Oxidation States:A Useful Tool in Describing Chemical Compounds

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1 Oxidation States:A Useful Tool in Describing Chemical Compounds The Oxidation state (O.S., aka oxidation number) of an element is a convenient bookkeeping technique for determining if an element is electron-rich or electron-poor in a compound o The O.S. does not correspond to a measurable property o O.S. are useful in naming compounds Rules for determining O.S.: 1.The O.S. of a free element = 0..The sum of the O.S. of all the atoms in a species (molecule or ion) = charge on the species. 3. The O.S. of a monatomic ion = charge on the ion. 4. In compounds, F always has an O.S 5.In compounds, oxygen usually has an O.S 6. In compounds, hydrogen usually has an O.S. = If there is no O or H in compound, the less electronegative element is assigned a positive O.S. Elements with lower electronegativities are to the left and below in the P.T. 11

2 O.S. of S in H SO 4 O.S. of S in Na S O 3 O.S. of S in S 4 O 6 O.S. of C in H C O 4 O.S. of Mo in Na MoO 8 1

3 The oxidation state (O.S.) can be described as a sometimes fictional charge. With monatomic ions, O.S. and charge are the same thing. For polyatomic ions and molecules, O.S. is fictional, but equal to what the charge would be if the compounds were entirely ionic. * 3-4 Oxidation States: A Useful Tool in Describing Chemical Compounds Most basic concepts in chemistry deal with measurable properties or phenomena. In a few instances, though, a concept has been devised more for convenience than because of any fundamental significance. This is the case with the oxidation state (oxidation number),* which is related to the number of electrons that an atom loses, gains, or otherwise appears to use in joining with other atoms in compounds. Consider NaCl. In this compound an Na atom, a metal, loses one electron to a Cl atom, a nonmetal. The compound consists of the ions Na + and Cl - (see Figure 3-4). Na + is in a +1 oxidation state and Cl - is in a -1 state. In MgCl an Mg atom loses two electrons to become Mg +,, and each Cl atom gains one electron to become Cl -. As in NaCl, the oxidation state of Cl is -1, but that of Mg is +. If we take the total of the oxidation states of all the atoms (ions) in a formula unit of MgCl, we get = 0. In the molecule Cl, the two Cl atoms are identical and should have the same oxidation state. But if their total is to be zero, each oxidation state must itself be 0. Thus, the oxidation state of an atom can vary, depending on the compounds in which it occurs. In the molecule H O, we arbitrarily assign H the oxidation state of +1. Then, because the total of the oxidation states of the atoms must be zero, the oxidation state of oxygen must be -. *Because oxidation state refers to a number, the term oxidation number is often used synonymously. We will use the two terms interchangeably.

4 3-4 Oxidation States: A Useful Tool in Describing Chemical Compounds 85 From these examples you can see that we need some conventions or rules for assigning oxidation states. The seven rules in Table 3. are sufficient to deal with most cases in this text, with this understanding: The rules must be applied in the numerical order listed, and whenever two rules appear to contradict each other (which they sometimes will), follow the lower numbered rule. Some examples are given for each rule, and the rules are applied in Example 3-7. TABLE 3. 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. [Examples: The O.S. of an isolated Cl atom is 0; the two Cl atoms in the molecule Cl both have an O.S. of 0.]. The total 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 is 0.] (b) an ion is equal to the charge on the ion. [Examples: The O.S. of Fe in is +3. The sum of the O.S. of all atoms in - MnO 4 is -l.] 3. In their compounds, the group 1 metals have an O.S. of +1 and the group metals have an O.S. of +. Fe 3+ [Examples: The O.S. of K is +1 in KCl and K CO 3 ; the O.S. of Mg is + in MgBr and Mg(NO 3 ).] 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 S, NH 3, and CH 4.] 6. In its compounds, oxygen usually has an O.S. of -. [Examples: The O.S. of O is - in H O, CO and KMnO 4.] 7. In binary (two-element) compounds with metals, group 17 elements have an O.S. of -1; group 16 elements, -; and group 15 elements, -3. [Examples: The O.S. of Br is -1 in MgBr ; the O.S. of S is - in Li S; and the O.S. of N is -3 in Li 3 N. ] The principal exceptions to rule 5 occur when H is bonded to metals, as in LiH, NaH, and CaH ; exceptions to rule 6 occur in compounds with O F bonds, such as OF, and in compounds where O atoms are bonded to one another, as in H O, and KO. * EXAMPLE 3-7 Assigning Oxidation States What is the oxidation state of the underlined element in (a) P (b) (c) MnO - 4 ; Al O 3 ; 4 ; (d) NaH ; (e) H O ; (f) Fe 3 O 4? Analyze Apply the rules in Table 3.. 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 O 3 : The total of the oxidation states of all the atoms in this formula unit is 0 (rule ). The O.S. of oxygen is - (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 total of the oxidation states of all the atoms in the ion is -1 (rule ). 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 +. This violates rule. Rules 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. (continued)

