Q.Explain why fluorine forms only one oxoacid, HOF.

Step 1: How the heavier halogens form multiple oxoacids.

Cl, Br and I form a whole series of oxoacids (e.g. $HOCl, HClO_2, HClO_3, HClO_4$) because they can access vacant d-orbitals to expand their octet and adopt positive oxidation states ($+1$ to $+7$) when bonded to the more electronegative oxygen.

Step 2: Fluorine cannot access positive oxidation states.

Fluorine belongs to period 2 (n = 2), so it has no d-orbitals available in its valence shell, and cannot expand its octet. Moreover, fluorine is the most electronegative element of all — more electronegative than oxygen itself — so in any F–O bond, oxygen (not fluorine) is the more electronegative partner, meaning fluorine cannot be assigned a positive oxidation state.

Step 3: The one oxoacid fluorine does form.

Fluorine forms only one binary compound with oxygen and hydrogen, HOF (called fluoric(I) acid or hypofluorous acid). Even here, since F is more electronegative than O, fluorine is still assigned the $-1$ oxidation state and oxygen is assigned $0$ — so fluorine never actually reaches a positive oxidation state even in its one oxoacid.

Step 4: Conclusion. …

Contrast why Cl/Br/I can form a series of oxoacids (d-orbital octet expansion, positive oxidation states towards O) with why F structurally and electronically cannot (no d-orbit …

• Do not say fluorine "doesn't react with oxygen" — HOF does exist; the point is fluorine is restricted to just this ONE compound and even there is not in a positive oxidation state.
• The reasoning must include BOTH factors — absence of d-orbitals AND fluorine's electronegativity exceeding oxygen's — a one-factor explanation is incomplete. …