Why fluorine does not exhibit any positive oxidation state?

Answer

The electronic configuration of F is `1s^(2) 2s^(2) 2p_(x)^(2) 2p_(y)^(2) 2p_(z)^(1)`. It has one unpaired electron. Since F is the most electronegative element known, therefore, the possibility of sharing its unpaired electron with an elemet more electronegative than itself does not arise and hence ti cannot show an oxidation state of `+l`. Further, F does not have d-orbital in its valence shell. Therefore, it cannot expand its valence shell and hence does not show higher positive oxidation of +3, +5 and +7. Thus, in nutshell, F cannot show any positive oxidation states.