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 changes, we will analyze each reaction step by step.

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

1. **Write the dissociation of reactants and products:**
- Sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) dissociates into:
\[ 2 \text{Na}^+ + \text{CO}_3^{2-} \]
- Sodium bicarbonate (\( \text{NaHCO}_3 \)) dissociates into:
\[ \text{Na}^+ + \text{HCO}_3^- \]

2. **Identify the change in oxidation states:**
- In \( \text{Na}_2\text{CO}_3 \), the carbonate ion (\( \text{CO}_3^{2-} \)) remains unchanged in terms of oxidation state.
- The sodium ions (\( \text{Na}^+ \)) remain unchanged as well.
- The change occurs in the carbonate ion converting to bicarbonate, where one hydrogen ion is added.

3. **Calculate n-factor:**
- The n-factor for \( \text{Na}_2\text{CO}_3 \) is 2 (from 2 Na ions).
- The n-factor for \( \text{NaHCO}_3 \) is 1 (from 1 Na ion).
- Therefore, the change in n-factor is:
\[ \text{n-factor change} = 1 - 2 = -1 \quad \text{(but we consider absolute values, so it is 1)} \]

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

1. **Write the dissociation of reactants and products:**
- Aluminum sulfate (\( \text{Al}_2(\text{SO}_4)_3 \)) dissociates into:
\[ 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
- Aluminum hydroxide (\( \text{Al(OH)}_3 \)) dissociates into:
\[ 2 \text{Al}^{3+} + 3 \text{OH}^- \]

2. **Identify the change in oxidation states:**
- The aluminum ions remain unchanged in oxidation state.
- The sulfate ions are converted to hydroxide ions.

3. **Calculate n-factor:**
- The n-factor for \( \text{Al}_2(\text{SO}_4)_3 \) is \( 2 \times 3 = 6 \) (from 3 sulfate ions).
- The n-factor for \( \text{Al(OH)}_3 \) is \( 3 \times 1 = 3 \) (from 3 hydroxide ions).
- Therefore, the change in n-factor is:
\[ \text{n-factor change} = 3 - 6 = -3 \quad \text{(but we consider absolute values, so it is 3)} \]

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

1. **Write the dissociation of reactants and products:**
- Sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)) dissociates into:
\[ 2 \text{Na}^+ + \text{S}_2\text{O}_3^{2-} \]
- Tetrathionate ion (\( \text{S}_4\text{O}_6^{2-} \)) is already in its ionic form.

2. **Identify the change in oxidation states:**
- For \( \text{S}_2\text{O}_3^{2-} \):
\[ 2X - 6 = -2 \implies 2X = 4 \implies X = +2 \]
- For \( \text{S}_4\text{O}_6^{2-} \):
\[ 4X - 12 = -2 \implies 4X = 10 \implies X = +2.5 \]

3. **Calculate n-factor:**
- The n-factor for \( \text{S}_2\text{O}_3^{2-} \) is \( 2 \times 2 = 4 \) (2 sulfur atoms).
- The n-factor for \( \text{S}_4\text{O}_6^{2-} \) is \( 4 \times 2.5 = 10 \) (4 sulfur atoms).
- Therefore, the change in n-factor is:
\[ \text{n-factor change} = 10 - 4 = 6 \]

### Summary of n-factors:
1. \( \text{Na}_2\text{CO}_3 \rightarrow \text{NaHCO}_3 \) : n-factor = 1
2. \( \text{Al}_2(\text{SO}_4)_3 \rightarrow \text{Al(OH)}_3 \) : n-factor = 3
3. \( \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{S}_4\text{O}_6^{2-} \) : n-factor = 6