from electron-half-equations!

Sours: https://www.chemguide.co.uk/inorganic/redox/oxidnstates.html

## Oxidation number sulfur

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