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Fluorine decomposes cold water to give:
[A] $4{{H}^{+}}\text{+4}{{\text{F}}^{-}}and\text{ }{{\text{O}}_{2}}$
[B] $HF\text{ and }{{\text{H}}_{2}}$
[C] HF only
[D] ${{H}_{2}}{{F}_{2}}\text{ and HF}{{\text{O}}_{4}}$

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Last updated date: 22nd Jun 2024
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Answer
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Hint: Fluorine is a strong oxidising agent. It reacts readily with group1 and group 2 elements. Fluorine will react readily with water and form hydrofluoric acid along with other oxygen products.

Complete answer:
We know that fluorine is a halogen and lies in group 17. All the halogens are oxidising agents but all of them do not react with water. Fluorine and chlorine are smaller in size and have higher reducing potential. Therefore, they can react with water.
In the question, we are asked about the decomposition of cold water with fluorine.
Fluorine is a very strong oxidising agent. It accepts an electron readily. When we add water to fluorine, it oxidises water and forms ozone and oxygen and forms hydrofluoric acid with the hydrogen atom. Ozone is present in very small amounts and we consider it as an impurity in this reaction. We can write the reaction as-

     \[\begin{align}
  & {{F}_{2}}+{{H}_{2}}O\to 2HF+[O] \\
 & 3{{F}_{2}}+3{{H}_{2}}O\to 6HF+{{O}_{3}} \\
 & 2{{F}_{2}}+2{{H}_{2}}O\to 4HF+{{O}_{2}} \\
\end{align}\]

As we can see from the above reactions, whether ozone, oxygen or nascent oxygen is formed depends on the number of moles of fluorine and water in the reaction.
However, if we carry out this reaction in ice cold water hypofluorous acid is also formed but in a very small amount. Hypofluorous acid is very unstable, its formula is HOF, it disproportionates readily into hydrofluoric acid, fluorine or water. This reaction takes place at a very low temperature of about -40 degrees Celsius. We can write the reaction as-

     \[\begin{align}
  & {{F}_{2}}+{{H}_{2}}O\to 2{{H}^{+}}+2{{F}^{-}}+\left[ O \right] \\
 & Or,\text{ 2}{{F}_{2}}+2{{H}_{2}}O\to 4{{H}^{+}}+4{{F}^{-}}+{{O}_{2}} \\
\end{align}\]
Therefore, the correct answer is option [A] $4{{H}^{+}}\text{+4}{{\text{F}}^{-}}and\text{ }{{\text{O}}_{2}}$.
So, the correct answer is “Option A”.

Note: Fluorine along with other halogens is a strong oxidising agent. As we know that halogens lie in group 17, which means there are seven electrons in their outermost shell. Gaining one electron will help them gain noble gas configuration. Therefore, they accept an electron readily and attain the stable configuration.