For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data
I. `2HI(g) hArr H_(2)(g)+I_(2)(g)`
II. `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)`
III. `2NO(g)hArr N_(2)(g)+O_(2)(g)`
IV. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

Options

A. I and III
B. II and IV
C. I and II
D. III and IV

Explanation

To determine for which of the given reactions the degree of dissociation cannot be calculated from the vapor density data, we need to analyze each reaction in terms of the change in the number of moles (Δn) during the reaction. The degree of dissociation can be calculated if Δn is not equal to zero.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- I. `2HI(g) ⇌ H₂(g) + I₂(g)`
- II. `2NH₃(g) ⇌ N₂(g) + 3H₂(g)`
- III. `2NO(g) ⇌ N₂(g) + O₂(g)`
- IV. `PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)`

2. **Calculate Δn for Each Reaction**:
- **Reaction I**:
- Reactants: 2 moles of HI
- Products: 1 mole of H₂ + 1 mole of I₂ = 2 moles
- Δn = moles of products - moles of reactants = 2 - 2 = 0

- **Reaction II**:
- Reactants: 2 moles of NH₃
- Products: 1 mole of N₂ + 3 moles of H₂ = 4 moles
- Δn = 4 - 2 = 2

- **Reaction III**:
- Reactants: 2 moles of NO
- Products: 1 mole of N₂ + 1 mole of O₂ = 2 moles
- Δn = 2 - 2 = 0

- **Reaction IV**:
- Reactants: 1 mole of PCl₅
- Products: 1 mole of PCl₃ + 1 mole of Cl₂ = 2 moles
- Δn = 2 - 1 = 1

3. **Determine Which Reactions Cannot Calculate Degree of Dissociation**:
- From the calculations:
- Reaction I: Δn = 0 (cannot calculate)
- Reaction II: Δn = 2 (can calculate)
- Reaction III: Δn = 0 (cannot calculate)
- Reaction IV: Δn = 1 (can calculate)

4. **Conclusion**:
- The reactions for which the degree of dissociation cannot be calculated from the vapor density data are:
- I. `2HI(g) ⇌ H₂(g) + I₂(g)`
- III. `2NO(g) ⇌ N₂(g) + O₂(g)`

### Final Answer:
The degree of dissociation cannot be calculated from the vapor density data for reactions I and III.