1. Oxidation Numbers (Ox #’s) What are they used for? Why do you need to learn them? to write chemical names and chemical formulas to balance redox equations for analytical, organic and inorganic chemistry What is an oxidation number? Ox # is a charge assigned to an ion or an atom. There are several cases to consider… 1.Elements, both monatomic, e.g., Cu and polyatomic, e.g., H 2, Cl 2, etc. 2.Ionic compounds, e.g., NaCl 3.Covalent compounds, e.g., HCl, H 2 O, etc. 1

2. 1.a) Ox # of a monatomic atom = 0 (because its net electric charge = 0) Consider a hydrogen atom, 1 H. It has one proton (+) and one electron (-). A hydrogen atom is neutral (-1 + 1 = 0). Its Ox # = 0. Consider a helium atom, 2 He. It has two protons (+2) and two electrons (-2). Its Ox# = 0. (-2 + 2 = 0) The same is true for all 118 monatomic atoms. They’re all neutral. They all have an Ox # = zero. 2 2 He 1H1H

3. 1.b) Ox # of a polyatomic element = 0 (because these atoms are neutral). Consider the diatomic hydrogen molecule, H 2. In the Lewis structure of H 2, the line drawn between the H atoms represents a covalent bond made of two shared e’s. e‘s are always shared equally between identical atoms. To determine H’s Ox #, divide the shared e’s and give one e’ to each atom. This produces two neutral H atoms, each owning one e’. Each H atom is now the same as its Lewis symbol. Recall that Lewis symbols represent neutral atoms by showing one dot for each e’ in the outermost shell (valence shell) of the atom. Since the atoms in H 2 are neutral, their Ox # = 0. 3 A line bond represents a shared pair of e’s in a covalent bond Lewis Symbols Ox # = 0

4. 4 The Ox # of a polyatomic element = 0 (continued). Consider the diatomic oxygen molecule, O 2. The Lewis structure of O 2 shows two shared e’ pairs (a double bond) represented by two lines. Dividing the shared e’s and giving two e’s to each atom, produces two neutral O atoms, each owning 6 valence e’s. Each O atom is now the same as its Lewis symbol. Since the atoms in O 2 are neutral, their Ox # = 0 in O 2. The same reasoning applies to diatomic nitrogen, N 2 Dividing its triple bond gives three e’s to each N atom, producing two neutral N atoms each owing 5 valence e’s N’s Ox # in N 2 = 0 The same is true for all polyatomic elements, e.g., P 4, S 8, F 2, Cl 2, Br 2, I 2, etc. Their atoms all have Ox # = 0. Lewis Symbols Ox # = 0

5. 5 2.In ionic compounds, the Ox # of an ion is the same as its electric charge Ox # = +2 Ox # = -1 Calcium chloride, CaCl 2, is an ionic compound, made of one calcium cation, Ca +2, and two chloride anions, 2Cl -. The Ox # of Ca +2 cation in CaCl 2 is the same as its charge, +2. The Ox # of each Cl - anion in CaCl 2 is the same as its charge, -1. Practice: State the Ox #’s of both atoms in Al 2 O 3 Answer: Since Al 2 O 3 is an ionic compound, the Ox #s of Al +3 and O -2 ions are the same as their electric charges, +3 and -2, respectively. Ox # = +3 Ox # = -2

6. 6 3.Calculating the Ox # of atoms in a covalent compound By definition, compounds are made of different types of atoms. Unlike ionic compounds (e.g., NaCl) that contain charged ions, covalent compounds (e.g., HCl) and covalent molecules (e.g. Cl 2 ) are contain no ions. In Cl 2, bonded e’s are shared equally, but in HCl, e’s are shared unequally because different atoms hold their electrons more or less strongly. In HCl, the shared e’s spend most of their time near Cl, as indicated by the larger e’-cloud around Cl and less time around H indicated by the smaller e’-cloud around H. NaCl, ionic compound doesn’t share e’s. HCl, polar covalent compound shares e’s unequally. Cl 2, nonpolar covalent molecule shares e’s equally.

