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With cold and dilute sodium hydroxide fluorine reacts to give:
a.) $NaF\ and \ OF_{2}$
b.) $NaF + O_{3}$
c.) $O_{2}\ and\ O_{3}$
d.) $NaF + O_{2}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: As we already know, sodium is metal, and fluoride is non-metal. So, the bond between them may be an ionic bond. To obtain the answer to the given question, we must keep in mind that sodium has one valence electron and fluorine is highly electronegative.

Complete step by step answer:
- As we are already aware of the fact that sodium is a metal with only one valence electron.
- Fluorine being small in size, is a highly electronegative element. And thus, the oxidation state is always -1.
- Sodium needs to lose electrons and fluorine needs to lose electrons in order to form a chemical bond. Thus, sodium donates its valence electron to fluoride, forming a cation, and fluorine accepts the electron, forming an anion. Thus, they distribute the difference in the charge between them.
- Now, due to its high electronegativity, fluoride does not form oxyacid with sodium hydroxide.
- So, it reacts with fluoride to form oxygen fluoride, and sodium fluoride which is a good conductor of electricity in a molten state along with water as a by-product. So, the correct answer is “Option A”.

Note: Disproportion reaction takes place between fluorine and cold and dilute sodium hydroxide. Here, fluorine is reduced to sodium fluoride as well as it is oxidized to oxygen difluoride. This is a type of redox reaction.