Question

To purify fluorine gas, fumes of HF are removed by
A. Solid $NaF$
B. ${H_2}$ gas
C. Solid $KH{F_2}$
D. None of these

Answer 1

Hint: Fluoride is a non-metal, as we well know, whereas sodium is a metal. As a result, the bond between them can be ionic. Remembering that fluorine is very electronegative and sodium has one valence electron can help us find the solution to the question.

Complete Step by Step Solution:
Fluorine is purified using solid $NaF$ by passing $HF$ vapours over it. It is eliminated as $NaF{H_2}$.

Since sodium is a metal with only one valence electron, we already know this. Due to its small size, fluorine is a very electronegative element. The oxidation state is always -1 as a result.

In order to create a chemical connection, fluorine and sodium both need to shed electrons. As a result, sodium gives fluoride its valence electron, forming a cation, and fluorine absorbs it, generating an anion. They split the charge difference between them as a result. Now, fluoride does not react with sodium hydroxide to produce oxyacid due to its high electronegativity.

As a result, it combines with fluoride to produce water as a by-product as well as sodium fluoride, which is a good conductor of electricity in a molten condition.
The right response is therefore "Option A."

Note: Fluorine and cold, diluted sodium hydroxide react in a disproportion reaction. Here, fluorine is both oxidised to oxygen difluoride and reduced to sodium fluoride. This kind of redox reaction exists.