Give the n factors for the following changes -
(i) `Na_(2) CO_(3) rarr NaHCO_(3)`
(ii) `Al_(2) (SO_(4))_(3) rarr Al(OH)_(3)`
(ii) `Na_(2) S_(2) O_(3) rarr S_(4) O_(6)^(-2)`

Options

A. 1,3,2
B. 1,3,1
C. 1,6,1
D. 2,6,1

Answer

To find the n factors for the given chemical changes, we will analyze each reaction step by step.

### (i) \( \text{Na}_2\text{CO}_3 \rightarrow \text{NaHCO}_3 \)

1. **Identify the oxidation states**:
- In \( \text{Na}_2\text{CO}_3 \):
- Sodium (Na) has an oxidation state of +1 (2 Na = +2).
- Carbon (C) in carbonate (CO₃²⁻) has an oxidation state of +4.
- Oxygen (O) has an oxidation state of -2 (3 O = -6).
- Overall charge = +2 + 4 - 6 = 0.
- In \( \text{NaHCO}_3 \):
- Sodium (Na) = +1.
- Hydrogen (H) = +1.
- Carbon (C) = +4.
- Oxygen (O) = -2 (3 O = -6).
- Overall charge = +1 + 1 + 4 - 6 = 0.

2. **Calculate the change in oxidation state**:
- Sodium remains +1 in both compounds.
- The only change occurs in the hydrogen ion, which is added to the carbonate ion.
- Therefore, the n factor is determined by the change in the number of moles of Na⁺ ions, which remains the same, but we consider the addition of H⁺.

3. **Conclusion**:
- The n factor for this reaction is **1** (due to the addition of one H⁺ ion).

### (ii) \( \text{Al}_2(\text{SO}_4)_3 \rightarrow \text{Al(OH)}_3 \)

1. **Identify the oxidation states**:
- In \( \text{Al}_2(\text{SO}_4)_3 \):
- Aluminum (Al) has an oxidation state of +3 (2 Al = +6).
- Sulfate (SO₄²⁻) has an oxidation state of -2 (3 SO₄ = -6).
- Overall charge = +6 - 6 = 0.
- In \( \text{Al(OH)}_3 \):
- Aluminum (Al) = +3.
- Hydroxide (OH) = -1 (3 OH = -3).
- Overall charge = +3 - 3 = 0.

2. **Calculate the change in oxidation state**:
- The oxidation state of Al remains +3 in both compounds.
- There is no change in the oxidation state of Al.
- The n factor is determined by the number of hydroxide ions produced, which is 3.

3. **Conclusion**:
- The n factor for this reaction is **3** (due to the formation of three hydroxide ions).

### (iii) \( \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{S}_4\text{O}_6^{2-} \)

1. **Identify the oxidation states**:
- In \( \text{Na}_2\text{S}_2\text{O}_3 \):
- Sodium (Na) = +1 (2 Na = +2).
- Sulfur (S) = +2 (2 S = +4).
- Oxygen (O) = -2 (3 O = -6).
- Overall charge = +2 + 4 - 6 = 0.
- In \( \text{S}_4\text{O}_6^{2-} \):
- Sulfur (S) = +2.5 (4 S = +10).
- Oxygen (O) = -2 (6 O = -12).
- Overall charge = +10 - 12 = -2.

2. **Calculate the change in oxidation state**:
- The change in oxidation state for sulfur is from +2 to +2.5, which is a change of +0.5.
- There are 2 sulfur atoms undergoing this change.

3. **Conclusion**:
- The n factor for this reaction is calculated as:
\[ \text{n factor} = \text{Change in oxidation state} \times \text{Number of atoms undergoing change} = 0.5 \times 2 = 1. \]

### Final Answers:
- (i) n factor = **1**
- (ii) n factor = **3**
- (iii) n factor = **1**