Question

When fluorine is reacted with cold dilute alkali, which of the following is obtained as a product ?
(A) ${O_{_2}}$
(B) ${O_2}{F_2}$
(C) $O{F_2}$
(D) ${H_2}$

Answer 1

$O{F_2}$ ( oxygen difluoride) is prepared by the reaction between fluorine and dilute aqueous solution of sodium hydroxide. It’s an oxygen halide.$NaOH$ , sodium hydroxide also called caustic soda, when it's hot and concentrated, reacts violently but when cold and dilute does not react violently.

Complete step by step answer:
The reaction between and cold and dilute $NaOH$ and Fluorine gas is given as ,

$4{F_2} + 6NaOH \to O{F_2} + 6NaF + 3{H_2}O + {O_2}$

$NaF$ ( sodium fluoride ) is formed as the byproduct .

Fluorine is available in two oxidation states , 0 and -1 i.e. ${F_2}$ and ${F^ - }$Here we are taking ${F_2}$ as the reactant.
This reaction is a disproportionation reaction, where fluorine is reduced to sodium fluoride as well as it is oxidized to oxygen difluoride.Such a reaction in which there is simultaneous oxidation and reduction of the same reactant is called a disproportionation reaction, a type of redox reaction.Now, when fluorine gas reacts with hot and concentrated sodium hydroxide ,

$2{F_2} + 4NaOH \to 4NaF + 2{H_2}O + {O_2}$

Oxygen is released without the formation of $O{F_2}$ gas .Also, ${O_2}{F_2}$ ( dioxygen difluoride) is formed when fluorine gas reacts with oxygen gas.

${O_2} + {F_2} \to {O_2}{F_2}$

Hydrogen gas is prepared in the laboratory by the action of dilute acid like hydrochloric acid or sulphuric acid on zinc metal.

$Zn + 2HCl \to ZnC{l_2} + {H_2} \uparrow$

$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2} \uparrow$

So, the correct answer is Option C .

Note: Oxygen is released during the decomposition of water

${H_2}O \to {H_2} + {O_2}$

And also during the reaction between fluorine gas and hot and concentrated sodium hydroxide.Oxygen difluoride, $O{F_2}$ is used as an oxidizer but is a hazardous, colorless gas.It’s a strong oxidizing agent.