See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

**Analysis of C-Cl Bond Length in Three Halides** **Step 1: Identify the hybridization of carbon in each compound** - (A) $CH_3-CH_2-Cl$: Carbon is $sp^3$ hybridized (saturated) - (B) $CH_2=CH-Cl$: Carbon is $sp^2$ hybridized (alkene) - (C) $HC\equiv C-Cl$: Carbon is $sp$ hybridized (alkyne) **Step 2: Recall the relationship between hybridization and bond length** As $s$-character increases in hybridization ($sp^3 < sp^2 < sp$), the orbitals become more compact and pull electrons closer to the nucleus. This increases orbital overlap and **decreases bond length**. $$\text{Bond length order: } sp < sp^2 < sp^3$$ **Step 3: Apply to C-Cl bonds** - (C) $sp$ C-Cl: **Shortest** (highest s-character = 50%) - (B) $sp^2$ C-Cl: **Medium** (s-character = 33%) - (A) $sp^3$ C-Cl: **Longest** (s-character = 25%) **Step 4: Write the decreasing order** $$\text{Bond length: } (B) > (C) > (A)$$ **Answer: (C)** is correct because as s-character increases, the C-Cl bond becomes shorter due to better orbital overlap and stronger attraction.