`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 basic medium, we can follow these steps:

### Step 1: Write the balanced half-reaction
In basic medium, the oxidation of \( NH_3 \) to \( NO \) can be represented as:
\[ 2 NH_3 + O_2 \rightarrow 2 NO + 2 H_2O \]

### Step 2: Determine the change in oxidation states
- In \( NH_3 \), nitrogen has an oxidation state of -3.
- In \( NO \), nitrogen has an oxidation state of +2.

### Step 3: Calculate the change in oxidation state for nitrogen
The change in oxidation state for one nitrogen atom from \( NH_3 \) to \( NO \) is:
\[ \text{Change} = +2 - (-3) = +5 \]

### Step 4: Calculate the total change for the reaction
Since 2 moles of \( NH_3 \) are oxidized to produce 2 moles of \( NO \), the total change in oxidation state for 2 moles of \( NH_3 \) is:
\[ 2 \times 5 = 10 \]

### Step 5: Determine the number of equivalents of \( NH_3 \)
The number of equivalents of \( NH_3 \) can be determined using the formula:
\[ \text{Number of equivalents} = \frac{\text{Total change in oxidation state}}{\text{Change in oxidation state per mole of substance}} \]
Since 1 mole of \( O_2 \) facilitates the oxidation of 2 moles of \( NH_3 \), we can find the number of equivalents of \( NH_3 \) oxidized by 1 mole of \( O_2 \):
\[ \text{Number of equivalents of } NH_3 = \frac{10}{5} = 2 \]

### Conclusion
Thus, the number of equivalents of \( NH_3 \) oxidized by 1 mole of \( O_2 \) is **2 equivalents**.

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