`NH_(3)` is oxidised to NO by `O_(2)` (air) in basic medium. Number of equivalent of `NH_(3)` oxidised by 1 mol of `O_(2)` is :

Explanation

To solve the problem of how many equivalents of \( NH_3 \) are oxidized by 1 mole of \( O_2 \) in a basic medium, we will follow these steps:

### Step 1: Write the balanced chemical equation
We start by writing the balanced equation for the oxidation of ammonia (\( NH_3 \)) by oxygen (\( O_2 \)). In basic medium, the reaction can be represented as:

\[ 4 NH_3 + 3 O_2 \rightarrow 2 NO + 6 H_2O \]

### Step 2: Determine the oxidation states
Next, we need to determine the oxidation states of nitrogen in \( NH_3 \) and \( NO \):

- In \( NH_3 \), the oxidation state of nitrogen (N) is \(-3\).
- In \( NO \), the oxidation state of nitrogen (N) is \(+2\).

### Step 3: Calculate the change in oxidation state
Now, we calculate the change in oxidation state for nitrogen:

- Change in oxidation state from \(-3\) to \(+2\) is:
\[ (+2) - (-3) = +5 \]

### Step 4: Calculate the total change in oxidation state for the reaction
Since 4 moles of \( NH_3 \) are oxidized, the total change in oxidation state for nitrogen in the reaction is:

\[ 4 \text{ moles} \times 5 = 20 \]

### Step 5: Determine the n-factor of \( NH_3 \)
The n-factor is defined as the total change in oxidation state per mole of the substance. Thus, for 4 moles of \( NH_3 \):

\[ \text{n-factor of } NH_3 = \frac{20}{4} = 5 \]

### Step 6: Calculate the number of equivalents
The number of equivalents of a substance is given by the formula:

\[ \text{Number of equivalents} = \frac{\text{number of moles} \times \text{n-factor}}{1} \]

Since we are considering 1 mole of \( O_2 \), and from the balanced equation, 3 moles of \( O_2 \) react with 4 moles of \( NH_3 \). Therefore, the number of equivalents of \( NH_3 \) oxidized by 1 mole of \( O_2 \) can be calculated as follows:

\[ \text{Equivalents of } NH_3 = \frac{4 \text{ moles of } NH_3}{3 \text{ moles of } O_2} \times 5 = \frac{20}{3} \approx 6.67 \]

Thus, the number of equivalents of \( NH_3 \) oxidized by 1 mole of \( O_2 \) is approximately \( 6.67 \).

### Final Answer
The number of equivalents of \( NH_3 \) oxidized by 1 mole of \( O_2 \) is approximately \( 6.67 \).

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