Which of the following is true for `N_(2)O_(5)`

Options

A. Paramagnetic
B. Anhydride of `HNO_(2)`
C. Brown gas
D. Exist in solid state In form of `[NO_(2)^(+)] [NO_(3)^(-)]`

Explanation

To determine which statement is true for \( N_2O_5 \), we will analyze each option step by step.

### Step 1: Analyze Option 1 - Paramagnetic
- **Calculation of Electrons**:
- Nitrogen (N) has an atomic number of 7, so each nitrogen atom has 7 electrons. For 2 nitrogen atoms:
\[ 2 \times 7 = 14 \text{ electrons} \]
- Oxygen (O) has an atomic number of 8, so each oxygen atom has 8 electrons. For 5 oxygen atoms:
\[ 5 \times 8 = 40 \text{ electrons} \]
- Total number of electrons in \( N_2O_5 \):
\[ 14 + 40 = 54 \text{ electrons} \]
- Since 54 is an even number, all electrons are paired, making \( N_2O_5 \ diamagnetic \) (not paramagnetic).
- **Conclusion**: Option 1 is **false**.

### Step 2: Analyze Option 2 - Anhydride of HNO2
- **Definition of Anhydrides**: An anhydride of an acid is a compound that can form that acid upon reaction with water.
- **Reaction with Water**:
- When \( N_2O_5 \) reacts with water, it produces nitric acid (HNO3):
\[ N_2O_5 + H_2O \rightarrow 2 HNO_3 \]
- Since \( N_2O_5 \) forms nitric acid (HNO3) and not nitrous acid (HNO2), this statement is incorrect.
- **Conclusion**: Option 2 is **false**.

### Step 3: Analyze Option 3 - Brown Gas
- **Physical State of \( N_2O_5 \)**:
- \( N_2O_5 \) is known to be a white solid at room temperature.
- **Conclusion**: Option 3 is **false**.

### Step 4: Analyze Option 4 - Ionic Form in Solid State
- **Ionic Form**:
- In the solid state, \( N_2O_5 \) exists as nitronium ion (\( NO_2^+ \)) and nitrate ion (\( NO_3^- \)).
- This can be represented as:
\[ N_2O_5 \rightarrow NO_2^+ + NO_3^- \]
- This statement is correct.
- **Conclusion**: Option 4 is **true**.

### Final Answer
The correct statement for \( N_2O_5 \) is **Option 4**: it exists in solid state in the form of \( NO_2^+ \) and \( NO_3^- \).

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