In the reaction`3ClO^(-)(Aq)to ClO3^(-)`+2CI^(-)` the equivalent mass of `ClO^(-)` is :

Options

A. `M//4`
B. `M//3`
C. `3M`
D. `3M//4`

Explanation

To find the equivalent mass of \( \text{ClO}^- \) in the reaction

\[ 3 \text{ClO}^- \rightarrow \text{ClO}_3^- + 2 \text{Cl}^- \]

we will follow these steps:

### Step 1: Determine the oxidation states
First, we need to calculate the oxidation states of chlorine in both \( \text{ClO}^- \) and \( \text{ClO}_3^- \).

1. For \( \text{ClO}^- \):
- Let the oxidation state of Cl be \( x \).
- The equation becomes:
\[ x + (-2) = -1 \implies x - 2 = -1 \implies x = +1 \]
So, the oxidation state of Cl in \( \text{ClO}^- \) is +1.

2. For \( \text{ClO}_3^- \):
- Let the oxidation state of Cl be \( y \).
- The equation becomes:
\[ y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5 \]
So, the oxidation state of Cl in \( \text{ClO}_3^- \) is +5.

### Step 2: Identify the type of reaction
In this reaction, \( \text{ClO}^- \) is being oxidized to \( \text{ClO}_3^- \) (oxidation state increases from +1 to +5) and reduced to \( \text{Cl}^- \) (oxidation state decreases from +1 to -1). This is a disproportionation reaction because the same species is both oxidized and reduced.

### Step 3: Calculate the n-factor
The n-factor is determined by the total change in oxidation states per mole of the substance.

1. For oxidation (from \( \text{ClO}^- \) to \( \text{ClO}_3^- \)):
- Change in oxidation state = \( +5 - (+1) = 4 \) (4 electrons lost)

2. For reduction (from \( \text{ClO}^- \) to \( \text{Cl}^- \)):
- Change in oxidation state = \( -1 - (+1) = -2 \) (2 electrons gained)

Using the formula for n-factor in disproportionation reactions:
\[ \frac{1}{n} = \frac{1}{n_{\text{oxidation}}} + \frac{1}{n_{\text{reduction}}} \]
where \( n_{\text{oxidation}} = 4 \) and \( n_{\text{reduction}} = 2 \):
\[ \frac{1}{n} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \]
Thus, \( n = \frac{4}{3} \).

### Step 4: Calculate equivalent mass
The equivalent mass is given by the formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{n} \]
Let the molar mass of \( \text{ClO}^- \) be \( M \). Therefore,
\[ \text{Equivalent mass} = \frac{M}{\frac{4}{3}} = \frac{3M}{4} \]

### Final Answer
The equivalent mass of \( \text{ClO}^- \) is \( \frac{3M}{4} \).

---