Oxidation States and Oxidation Numbers Notes

Oxidation States or Oxidation Number

Introduction

• Last Packet!!!!!!
• Oxidation/Reduction (REDOX)
• Reference Table: Periodic Table & J

Before We Start, Recall…

• Each box on the periodic table has one or multiple Oxidation States in the upper right-hand corner of the box.
• The oxidation number is the charge that an atom takes on after either losing or gaining one or more electrons on the outermost shell in order to most efficiently stabilize its amount of energy.
• Any free element (not combined with any other element) has an oxidation number of 0.
  • Ex: $"Na = Na^0$

Oxidation

• Is the loss of electrons by a molecule, atom, or ion.
• Metals lose electrons to have a complete outer shell (more stable).
• For example, Na will lose one electron to become $Na^+$.
• Therefore, Na is Oxidized.
• When any atom loses electrons its oxidation number increases.
  • Example: $Na \rightarrow Na^+ + e^-$
  • Its oxidation number increases from 0 to +1.

Reduction

• Is the gain of electrons by a molecule, atom, or ion.
• Nonmetals gain electrons to have a complete outer shell.
• Phosphorous will gain three electrons to have 8 in its outer shell.
• Because P gains 3 electrons, P is Reduced.
• When any atom gains electrons (negative particles) the oxidation number decreases.
  • Example: $P + 3e^- \rightarrow P^{3-}$
  • Its oxidation # decreases from 0 to -3.

OIL RIG

• JUST REMEMBER…
• "OIL RIG" (oxidation is losing, reduction is gaining)

Oxidation States

• Oxidation States are assigned to atoms to identify how many electrons are either gained or lost by an atom to form an ion.
• For example metals in Group 1 (like Na) have a +1 oxidation state. Metals in Group 2 (like Ca) have a +2 oxidation state. Therefore metals in Group 1 loses one electron when they form compounds and Group 2 loses two electrons when they form compounds.

Changes in Oxidation Number

• Changes in Oxidation Number indicate that a redox reaction has occurred.
• It is important to learn the rules for assigning oxidation states to atoms in order to determine whether oxidation or reduction has occurred.

Oxidizing Agent

• Oxidizing Agent: an electron acceptor

Oxidation State Rules

1. Free elements (not combined with any other element) have an oxidation number of Zero.

   • Ex: Na, $O<em>2$, $H</em>2$
   • $Na = 0$
   • $O_2 = 0$
   • $H_2 = 0$

2. All metals in group 1 have an oxidation number of +1.

   • $Na^+ = +1$
   • $Li^+ = +1$
   • $K^+ = +1$

3. All metals in group 2 have an oxidation number of +2.

   • $Mg^{2+} = +2$
   • $Ca^{2+} = +2$
   • $Sr^{2+} = +2$
   • Memorize

4. F (fluorine) always has an oxidation of -1 as an ion

   • Ex: $Mg$ has an oxidation number of +2.
   • $F^- = -1$

5. The oxidation of simple ions is equal to the Charge on the ion

   • $Na^+ = +1$
   • $Cl^- = -1$
   • $Mg^{2+} = +2$
   • Examples: $NaCl$ & $MgCl_2$
   • $NaCl \rightarrow Na^+ Ion + Cl^- Ion = 0$
   • $Mg Cl_2 \rightarrow Mg = +2, Cl = -1$

6. The Sum of the oxidation numbers must equal 0 in a compound

7. In Polyatomic ions, the Sum of the oxidation numbers of all the atoms must equal the Charge of the ion

   • Example: sulfate ion $SO_4$. Oxygen has an oxidation of -2, and therefore $(-2)x(+4)=-2$ Remember that the overall charge of this ion has to be -2. What is the oxidation # of sulfur?
   • $SO_4^{-2}$
   • $S + 4(-2) = -2$
   • $S + (-8) = -2$
   • $S = 6$

8. In general, Oxygen has an oxidation number of -2. Oxygen has an oxidation number of -1 in Peroxides ($O<em>2^{-2}$) Example: $H</em>2O<em>2$. Oxygen has an oxidation number +2 in compounds with flourine. Example: $OF</em>2$

9. Hydrogen has an oxidation number of +1 in all compounds combined with a non-metal. Hydrogen has an oxidation number of -1 when it is a Metal hydrides. Example: $LiH$, and $CaH_2$

   • Metal + Hydrogen

Assigning Oxidation Numbers

• When you have a binary compound or polyatomic (ternary) compound, assigning oxidation numbers is a little easier than when you have to assign oxidation numbers to compounds with more than two elements.
• Examples
  • Hydrogen Peroxide
    • $H<em>2O</em>2$
    • $2H + 2O = 0$
  • $NaCl$
    • $Na = +1$
    • $Cl = -1$
    • $(+1) + 1(Cl) = 0$
    • $Cl = -1$
  • $CaO$
    • $Ca Ion + 1 O Ion = 0$
    • $+2 + 1O = 0$
    • $O = -2$
  • $CO_2$
    • $1C + 2O = 0$
    • $1C + 2 (-2) = 0$
    • $1C = +4$
    • $C = +4$
    • $O = -2$
  • $Li_3N$
    • $3 Li Ion + 1N = 0$
    • $3 (+1) + 1N = 0$
    • $3 + N = 0$
    • $N = -3$

Assigning Oxidation Numbers to Compounds with More Than Two Elements

• Dr. McGuiness' "bookend" technique to assigning oxidation numbers to compounds with more than two elements.
  • a) Identify the oxidation # of the last element (overall charge)
  • b) Identify the oxidation # of the first element (overall charge)
  • c) If there is no charge to the Compound, then the overall charge must be 0, therefore you can determine the oxidation # of the element in the middle.
  • Example:
    • $Li(MnO_4)$
    • $(+1) + (?) + (-8) = 0$