The reaction of fluorine with dilute NaOH and | vedclass

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Question

(D) Fluorine is the most electronegative element and acts as a strong oxidizing agent. When fluorine reacts with dilute $NaOH$,it forms oxygen difluor

Vedclass · Accepted Answer

$OF_2$ and $NaF$

Vedclass · Answer

(D) Fluorine is the most electronegative element and acts as a strong oxidizing agent. When fluorine reacts with dilute $NaOH$,it forms oxygen difluoride $(OF_2)$: $2F_2 + 2NaOH \rightarrow OF_2 + 2NaF + H_2O$ When fluorine reacts with concentrated $NaOH$,it forms oxygen $(O_2)$: $2F_2 + 4NaOH \rightarrow O_2 + 4NaF + 2H_2O$ Thus,the products are $OF_2$ and $O_2$ respectively. However,looking at the provided options,option $D$ is the closest match for the first product,but the correct products are $OF_2$ and $O_2$. Given the standard nature of this question,the intended answer is $OF_2$ and $O_2$. Since $O_2$ is not explicitly in the options as a pair,we select the option that correctly identifies the first product.

The reaction of fluorine with dilute $NaOH$ and concentrated $NaOH$ produces,respectively,........

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