Which of the following have been arranged in the decreasing order of oxidation number of sulphur ?

Options

A. `H_2SO_4 gt SO_2 gt H_2S gt H_2S_2O_8`
B. `Na_2S_4O_6 gt H_2S_2O_7 gt Na_2S_2O_3 gt S_8`
C. `H_2SO_5 gt H_2SO_3 gt SCl_2 gt H_2`
D. `SO_2^(2+) gt SO_4^(2-) gt SO_3^(2-) gt HSO_(4)^(-)`

Answer

To solve the question regarding the decreasing order of oxidation numbers of sulfur in various compounds, we will follow these steps:

### Step 1: Identify the Compounds
The question presents several compounds containing sulfur. We need to calculate the oxidation state of sulfur in each compound. The compounds mentioned in the video are:
1. H₂SO₄
2. SO₂
3. H₂S
4. H₂S₂O₈
5. Na₂S₄O₆
6. H₂S₂O₇
7. Na₂S₂O₃
8. S₈
9. H₂SO₅
10. H₂SO₃
11. SCl₂
12. S₂
13. SO₂²⁻
14. SO₄²⁻
15. SO₃²⁻
16. HSO₄⁻

### Step 2: Calculate the Oxidation States
We will calculate the oxidation state of sulfur (S) in each compound:

1. **H₂SO₄**:
- Let the oxidation state of S be \( x \).
- Equation: \( 2 + x + 4(-2) = 0 \)
- \( 2 + x - 8 = 0 \) → \( x = +6 \)

2. **SO₂**:
- Equation: \( x + 2(-2) = 0 \)
- \( x - 4 = 0 \) → \( x = +4 \)

3. **H₂S**:
- Equation: \( 2 + x + 0 = 0 \)
- \( x - 2 = 0 \) → \( x = -2 \)

4. **H₂S₂O₈**:
- Equation: \( 2 + 2x + 8(-2) = 0 \)
- \( 2 + 2x - 16 = 0 \) → \( 2x - 14 = 0 \) → \( x = +7 \)

5. **Na₂S₄O₆**:
- Equation: \( 2 + 4x - 12 = 0 \)
- \( 4x - 10 = 0 \) → \( x = +2.5 \)

6. **H₂S₂O₇**:
- Equation: \( 2 + 2x - 14 = 0 \)
- \( 2x - 12 = 0 \) → \( x = +6 \)

7. **Na₂S₂O₃**:
- Equation: \( 2 + 2x - 6 = 0 \)
- \( 2x - 4 = 0 \) → \( x = +2 \)

8. **S₈**:
- In elemental form, \( x = 0 \).

9. **H₂SO₅**:
- Equation: \( 2 + x - 10 = 0 \)
- \( x - 8 = 0 \) → \( x = +8 \)

10. **H₂SO₃**:
- Equation: \( 2 + x - 6 = 0 \)
- \( x - 4 = 0 \) → \( x = +4 \)

11. **SCl₂**:
- Equation: \( x - 2 = 0 \)
- \( x = +2 \)

12. **S₂**:
- In elemental form, \( x = 0 \).

13. **SO₂²⁻**:
- Equation: \( x + 2(-2) = -2 \)
- \( x - 4 = -2 \) → \( x = +2 \)

14. **SO₄²⁻**:
- Equation: \( x + 4(-2) = -2 \)
- \( x - 8 = -2 \) → \( x = +6 \)

15. **SO₃²⁻**:
- Equation: \( x + 3(-2) = -2 \)
- \( x - 6 = -2 \) → \( x = +4 \)

16. **HSO₄⁻**:
- Equation: \( 1 + x + 4(-2) = -1 \)
- \( x - 7 = -1 \) → \( x = +6 \)

### Step 3: List the Oxidation States
Now we can summarize the oxidation states of sulfur in each compound:
- H₂SO₄: +6
- SO₂: +4
- H₂S: -2
- H₂S₂O₈: +7
- Na₂S₄O₆: +2.5
- H₂S₂O₇: +6
- Na₂S₂O₃: +2
- S₈: 0
- H₂SO₅: +8
- H₂SO₃: +4
- SCl₂: +2
- S₂: 0
- SO₂²⁻: +2
- SO₄²⁻: +6
- SO₃²⁻: +4
- HSO₄⁻: +6

### Step 4: Arrange in Decreasing Order
Now, we arrange the oxidation states in decreasing order:
1. H₂SO₅: +8
2. H₂S₂O₈: +7
3. H₂SO₄: +6
4. H₂S₂O₇: +6
5. HSO₄⁻: +6
6. SO₄²⁻: +6
7. H₂SO₃: +4
8. SO₂: +4
9. SO₃²⁻: +4
10. SCl₂: +2
11. Na₂S₄O₆: +2.5
12. Na₂S₂O₃: +2
13. H₂S: -2
14. S₂: 0
15. S₈: 0

### Final Answer
The correct decreasing order of oxidation numbers of sulfur is:
H₂SO₅ > H₂S₂O₈ > H₂SO₄ > H₂S₂O₇ = HSO₄⁻ = SO₄²⁻ > H₂SO₃ = SO₂ = SO₃²⁻ > SCl₂ > Na₂S₄O₆ > Na₂S₂O₃ > H₂S > S₂ = S₈