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Oxidation Numbers, or Oxidation States, Tutorial

Key Concepts

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How to Apply the Rules to Find the Oxidation Number (State) of an Element

    general example
Step 1: Write the formula for the species (molecule, ion or compound). MX2
Step 2: Assign a symbol to the overall charge on the species, and include its value if known. ON(MX2) = 0
Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present. 1 × M
2 × X
Step 4: Assign a symbol to represent the oxidation state of each element in the species. ON(M)
ON(X)
Step 5: Write in any known values for the oxidation state (number) for the elements above. ON(M) = ?
ON(X) = ?
Step 6: Write an equation using these symbols to show that the sum of the oxidation state (number) for "atoms" of each element present is equal to the overall charge on the species. ON(MX2)=
ON(M)+[2×ON(X)]
Step 7: Substitute in the known values and solve to find the unknown value. 0=ON(M)+[2×ON(X)]

The best way of learning how to do this is to do some examples .....

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Worked Examples: Calculating Oxidation Numbers for Elements in Electrically Neutral Species

Question 1: Find the oxidation number (oxidation state) for fluorine in F2.

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the oxidation number (oxidation state) of fluorine

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    formula of a molecule: F2(g)
  3. What is the relationship between what you know and what you need to find out?

    Step 1: Write the formula for the species (molecule, ion or compound).

    F2

    Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

    ON(F2) = 0
    The F2 molecule is electrically neutral, it has no overall charge.

    Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

    2 atoms of fluorine = 2 × F

    Step 4: Assign a symbol to represent the oxidation state of each element in the species.

    Let ON(F) = oxidation number of each atom of fluorine in the molecule

    Step 5: Write in any known values for the oxidation state (number) for the elements above.

    ON(F) = ?

    Step 6: Write an equation using these symbols to show that the sum of the oxidation state (number) for "atoms" of each element present is equal to the overall charge on the species.

    ON(F2) = 2 × ON(F)
    Note that the F2 molecule is made up of 2 F atoms and each atom must be assigned an oxidation number (state).

    Step 7:Substitute the known values into the equation:

    0 = 2 × ON(F)

  4. Solve the equation to find the oxidation number (oxidation state) of fluorine

    0 = 2 × ON(F)
    Divide both sides of the equation by 2:
    0 ÷ 2 = [2 × ON(F)] ÷ 2
    0 = ON(F)
    Oxidation number for fluorine = 0

  5. Is your answer plausible?

    One of the Rules for Assigning Oxidation Numbers (States), Rule 1, states that the,

    "Oxidation number (oxidation state) of the atoms of an element in the free state, including its standard state state, is 0"

    F2(g) represents the free state (and standard state) for fluorine.
    So the oxidation state (number) of fluorine in F2 must be 0

    Since this agrees with the value we calculated above, we are confident that our answer is plausible.

  6. State your solution to the problem "find the oxidation number (state) for fluorine":

    Oxidation number (oxidation state) of fluorine in F2 is 0

Question 2: Find the oxidation number (oxidation state) of carbon in CO2

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate oxidation state of carbon (C)

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    Formula of a compound: CO2
  3. What is the relationship between what you know and what you need to find out?

    Step 1: Write the formula for the species (molecule, ion or compound).

    CO2

    Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

    ON(CO2) = 0
    The CO2 molecule is electrically neutral, it has no overall charge.

    Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

    1 atom of carbon = 1 × C

    2 atoms of oxygen = 2 × O

    Step 4: Assign a symbol to represent the oxidation state of each element in the species.

    Let the oxidation number of carbon be ON(C)

    Let the oxidation number of oxygen be ON(O)

    Step 5: Write in any known values for the oxidation state (number) for the elements above.

    ON(C) = ?

    ON(O) = -2
    (from the Rules, oxidation state of O is always -2 except in peroxides, superoxides and oxygen fluorides)

    Step 6: Write an equation using these symbols to show that the sum of the oxidation state (number) for "atoms" of each element present is equal to the overall charge on the species.

    ON(CO2) = ON(C) + [2 × ON(O)]
    Note that the CO2 molecule is made up of 1 C atom and 2 O atoms and each atom must be assigned an oxidation number (state).