5 86 Chapter 3 Chemical Compounds (e) H O : 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 -). The sum of the oxidation states of the two H atoms is + and that of the two O atoms must be -. 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 total must be The O.S. per Fe atom is or 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 ), 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 (-) = -1, which is equal to the charge on the MnO 4 ion. PRACTICE EXAMPLE A: What is the oxidation state of the underlined element in each of the following: S 8 ; Cr O - 7 ; Cl O; KO? PRACTICE EXAMPLE B: What is the oxidation state of the underlined element in each of the following: S O - 3 ; Hg Cl ; KMnO 4 ; H CO? In part (f) of Example 3-7, we got the somewhat surprising answer of + 3 for the oxidation state of the iron atoms in Fe 3 O 4. Prior to that, we saw only integral values for oxidation states. How does this fractional value come about? Generally, it comes from the assumption that all the atoms of an element have the same oxidation state in a given compound. Usually they do, but not always. Fe 3 O 4, for example, is probably better represented as FeO # Fe O 3, that is, through a combination of two simpler formula units. In FeO, the O.S. of the Fe atom is +. In Fe O 3, the O.S. of each of two Fe atoms is +3. When we average the oxidation states over all three Fe atoms, we get a nonintegral value: >3 = 8 3 = 3. Also, we may at times need to fragment a formula into its constituent parts before assigning oxidation states. The ionic compound NH 4 NO 3, for + instance, consists of the ions NH and NO The oxidation state of N in + NH is -3, and in NO - 4 3, +5, and we do not want to average them. It is far more useful to know the oxidation states of the individual N atoms than it is to deal with an average oxidation state of +1 for the two N atoms.

6 7A 7B (E) S 8 For an atom of a free element, the oxidation state is 0 (rule 1). CrO 7 The sum of all the oxidation numbers in the ion is (rule ). The O.S. of each oxygen is (rule 6). Thus, the total for all seven oxygens is 14. The total for both chromiums must be +1. Thus, each Cr has an O.S. = +6. ClO The sum of all oxidation numbers in the compound is 0 (rule ). The O.S. of oxygen is (rule 6). The total for the two chlorines must be +. Thus, each chlorine must have O.S. = +1. KO The sum for all the oxidation numbers in the compound is 0 (rule ). The O.S. of potassium is +1 (rule 3). The sum of the oxidation numbers of the two oxygens must be 1. Thus, each oxygen must have O.S. = 1/. (E) SO 3 The sum of all the oxidation numbers in the ion is (rule ). The O.S. of oxygen is (rule 6). Thus, the total for three oxygens must be 6. The total for both sulfurs must be +4. Thus, each S has an O.S. = +. Hg Cl The O.S. of each Cl is 1 (rule 7). The sum of all O.S. is 0 (rule ). Thus, the total for two Hg is + and each Hg has O.S. = +1. KMnO 4 The O.S. of each O is (rule 6). Thus, the total for 4 oxygens must be 8. The K has O.S. = +1 (rule 3). The total of all O.S. is 0 (rule ). Thus, the O.S. of Mn is +7. HCO The O.S. of each H is +1 (rule 5), producing a total for both hydrogens of +. The O.S. of O is (rule 6). Thus, the O.S. of C is 0, because the total of all O.Ss. is 0 (rule ).

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