Question 3.7: Assigning Oxidation States. What is the oxidation state of t......
Assigning Oxidation States
What is the oxidation state of the underlined element in (a) \underline{P}_{4}; (b) \underline{Al}_{2}O_{3}; (c) \underline{Mn}O_{4}^{- }; (d) NaH; (e) H_{2}\underline{O}_{2}; (f) \underline{Fe}_{3}O_{4}?
Analyze
Apply the rules in Table 3.2.
Solve
(a) P_{4}: This formula represents a molecule of elemental phosphorus. For an atom of a free element, the O.S. = 0 (rule 1). The O.S. of P in P_{4} is 0.
(b) Al_{2}O_{3}: The total of the oxidation states of all the atoms in this formula unit is 0 (rule 2). The O.S. of oxygen is −2 (rule 6). The total for three O atoms is −6. The total for two Al atoms is +6. The O.S. of Al is +3.
(c) MnO_{4}^{−}: This is the formula for permanganate ion. The sum of the oxidation states of all the atoms in the ion is −1 (rule 2). The total for the four O atoms is −8. The O.S. of Mn is +7.
(d) NaH: This is a formula unit of the ionic compound sodium hydride. Rule 3 states that the O.S. of Na is +1. Rule 5 indicates that H should also have an O.S. of +1. If both atoms had an O.S. of +1, the total for the formula unit would be +2. This violates rule 2. Rules 2 and 3 take precedence over rule 5. Na has an O.S. of +1; the total for the formula unit is 0; and the O.S. of H must be −1.
(e) H_{2}O_{2}: This is hydrogen peroxide. Rule 5, stating that H has an O.S. of +1, takes precedence over rule 6 (which says that oxygen has an O.S. of −2). The sum of the oxidation states of the two H atoms is +2 and that of the two O atoms must be −2. The O.S. of O must be −1.
(f) Fe_{3}O_{4}: The total of the oxidation states of four O atoms is −8. For three Fe atoms, the sum of the oxidation states must be +8. The O.S. per Fe atom is \frac{8}{3} or +2\frac{2}{3}.
Assess
With practice, you should be able to do the arithmetic associated with assigning oxidation states in your head, that is, without writing down any arithmetic expressions. Also, when you have determined an oxidation state by using arithmetic (as required by rule 2), check your result by making sure the sum of the oxidation states is equal to the charge on the atom, molecule, or ion. For example, in part (c), we determined that the oxidation state of Mn in MnO_{4}^{−} is +7. We know this result is correct because the sum of the oxidation states is 7 + 4(−2) = −1, which is equal to the charge on the MnO_{4}^{−} ion.
| TABLE 3.2 Rules for Assigning Oxidation States | |||||||
| 1. The oxidation state (O.S.) of an individual atom in a free element (uncombined with other elements) is 0. | |||||||
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[Examples: The O.S. of an isolated Cl atom is 0; the two Cl atoms in the molecule Cl_{2} both have an O.S. of 0.] |
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| 2. The sum of the O.S. of all the atoms in | |||||||
| (a) neutral species, such as isolated atoms, molecules, and formula units, is 0;
[Examples: The sum of the O.S. of all the atoms in CH_{3}OH and of all the ions in MgCl_{2} is 0.] |
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| (b) an ion is equal to the charge on the ion. | |||||||
| [Examples: The O.S. of Fe in Fe^{3+} is +3. The sum of the O.S. of all atoms in MnO_{4}^{-} is -1.] | |||||||
| 3. In their compounds, the group 1 metals have an O.S. of +1 and the group 2 metals have an O.S. of +2. | |||||||
| [Examples: The O.S. of K is +1 in KCl and K_{2}CO_{3}, the O.S. of Mg is +2 in MgBr_{2} and Mg(NO_{3})_{2}.] | |||||||
| 4. In its compounds, the O.S. of fluorine is -1. | |||||||
| [Examples: The O.S. of F is -1 in HF, ClF_{3}, and SF_{6}.] | |||||||
| 5. In its compounds, hydrogen usually has an O.S. of +1. | |||||||
| [Examples: The O.S. of H is +1 in HI, H_{2}S, NH_{3}, and CH_{4}.] | |||||||
| 6. In its compounds, oxygen usually has an O.S. of -2. | |||||||
| [Examples: The O.S. of O is -2 in H_{2}O, CO_{2} and KMnO_{4}.] | |||||||
| 7. In binary (two-element) compounds with metals, group 17 elements have an O.S. of -1; group 16 elements, -2, and group 15 elements, -3. | |||||||
| [Examples: The O.S. of Br is -1 in MgBr_{2}; the O.S. of S is -2 in Li_{2}S; and the O.S. of N is -3 in Li_{3}N.] |