1) The document discusses naming compounds and balancing chemical equations using oxidation numbers. It provides examples of naming metal and non-metal compounds as well as oxide compounds. 2) Methods for determining oxidation numbers from compound names and formulas are explained. Examples are given for nitrate compounds and working out formulas from compound names. 3) The document demonstrates balancing chemical equations using oxidation numbers, giving examples of copper (II) oxide reacting with ammonia and manganate (VII) ions reacting with iron (II) ions.

1. Naming compounds
Balancing chemical equations
using oxidation numbers
Chapter 7
Part 3
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2. Naming compounds:
Species of metals and non-metals
1- Name the metal part of the compound
2- Writer the oxidation number of metal between two brackets
3- Name the non-metal part
Example:
FeCl2
Iron (II) chloride
But how did we decide that Fe oxidation number is +2 ?
Cl=-2 and FeCl2=0
Fe + (-2) =0
Fe= +2
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3. Oxides of nitrogen:
• Naming oxygen compounds:
1- Name the first part of the compound
2- Put oxidation number between two brackets
3- Add oxide word to express the presence of oxygen in the compound
• N2O
Nitrogen (I) oxide (oxygen =-2 , N2O=0 , and 2N + (-2) = 0 so N=+1
NO2
Nitrogen (IV) oxide
Oxygen = -2 , 2O= -4 , NO2 = 0
N + (-4) =0 , so N = +4
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4. Nitrate ions:
Na+NO2
-
Oxidation number of Na=+1 , 2O = -2X2 = -4 ,
NO2 = -1 N + (-4) = -1 N = +3
Soduim nitrate (III)
Na+NO3
-
Na = +1 NO3 = -1
N + (3 X O) = -1
N + (3X-2) = -1
N = +5
Sodium nitrate (V)
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5. How to work out formula from compound
name?
• What is the formula of sodium chlorate(V)?
• Sodium = +1
• Oxygen = -2
• Chlorine oxi number is +5 from (V)
• Na + O + Cl = 0
(+1) + n(-2) + (+5) = 0
n= 3
Na ClO3
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6. Exercise:
• sodium chlorate(I)
• Sodium = +1
• Chlorine = +1
• Oxygen = -2
Na + Cl + nO =0
+1 +1 + nX-2 = 0
n = 1
Na ClO
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7. Balancing chemical equations using oxidation
numbers:
• Copper(II) oxide (CuO) reacts with ammonia (NH3 ) to form copper,
nitrogen (N2) and water.
1- write unbalanced chemical equation
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8. 2- Deduce the ox. no. changes:
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9. 3- Balance the ox. no. changes
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4- Balance the atoms:

10. Exercise:
• Manganate (VII) ions (MnO4 – ) react with Fe2+ ions in the presence of
acid (H+ ) to form Mn2+ ions, Fe3+ ions and water.
1- write unbalanced chemical equation
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11. 2- Deduce the ox. no. changes:
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12. 3- Balance the ox. no. changes
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13. 4- Balance the atoms:
The total charge on the other reactants is: (1–)(from MnO4 – ) + (5 ×
2+)(from 5Fe2+) = 9+
■ The total charge on the products is: (2+)(from Mn2+) + (5 × 3+)(from
5Fe3+) = 17+
■ To balance the charges we need 8 H+ ions on the left.
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14. End of Redox Reactions
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