JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 14)
The correct order of bond dissociation enthalpy of halogens is :
F2 > Cl2 > Br2 > I2
Cl2 > Br2 > F2 > I2
I2 > Br2 > Cl2 > F2
Cl2 > F2 > Br2 > I2
Explanation
Among halogens (F2, Cl2, Br2 and I2), bond dissociation enthalpy ($$\Delta$$dissH$$^\circ$$) of I2, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.
Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F2 molecule. As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.
Thus, the order of $$\Delta$$dissH$$^\circ$$ (in kJ mol$$-$$1) is
$$\mathop {Cl - Cl}\limits_{(242.6)} > \mathop {Br - Br}\limits_{(192.3)} > \mathop {F - F}\limits_{(158.8)\,Electrostatic\,repulsion} > \mathop {I - I}\limits_{(151.1)\,Steric\,repulsion} $$
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