[Sameer16092005]

Hello everyone
Request to help me understand
1- How fluorine has the oxidation number +1 in HOF compound, even though in all other compound it has oxidation number -1.
2. Metals have the capability to loose electron and form positive ions, but in HOF compound Fluorine has an oxidation number +1, while fluorine is a non metal, how does it loses electron and gain +ve charge ?

[AWK]

IUPAC 2016 definition:
The oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

Allen electronegativities should be used.

[Sameer16092005]

Hi
Request if someone can clarify my queries.

rgds
Sameer

[Enthalpy]

https://en.wikipedia.org/wiki/Electronegativity#Allen_electronegativity
4.193 F
3.610 O
2.300 H

The book "F Fluorine: Compounds with Oxygen and Nitrogen" on page 142 reports "polemics on the oxidation number of F in HOF".

[Borek]

How fluorine has the oxidation number +1 in HOF compound

Does it? I would assign -1.

Please note ON are just an approximation, an accounting device, apart from some highly ionic compounds they don't reflect the real nature of atoms in compounds. Many people think about ON as equivalent to "charge on atom" but in fact typically there is no such charge nor other physical property of the atom that can be measured and related to the ON. They do help counting electrons in redox reactions, that's all.

[MNIO]

Where on Earth did you get the idea that F was +1 in hypofluorous acid?

Oxidation state (number) is based entirely on electronegativities and assumes complete transfer of electrons.  It's not reality but an accounting term (as Borek mentioned) useful in balancing REDOX reactions. 
  EN H = 2.20
  EN O = 3.44
  EN F = 3.98

in HOF
  H = +1     (O with the higher EN gets an electron from H making H = +1)
  F = -1      (F with the higher EN takes an electron from O making F = -1
  O = 0       (O takes an electron and gives one away  or if you prefer,
                   overall HOF has 0 net charge   +1 (H) + x (O) - 1(F) = 0    x=0)