The order of reactivity of halogens towards hydrogen is — Hydrogen Chemistry Question

💡 Solution & Explanation

# Reactivity of Halogens with Hydrogen **Step 1: Identify the reaction** Halogens react with hydrogen to form hydrogen halides: $$X_2 + H_2 \rightarrow 2HX$$ **Step 2: Determine the order based on electronegativity and bond strength** The reactivity depends on: - **Electronegativity** of the halogen (higher = more reactive) - **Bond dissociation energy** of $X-X$ (lower = easier to break, more reactive) | Halogen | Electronegativity | X-X Bond Energy | |---------|-------------------|-----------------| | F | 4.0 (highest) | 158 | | Cl | 3.0 | 243 | | Br | 2.8 | 193 | | I | 2.5 (lowest) | 151 | **Step 3: Apply the trend** - **Fluorine ($F_2$)** is most reactive: highest electronegativity pulls electron density, making the $F-F$ bond weaker relative to reactivity - **Chlorine ($Cl_2$)** is next: good balance of electronegativity and reasonable bond energy - **Bromine ($Br_2$)**: moderate reactivity - **Iodine ($I_2$)** is least reactive: lowest electronegativity and weakest driving force **Answer: The reactivity order is** $$F_2 > Cl_2 > Br_2 > I_2$$ Fluorine reacts explosively with $H_2$ even in darkness, while iodine requires heat and a catalyst.