Fluorine exhibits only -1 oxidation state whereas other halogens show +1, +3, +5 and +7 oxidation states also. Explain.

Explanation

Fluorine, like other halogens, has seven electrons in the valence shell and thus accepts one more electron to acquire the stable electronic configuration of the nearest noble gas. Therefore, like other halogens, it also exhibits and oxidation state of -1.
Further, since fluorine is the most electronegative element known, it cannot share its unpaired electron with an element more electronegative than itself and hence cannot show an oxidation state of +1 (except in HOF). However, other halogens (Cl, Br, I) can share their respective unpaired electron with more electronegative elements such as F and O and hence can show positive oxidation state of +1.
Further, fluorine does not have d-orbitals in its valence shell. Therefore, it cannot expand its valence shell and hence does not show higher positive oxidation states of +3, +5 and +7. On the other hand, other halogens have vacant d-orbitals and, therefore, can expand their octets by exciting one, two or even three electrons form ns and np orbitals to nd-orbitals. As a result, 3, 5 and 7 half-filled orbitals are made available for bond formation and hence, Cl, Br and I also show higher positive oxidation states of +3, +5 and +7. Thus, form the above discussion, it follows that fluorine shows only -1 oxidation state while other halogens show positive oxidation states of +1, +3, +5 and +7.