Match the List-I and the List-II and select the correct answer using the codes given below the lists :
`{:(,"List-I" ,, "List -II" ) ,(,"(Compounds of Nitrogen)" ,, "(Valency)"), (A., N_(2)O , 1. , 1), (B., NO, 2., 2), (C., N_(2)O_(5) , 3., 3), (D., NO_(2) , 4., 4), (,,5., 5):}`

Options

A. `{:(,"A" , "B", "C", "D"), ((a), 1,4,5,2):}`
B. `{:(,"A" , "B", "C", "D"), ((b), 5,2,3,4):}`
C. `{:(,"A" , "B", "C", "D"), ((c), 1,2,5,4):}`
D. `{:(,"A" , "B", "C", "D"), ((d), 5,4,3,2):}`

Answer

To solve the question, we need to match the compounds of nitrogen in List-I with their corresponding valencies in List-II. Let's go through each compound step by step.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Valencies:**
- We have four nitrogen compounds: N₂O, NO, N₂O₅, and NO₂.
- We need to calculate the oxidation state (valency) of nitrogen in each compound.

2. **Calculate the Valency of N₂O:**
- The formula for N₂O can be expressed as:
\[ 2x + (-2) = 0 \]
where \(x\) is the oxidation state of nitrogen.
- Rearranging gives:
\[ 2x = 2 \implies x = 1 \]
- Therefore, the valency of N₂O is **1**.

3. **Calculate the Valency of NO:**
- The formula for NO can be expressed as:
\[ x + (-2) = 0 \]
- Rearranging gives:
\[ x = 2 \]
- Therefore, the valency of NO is **2**.

4. **Calculate the Valency of N₂O₅:**
- The formula for N₂O₅ can be expressed as:
\[ 2x + (-10) = 0 \]
- Rearranging gives:
\[ 2x = 10 \implies x = 5 \]
- Therefore, the valency of N₂O₅ is **5**.

5. **Calculate the Valency of NO₂:**
- The formula for NO₂ can be expressed as:
\[ x + (-4) = 0 \]
- Rearranging gives:
\[ x = 4 \]
- Therefore, the valency of NO₂ is **4**.

6. **Match the Compounds with Their Valencies:**
- Now we can summarize the findings:
- A. N₂O → 1
- B. NO → 2
- C. N₂O₅ → 5
- D. NO₂ → 4

7. **Final Matching:**
- Based on the calculations:
- A matches with 1
- B matches with 2
- C matches with 5
- D matches with 4

### Conclusion:
The correct matching of List-I with List-II is:
- A → 1
- B → 2
- C → 5
- D → 4

Thus, the correct answer is **Option C**.