In which of the following pairs in there the greatest difference in the oxidation numbers of the underlined elements?

Options

A. `ul(N)O_(2)` and `ul(N_(2))O_(4)`
B. `ulP_(2)O_(5)` and `ul(P_(4))O_(10)`
C. `ul(N_(2))O` and `ul(N)O`
D. `ul(S)O_(2)` and `ul(S)O_(3)`

Explanation

To solve the problem of finding the greatest difference in oxidation numbers of the underlined elements in the given pairs, we will calculate the oxidation states for each pair and then determine the differences.

### Step-by-Step Solution:

1. **Identify the pairs and the underlined elements:**
- The pairs given are NO2 and N2O4, P2O5 and P4O10, N2O and NO, SO2 and SO3.

2. **Calculate the oxidation number for the first pair (NO2 and N2O4):**
- For NO2:
- Let the oxidation number of N be \( X \).
- The equation is: \( X + 2(-2) = 0 \)
- Simplifying gives: \( X - 4 = 0 \) → \( X = +4 \)

- For N2O4:
- Let the oxidation number of N be \( Y \).
- The equation is: \( 2Y + 4(-2) = 0 \)
- Simplifying gives: \( 2Y - 8 = 0 \) → \( 2Y = 8 \) → \( Y = +4 \)

- **Difference**: \( |4 - 4| = 0 \)

3. **Calculate the oxidation number for the second pair (P2O5 and P4O10):**
- For P2O5:
- Let the oxidation number of P be \( A \).
- The equation is: \( 2A + 5(-2) = 0 \)
- Simplifying gives: \( 2A - 10 = 0 \) → \( 2A = 10 \) → \( A = +5 \)

- For P4O10:
- Let the oxidation number of P be \( B \).
- The equation is: \( 4B + 10(-2) = 0 \)
- Simplifying gives: \( 4B - 20 = 0 \) → \( 4B = 20 \) → \( B = +5 \)

- **Difference**: \( |5 - 5| = 0 \)

4. **Calculate the oxidation number for the third pair (N2O and NO):**
- For N2O:
- Let the oxidation number of N be \( C \).
- The equation is: \( 2C + (-2) = 0 \)
- Simplifying gives: \( 2C - 2 = 0 \) → \( 2C = 2 \) → \( C = +1 \)

- For NO:
- Let the oxidation number of N be \( D \).
- The equation is: \( D + (-2) = 0 \)
- Simplifying gives: \( D - 2 = 0 \) → \( D = +2 \)

- **Difference**: \( |1 - 2| = 1 \)

5. **Calculate the oxidation number for the fourth pair (SO2 and SO3):**
- For SO2:
- Let the oxidation number of S be \( E \).
- The equation is: \( E + 2(-2) = 0 \)
- Simplifying gives: \( E - 4 = 0 \) → \( E = +4 \)

- For SO3:
- Let the oxidation number of S be \( F \).
- The equation is: \( F + 3(-2) = 0 \)
- Simplifying gives: \( F - 6 = 0 \) → \( F = +6 \)

- **Difference**: \( |4 - 6| = 2 \)

6. **Compare the differences calculated:**
- Pair 1 (NO2 and N2O4): Difference = 0
- Pair 2 (P2O5 and P4O10): Difference = 0
- Pair 3 (N2O and NO): Difference = 1
- Pair 4 (SO2 and SO3): Difference = 2

7. **Conclusion:**
- The greatest difference in oxidation numbers is found in the pair SO2 and SO3, which is 2.

### Final Answer:
The correct option is D (SO2 and SO3).