Q.Explain why fluorine always exhibit an oxidation state of -1?

Step 1. Note fluorine's electron configuration.

Fluorine is 1s2 2s2 2p5 — one electron short of the stable, fully-filled noble-gas configuration of neon.

Step 2. Note fluorine's small size and highest electronegativity.

Fluorine is the smallest halogen and the most electronegative element in the entire periodic table, so in any bond it forms, it always attracts the shared electron pair(s) towards itself, effectively behaving as though it has gained an electron (oxidation state -1) rather than shared one on equal terms.

Step 3. Note fluorine's lack of d orbitals.

Because fluorine is a period-2 element, it has no accessible d orbitals in its valence shell. This means, unlike chlorine, bromine and iodine (which can use empty d orbitals to expand their octet and form species like ClF3, BrF5 or IF7, formally showing positive oxidation states such as +3, +5, +7), fluorine can never expand beyond a single covalent bond, and can never itself be forced into a positive oxidation state by an even more electronegative partner, since no element is more electronegative than fluorine.

Step 4. Combine to explain the fixed -1 state.

Taken together — highest electronegativity (it never loses electron density to any partner) and no valence d orbitals (it can never share more than one electron pair or reach a higher covalency) — fluorine is restricted to gaining exactly one electron to complete its octet, fixing its oxidation state permanently at -1 in every one of its compounds.

Combine three facts about fluorine — its near-complete octet (2p5), its position as the most electronegative element, and its complete lack of valence d orbitals — to show why -1 is its only possible oxidation state.

Citing only 'fluorine is very electronegative' without also explaining the second, equally necessary reason — the absence of d orbitals — which is what actually prevents fluorine from ever showing the positive oxidation states seen for the other halogens.