Question

The oxidation state of oxygen in ${{\text{F}}_{\text{2}}}{\text{O}}$ is:
A. $ + 1$
B. $ - 1$
C. $ + 2$
D. $ - 2$

Answer 1

Hint: The number of electrons gained or lost by an atom in order to form a bond with another atom is known as the oxidation state of that particular atom.
In ${{\text{F}}_{\text{2}}}{\text{O}}$, the most electronegative atom is ${\text{F}}$. Thus, the charge on ${\text{F}}$ is negative. Also, ${{\text{F}}_{\text{2}}}{\text{O}}$ is a neutral compound. Thus, the net charge on the compound is $0$.

Complete step by step answer:
Determine the most electronegative atom as follows:
In ${{\text{F}}_{\text{2}}}{\text{O}}$, the most electronegative atom is ${\text{F}}$.
Thus, the charge on ${\text{F}}$ is negative. And the charge on ${\text{O}}$ is positive.
Step 2:
Determine the oxidation state of oxygen in ${{\text{F}}_{\text{2}}}{\text{O}}$ as follows:
${\text{F}}$ has one unpaired electron. Thus, the oxidation state of ${\text{F}}$ is $ - 1$.
The net charge on ${{\text{F}}_{\text{2}}}{\text{O}}$ is $0$.
Thus,
$\left( {2 \times {\text{Oxidation state of F}}} \right) + \left( {{\text{1}} \times {\text{Oxidation state of O}}} \right) = {\text{Net charge on }}{{\text{F}}_2}{\text{O}}$
$\left( {2 \times - 1} \right) + \left( {{\text{1}} \times {\text{Oxidation state of O}}} \right) = 0$
$ - 2 + \left( {{\text{1}} \times {\text{Oxidation state of O}}} \right) = 0$
 ${\text{Oxidation state of O}} = + 2$
Thus, the oxidation state of oxygen is ${{\text{F}}_{\text{2}}}{\text{O}}$ is $ + 2$.

So, the correct answer is Option A.

Note:
The atomic number of fluorine is $9$. Thus, fluorine has $9$ electrons.
The electronic configuration of fluorine is $1{s^2}\,2{s^2}\,2{p^5}$.
Thus, fluorine has only one unpaired electron.
The element is said to complete its octet when it has eight electrons in its valence shell. Fluorine has five electrons in its valence ${\text{2p}}$ shell.
Fluorine requires three more electrons to complete its octet. But only one electron can be filled in the valence ${\text{2p}}$ orbital. The octet cannot be expanded as there is no $d$ orbital. Thus, fluorine cannot have a positive oxidation state.