Match List - I (compound) with list - II (Oxidation state of N) and select the correct answer using the codes given below the list :-
`{:(" List - I"," List - II"),((A)KNO_(3),(a)-1//3),((B)HNO_(2),(b)-3),((C)NH_(4)Cl,(c)0),((D)NaN_(3),(d)+3),(,(e) +5):}`

Options

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

Answer

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

### Step 1: Determine the oxidation state of nitrogen in KNO₃

1. **Write the equation for oxidation states**:
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2 (and there are 3 oxygen atoms).
- Let the oxidation state of nitrogen be \( X \).

The equation becomes:
\[ +1 + X + 3(-2) = 0 \]
Simplifying this:
\[ +1 + X - 6 = 0 \implies X - 5 = 0 \implies X = +5 \]

2. **Conclusion**:
- The oxidation state of nitrogen in KNO₃ is +5.

### Step 2: Determine the oxidation state of nitrogen in HNO₂

1. **Write the equation for oxidation states**:
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2 (and there are 2 oxygen atoms).
- Let the oxidation state of nitrogen be \( X \).

The equation becomes:
\[ +1 + X + 2(-2) = 0 \]
Simplifying this:
\[ +1 + X - 4 = 0 \implies X - 3 = 0 \implies X = +3 \]

2. **Conclusion**:
- The oxidation state of nitrogen in HNO₂ is +3.

### Step 3: Determine the oxidation state of nitrogen in NH₄Cl

1. **Write the equation for oxidation states**:
- Hydrogen (H) has an oxidation state of +1 (and there are 4 hydrogen atoms).
- Chlorine (Cl) has an oxidation state of -1.
- Let the oxidation state of nitrogen be \( X \).

The equation becomes:
\[ 4(+1) + X - 1 = 0 \]
Simplifying this:
\[ 4 + X - 1 = 0 \implies X + 3 = 0 \implies X = -3 \]

2. **Conclusion**:
- The oxidation state of nitrogen in NH₄Cl is -3.

### Step 4: Determine the oxidation state of nitrogen in NaN₃

1. **Write the equation for oxidation states**:
- Sodium (Na) has an oxidation state of +1.
- Let the oxidation state of nitrogen be \( X \) (and there are 3 nitrogen atoms).

The equation becomes:
\[ +1 + 3X = 0 \]
Simplifying this:
\[ 3X = -1 \implies X = -\frac{1}{3} \]

2. **Conclusion**:
- The oxidation state of nitrogen in NaN₃ is -1/3.

### Final Matching

Now we can match the compounds with their oxidation states:

- **KNO₃**: +5 (matches with E)
- **HNO₂**: +3 (matches with D)
- **NH₄Cl**: -3 (matches with B)
- **NaN₃**: -1/3 (matches with A)

### Summary of Matches
- A: KNO₃ → +5 (E)
- B: HNO₂ → +3 (D)
- C: NH₄Cl → -3 (B)
- D: NaN₃ → -1/3 (A)

### Answer Code
The correct answer code is: **(A)E, (B)D, (C)B, (D)A**.