Match list -1 ( compound ) with list -II ( Oxidation state of nitrogen ) and select the correct answer using the codes given below the lists -
`{:(,"List-I",,"List-II"),((A),NaN_(3),(a),+5),((B),N_(2)H_(4),(b) ,+ 2),((C ) , NO, ( c ) , - 1//3),((D ) ,N_(2) O_(5), (d) , -2):}`
Codes `:-`

Options

A. `{:(A,B,C,D),(c,d,b,a):}`
B. `{:(A,B,C,D),(d,c,b,a):}`
C. `{:(A,B,C,D),(c,d,a,b):}`
D. `{:(A,B,C,D),(d,c,a,b):}`

Answer

To solve the problem of matching compounds from List-I with their corresponding oxidation states of nitrogen from List-II, we will calculate the oxidation state of nitrogen in each compound step by step.

### Step 1: Determine the oxidation state of nitrogen in NaN₃ (A)
1. The formula for sodium azide is NaN₃.
2. Sodium (Na) has an oxidation state of +1.
3. Let the oxidation state of nitrogen be \( x \). Since there are three nitrogen atoms, the total contribution from nitrogen is \( 3x \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ +1 + 3x = 0 \]
5. Solving for \( x \):
\[ 3x = -1 \implies x = -\frac{1}{3} \]
6. Therefore, the oxidation state of nitrogen in NaN₃ is **-1/3**.

### Step 2: Determine the oxidation state of nitrogen in N₂H₄ (B)
1. The formula for hydrazine is N₂H₄.
2. Each hydrogen (H) has an oxidation state of +1. Since there are four hydrogen atoms, their total contribution is \( +4 \).
3. Let the oxidation state of nitrogen be \( x \). There are two nitrogen atoms, so the total contribution from nitrogen is \( 2x \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ 2x + 4 = 0 \]
5. Solving for \( x \):
\[ 2x = -4 \implies x = -2 \]
6. Therefore, the oxidation state of nitrogen in N₂H₄ is **-2**.

### Step 3: Determine the oxidation state of nitrogen in NO (C)
1. The formula for nitric oxide is NO.
2. Oxygen (O) has an oxidation state of -2.
3. Let the oxidation state of nitrogen be \( x \). The total contribution from nitrogen is \( x \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ x - 2 = 0 \]
5. Solving for \( x \):
\[ x = +2 \]
6. Therefore, the oxidation state of nitrogen in NO is **+2**.

### Step 4: Determine the oxidation state of nitrogen in N₂O₅ (D)
1. The formula for dinitrogen pentoxide is N₂O₅.
2. Each oxygen (O) has an oxidation state of -2. Since there are five oxygen atoms, their total contribution is \( 5 \times -2 = -10 \).
3. Let the oxidation state of nitrogen be \( x \). There are two nitrogen atoms, so the total contribution from nitrogen is \( 2x \).
4. The overall charge of the compound is neutral (0), so we can set up the equation:
\[ 2x - 10 = 0 \]
5. Solving for \( x \):
\[ 2x = 10 \implies x = +5 \]
6. Therefore, the oxidation state of nitrogen in N₂O₅ is **+5**.

### Summary of Results
- A: NaN₃ → Oxidation state of nitrogen = **-1/3** (C)
- B: N₂H₄ → Oxidation state of nitrogen = **-2** (D)
- C: NO → Oxidation state of nitrogen = **+2** (B)
- D: N₂O₅ → Oxidation state of nitrogen = **+5** (A)

### Final Matching
- A → C
- B → D
- C → B
- D → A

Thus, the correct answer is **CDBA**.