STATEMENT - 1 : `NaNO_(2)` and `NaNO_(3)` gives brown ring test.
STATEMENT - 2 : Both gives blue coloured gas with conc. `H_(2)SO_(4)`.
STATEMENT - 3 : `NO_(3)^(-)` gives ammonia with `NaOH` in presence of `Cu`.

Options

A. TFT
B. TTT
C. FTF
D. TTF

Answer

To determine the validity of the statements provided, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement 1:** NaNO₂ and NaNO₃ give a brown ring test.

- The brown ring test is a qualitative test for the presence of nitrites (NO₂⁻) and nitrates (NO₃⁻).
- When nitrites or nitrates react with concentrated sulfuric acid (H₂SO₄) and iron(II) sulfate (FeSO₄), a complex is formed that results in a brown ring.
- The reaction involves the formation of nitric oxide (NO) and the subsequent formation of the brown ring complex [Fe(H₂O)₆]²⁺-NO.
- Therefore, **Statement 1 is true.**

### Step 2: Analyze Statement 2
**Statement 2:** Both give blue-colored gas with concentrated H₂SO₄.

- When nitrites (NaNO₂) react with concentrated H₂SO₄, they form nitrous acid (HNO₂), which then disproportionates to form nitric acid (HNO₃) and nitric oxide (NO).
- Nitrate (NaNO₃) reacts similarly, producing nitric oxide (NO) as well.
- Nitric oxide (NO) is a blue-colored gas.
- Therefore, **Statement 2 is true.**

### Step 3: Analyze Statement 3
**Statement 3:** NO₃⁻ gives ammonia with NaOH in presence of Cu.

- When nitrate ions (NO₃⁻) react with sodium hydroxide (NaOH) in the presence of copper (Cu), a reduction reaction occurs.
- The nitrate ion (with nitrogen in +5 oxidation state) is reduced to ammonia (NH₃, with nitrogen in -3 oxidation state).
- This reaction also produces copper hydroxide (Cu(OH)₂), which is a blue precipitate.
- Therefore, **Statement 3 is true.**

### Conclusion
All three statements are true:
- **Statement 1:** True
- **Statement 2:** True
- **Statement 3:** True

### Final Answer
All statements are true.

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