When an oxide `M_(2)O_(3)` is oxidised to `M_(2)O_(5)` its equivalent is

Options

A. M. w/1
B. M. w/2
C. M. w/4
D. M. w/8

Explanation

To find the equivalent weight of the oxide \( M_2O_3 \) when it is oxidized to \( M_2O_5 \), we can follow these steps:

### Step 1: Determine the oxidation states of the metal in both oxides.
- In \( M_2O_3 \):
- Let the oxidation state of \( M \) be \( x \).
- The oxidation state of oxygen is \(-2\).
- The equation for neutrality is:
\[ 2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = +3 \]

- In \( M_2O_5 \):
- The oxidation state of \( M \) is still \( x \).
- The equation for neutrality is:
\[ 2x + 5(-2) = 0 \implies 2x - 10 = 0 \implies 2x = 10 \implies x = +5 \]

### Step 2: Calculate the change in oxidation state.
- The change in oxidation state for one atom of \( M \) is:
\[ +5 - (+3) = +2 \]
- Since there are 2 atoms of \( M \) in \( M_2O_3 \), the total change in oxidation state is:
\[ 2 \times 2 = 4 \]

### Step 3: Calculate the equivalent weight.
- The formula for equivalent weight is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{N factor}} \]
- Here, the N factor (change in oxidation state) is 4.
- Therefore, the equivalent weight is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{4} \]

### Final Answer:
The equivalent weight of the oxide \( M_2O_3 \) when oxidized to \( M_2O_5 \) is \( \frac{\text{Molecular weight}}{4} \).

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