`Cr_(2)O_(7)^(2-)overset(H^(+))rarrCr^(3+)`, Eq. wt. of `Cr_(2)O_(7)^(2-)` is :-

Options

A. mol wt/6
B. mol wt/3
C. mol, wt/4
D. mol, wt/2

Answer

To find the equivalent weight of \( \text{Cr}_2\text{O}_7^{2-} \), we will follow these steps:

### Step 1: Determine the Molecular Weight of \( \text{Cr}_2\text{O}_7^{2-} \)

The molecular weight can be calculated using the atomic weights of chromium (Cr) and oxygen (O):
- Atomic weight of Cr = 52 g/mol
- Atomic weight of O = 16 g/mol

The molecular weight of \( \text{Cr}_2\text{O}_7^{2-} \) is calculated as follows:
\[ \text{Molecular weight} = 2 \times \text{Atomic weight of Cr} + 7 \times \text{Atomic weight of O} \]
\[ \text{Molecular weight} = 2 \times 52 + 7 \times 16 = 104 + 112 = 216 \text{ g/mol} \]

### Step 2: Determine the Change in Oxidation State

In the conversion from \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \):
- The oxidation state of Cr in \( \text{Cr}_2\text{O}_7^{2-} \) is +6.
- The oxidation state of Cr in \( \text{Cr}^{3+} \) is +3.

The change in oxidation state for each chromium atom is:
\[ +6 \text{ to } +3 \implies \text{decrease of } 3 \]
Since there are 2 chromium atoms, the total decrease in oxidation state is:
\[ 2 \times 3 = 6 \]
This indicates that 6 electrons are gained in total during the reaction.

### Step 3: Calculate the n-factor

The n-factor represents the number of electrons gained or lost per formula unit in the reaction. Here, since the oxidation state decreases (indicating a gain of electrons), the n-factor is:
\[ n = 6 \]

### Step 4: Calculate the Equivalent Weight

The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
Substituting the values we found:
\[ \text{Equivalent weight} = \frac{216 \text{ g/mol}}{6} = 36 \text{ g/equiv} \]

### Final Answer

The equivalent weight of \( \text{Cr}_2\text{O}_7^{2-} \) is \( 36 \text{ g/equiv} \).

---