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## Q. 4.7

Assigning Oxidation Numbers Use the rules from Table 4.5 to obtain the oxidation number of the chlorine atom in each of the following: (a) $HClO_4$ (perchloric acid), (b) $ClO_3^-$ (chlorate ion).

PROBLEM STRATEGY

In each case, write the expression for the sum of the oxidation numbers, equating this to zero for a compound or to the charge for an ion (Rule 6). Now, use Rules 2 to 5 to substitute oxidation numbers for particular atoms, such as -2 for oxygen and +1 for hydrogen, and solve for the unknown oxidation number (Cl in this example).

Table 4.5

 Rules for Assigning Oxidation Numbers Rule Applies to Statement 1 Elements The oxidation number of an atom in an element is zero. 2 Monatomic ions The oxidation number of an atom in a monatomic ion equals the charge on the ion. 3 Oxygen The oxidation number of oxygen is  -2 in most of its compounds. (An exception is O in $H_2O_2$ and other 4 Hydrogen The oxidation number of hydrogen is + 1 in most of its compounds. (The oxidation number of hydrogen is -1 in binary compounds with a metal, such as $CaH_2$.) 5 Halogens The oxidation number of fluorine is  -1 in all of its compounds. Each of the other halogens (Cl, Br, I) has an oxidation number of  -1 in binary compounds, except when the other element is another halogen above it in the periodic table or the other element is oxygen. 6 Compounds and ions The sum of the oxidation numbers of the atoms in a compound is zero. The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.

## Verified Solution

a. For perchloric acid, Rule 6 gives the equation

(Oxidation number of H) + (oxidation number of Cl) + 4 × (oxidation number of O) = 0

Using Rules 3 and 4, you obtain

(+1) + (oxidation number of Cl) + 4 × (-2) = 0

Therefore,

Oxidation number of Cl (in $HClO_4$) = -(+1) – 4 × (-2) = +7

b. For the chlorate ion, Rule 6 gives the equation

(Oxidation number of Cl) + 3 × (oxidation number of O) = -1

Using Rule 3, you obtain

(Oxidation number of Cl) + 3 × (-2) = -1

Therefore,

Oxidation number of Cl (in $ClO_3^-$) = -1 – 3 × (-2) = +5