Equivalent weight of `N_(2)` in the change
`N_(2) rarr NH_(3)` is

Options

A. `28//6`
B. `28`
C. `28//2`
D. `28//3`

Explanation

To find the equivalent weight of \( N_2 \) in the reaction \( N_2 \rightarrow NH_3 \), we will follow these steps:

### Step 1: Determine the Molecular Weight of \( N_2 \)
The molecular weight of nitrogen gas \( N_2 \) can be calculated as follows:
- The atomic weight of nitrogen (N) is approximately 14 g/mol.
- Since \( N_2 \) consists of two nitrogen atoms, the molecular weight is:
\[ \text{Molecular weight of } N_2 = 2 \times 14 = 28 \text{ g/mol} \]

### Step 2: Identify the Change in Oxidation State
Next, we need to determine the change in oxidation state of nitrogen in this reaction:
- In \( N_2 \), nitrogen has an oxidation state of 0.
- In \( NH_3 \), nitrogen has an oxidation state of -3.
- The change in oxidation state for one nitrogen atom is:
\[ 0 \rightarrow -3 \quad \text{(a change of 3)} \]
- Since there are 2 nitrogen atoms in \( N_2 \), the total change in oxidation state is:
\[ 2 \times 3 = 6 \]

### Step 3: Determine the Valency Factor
The valency factor is defined as the total number of electrons gained or lost per molecule in the reaction. In this case, since \( N_2 \) is reduced to \( NH_3 \), the valency factor is:
\[ \text{Valency factor} = 6 \]

### Step 4: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Valency factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight} = \frac{28 \text{ g/mol}}{6} = \frac{28}{6} \text{ g/equiv} \]

### Final Answer
Thus, the equivalent weight of \( N_2 \) in the reaction \( N_2 \rightarrow NH_3 \) is:
\[ \frac{28}{6} \text{ g/equiv} \]