7. Linus Pauling’s Table of Electronegativities (EN) More EN atoms hold e’s more strongly than less EN atoms. More EN atoms have higher EN values, nonmetals being the highest and metals being the lowest. H 2.1 Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Na 1.0 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.0 K 0.9 Ca 1.0 Sc 1.3 Ti 1.4 V 1.5 Cr 1.6 Mn 1.6 Fe 1.7 Co 1.7 Ni 1.8 Cu 1.8 Zn 1.6 Ga 1.7 Ge 1.9 As 2.1 Se 2.4 Br 2.8 Rb 0.9 Sr 1.0 Y 1.2 Zr 1.3 Nb 1.5 Mo 1.6 Tc 1.7 Ru 1.8 Rh 1.8 Pd 1.8 Ag 1.6 Cd 1.6 In 1.6 Sn 1.8 Sb 1.9 Te 2.1 I 2.5 Cs 0.8 Ba 1.0 La 1.1 Hf 1.3 Ta 1.4 W 1.5 Re 1.7 Os 1.9 Ir 1.9 Pt 1.8 Au 1.9 Hg 1.7 Tl 1.6 Pb 1.7 Bi 1.8 Po 1.9 At 2.1 7 EN increases

8. 8 To calculate the Ox # of atoms in covalent compounds, all shared electrons (in covalent bonds) are assigned to the more electronegative (EN) atom (and taken away from the less EN atom). Calculating the Ox # of atoms in a covalent compound (continued) Consider HCl Since Cl (EN = 3.0) is more EN than H (EN = 2.1), we assign both shared e’s in the covalent bond to Cl. We ascribe all 8 e’s in the structure to Cl, thus giving Cl one more e’ than its neutral atom has (see the Lewis symbol of Cl) Thus Cl in HCl has Ox # = -1. This leaves the H atom in the structure without any e’s, one less e’s than its neutral atom has (see the Lewis symbol of H). Thus H in HCl has Ox # = +1. Be Careful. No ions are actually present in HCl. The Ox #’s do not represent charges in covalent compounds. H owns 0 e’s Ox # = +1 Cl owns 8 e’s Ox # = -1 Lewis Symbols

9. Look at the Lewis structure of a H 2 O molecule. O has two pairs of unshared (nonbonded) e’s. Nonbonded e’s belong solely to the O atom. There are also two shared e’ pairs (the covalently bonded e’s) These 4 bonded e’s are not shared equally. They spend more time around O, because O (EN=3.5) is more EN than H (EN=2.1) 9 So all 8 valence e’s are assigned to O when calculating its Ox #. Thus O is assigned 8 e’s and the H’s are not assigned any e’s. Each H in H 2 O has one less e’ than its Lewis symbol, so H is assigned an Ox # of +1 The O in H 2 O has two more e’s than its Lewis symbol, so O is assigned an Ox # of -2. EN O = 3.5 EN H = 2.1 O owns 8 e’s Ox # = -2 H owns 0 e’s Ox # = +1 Calculating the Ox # of atoms in a covalent compound (continued) Lewis Symbols

10. Oxidation Numbers Note that in almost all its compounds, O has Ox # = -2. But recall that H and O are not truly charged in H 2 O. H 2 O is not ionic, but is considered ionic only to calculate Ox #’s. Peroxides are exceptions. Peroxides have one more O in their formula than is normally present. Hydrogen peroxide, H 2 O 2, is a simple example of a peroxide. The O atoms in peroxides have Ox # = -1, because they are assigned 7 e’s (one more than a neutral O atom). H can also have an Ox # of -1, when it is bonded to a less EN element, such as a reactive metal. Lithium hydride, LiH, is an example. In LiH, H owns one more e’ than its Lewis symbol so its Ox # = -1. Note that the ‘ide’ ending in hydride indicates that H is more EN and is named as an anion. 10 Ox # = -2 Ox # = +1 Ox # = -1 Ox # = +1 Ox # = -1 Ox # = +1

11. 11 Oxidation Numbers Now that you understand how Ox #’s are determined, you can relax in the knowledge that you don’t have to draw Lewis structures every time you need to find an Ox #. There is a simple method for learning Ox #’s, that will require a little memory and a little understanding of the periodic table. Group 1A metals (with 1 valence e’) are always Ox # +1 (so is Ag+) Group 2A metals (with 2 valence e’s) are always Ox # +2 (so are Zn +2 and Cd +2 ) Group 3A elements (with 3 valence e’s) B, Al and Ga are always +3. Recall that the A-Group metals tend to lose all their valence e’s to become isoelectronic with the nearest noble gas. So if you know the group number of an A-Group metal, you usually know its Ox #. Lewis Symbols

