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Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?

  • a)
    Bond dissociation energy

  • b)
    Ionization enthalpy

  • c)
    Electron affinity

  • d)
    Hydration enthalpy 

Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Which among the following factors is the most important in making fluo...
The fluorine has low dissociation energy of F - F bond and reaction of atomic fluorine is exothermic in nature
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Which among the following factors is the most important in making fluo...
Bond dissociation energy
The bond dissociation energy is the energy required to break a bond in a molecule. In the case of fluorine, the bond dissociation energy is very high due to the small size of the fluorine atom and the strong electronegativity of fluorine. This means that the fluorine atom has a strong hold on its valence electrons and is less likely to share them with other atoms. As a result, fluorine is able to easily gain electrons from other atoms, making it a strong oxidizing agent.

Ionization enthalpy
Ionization enthalpy is the energy required to remove an electron from an atom or ion. In the case of fluorine, the ionization enthalpy is also very high due to the strong hold of fluorine on its valence electrons. This high ionization enthalpy means that it is difficult to remove an electron from a fluorine atom, making it less likely to accept additional electrons and therefore less likely to act as a strong oxidizing agent.

Electron affinity
Electron affinity is the energy change that occurs when an atom gains an electron to form an anion. Fluorine has a high electron affinity, meaning that it has a strong attraction for additional electrons. This high electron affinity allows fluorine to easily accept electrons from other atoms, making it a strong oxidizing agent.

Hydration enthalpy
Hydration enthalpy is the energy change that occurs when an ion is hydrated, or surrounded by water molecules. In the case of fluorine, the hydration enthalpy is very high due to the small size of the fluorine ion and its strong charge density. This high hydration enthalpy means that the fluorine ion is strongly attracted to water molecules, making it less likely to attract electrons from other atoms and therefore less likely to act as a strong oxidizing agent.

Among these factors, the hydration enthalpy is the most important in making fluorine the strongest oxidizing halogen. The high hydration enthalpy of fluorine makes it less likely to attract electrons from other atoms, reducing its oxidizing power.
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Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?a)Bond dissociation energyb)Ionization enthalpyc)Electron affinityd)Hydration enthalpyCorrect answer is option 'D'. Can you explain this answer?
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Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?a)Bond dissociation energyb)Ionization enthalpyc)Electron affinityd)Hydration enthalpyCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?a)Bond dissociation energyb)Ionization enthalpyc)Electron affinityd)Hydration enthalpyCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ?a)Bond dissociation energyb)Ionization enthalpyc)Electron affinityd)Hydration enthalpyCorrect answer is option 'D'. Can you explain this answer?.
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