    Step 7:Substitute the known values into the equation:

    0 = ON(C) + (2 × -2)

  4. Solve the equation to calculate the oxidation number (state) of carbon

    ON(C) = 0 - (2 × -2)
          = 0 - (-4)
          = +4

  5. Is your answer plausible?

    Oxygen is more electronegative than carbon, that is, oxygen has a greater power to attract the shared negatively charged electrons in the covalent bond towards itself than a carbon atom has, so we expect oxygen to have a negative value for its oxidation number while carbon will therefore have a positive value for its oxidation number.
    In this respect our answer is plausible.

    Work backwards by using your calculated value for the oxidation state of carbon, and your value for the oxidation state of oxygen as given by the "rules" to calculate the overall charge on the CO2 molecule.
    Let ON(CO2) = charge on the molecule
    and ON(O) = oxidation number of oxygen = -2
    and ON(C) = oxidation number of carbon = +4
    Then:
    ON(CO2) = ON(C) + [2 × ON(O)]
    ON(CO2) = +4 + (2 × -2) = +4 + (-4) = +4 - 4 = 0
    So CO2 is found to be electrically neutral.
    We are therefore reasonably confident that our answer is correct.

  6. State your solution to the problem "find the oxidation number (state) for carbon":

    Oxidation number (oxidation state) of carbon is +4
    ( or +IV if you are using the roman numeral convention)

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Worked Examples: Calculating Oxidation Numbers (States) for Elements in Charged Species (Ions)

Question 1: Find the oxidation state (oxidation number) of sulfur in S2-.

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate oxidation number (oxidation state) for sulfur

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    formula of the ion: S2-
  3. What is the relationship between what you know and what you need to find out?

    Step 1: Write the formula for the species (molecule, ion or compound).

    S2-

    Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

    Let ON(S2-) = charge on ion

    ON(S2-) = -2

    Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

    1 atom of sulfur = 1 × S

    Step 4: Assign a symbol to represent the oxidation state of each element in the species.

    Let ON(S) = oxidation number of sulfur in the ion

    Step 5: Write in any known values for the oxidation state (number) for the elements above.

    ON(S) = ?

    Step 6: Write an equation using these symbols to show that the sum of the oxidation state (number) for "atoms" of each element present is equal to the overall charge on the species.

    ON(S2-) = ON(S)

    Step 7:Substitute the known values into the equation:

    -2 = ON(S)

  4. Solve the equation to calculate the oxidation number (state)

    ON(S) = -2

  5. Is your answer plausible?

    Rule 2 states that "the oxidation number (or oxidation state) for the ion of an element is equal to the charge on the ion."
    S2- is the sulfide ion, the ion of the element sulfur.
    So, the oxidation number (oxidation state) of S2- will be equal to the charge on the ion.
    Charge on S2- is 2-
    Therefore oxidation state (oxidation number) of sulfur in S2- must be -2

    Since this value agrees with that which we calculated above, we are reasonably confident that our answer is plausible.

  6. State your solution to the problem "find the oxidation number (state) for sulfur":

    Oxidation number (oxidation state) for sulfur in S2- is -2

Question 2: Find the oxidation number (oxidation state) for chromium in Cr2O72-

Solution:

(Based on the StoPGoPS approach to problem solving.)

  1. What is the question asking you to do?

    Calculate the oxidation number (oxidation state) of chromium

  2. What data (information) have you been given in the question?

    Extract the data from the question:

    Formula of the polyatomic ion: Cr2O72-
  3. What is the relationship between what you know and what you need to find out?

    Step 1: Write the formula for the species (molecule, ion or compound).

    Cr2O72-

    Step 2: Assign a symbol to the overall charge on the species, and include its value if known.

    Let ON(Cr2O72-) be the charge on the polyatomic ion
    ON(Cr2O72-) = -2

    Step 3: List the elements making up the species (molecule, ion or compound) and note how many atoms of this element are present.

    2 atoms of chromium = 2 × Cr
    7 atoms of oxygen = 7 × O

    Step 4: Assign a symbol to represent the oxidation state of each element in the species.

    Let ON(Cr) = oxidation number of each chromium atom

    Let ON(O) = oxidation number of each oxygen atom

    Step 5: Write in any known values for the oxidation state (number) for the elements above.

    ON(Cr) = ?

