Question

The decreasing order of C--X bond length in \(CH_3-X\) is

• A. \(CH_3I > CH_3Br > CH_3Cl > CH_3F\)
• B. \(CH_3F > CH_3Cl > CH_3Br > CH_3I\)
• C. \(CH_3F > CH_3Cl > CH_3I > CH_3Br\)
• D. \(CH_3I > CH_3Cl > CH_3F > CH_3Br\)

Hint:

Bond length increases with increasing atomic size. Since iodine is the largest halogen, C--I bond is the longest among methyl halides.

Answer 1

The Correct Option is A

Concept: Bond length depends mainly upon the size of the bonded atoms. As atomic radius increases, bond length also increases. The halogen atomic radii follow the order: \[ F < Cl < Br < I \] Therefore, the carbon-halogen bond length increases as the size of halogen increases.

Step 1: Compare the sizes of halogens. The atomic sizes are: \[ F < Cl < Br < I \] Iodine is the largest halogen while fluorine is the smallest.

Step 2: Relate size to bond length. Larger halogen atoms form longer bonds with carbon. Hence: \[ C-I > C-Br > C-Cl > C-F \]

Step 3: Write the decreasing order. Thus, \[ CH_3I > CH_3Br > CH_3Cl > CH_3F \] Therefore the correct option is \[ \boxed{(a)} \]