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Oxidation State

Oxidation-Reduction (redox) reactions take place in the world at every moment. In fact, they are directly related to the origin of life. For instance, oxidation of nutrients forms energy and enables human beings, animals, and plants to thrive. If elements or compounds were exposed to oxygen, after a series of reactions the oxygen will be converted into carbon dioxide or water (combustion). To fully understand redox and combustion reactions, we must first learn about oxidation states (OS).

Introduction

An oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element. The terms “oxidation state” and “oxidation number” are often used interchangeably. The transfer of electrons is described by the oxidation state of the molecule. One might mistaken formal charge for oxidation state but they are different. Oxidation state is commonly used to determine the changes in redox reactions and is numerically similar to valence electrons, but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.

Oxidation State Rules

To effectively assign oxidation states to a compound, the seven basic rules must be followed in order. Remember to use the rule that comes first if two rules conflict with each other.  These rules hold true for most compounds..

 

  • RULE 1: Any individual atom uncombined with other elements has the oxidation state of 0 (zero).

 Ex.) The OS for Ag is 0. The oxidation state for O (oxygen) or O2 is 0 as long as it is uncombined with any other element.

 

  • RULE 2: The total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species.

a.) In neutral species, the total sum of the oxidation state of all atoms is 0.  

    Ex.) The sum of OS for NaCl is 0 since the OS of Na = +1 and the OS of Cl = -1, therefore NaCl total OS = 0

b.) In ions, the total sum of the oxidation state is the charge of the ion.  

    Ex.) The OS of Ca2+ (Calcium) is = +2. The total sum of the OS of all atoms in CrO42−  (Chromate ion) is -2. The total sum of the oxidation states of all atoms in CH3COO- (Acetate ion) is -1.

 

  • RULE 3: In a compound, the Oxidation state for Group 1(1A) metal is +1 and for Group 2(2A) metal, the oxidation state is +

Ex.) In NaCl, Na has the oxidation state of +1 since it is a Group 1 Alkali metal. Cl would have an oxidation number of -1 to make the sum of the oxidation states 0 (Rule 2). In MgCl2, Mg has the oxidation state of +2, since it is a Group 2 Alkaline Earth metal. Cl would have an oxidation state of -1, and since there are 2 Cl atoms, the overall charge of the species would again be 0 (Rule 2).

 

  • RULE 4: The oxidation state of FLOURINE is -1 in a compound. 

Ex.) OS of F is -1 in HF, SF6

 

  • RULE 5: The oxidation state of HYDROGEN is +1 in a compound.

Ex.) OS of H is +1 in HI, CH4, NH4+

 

  • RULE 6: The oxidation state of OXYGEN is -2 in a compound. 

 Ex.) OS of Oxyen is -2 in OH-, H2O, CO32-

 

  • RULE 7: In two-element compounds with metals, Group 15(3A) elements have the oxidation state of -3, Group 16(6A) elements have the oxidation state of -2, and Group 17(7A) elements have the oxidation state of -1. 

Ex.) In MgBr2, Br has the oxidation state of -1, since it is a Group 17 element. In Li2S, S has the oxidation state of -2, since it is a Group 16.

 

Example 1

Find the oxidation state of Cr in CrO4-2

SOLUTION

  1. From Rule #2, we know that the sum of OS for this compound is -2. 
  2. From Rule #6, we know that Oxygen in a compound is -2. Since we have four oxygen, -2 x 4 = -8.
  3. With those information, set up an short equation

\[Cr + (-8) = -2\]

So the oxidation state of Cr is +6 

Example 2

Find the oxidation state of C in C2H3O2-

SOLUTION

  1. From Rule #2, we know that the sum of OS for this compound is -1
  2. From Rule #5, we know that Hydrogen in a compound is +1. Since we have three Hydrogen, +1 x 3 = +3
  3. From Rule #6, we know that Oxygen in a compound is -2. Since we have two oxygen, -2 x 2 = -4
  4. With those information, set up an equation 

\[2C + (+3) + (-4) = -1\]

Rewrite as

\[2C + 3 - 4 = -1\]

\[2C = 0 \]

\[C = 0/2 = 0 \]

The oxidation state for C is 0.

Example 3

Find the oxidation state of S in S2O32-

SOLUTION

  1. From Rule #2, we know that the sum of OS for this compound is -2
  2. From Rule #6, we know that Oxygen in a compound is -2. Since we have three oxygen, -2 x 3 = -6
  3. With those information, set up an equation  2S + (-6) = -2

Rewrite as

\[2S - 6= -2\]

\[2S = 4\]

\[S = 4/2 = 2\]

The oxidation state for S is 2.

