SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 4

Chemical Formulas

OXIDATION STATES AND FORMULA WRITING

To keep track of the transfer of electrons in all formulas, chemists have devised a system of electron bookkeeping called oxidation states (or oxidation numbers). In this method, an oxidation state is assigned to each member of a formula or polyatomic ion. It is designated by a small, whole-number superscript preceded by a plus or minus sign. This is not to be confused with the ionic charges we have been using thus far that are used to the right of ionic charge. These charges are directly related to the bonding that occurs in compounds and was discussed in the previous chapter. Oxidation states are also used to track electron transfers. This is discussed in the chapter about oxidation and reduction.

The Rules for Assigning an Oxidation State

Below are the basic rules for assigning an oxidation state to each element. By applying simple rules, oxidation states can be assigned to most elements or compounds. To apply these rules, remember that the sum of the oxidation states must be zero for an electrically neutral compound. For an ion, the sum of the oxidation states must equal the charge on the ion.

  1. The oxidation state of an atom in an element is zero. Examples: 0 for Na(s), O2(g), and H(l)

  2. The oxidation state of a monoatomic ion is the same as its charge. Examples: Na+1, Cl−1

  3. The oxidation state of fluorine is −1 in its compounds. Examples: HF as 1H(+1) + 1F(−1) = 0; PFas 1P(+3) + 3F(−1) = 0

  4. The oxidation state of oxygen is usually −2 in its compounds. Example: H2O where 2H(+1) + 1O(−2) = 0 (Exceptions occur when the oxygen is bonded to fluorine and the oxidation state of fluorine takes precedence and in peroxide compounds where the oxidation state is assigned the value of −1.)

  5. The oxidation state of hydrogen in most compounds is +1. Examples: H2O, HCl, NH(In hydrides, there is an exception. Oxygen is assigned the value of −1.)

Some Examples of Using Oxidation States in Formulas

EXAMPLE 1: In Na2SO4, what is the oxidation state of sulfur?

Since the positive sum plus the negative sum must equal 0,

(+2) + x + (−8) = 0

The sulfur must have a +6 oxidation state.

EXAMPLE 2: What is the oxidation state of chromium in K2Cr2O7?

K = 2 × (+1) = +2
Cr = 2 × (x) = 2x
O = 7 × (−2) = −14
(+2) + (2x) + (−14) = 0
2x = +12
x = +6

The chromium has a +6 oxidation state.

    In a polyatomic ion, the algebraic sum of the positive and negative oxidation states of all the atoms in the formula must be equal to the charge of the ion.

EXAMPLE 3:  What is the oxidation state of sulfur in SO42−?

S = 1 × (x) = x
O = 4 × (−2) = −8

Since they must equal −2 (the charge of the ion),

x + (−8) = −2
x = +6

Sulfur must have a +6 oxidation state in the ion.