A 0.1 mole of a metal is burnt in air to form oxide. The same oxide is then reduced by 0.05 M, 4 litres `S_(2)O_(3)^(2-)` (acidic medium) to +3 oxidation state of metal. What is the oxidation state of metal in oxide ?

Answer

To find the oxidation state of the metal in the metal oxide, we can follow these steps:

### Step 1: Understand the Reaction
We start with 0.1 moles of a metal (let's denote it as M) that is burnt in air to form a metal oxide (M_xO_y). The metal oxide is then reduced by thiosulfate ions (S2O3^2-) in an acidic medium to produce the metal in the +3 oxidation state.

### Step 2: Determine the Moles of Thiosulfate
We need to calculate the number of moles of thiosulfate used in the reaction. The concentration of thiosulfate is given as 0.05 M and the volume is 4 liters.

\[ \text{Number of moles of } S2O3^{2-} = \text{Molarity} \times \text{Volume} = 0.05 \, \text{mol/L} \times 4 \, \text{L} = 0.2 \, \text{mol} \]

### Step 3: Calculate the n-factor for Thiosulfate
The oxidation state of sulfur in thiosulfate (S2O3^2-) can be calculated as follows:
Let the oxidation state of sulfur be \( x \):
\[ 2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = 2 \]
Thus, the oxidation state of sulfur in thiosulfate is +2.

When thiosulfate is oxidized to tetrathionate (S4O6^2-), the oxidation state of sulfur in tetrathionate can be calculated similarly:
Let the oxidation state of sulfur in tetrathionate be \( y \):
\[ 4y + 6(-2) = -2 \implies 4y - 12 = -2 \implies 4y = 10 \implies y = 2.5 \]
Thus, the oxidation state of sulfur in tetrathionate is +2.5.

### Step 4: Calculate the Change in Oxidation State
The change in oxidation state for sulfur from thiosulfate to tetrathionate is:
\[ \Delta \text{Oxidation State} = 2.5 - 2 = 0.5 \]
Since there are 2 sulfur atoms in thiosulfate, the total change in oxidation state (n-factor) for thiosulfate is:
\[ \text{n-factor} = 2 \times 0.5 = 1 \]

### Step 5: Relate the n-factor of Metal Oxide to Thiosulfate
The number of equivalents of thiosulfate used is equal to the number of equivalents of the metal oxide reduced:
\[ \text{Equivalents of } S2O3^{2-} = \text{Number of moles} \times \text{n-factor} = 0.2 \, \text{mol} \times 1 = 0.2 \, \text{equivalents} \]

### Step 6: Calculate the n-factor for Metal Oxide
Let the n-factor of the metal oxide be \( n \). We know that:
\[ \text{Equivalents of Metal Oxide} = \text{Number of moles of Metal} \times n \]
Given that we have 0.1 moles of metal:
\[ 0.1 \, \text{mol} \times n = 0.2 \implies n = \frac{0.2}{0.1} = 2 \]

### Step 7: Relate the n-factor to the Change in Oxidation State
The n-factor of the metal oxide is equal to the total change in oxidation state. If the oxidation state of the metal in the oxide is \( x \) and it is reduced to +3, we have:
\[ x - 3 = 2 \implies x = 2 + 3 = 5 \]

### Conclusion
The oxidation state of the metal in the metal oxide is +5.