Chemistry Hydrocarbons NUMS 2019
PMDC Verified Question 27 of 65
Order of reactivity of halogen toward alkane is
A
\( \text{F}_2 > \text{I}_2 > \text{Br}_2 > \text{Cl}_2 \)
B
\( \text{F}_2 > \text{Br}_2 > \text{Cl}_2 > \text{I}_2 \)
C
\( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \)
D
\( \text{F}_2 > \text{Cl}_2 > \text{I}_2 > \text{Br}_2 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \)
Concept:

The reactivity of halogens in free-radical substitution of alkanes is dictated by the electronegativity of the halogen, the weakness of the X-X bond, and the high exothermicity of forming the H-X bond.

Solution:

  • Fluorine is incredibly reactive due to a weak F-F bond and the highly exothermic nature of the reaction. It often reacts explosively.


  • Chlorine reacts rapidly under UV light.


  • Bromine is less reactive and highly selective, reacting slower.


  • Iodine is the least reactive; its reaction is endothermic, slow, and highly reversible.


  • Therefore, the strictly descending order of reactivity follows the periodic group: \( \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2 \).


Why other options are incorrect:

  • Any other sequence breaks the periodic trend correlating reactivity with electronegativity and bond energies of the halogens.

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