Question

Match List-I with List-II:

Choose the correct answer from the options given below:

• (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

Assigning oxidation states is a highly reliable way to solve redox-matching questions.
Write out the basic formulas: \( \text{SO}_3^{2-} (+4) \rightarrow \text{S}^{2-} (-2) \) involves a 6-electron shift, which instantly connects (D) to (I).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To determine the stoichiometry of electrons required for any inorganic reduction reaction, we calculate the difference between the formal oxidation states of the key atom (nitrogen or sulfur) in the starting reactant and the final product.

Step 2: Detailed Explanation:
Let us compute the electron requirements for each of the four reductive reactions:
1. Reduction of sulfite to sulfide (D):
- In the sulfite ion (\( \text{SO}_3^{2-} \)), oxygen is assigned its standard oxidation state of \( -2 \).
Thus, for the neutral ion: \( S + 3(-2) = -2 \Rightarrow S = +4 \).
- In the sulfide ion (\( \text{S}^{2-} \)), the oxidation state of sulfur is \( -2 \).
- The change in oxidation state represents the number of electrons required:
\[ (+4) - (-2) = 6 \text{ electrons} \] Thus, (D) matches with (I).
2. Reduction of nitrate to nitrite (C):
- In the nitrate ion (\( \text{NO}_3^- \)), we have: \( N + 3(-2) = -1 \Rightarrow N = +5 \).
- In the nitrite ion (\( \text{NO}_2^- \)), we have: \( N + 2(-2) = -1 \Rightarrow N = +3 \).
- The difference in oxidation state is:
\[ (+5) - (+3) = 2 \text{ electrons} \] Thus, (C) matches with (II).
3. Reduction of nitrate to hydrazine (B):
- In the starting nitrate ion (\( \text{NO}_3^- \)), the nitrogen is in the \( +5 \) state.
- In hydrazine (\( \text{N}_2\text{H}_4 \)), hydrogen has its standard oxidation state of \( +1 \).
Thus: \( 2N + 4(+1) = 0 \Rightarrow 2N = -4 \Rightarrow N = -2 \).
- Reducing a single nitrogen atom from \( +5 \) to \( -2 \) requires:
\[ (+5) - (-2) = 7 \text{ electrons} \] Thus, (B) matches with (IV).
4. Conversion of nitrogen to ammonia and hydrogen (A):
- In biological nitrogen fixation, dinitrogen gas (\( \text{N}_2 \)) is reduced to generate ammonia and hydrogen gas.
- The stoichiometric equation is:
\[ \text{N}_2 + 8\text{H}^+ + 8e^- \rightarrow 2\text{NH}_3 + \text{H}_2 \] - This complete reaction requires a total of 8 electrons.
Thus, (A) matches with (III).

Step 3: Final Answer:
The matching pairs are (A)-(III), (B)-(IV), (C)-(II), and (D)-(I), which corresponds to option (C).