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# Oxidation number of fluorine in ${{F}_{2}}O$ is:(a) +1(b) +2(c) -1(d) -2

Last updated date: 24th Jun 2024
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Hint: Oxidation number is the charge of the atom it appears when present in the combined state. The element which has more electronegativity will have a more negative charge and the atom with less electronegativity will have a positive oxidation number.

The oxidation number of an element may be defined as the charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms. The oxidation number is also known as the oxidation state.
The oxidation state of halogens is (Fluorine, Chlorine, Bromine, Iodine) is -1. Because they have 7 electrons in their outermost valence shell and they require only one electron to complete their shell. Therefore, they gain 1 electron and acquire an -1 oxidation state.
In compounds formed by the union of different elements, the more electronegative atom will have a negative oxidation state, and the less electronegative atom will have a positive oxidation state.
In ${{F}_{2}}O$ the fluorine atom is more electronegative than the oxygen atom, hence it will have an -1 oxidation state.
Oxygen being less electronegative atom will have a positive oxidation state.
The oxidation state of oxygen is:
2(-1) + x = 0
x =+2
Hence, the oxidation state of fluorine in ${{F}_{2}}O$ is -1 and that of oxygen is +2.

So, the correct answer is an option (c) -1

Note: You may get confused that the oxygen always has a -2 oxidation state and you would solve accordingly, then you would get the answer +1. So, you may get confused between option (a). electronegativity is the factor that must not be ignored.