A. Define the terms 'oxidation' and 'reduction.

B. The oxidation number of S

inS2O32 - is:

C. The oxidation number of S

inS4O62 - is:

How were the assignations made?

A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.

B. The oxidation number of S in S₂O₃²⁻ is + 2.

C. The oxidation number of S in S₄O₆²⁻ is + 5/2.

Explanation:

A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.

The overall charge on a molecule is equal to the sum of the oxidation states of all the atoms in a molecule.

B. S₂O₃²⁻ : The oxidation state of oxygen (O) is - 2 and overall charge on molecule is - 2.

Therefore, the oxidation state of sulfur (x) can be calculated by

2x + 3 (-2) = - 2

2x + (-6) = - 2

2x = - 2 + 6 = 4

x = 4 : 2 = + 2

Therefore, the oxidation number of S is + 2.

C. S₄O₆²⁻ : The oxidation state of oxygen (O) is - 2 and overall charge on molecule is - 2.

Therefore, the oxidation state of sulfur (x) can be calculated by

4x + 6 (-2) = - 2

4x + (-12) = - 2

4x = - 2 + 12 = 10

x = 10 : 4 = + 5/2

Therefore, the oxidation number of S is + 5/2.