Question

Why would fluorine have a positive oxidation state in HOF even though it is more electronegative?

Answer 1

Hint: We acquaint oxidation states with sorting out our pondering oxidation–decrease responses and electrochemical cells. At the point when we characterize oxidation states, we make a bunch of rules for distributing the electrons in a particle or particle to the individual atoms that make it up.

Complete answer:
We have to know that, an electronegativity, represented as $X$ , is the propensity of an atom to attract shared electrons to itself. An atom's electronegativity is influenced by the two its nuclear number and the distance at which its valence electrons dwell from the charged core. The higher the related electronegativity, the more a particle or a substituent bunch draws in electrons. Something contrary to electronegativity is electro-positivity: a proportion of a component's capacity to give valence electrons.
As fluorine is the most electronegative component in the intermittent table, it ought to be $ - 1$ . Fluorine is more electronegative than oxygen so it will procure $ - 1$ charge and $OH$ would gain $ + 1$ charge. In $O{H^ + }$ , oxygen happens in nonpartisan charge arrangement and in this manner, hydrogen gains $ + 1$ charge. $H - O - F$ the oxidation number of hydrogen, oxygen, and fluorine are $ + 1$ , $0$ and $ - 1$ individually. For fluorine to display any certain oxidation state it ought to be clung to a negative component much higher than it. Which is beyond the realm of imagination, along these lines fluorine doesn't show any sure oxidation state.

Note:
In natural science, electronegativity is connected more with unexpected utilitarian gatherings in comparison to with singular molecules. The terms bunch electronegativity and substituent electronegativity are utilized interchangeably. Notwithstanding, it is entirely expected to recognize the inductive impact and the reverberation impact.