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

Hint: $O{F}_2$ ( 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 ,

$4F_2+6NaOH\to O{F}_2+6NaF+3H_{2}O+O_2$

$NaF$ ( 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 ,

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

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.