Match the entires of Column I with appropiate entires of column II and choose the correct option out of the four option (a) ,(b) ,(c ) (d) given at the end of each question
`{:(,"column I (Compound)" ,, "column II (Oxidation numbers of S atoms)"),((A) ,Na_(2)S_(2) , (p), +6),((B), Na_(2)S_(2)O_(3) , (q), -1-1),((C ) ,Na_(2)S(2)O_(7) , (r),+6+6),((D),H_(2)SO_(4) , (s),-2+4):}`

Options

A. A-r,B-p,C-s,D-q
B. A-p,B-s,c-q,D-r
C. A-q,B-s,c-r,D-p
D. A-s,B-r,c-p,D-q

Answer

To solve the problem of matching the compounds in Column I with their corresponding oxidation numbers of sulfur atoms in Column II, we will calculate the oxidation state of sulfur in each compound step by step.

### Step-by-Step Solution:

**Step 1: Calculate the oxidation number of sulfur in Na₂S₂ (Compound A)**

1. **Identify the oxidation state of sodium (Na)**: Sodium has an oxidation state of +1.
2. **Set up the equation**: For Na₂S₂, we have:
\[ 2(+1) + 2x = 0 \]
where \(x\) is the oxidation state of sulfur.
3. **Solve for x**:
\[ 2 + 2x = 0 \implies 2x = -2 \implies x = -1 \]
4. **Conclusion**: The oxidation number of sulfur in Na₂S₂ is -1.

**Step 2: Calculate the oxidation number of sulfur in Na₂S₂O₃ (Compound B)**

1. **Identify the oxidation states**: Sodium (Na) is +1, and oxygen (O) is -2.
2. **Set up the equation**: For Na₂S₂O₃, we have:
\[ 2(+1) + 2x + 3(-2) = 0 \]
3. **Solve for x**:
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]
4. **Conclusion**: The oxidation number of sulfur in Na₂S₂O₃ is +2.

**Step 3: Calculate the oxidation number of sulfur in Na₂S₂O₇ (Compound C)**

1. **Identify the oxidation states**: Sodium (Na) is +1, and oxygen (O) is -2.
2. **Set up the equation**: For Na₂S₂O₇, we have:
\[ 2(+1) + 2x + 7(-2) = 0 \]
3. **Solve for x**:
\[ 2 + 2x - 14 = 0 \implies 2x - 12 = 0 \implies 2x = 12 \implies x = +6 \]
4. **Conclusion**: The oxidation number of sulfur in Na₂S₂O₇ is +6.

**Step 4: Calculate the oxidation number of sulfur in H₂SO₄ (Compound D)**

1. **Identify the oxidation states**: Hydrogen (H) is +1, and oxygen (O) is -2.
2. **Set up the equation**: For H₂SO₄, we have:
\[ 2(+1) + x + 4(-2) = 0 \]
3. **Solve for x**:
\[ 2 + x - 8 = 0 \implies x - 6 = 0 \implies x = +6 \]
4. **Conclusion**: The oxidation number of sulfur in H₂SO₄ is +6.

### Summary of Results:
- A: Na₂S₂ → -1 (q)
- B: Na₂S₂O₃ → +2 (not listed)
- C: Na₂S₂O₇ → +6 (r)
- D: H₂SO₄ → +6 (p)

### Final Matching:
- A: (q) -1
- B: (not listed)
- C: (r) +6
- D: (p) +6