Question

For the compounds $CH_3Cl, \, CH_3Br,\,CH_3I$ and $CH_3F,$ the correct order of increasing $C$-halogen bond length is :

• A. $CH_3F < CH_3Cl < CH_3C < CH_3I$
• B. $CH_3F < CH_3Br < CH_3Cl < CH_3I$
• C. $CH_3F < CH_3Cl < CH_3Br < CH_3I$
• D. $CH_3Cl < CH_3Br < CH_3F < CH_3I$

Answer 1

The Correct Option is C

The question asks for the order of increasing bond length for C-halogen bonds in the molecules \(CH_3Cl, \, CH_3Br,\,CH_3I\), and \(CH_3F\). To solve this, we need to understand the relationship between bond length and the atomic size of halogens. 

1. Bond length is typically influenced by the size of the atoms involved. As the atomic size of the halogen increases, the bond length increases. Halogens are a group in the periodic table, ordered by increasing atomic number and size as: F < Cl < Br < I.
2. In the molecules given, the C-halogen bond length will follow the order of the atomic size of the halogen:
   • \(CF\) bond will be the shortest because fluorine is the smallest halogen.
   • \(CCl\) bond will be longer than \(CF\) because chlorine is larger than fluorine.
   • \(CBr\) bond will be longer than \(CCl\) since bromine is larger than chlorine.
   • \(CI\) bond will be the longest as iodine is the largest of the halogens listed.
3. This reasoning leads to the order of increasing C-halogen bond length as: \(CH_3F < CH_3Cl < CH_3Br < CH_3I\).

This order matches the given correct option: \(CH_3F < CH_3Cl < CH_3Br < CH_3I\).