Statement-1. Electrolysis of NaH in the fused state liberates `H_(2)` at the anode.
Statement -2. NaH contains `H^(-)` ions.

Answer

To solve the question regarding the electrolysis of sodium hydride (NaH) in the fused state, we will analyze both statements step by step.

### Step 1: Understanding Electrolysis of NaH
Electrolysis is a process that uses an electric current to drive a chemical reaction. In the case of NaH, when it is fused (melted), it dissociates into its constituent ions.

### Step 2: Dissociation of NaH
When NaH is fused, it dissociates into sodium ions (Na⁺) and hydride ions (H⁻):
\[ \text{NaH} \rightarrow \text{Na}^+ + \text{H}^- \]

### Step 3: Identifying the Anode and Cathode Reactions
In electrolysis, the anode is the electrode where oxidation occurs, and the cathode is where reduction occurs.

- **At the Cathode**: The sodium ions (Na⁺) gain electrons (reduction) to form sodium metal:
\[ \text{Na}^+ + e^- \rightarrow \text{Na} \]

- **At the Anode**: The hydride ions (H⁻) lose electrons (oxidation) to form hydrogen gas:
\[ \text{H}^- \rightarrow \text{H} + e^- \]
Two hydrogen atoms will combine to form hydrogen gas (H₂):
\[ 2\text{H} \rightarrow \text{H}_2 \]

### Step 4: Evaluating Statement 1
Statement 1 claims that the electrolysis of NaH in the fused state liberates H₂ at the anode. Based on our analysis, this statement is correct because hydrogen gas is indeed produced at the anode.

### Step 5: Evaluating Statement 2
Statement 2 states that NaH contains H⁻ ions. From the dissociation process we discussed, we see that NaH does indeed dissociate into Na⁺ and H⁻ ions. Therefore, this statement is also correct.

### Conclusion
Both statements are correct:
- Statement 1 is correct: Electrolysis of NaH in the fused state liberates H₂ at the anode.
- Statement 2 is correct: NaH contains H⁻ ions.

### Final Answer
Both Statement 1 and Statement 2 are correct.

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