JEE CHEMISTRY MCQ: CHEMICAL BONDING & MOLECULAR STRUCTURE

Solution approach

1. Correct Answer + Core Reason: ${\mathrm{C}_2}{\mathrm{H}_6}$. Single bonds are longer than double or triple bonds.
   
   Step-by-Step Solution:
   Step 1: Identify the compounds: Ethane (${\mathrm{C}_2}{\mathrm{H}_6}$), Ethyne (${\mathrm{C}_2}{\mathrm{H}_2}$), Dihaloalkane (${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$), Ethene (${\mathrm{C}_2}{\mathrm{H}_4}$).
   Step 2: Determine the hybridization of carbon in each compound: Ethane ($s{p^3}$), Ethyne ($sp$), Dihaloalkane (${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$), Ethene ($s{p^2}$).
   Step 3: Relate bond length to hybridization: Higher s-character leads to shorter bond length. The order of s-character is $sp > s{p^2} > s{p^3}$.
   Step 4: Determine the order of C-H bond length: ${\mathrm{C}_2}{\mathrm{H}_6} > {\mathrm{C}_2}{\mathrm{H}_4} > {\mathrm{C}_2}{\mathrm{H}_2}$.
   
   Why other options are wrong:
   - Option B: ${\mathrm{C}_2}{\mathrm{H}_2}$ has a triple bond, which is shorter than a single bond.
   - Option C: ${\mathrm{C}_2}{\mathrm{H}_2}\mathrm{B}{\mathrm{r}_2}$ contains a triple bond, which is shorter than a single bond.
   - Option D: ${\mathrm{C}_2}{\mathrm{H}_4}$ has a double bond, which is shorter than a single bond.
   
   Key Concepts:
   - Hybridization
   - Bond length
   - s-character
   
   Memory Aids:
   - More s-character, shorter the bond.
   
   Quick Tips:
   - Remember the relationship between hybridization and bond length.