Medium MCQ +4 / -1 PYQ · JEE Mains 2021

The correct order of bond dissociation enthalpy of halogens is :

  1. A F<sub>2</sub> &gt; Cl<sub>2</sub> &gt; Br<sub>2</sub> &gt; I<sub>2</sub>
  2. B Cl<sub>2</sub> &gt; Br<sub>2</sub> &gt; F<sub>2</sub> &gt; I<sub>2</sub> Correct answer
  3. C I<sub>2</sub> &gt; Br<sub>2</sub> &gt; Cl<sub>2</sub> &gt; F<sub>2</sub>
  4. D Cl<sub>2</sub> &gt; F<sub>2</sub> &gt; Br<sub>2</sub> &gt; I<sub>2</sub>

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

This question is part of PrepWiser's free JEE Main question bank. 131 more solved questions on Periodic Table and Periodicity are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →