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When fluorine is reacted with cold dilute alkali, which of the following is obtained as a product ?
(A) O2
(B) O2F2
(C) OF2
(D) H2

Answer
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Hint: OF2 ( 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 ,

4F2+6NaOHOF2+6NaF+3H2O+O2

NaF ( sodium fluoride ) is formed as the byproduct .

Fluorine is available in two oxidation states , 0 and -1 i.e. F2 and FHere we are taking F2 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 ,

2F2+4NaOH4NaF+2H2O+O2

Oxygen is released without the formation of OF2 gas .Also, O2F2 ( dioxygen difluoride) is formed when fluorine gas reacts with oxygen gas.

O2+F2O2F2

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

Zn+2HClZnCl2+H2

Zn+H2SO4ZnSO4+H2

So, the correct answer is Option C .

Note: Oxygen is released during the decomposition of water

H2OH2+O2

And also during the reaction between fluorine gas and hot and concentrated sodium hydroxide.Oxygen difluoride, OF2 is used as an oxidizer but is a hazardous, colorless gas.It’s a strong oxidizing agent.

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