The correct decreasing order of Vanderwaal constant 'a' for `H_(2) , CH_(4) , O_(2)` , gases is

Options

A. `H_(2) lt CH_(4) lt O_(2)`
B. `CH_(4) lt O_(2) lt H_(2)`
C. `CH_(4) gt O_(2) gt H_(2)`
D. `H_(2) gt CH_(4) gt O_(2)`

Explanation

To determine the correct decreasing order of the Van der Waals constant 'a' for the gases H₂, CH₄, and O₂, we can follow these steps:

### Step 1: Understand the Van der Waals constant 'a'
The Van der Waals constant 'a' is a measure of the attraction between particles in a gas. A higher 'a' value indicates stronger intermolecular forces.

### Step 2: Identify the types of intermolecular forces
- **Permanent Dipole**: Molecules with permanent dipoles have stronger attractions.
- **Instantaneous Dipole**: Non-polar molecules can have temporary dipoles, but this is generally weaker.
- **Surface Area**: Larger surface area allows for more significant interactions between molecules.

### Step 3: Analyze the given gases
1. **H₂ (Hydrogen)**:
- Diatomic, non-polar, very small surface area, and low molar mass.
- Weakest intermolecular forces.

2. **O₂ (Oxygen)**:
- Diatomic, non-polar, larger surface area than H₂, and higher molar mass than H₂.
- Weaker than CH₄ but stronger than H₂.

3. **CH₄ (Methane)**:
- Tetrahedral geometry, non-polar, but has a larger surface area compared to H₂ and O₂.
- Higher molar mass than both H₂ and O₂, leading to stronger intermolecular forces.

### Step 4: Compare the values of 'a'
- CH₄ has the highest value of 'a' due to its larger surface area and higher molar mass.
- O₂ comes next because it has a higher molar mass than H₂ and a larger surface area than H₂.
- H₂ has the lowest value of 'a' due to its very small size and low molar mass.

### Step 5: Write the decreasing order
The decreasing order of the Van der Waals constant 'a' for the gases is:
**CH₄ > O₂ > H₂**

### Final Answer
The correct decreasing order of Van der Waals constant 'a' for H₂, CH₄, and O₂ is:
**CH₄ > O₂ > H₂**

---