Question

Fluorine with dilute $NaOH$ gives
(A) $O{{F}_{2}}$
(B) ${{O}_{3}}$
(C) ${{O}_{2}}$
(D) $HF$ and ${{O}_{2}}$

Answer 1

Hint: In this question, we have to give the products by the reaction of fluorine with dilute NaOH. We know that sodium exists as a metal and fluorine exists as a non- metal. By analysing the properties of fluorine and sodium, we are able to choose the correct option.

Complete Step by Step Solution:
We are asked to give the product by the reaction of fluorine with dilute $NaOH$
Fluorine is an element which is in the 7th period of the periodic table, it is the first element of the group. As it is the most electronegative element, It means that fluorine reacts with anything around it. It is a powerful oxidizer and oxidation state is always -1. As we know, sodium exists as a metal with only one valence electron.

 Sodium needs to lose electrons and fluorine needs to lose electrons in order to form a stable chemical bond. Thus, sodium gives its valence electron to fluorine to form a cation, and fluorine accepts the electron to form an anion. Thus, they spread out the difference in the charge between them.

 Now, because of 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.
${{F}_{2}}+(dil)NaOH\to 2NaF+O{{F}_{2}}+{{H}_{2}}O$
Thus, Option (A) is correct.

Note: As to answer this type of questions, students must have the proper knowledge of the properties of sodium and fluorine. Only then will I be able to tick the correct answer. Students must revise the topics like electron affinity, ionisation energy etc so that there will be no confusion in their mind related to these topics.