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CHAP. 14]                        OXIDATION AND REDUCTION                              203


               7. The oxidation number of every halogen atom in its compounds is −1 except for a chlorine, bromine, or
                   iodine atom combined with oxygen or a halogen atom higher in the periodic table. For example, the chlorine
                   atoms in each of the following compounds have oxidation numbers of −1:

                                            SnCl 2  SiCl 4   KCl    PCl 3   HCl
                   The chlorine atom in each of the following has an oxidation number different from −1:
                                                 −
                                    Cl 2 O 3  ClO 2   ClF 3 (F has an oxidation number of −1.)
                   With these rules, we can quickly and easily calculate the oxidation numbers of an element most of the time
               from the formulas of its compounds.

               EXAMPLE 14.1. What are the oxidation numbers of fluorine and iodine in IF 7 ?
                Ans.  F has an oxidation state of −1 (rule 7). I has an oxidation number of +7 (rule 1). (I is not −1, because it is combined
                     with a halogen higher in the periodic table.)

               EXAMPLE 14.2. Calculate the oxidation number of S in (a)SO 2 ,(b)SO 3 , and (c)SO 3 .
                                                                      2−
                Ans.  Let x = oxidation number of S in each case:
                     (a)             x + 2(−2) = 0    x = +4   (b)              x + 3(−2) = −2  x = +4


                          Oxidation  Two   Oxidation   Total        Oxidation  Three  Oxidation  Total
                          number of  atoms  number of  charge on    number of  atoms  number of  charge
                            sulfur          oxygen   molecule        sulfur          oxygen   on ion



                      (c)  x + 3(−2) = 0  x = +6

                                                                                 2−
                                                                  2−  and (b)SinS 2 O 3 .
               EXAMPLE 14.3. Calculate the oxidation number of (a)CrinCr 2 O 7
                Ans.  (a)                 2x + 7(−2) = −2     x = +6

                            Two      Oxidation  Seven   Oxidation  Total
                          chromium   number of  oxygen  number of  charge
                           atoms     chromium   atoms    oxygen    on ion


                     (b)  2x + 3(−2) = −2   x = +2

                   Oxidation numbers are most often, but not always, integers. If there are nonintegral oxidation numbers, there
               must be multiple atoms so that the number of electrons is an integer.

               EXAMPLE 14.4. Calculate the oxidation number of N in NaN 3 , sodium azide.
                Ans.  Na has an oxidation number of +1. Therefore, the total charge on the three nitrogen atoms is 1−, and the average
                                                                                  1
                                                         1
                     charge, which is equal to the oxidation number, is −. (Three nitrogen atoms, times − each, have a total of oxidation
                                                         3                        3
                     numbers equal to an integer, −1.)
               14.3. PERIODIC RELATIONSHIPS OF OXIDATION NUMBERS
                   Oxidation numbers are very useful in correlating and systematizing a lot of inorganic chemistry. For example,
               the metals in very high oxidation states behave as nonmetals. They form oxyanions like MnO 4 , but do not form
                                                                                          −
               highly charged monatomic ions, for example. A few simple rules allow the prediction of the formulas of covalent
               compounds using oxidation numbers, just as predictions were made for ionic compounds in Chap. 5 by using
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