Excess of ammonia with sodium hypochloride solution in the presence of glue or gelatine gives

Options

A. `NaNH_(2)`
B. `NH_2NH_2`
C. `N_2`
D. `NH_4Cl`

Answer

To solve the question regarding the reaction of excess ammonia with sodium hypochlorite in the presence of glue or gelatine, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Ammonia (NH₃)
- Sodium hypochlorite (NaOCl)

### Step 2: Understand the Role of Glue or Gelatine
Glue or gelatine acts as a stabilizing agent in the reaction, which can help in the formation of the product by providing a medium for the reaction to occur.

### Step 3: Write the Reaction
When ammonia reacts with sodium hypochlorite, the following reaction occurs:
\[ \text{NH}_3 + \text{NaOCl} \rightarrow \text{N}_2\text{H}_4 + \text{NaCl} + \text{H}_2\text{O} \]
In this reaction, ammonia (NH₃) is converted into hydrazine (N₂H₄).

### Step 4: Identify the Product
The main product formed from this reaction is hydrazine (N₂H₄), which can also be represented as NH₂-NH₂.

### Step 5: Conclusion
Thus, the excess of ammonia with sodium hypochlorite solution in the presence of glue or gelatine gives hydrazine (N₂H₄).

### Final Answer
The answer is: **Hydrazine (N₂H₄)**.

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