Fractional Values for Oxidation States 

The oxidation states are usually in whole numbers, but in some cases, they are in fractional numbers. Consider Fe3O4

  • Using rule #2, the sum of the oxidation state of all atoms is 0
  • Using rule #6, the O has the oxidation state of -2.  4x(-2) = -8
  • Equation: 3Fe + (-8) =  0 
  • 3Fe = 8
  • Fe = 8/3 

In most cases, atoms of the same element in a given compound have the same oxidation states, but each atom can have a different state than the other. For Fe3O4, two Fe atoms have an oxidation state of +3 and one of +2, which makes the total oxidation state of Fe= 8/3.

Instead of averaging the oxidation states, we must keep them separated. This is called fragmenting, which occurs if there is an ionic compound, and the ions can be separated. The oxidation state of the individual atoms in the different ions can be determined. 

Practice Problems

Determine the oxidation states of the underlined elements in

  1) Cl2;    2) NaH;    3) H2CO;    4) S2O3 2-;    5) KMnO4.; 6)FeCl3;   7)N2;    8)H2SO4;    9)HClO2;    10)CuSO4;

Transition metal complex

 11) [Co(NH3)5CO3]Br

Solutions

  1. Using rule #1, Cl isn't combine with any other element other than itself, therefore the oxidation state for Cl is 0.
  2. Using rule #2, we know that the sum of the oxidation states for Na and H is 0 because NaH is a neutral compound.  Now, we might be tempted to say H has the oxidation state of +1 because of rule #5.  However, recall that one must use the rules in order.  Because Na is in Group 1A, according to rule #3, the oxidation state of Na is +1.  Thus, H must have the oxidation state of -1 so that the sum of the oxidation state for Na and H is 0.
  3. Using rule #2, the sum of the oxidation state is 0. Using rule #5, H has the oxidation state of +1, and using rule #6, O has the oxidation state of -2.  The equation for H2CO:  2(+1) + C + (-2) = 0. -----> 2 + C - 2 = 0  --------> C = 0
  4. Using rule #2, the sum of the oxidation state is -2.  Using rule #6, O has the oxidation state of -2. We want to know the oxidation of S, so the equation forS2O3 2-: 2S + 3(-2) = -2  ------> 2S -6 = -2 ------>   2S = +4  ------> S = 4/2 -----> S = 2
  5. Using rule #2, the sum of the oxidation state is 0, and using rule #3, the oxidation state for K is +1.  Using rule #6, O has the oxidation state of -2. The equation for KMnO4: 1 + Mn + 4(-2) = 0 -----> Mn = -1 + 8 -----> Mn = 7
  6. Using rule #2, the sume of oxidation state is 0. Using rule #7, Cl has a -1 charge. The equation for FeCl3 Fe + 3(-1) = 0 -----> Fe -3 = 0 -----> Fe = 3
  7. Using rule #1, N isn't combine with any other element other than itself, therefore the oxidation state for N is 0.
  8. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1.  Using rule #6, O has the oxidation state of -2. The equation for H2SO4: 2(1) + S + 4(-2) = 0 -----> S = -2 + 8 -----> S= 6
  9. Using rule #2, the sum of the oxidation state is 0, and using rule #5, the oxidation state for H is +1.  Using rule #6, O has the oxidation state of -2. The equation for HClO2: 1 + Cl + 2(-2) = 0 -----> Cl = -1 + 4 -----> Cl = 3
  10. Using rule #2, the sum of the oxidation state is 0.  Using rule #6, O has the oxidation state of -2. Cu only have one charge which is +2. The equation for CuSO4: 2 + S + 4(-2) = 0 -----> S = -2 + 8 -----> S = 6
  11. [Co(NH3)5CO3]Br is a transition metal complex, whatever is inside the bracket [  ] has to balance out with the outside, which is the Br in this case to make the whole complex = 0.
  • Lets start off with the outside (Br) first, the OS of Br is -1 (Rule #7).
  • Everything INSIDE of the bracket must equal to +1. Since Br is -1, the whole bracket is +1 = 0.
  • We know that NH3 have a 0 charge.
  • The charge of CO3 is -2.
  • The equation for the INSIDE bracket is Co + 0(5) + (-2) = +1 
  • Co -2 = +1 ----> oxidation state for Co = +3

References

  1. Petrucci, Ralph. Harwood, William. Herring, Geoffrey. Madura, Jeffery. GENERAL CHEMISTRY Principles and Modern Applications 9th Edition. Macmillan Publishing co, New Jersey. 1989.
  2. Dumas, Philip. Fikar, Ronald. Samples, Jerry. Templin, Jay. The Best Test Preparation for the Advanced Placement Examination: CHEMISTRY. Research and Education Association, New Jersey. 2003. Print.
  3. "oxidation number." Encyclopædia Britannica. 2010. Encyclopædia Britannica Online. 21 Nov. 2010 <http://www.britannica.com/EBchecked/topic/436613/oxidation-number>.
  4. http://www.khanacademy.org/about
  5. Mascetta, Joseph A. Chemistry The Easy Way. 3rd ed. New York: Barron's Educational Series, 1996.

Contributors

  • Han Nguyen (UCD), Luvleen Brar (UCD)