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SS1: CHEMISTRY - 1ST TERM

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  1. Introduction to Chemistry and Laboratory Apparatus | Week 1
    5Topics
    |
    1 Quiz
  2. Nature of Matter | Week 2
    3Topics
    |
    1 Quiz
  3. Separation Techniques I | Week 3
    1Topic
    |
    1 Quiz
  4. Separation Techniques II | Week 4
    5Topics
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    1 Quiz
  5. Particulate Nature of Matter I | Week 5
    5Topics
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    1 Quiz
  6. Particulate Nature of Matter II | Week 6
    9Topics
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    1 Quiz
  7. Symbols, Formulae & Oxidation Number | Week 7
    7Topics
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    1 Quiz
  8. Laws of Chemical Combination | Week 8
    4Topics
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    1 Quiz
  9. Chemical Equation & Chemical Combination (Chemical Bonding) I | Week 9
    4Topics
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    1 Quiz
  10. Chemical Combination (Chemical Bonding) II | Week 10
    4Topics
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    1 Quiz
  11. Chemical Combination (Chemical Bonding) III & Shapes of Covalent Molecules | Week 11
    3Topics
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    1 Quiz
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In order to understand the basic principles underlying the IUPAC system of naming (International Union of Pure and Applied Chemistry), a good knowledge of the concept of oxidation numbers to the atoms of the element in a molecule or ion is very essential.

Definition of Oxidation Number:

Oxidation number can be defined as the charge on an ion or it is the number of electron(s) on the valence shell of an atom or it is the number of electrons required by an atom to acquire a filled electronic structure.

Uses of Oxidation Numbers:

1. The oxidation number is used in the I.U.P.A.C system of naming chemical compounds.

2. Used in classifying elements into a particular group (family) on the periodic table.

3. Used to know which atom is oxidized or reduced in a redox reaction

4. Used in balancing half-cell equations.

Rules for Determining Oxidation Numbers of Compounds and Elements

1. The Oxidation number of the non-combined element is zero. Example: O2, H2, Mg, Hg, Cu, Ag, etc.

2. Positive oxidation states are assigned to more electropositive elements and negative oxidation states are assigned to more electronegative elements. Example, (Na+ Cl)

3. In a compound, the sum of the positive oxidation state and negative oxidation state is zero

4. For any ion, the sum of the oxidation numbers is equal to the charge on the ion. Example: Cl = -1, Fe3+ = +3, etc

5. The oxidation number of hydrogen is +1 except in metal hydride where it is -1. Example: (NaH, Na+1, H-1)

6. The oxidation number of oxygen is -2, except in peroxide where it is -1 e.g Na2O2.

7. In any Radical, the sum of the oxidation number of all the atoms equals the charge on its ions.

Example I:

Find the oxidation number of copper in Cu2O.

Solution:

Oxidation Number of Oxygen is -2 (Rule 6)

In a compound, the sum of positive oxidation state and negative oxidation state is zero (Rule3). 

Therefore, Cu2O

2Cu +(-2) = 0 (Rule 3)

2Cu – 2 = 0

2Cu  =  2

Cu = \( \frac{2}{2} \)

Cu = + 1

Oxidation number of Cu in Cu2O = +1

Example II

What is the oxidation number of sulphur in SO2?

Solution:

S + (-2 x 2) = 0

S – 4 = 0

S = +4

Oxidation Number of S in SO2 = +4

Example III

What is the oxidation number of Phosphorus in PH3?

Solution:

Each hydrogen atom has oxidation Number of +1 (Rule 5)

P + (1 x 3) = 0

P + 3 = 0

P = -3

Oxidation Number of Phosphorus in PH3 = -3

Example IV

Find the oxidation number of sulphur in H2SO4.

Solution:

In H2SO4, the oxidation number of H is +1 and Oxygen is -2.

(1 x 2)  +  S + (-2 x 4) = 0

2 + S – 8 = 0

S = -2 + 8

S = +6

Oxidation Number of sulphur in H2SO4 = +6

Example V

Calculate the oxidation number of chromium in Cr2O72-

Solution:

Oxidation Number of Oxygen = -2 (Rule 6)

Net charge on the Ion = -2 (Rule 7)

Therefore, Cr in Cr2O72-

2Cr + (-2 x 7) = -2

2Cr – 14 = -2

2Cr = + 14 – 2

2Cr = +12

Cr = \( \frac{12}{2} \)

Cr = +6

Oxidation Number of Chromium in Cr2O72- is +6

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