Question

Which one of the following oxyacids of fluorine exists?
A.${\text{HOF}}$
B.${\text{HF}}{{\text{O}}_{\text{3}}}$
C.${\text{HF}}{{\text{O}}_{\text{4}}}$
D.${\text{HF}}{{\text{O}}_{\text{2}}}$

Answer 1

Oxyacids are acids which contain oxygen. The halogens can form four series of oxyacids. These are hypohalous acids, halous acids, halic acids and perhalic acids. In hypohalous acid, the halogen is in +1 oxidation state, while in halous acids, halic acids and perhalic acids, the halogen is in +3, +5 and +7 oxidation states respectively. Fluorine has only one positive oxidation state which is +1.

Complete step by step answer:
-Fluorine has very high electronegativity and is smaller in size than the other members of the halogen family. It differs from other orbitals in the aspect that it does not possess d – orbitals. All other halogens have vacant d – orbitals in their respective valence shells due to which one, two and three electrons from np and ns orbitals can be excited into the vacant d orbitals. Thus, 3, 5 or 7 half – filled orbitals are made available for bond formation. So, apart from +1 oxidation state, chlorine, bromine and iodine also show +3, +5 and +7 positive oxidation states. Thus, due to the absence of vacant d – orbitals, fluorine cannot show higher positive oxidation states of +3, +5 and +7 and shows only +1 positive oxidation state.
-In ${\text{HOF}}$, fluorine is in +1 oxidation state which is acceptable and hence ${\text{HOF}}$ exists. It is called hypofluorous acid and it is the only oxyacid of fluorine which exists. So, option A is correct.
-In ${\text{HF}}{{\text{O}}_{\text{3}}}$ , fluorine is in +5 oxidation state which is not acceptable and hence ${\text{HF}}{{\text{O}}_{\text{3}}}$ does not exist. In ${\text{HF}}{{\text{O}}_{\text{4}}}$ , fluorine is in +7 oxidation state which is not acceptable and hence it does not exist. In ${\text{HF}}{{\text{O}}_{\text{2}}}$ , fluorine is in +3 oxidation state which is not acceptable and hence it does not exist.

So, option A is correct.

Note:
The oxygen atom in ${\text{HOF}}$ is $sp^{3}$ hybridized like that in ${\text{HOCl}}$ . But due to the higher electronegativity of fluorine as compared to chlorine, the bond pair-bond pair repulsions in ${\text{HOF}}$ are weaker than in ${\text{HOCl}}$ . So, the ${\text{HOF}}$ bond angle is slightly smaller than the ${\text{HOCl}}$ bond angle. Also, the ${\text{O - Cl}}$ bond length is longer than the ${\text{O - F}}$ bond length due to the bigger size of chlorine over fluorine.