12. 12 Oxidation Numbers In Group 7A, F always has Ox # = -1. The other halogens may have ⊕ Ox #’s but will always be -1 when they are the more EN atom in a compound, e.g., HCl (Cl is more EN), NaBr (Br is more EN), KI (I is more EN). So Cl, Br and I all have Ox #’s = -1 in these compounds. In Group 6A, Ox # of O is always -2 (except in peroxides). The other chalcogens will also have Ox # of -2 when they are the more EN atom in a compound, e.g., H 2 S. (Ox # S = -2) The Group 5A atoms will always have Ox # of -3 when they are the more EN atom in a compound, e.g., NH 3 (Ox # N = -3) Recall that the A-Group nonmetals tend to gain enough e’s to fill their valence shell to become isoelectronic with the nearest noble gas. Lewis Symbols

13. 13 Atoms with fixed Ox #’s are shown. Metals cannot have ⊖ Ox #’s (they never gain e’s). Most nonmetals can have ⊖ or ⊕ Ox #’s, e.g., ICl (Ox # Cl = -1), ClF (Ox # Cl = +1, Ox # F = -1) Note that the nonmetal with higher EN always uses its ⊖ Ox # 7e + 1e = 8 6e + 2e = 8 5e + 3e = 8 Fixed Ox #’s of Atoms

14. 14 Oxidation Numbers Once you memorize the Ox #’s of those elements that have fixed Ox #’s, then all other Ox #’s are calculated from the chemical formula. Important Rule: The sum of all the Ox #’s in a formula equals the total charge of the formula. Examples: The charge on hydroxide, OH -, is -1, so the sum of the Ox #’s of O and H must = -1. The charge on H 2 O is 0. A charge of zero is never written in a formula. The sum of Ox #’s of O and H must = 0 The charge on ammonium ion, NH 4 +, is +1. The sum of Ox #’s of N and H must = +1, so Ox # N = -3 -2 -3

15. 15 Oxidation Numbers Example 1: Calculate the Ox # of the manganese atom in MnO 2 Process: Use the known Ox # of O to calculate the unknown Ox # of Mn Ox # of ‘O’ = -2. Multiply this by 2 because there are 2 ‘O’ atoms in the formula: (2 × -2 = -4) Since the formula shows no charge, you know it is neutral. This means that the Mn atom must have an Ox # = +4. You can think of it as a simple math equation, where the sum of all Ox #’s = total formula charge Mn + (-4) = 0  Mn = 0- (-4)  Mn = +4 This compound is named manganese(IV) oxide, where ‘IV’ is the Ox # of Mn in Roman numerals Note that the Ox # of an atom is stated in a name only when the atom can have > 1 Ox #

16. 16 Oxidation Numbers Example 2: Calculate the Ox # of the manganese in Mn 2 O 7 Process: Use the known Ox #’s of O to calculate the unknown Ox # of Mn Ox # of ‘O’ = -2. Multiply this by 7 because there are 7 ‘O’ atoms in the formula: (7 × -2 = -14) Since the formula shows no charge, you know it is neutral. This means that both Mn atoms together must contribute a charge of +14. Divide this by 2, to find the Ox # of each Mn atom. You can think of it as a simple math equation, where the sum of all Ox #’s = total formula charge 2Mn + (-14) = 0  2Mn = 0- (-14)  2Mn = +14  Mn = +14/2 = +7 This compound is named manganese(VII) oxide, where ‘VII’ is the Ox # of Mn in Roman numerals

17. 17 Oxidation Numbers Practice: Calculate the Oxid # of the underlined atom in each formula N 2 O N 2 O 5 SO 3 -2 Cd(BrO 3 ) 2 SiF 4 Na 2 CrO 4 P 2 O 7 -4 HCO 3 - Cr 2 S 3 Fe 3 P 2 N= +1 N= +5 S= +4 Br= +5 Si= +4 Cr= +6 P= +5 C= +4 Cr= +3 Fe= +2 2N + (-2) = 0 2N + (-10) = 0 S + (-6) = -2 2 + 2(Br -6) = 0 (Br -6) = -1 Si -4 = 0 2 + Cr -8 = 0 2P + -14 = -4 1 + C -6 = -1 2Cr +3(-2) = 0 3Fe +2(-3) = 0 2P = +10 C = -1 -1 +6 2Cr = 6 3Fe = 6

18. 18 Oxidation Number Summary: exampleelectric chargeOx # Monatomic elementCu, Al, Zn, etc.0Ox # = 0 Polyatomic elementH 2, O 2, P 4, S 8, etc.0Ox # = 0 Ionic compoundsNaCl, CaCl 2, Al 2 O 3 variable charge but ≠ 0 Ox # = ion charge Covalent compoundsHCl, CO 2 no chargesShared e’s assigned to more EN atom Ox # ≠ charge

19. 19 A List of Common Ox #’s