For which of the following gases should the correction for the molecular volume be largest :
`CO,CO_(2),NH_(3) " or " SF_(6)` ?

Options

A. CO
B. `CO_(2)`
C. `NH_(3)`
D. `SF_(6)`

Explanation

To determine which gas requires the largest correction for molecular volume, we can analyze the gases provided: CO, CO₂, NH₃, and SF₆, using the Van der Waals equation.

### Step-by-Step Solution:

1. **Understand the Van der Waals Equation**:
The Van der Waals equation is given by:
\[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \]
Here, \(a\) accounts for intermolecular forces, and \(b\) accounts for the volume occupied by the gas molecules themselves.

2. **Identify the Role of \(b\)**:
The parameter \(b\) in the Van der Waals equation represents the volume correction for the finite size of the molecules. The larger the size of the molecules, the larger the value of \(b\).

3. **Compare the Molecular Sizes**:
- **CO (Carbon Monoxide)**: Composed of one carbon and one oxygen atom.
- **CO₂ (Carbon Dioxide)**: Composed of one carbon and two oxygen atoms.
- **NH₃ (Ammonia)**: Composed of one nitrogen and three hydrogen atoms.
- **SF₆ (Sulfur Hexafluoride)**: Composed of one sulfur atom and six fluorine atoms.

4. **Determine the Largest Molecular Size**:
Among the given gases, SF₆ has the largest molecular size due to the presence of sulfur (which is a larger atom) and six fluorine atoms, making it a larger molecule overall compared to CO, CO₂, and NH₃.

5. **Conclusion**:
Since the value of \(b\) is directly proportional to the size of the molecules, the gas with the largest molecular volume correction will be SF₆. Therefore, the correction for molecular volume will be largest for SF₆.

### Final Answer:
The gas for which the correction for the molecular volume should be largest is **SF₆**.

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