Which of the following statements is incorrect about fluorine (F)?

Fluorine (F) is the most electronegative element in the periodic table and exhibits several unique properties that differentiate it from other halogens. Understanding its chemical behavior is important in inorganic chemistry.

Oxidation State:
Fluorine is highly electronegative and always exhibits a –1 oxidation state in its compounds. It cannot attain positive oxidation states because of its strong tendency to attract electrons, making it highly reactive and a strong oxidizing agent.

Octet Rule:
Being a very small atom, fluorine cannot expand its valence shell beyond eight electrons. Unlike heavier halogens such as chlorine, bromine, or iodine, fluorine cannot follow the extended octet rule. Therefore, any statement suggesting that fluorine can have more than eight valence electrons in compounds is incorrect. Its chemical compounds strictly satisfy the octet rule, limiting fluorine to one unpaired electron in its ground-state configuration that participates in bonding.

Bond Dissociation Energy:
Fluorine has the strongest F–F bond among halogens, not the weakest. Its small atomic size and high electronegativity result in a very strong bond. Therefore, statement C claiming the lowest dissociation energy is incorrect. This high bond energy also explains why fluorine reacts explosively with many substances despite being diatomic.

Summary:
Fluorine always has a –1 oxidation state, strictly follows the octet rule, and possesses the highest bond dissociation energy among halogens. Statements suggesting that it follows the extended octet or has the lowest dissociation energy are incorrect.

In conclusion, the incorrect statements about fluorine are both B and C, making option 4 the correct answer. This highlights fluorine’s unique chemistry, including its high electronegativity, strict octet adherence, and exceptionally strong F–F bond, which are critical for predicting its reactivity and compound formation