Equivalent weight of `(NH_(4))_(2)Cr_(2)O_(7)` in the changes is:
`(NH)_(4)Cr_(2)O_(7)rarrN_(2)+Cr_(2)O_(3)+4H_(2)O`

Options

A. `M//6`
B. `M//8`
C. `M//2`
D. `M//3`

Explanation

To find the equivalent weight of \((NH_4)_2Cr_2O_7\) in the given redox reaction, we will follow these steps:

### Step 1: Write the balanced redox reaction
The reaction provided is:
\[ (NH_4)_2Cr_2O_7 \rightarrow N_2 + Cr_2O_3 + 4H_2O \]

### Step 2: Determine the oxidation states
- In \((NH_4)_2Cr_2O_7\):
- Nitrogen in \((NH_4)^+\) has an oxidation state of -3.
- Chromium in \((Cr_2O_7)^{2-}\) has an oxidation state of +6.
- Oxygen has an oxidation state of -2.

- In the products:
- Nitrogen in \(N_2\) has an oxidation state of 0.
- Chromium in \(Cr_2O_3\) has an oxidation state of +3.
- Oxygen in \(H_2O\) remains at -2.

### Step 3: Identify the changes in oxidation states
- For Nitrogen:
- From -3 in \((NH_4)_2Cr_2O_7\) to 0 in \(N_2\):
- Change = \(0 - (-3) = +3\) (each nitrogen atom increases by 3).
- Since there are 2 nitrogen atoms, total change = \(2 \times 3 = 6\).

- For Chromium:
- From +6 in \((Cr_2O_7)^{2-}\) to +3 in \(Cr_2O_3\):
- Change = \(3 - 6 = -3\) (each chromium atom decreases by 3).
- Since there are 2 chromium atoms, total change = \(2 \times (-3) = -6\).

### Step 4: Calculate the total number of electrons transferred
- Total electrons gained by Cr = 6 (as Cr is reduced).
- Total electrons lost by N = 6 (as N is oxidized).

### Step 5: Calculate the equivalent weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valency Factor}} \]
Where the valency factor is the total number of electrons transferred in the reaction.

### Step 6: Calculate the molecular weight of \((NH_4)_2Cr_2O_7\)
- Molecular weight of \(N = 14 \, \text{g/mol}\)
- Molecular weight of \(H = 1 \, \text{g/mol}\)
- Molecular weight of \(Cr = 52 \, \text{g/mol}\)
- Molecular weight of \(O = 16 \, \text{g/mol}\)

Calculating:
\[ \text{Molecular Weight} = 2 \times (14 + 4 \times 1) + 2 \times 52 + 7 \times 16 \]
\[ = 2 \times (14 + 4) + 104 + 112 \]
\[ = 2 \times 18 + 104 + 112 = 36 + 104 + 112 = 252 \, \text{g/mol} \]

### Step 7: Substitute into the equivalent weight formula
\[ \text{Equivalent Weight} = \frac{252}{6} = 42 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \((NH_4)_2Cr_2O_7\) in the reaction is \(42 \, \text{g/equiv}\).

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