Valency and oxidation number of nitrogen in `N_(2)O_(5)`

Options

A. `5 , +5`
B. `4 , +5`
C. `3 , +3`
D. `3 , +5`

Answer

To determine the valency and oxidation number of nitrogen in \( N_2O_5 \), we can follow these steps:

### Step 1: Assign oxidation states
Let the oxidation number of nitrogen be \( x \). In \( N_2O_5 \), there are 2 nitrogen atoms, so the total contribution from nitrogen will be \( 2x \).

### Step 2: Determine the oxidation state of oxygen
Oxygen typically has an oxidation state of \(-2\). Since there are 5 oxygen atoms in \( N_2O_5 \), the total contribution from oxygen will be \( 5 \times (-2) = -10 \).

### Step 3: Set up the equation
The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can set up the equation:
\[ 2x + (-10) = 0 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ 2x - 10 = 0 \\ 2x = 10 \\ x = \frac{10}{2} = 5 \]
Thus, the oxidation number of nitrogen in \( N_2O_5 \) is \( +5 \).

### Step 5: Determine the valency of nitrogen
Valency refers to the ability of an atom to bond with other atoms. In \( N_2O_5 \), nitrogen forms 4 covalent bonds (2 nitrogen atoms each forming 4 bonds with oxygen). Therefore, the valency of nitrogen in \( N_2O_5 \) is 4.

### Final Answer
- **Oxidation Number of Nitrogen:** \( +5 \)
- **Valency of Nitrogen:** \( 4 \)

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