    ON(O) = -2
    (from the Rules, oxidation state of O is always -2 except in peroxides, superoxides and oxygen fluorides)

    Step 6: Write an equation using these symbols to show that the sum of the oxidation state (number) for "atoms" of each element present is equal to the overall charge on the species.

    ON(Cr2O72-) = [2 × ON(Cr)] + [7 × ON(O)]
    remember, each molecule of Cr2O72- is made up of 2 Cr atoms and 7 O atoms and each atom has an oxidation state (oxidation number)

    Step 7:Substitute the known values into the equation:

    -2 = [2 × ON(Cr)] + [7 × -2]

  4. Solve the equation to calculate the oxidation number (state)

    -2 = [2 × ON(Cr)] + [7 × -2]
    -2 = 2ON(Cr) -14
    Collect like terms:
    -2 + 14 = 2ON(Cr)
    +12 = 2ON(Cr)
    Divide both sides of the equation by 2:
    +12 ÷ 2 = 2ON(Cr) ÷ 2
    +6 = ON(Cr)

  5. Is your answer plausible?

    Chromium is a transition metal so we expect it to be less electronegative than oxygen, therefore we expect chromium to have a positive oxidation number and for oxygen to have a negative oxidation number.
    On this basis, our answer is plausible.

    Work backwards by using our calculated value for the oxidation state of chromium and the known oxidation state for oxygen to see if we arrive at a charge of 2- for the polyatomic ion given in the question:
    ON(Cr) = +6
    ON(O) = -2
    ON(polyatomic ion) = (2 × +6) + (7 × -2)
        = +12 + -14 = -2

    Since this is the charge on the polyatomic ion given in the question, we are confident our answer is plausible.

  6. State your solution to the problem "find the oxidation number (state) for chromium":

    Oxidation state (oxidation number) of chromium in Cr2O72- is +6

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1. Note that oxidation state and "charge" are NOT the same thing, they are infact two different concepts.
The charge on an ion is real and measurable, caused by a transfer of electons that results in a deficiency or excess of electrons:
For example, a highly reactive sodium atom, Na, loses an electron to form a very stable sodium ion, Na+.
The sodium "atom" is now charged because there is 1 more positively charged proton in the nucleus than there are negatively charged electrons orbiting around the nucleus.
The charge on the sodium ion is 1+.

An oxidation state is more of an "account keeping", helping us to understand how charge is distributed within a molecule.
CO is a covalent molecule, it is made up of a carbon atom and an oxygen atom sharing electrons to make a covalent bond, but because of the difference in the electronegativities of carbon and oxygen, this sharing of electrons is unequal.
Oxygen, being more electronegative than carbon, has the greater attraction for the shared electrons than does carbon.
We can use oxidation states to give us a feel for how unequal this share of electrons is.
In the CO compound, carbon has an oxidation number of +2, and oxygen has an oxidation state of -2.
Because these oxidation states are NOT charges on ions, most people would place the sign (+ or -) to the left of the number.
So for the oxygen ion, the charge is 2- (sign to the right of the number)
BUT the oxidation state of oxygen in a compound is written as -2 (sign to the left of the number)
However, this "rule" is not always obeyed, and you should follow the example of your teacher.

There are also people who advocate the use of roman numerals for the use of oxidation states (oxidation numbers).
In this case, hydrogen with an oxidation number (state) +1 would be represented as +I
and oxygen with an oxidation number (state) of -2 would be represented as -II
Once again, use the convention your teacher uses.

2. The standard state of an element is the form it takes at a temperature of 25°C and 100 kPa pressure.
(I) Metals are solids (M(s)) EXCEPT mercury which is a liquid. (Hg(l))
(II) Non-metals will be solids with a range of molecular formulae, EXCEPT for:
    (i) bromine is liquid (Br2(l))
    (ii) 11 gaseous elements:

(a) ALL 6 of the Noble Gases (Group 18) are monatomic gases (He(g), Ne(g), Ar(g), Kr(g), Xe(g), Ra(g))
(b) 5 diatomic gases, F2(g), Cl2(g), O2(g), N2g, H2(g)

3. Chemists use the word "neutral" in different contexts to mean different things.
The compound HCl is electrically neutral because it has no overall charge.
But HCl is a proton donor, so it is acidic, not neutral, when we talk about its acidity.