(i) Occurrence of fluorine : Fluorine does not occur free in nature but occurs mostly as fluorspar CaF₂ , cryolite, Na₃AlF₆ and fluorapatite, CaF₂.3Ca₃(PO₄)₂. Traces of fluoride occur in sea water, bones, teeth, blood, milk etc.

(ii) Difficulties encountered during its isolation : (a) F₂ attacks all the materials of the apparatus such as glass, platinum, carbon and other metals, (b) F₂ is the strongest oxidising agent and hence no oxidising agent can oxidise F^– ions to F₂ . (c) F₂ cannot be prepared even by electrolysis of an aqueous solution of HF because F₂ formed reacts violently with water. If also cannot be prepared by electrolysis of anhydrous HF because it is not only poisonous, corrosive and volatile but also is a bad conductor of electricity.

(iii) Preparation: F₂ is now prepared by electrolysis of a solution of KHF₂ (1 part) in anyhydrous HF (5 parts) in a vessel (modern method) made of Ni — Cu alloy or Ni — Cu — Fe alloy called the monel metal using carbon electrodes. During the electrolysis following reactions occur.

KHF₂ ————> KF + HF; KF ———> K⁺ + F^–;

At cathode : K⁺ + e^– ———> K; 2K + 2HF ———> 2 KF + H₂

At anode : F^– ————> F + e^– ; F + F ———> F₂

(iv) Properties: It is the most reactive of all the halogens. It Combines with metals as well as non-metals to form fluorides. It decomposes water forming O₂ and O₃ and reacts vigorously with hydrogen of hydrocarbons leaving behind fluorinated hydrocarbons.

(HF being a volatile liquid fumes in air)

It is a strong oxidizing agent and oxidizes KClO₃ to KClO₄, KIO₃ to KIO₄ and bisulfates to peroxy sulfates.

KCIO₃ + F₂ + H₂O ———> KCIO₄ + H₂F₂

2NaHSO₄ + F₂ ———> Na₂S₂O₆ + 2HF

It reacts with to form nitrogen and with H₂S forming SF₆.

2NH₃ + 3F₂ ——> N₂ + 6 HF (oxidation reaction)

H₂S + 4F₂ ——> SF₆ + 2 HF

Fluorine reacts with cold and dilute sodium hydroxide solution to give oxygen difluoride (OF₂)

2F₂ + 2NaOH (cold, dil) ——> 2NaF + H₂O + OF₂

However, with hot and concentrated sodium hydroxide solution it gives oxygen

2F₂ + 4NaOH (hot, conc.) ——> 4NaF + 2H₂O + O₂

Since F₂ is the strongest oxidizing agent, it is always reduced and hence does not show disproportionation reactions while others halogens do.

F₂ oxidizes all other halide ions to the corresponding halogens (F₂ + 2 X^– ———> 2F^– + X₂) ; (X = Cl, Br or I)