The reaction between `NH_2^(Ө)` and `N_2 O` gives

Options

A. `NO`
B. `N_2 O_5`
C. `NH_2 NH_2`
D. `N_3^(Ө)`

Explanation

To solve the question regarding the reaction between \( NH_2^- \) (the amide ion) and \( N_2O \) (nitrous oxide), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- \( NH_2^- \) (amide ion)
- \( N_2O \) (nitrous oxide)

### Step 2: Write the Reaction
When \( NH_2^- \) reacts with \( N_2O \), we can set up the reaction as follows:
\[ NH_2^- + N_2O \rightarrow ? \]

### Step 3: Analyze the Reaction
The amide ion \( NH_2^- \) acts as a nucleophile and can react with \( N_2O \). The reaction typically leads to the formation of the azide ion \( N_3^- \) and water.

### Step 4: Write the Balanced Equation
The balanced chemical equation for the reaction can be written as:
\[ NH_2^- + N_2O \rightarrow N_3^- + H_2O \]

### Step 5: Count Atoms and Charges
- On the left side (reactants):
- Nitrogen: 3 (1 from \( NH_2^- \) and 2 from \( N_2O \))
- Hydrogen: 2 (from \( NH_2^- \))
- Oxygen: 1 (from \( N_2O \))

- On the right side (products):
- Nitrogen: 3 (from \( N_3^- \))
- Hydrogen: 2 (from \( H_2O \))
- Oxygen: 1 (from \( H_2O \))

The charges also balance out: \( -1 \) from \( NH_2^- \) and \( 0 \) from \( N_2O \) gives \( -1 \) on the left, while \( -1 \) from \( N_3^- \) gives \( -1 \) on the right.

### Step 6: Conclusion
The product formed when \( NH_2^- \) reacts with \( N_2O \) is the azide ion \( N_3^- \).

### Final Answer
The correct answer is \( N_3^- \) (the azide ion).

---