Plz answer as soonas possible

1.	Plz answer as soonas possible
Answer» Oxidation state of 'S' in S₂O₃²⁻ : 2x + 3(-2) = -2 2x = -2 + 6 2x = 4 x = +2 Oxidation state of 'S' in S₄O₆²⁻ : 4x + 6(-2) = -2 4x - 12 = -2 4x = 10 x = +2.5 So, In the above reaction, the oxidation state of sulphur CHANGES from +2 to +2.5. Thus the change in oxidation state = 2.5 - 2 = 0.5 → But the change occurs in TWO sulphur atoms, hence the total change in oxidation state is 2 x 0.5 = 1 Thus, n-factor of Na₂S₂O₃ = 1 So, EQUIVALENT weight = molecular weight / 1 = Molecular weight Hope it helps!

Answer»

Oxidation state of 'S' in S₂O₃²⁻ :

2x + 3(-2) = -2

2x = -2 + 6

2x = 4

x = +2

Oxidation state of 'S' in S₄O₆²⁻ :

4x + 6(-2) = -2

4x - 12 = -2

4x = 10

x = +2.5

So,

In the above reaction, the oxidation state of sulphur CHANGES from +2 to +2.5.

Thus the change in oxidation state = 2.5 - 2 = 0.5

→ But the change occurs in TWO sulphur atoms, hence the total change in oxidation state is 2 x 0.5 = 1

Thus, n-factor of Na₂S₂O₃ = 1

So, EQUIVALENT weight = molecular weight / 1 = Molecular weight

Hope it helps!

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