Why does fluorine not form a positive oxidation state?

Fluorine does not form a positive oxidation state because it has the highest electronegativity of all elements.

Fluorine, with an atomic number of 9, is the most electronegative element on the periodic table. Electronegativity is a measure of the ability of an atom to attract electrons towards itself. The higher the electronegativity, the greater the atom's attraction for bonding electrons. Fluorine's electronegativity value is 4.0 on the Pauling scale, the highest of all elements. This means that fluorine has a very strong tendency to attract electrons and form negative ions, rather than losing electrons and forming positive ions.

The electronic configuration of fluorine is 1s² 2s² 2p⁵. This means that it has seven electrons in its outermost shell and needs only one more electron to achieve a stable, full outer shell of eight electrons, following the octet rule. This makes fluorine highly reactive, as it will readily gain an electron to achieve this stable configuration. When fluorine gains an electron, it forms a negative ion (F-) and achieves an oxidation state of -1.

In contrast, for fluorine to form a positive oxidation state, it would need to lose electrons. However, this is energetically unfavourable. The energy required to remove an electron from fluorine is very high due to its high electronegativity and the stability of its full outer shell once it has gained an electron. Therefore, fluorine does not form a positive oxidation state.

Furthermore, the concept of oxidation states is a way of keeping track of electrons in chemical reactions. In this system, fluorine is always assigned an oxidation state of -1 in its compounds. This is because, in any compound, fluorine will always be the element that attracts electrons most strongly due to its high electronegativity. Therefore, it will always gain electrons and form a negative oxidation state, rather than losing electrons and forming a positive oxidation state.