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

### Step 1: Determine the structure of \( N_2O_5 \)
- \( N_2O_5 \) consists of two nitrogen atoms and five oxygen atoms.
- The structure can be visualized as two nitrogen atoms bonded to five oxygen atoms, where one nitrogen forms a double bond with one oxygen and single bonds with the other oxygen atoms.

### Step 2: Calculate the valency of nitrogen
- Valency is determined by the number of bonds an atom can form.
- In \( N_2O_5 \), each nitrogen atom forms four bonds (one double bond with one oxygen and two single bonds with two other oxygen atoms).
- Therefore, the valency of nitrogen in \( N_2O_5 \) is 4.

### Step 3: Calculate the oxidation number of nitrogen
- Let the oxidation number of nitrogen be \( x \).
- The oxidation number of oxygen is typically \(-2\).
- In \( N_2O_5 \), we have two nitrogen atoms and five oxygen atoms. The equation for the total oxidation state can be set up as follows:

\[ 2x + 5(-2) = 0 \]

### Step 4: Solve the equation
- Simplifying the equation:

\[ 2x - 10 = 0 \]

- Adding 10 to both sides:

\[ 2x = 10 \]

- Dividing both sides by 2:

\[ x = 5 \]

### Conclusion
- The oxidation number of nitrogen in \( N_2O_5 \) is \( +5 \).

### Final Answer
- Valency of nitrogen: 4
- Oxidation number of nitrogen: +5