Match the hydrocarbons in column I with the boiling points given in column II.
`{:("Column I","Column II"),((i)." n-Pentane",(a)." 282.5 K"),((ii)."iso-Pentane",(b)." 309 K"),((iii)." neo-Pentane",(c)." 301 K"):}`

Explanation

To solve the problem of matching the hydrocarbons in Column I with their respective boiling points in Column II, we need to analyze the structure of each hydrocarbon and understand how branching affects boiling points.

### Step-by-Step Solution:

1. **Identify the Hydrocarbons**:
- (i) n-Pentane
- (ii) iso-Pentane
- (iii) neo-Pentane

2. **Understand the Structure**:
- **n-Pentane**: This is a straight-chain hydrocarbon with the formula \( C_5H_{12} \). It has no branches, which means it has a larger surface area.
- **iso-Pentane**: This is a branched-chain hydrocarbon with the formula \( C_5H_{12} \) but has one branch. Its structure is less linear than n-pentane.
- **neo-Pentane**: This is a highly branched hydrocarbon with the formula \( C_5H_{12} \) and has two branches. It has the least surface area among the three.

3. **Determine the Boiling Points**:
- The boiling point of hydrocarbons generally increases with increasing surface area. Thus, the order of boiling points will be influenced by the degree of branching:
- **n-Pentane** has the highest boiling point because it has the longest chain and the most surface area.
- **iso-Pentane** has a lower boiling point than n-pentane but higher than neo-pentane due to its single branch.
- **neo-Pentane** has the lowest boiling point because it is the most branched.

4. **Match the Hydrocarbons with Boiling Points**:
- n-Pentane: (b) 309 K
- iso-Pentane: (c) 301 K
- neo-Pentane: (a) 282.5 K

### Final Matching:
- (i) n-Pentane → (b) 309 K
- (ii) iso-Pentane → (c) 301 K
- (iii) neo-Pentane → (a) 282.5 K