The correct order of bond dissociation enthalpy of halogens is :
Solution
<p>Among halogens (F<sub>2</sub>, Cl<sub>2</sub>, Br<sub>2</sub> and I<sub>2</sub>), bond dissociation enthalpy ($\Delta$<sub>diss</sub>H$^\circ$) of I<sub>2</sub>, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.</p>
<p>Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F<sub>2</sub> molecule. As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.</p>
<p>Thus, the order of $\Delta$<sub>diss</sub>H$^\circ$ (in kJ mol<sup>$-$1</sup>) is</p>
<p>$$\mathop {Cl - Cl}\limits_{(242.6)} > \mathop {Br - Br}\limits_{(192.3)} > \mathop {F - F}\limits_{(158.8)\,Electrostatic\,repulsion} > \mathop {I - I}\limits_{(151.1)\,Steric\,repulsion} $$</p>
About this question
Subject: Chemistry · Chapter: Periodic Table and Periodicity · Topic: Periodic Trends
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