Question
Question: The oxidation state of oxygen in \({{\text{F}}_{\text{2}}}{\text{O}}\) is: A. \( + 1\) B. \( - 1...
The oxidation state of oxygen in F2O is:
A. +1
B. −1
C. +2
D. −2
Solution
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 F2O, the most electronegative atom is F. Thus, the charge on F is negative. Also, F2O 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 F2O, the most electronegative atom is F.
Thus, the charge on F is negative. And the charge on O is positive.
Step 2:
Determine the oxidation state of oxygen in F2O as follows:
F has one unpaired electron. Thus, the oxidation state of F is −1.
The net charge on F2O is 0.
Thus,
(2×Oxidation state of F)+(1×Oxidation state of O)=Net charge on F2O
(2×−1)+(1×Oxidation state of O)=0
−2+(1×Oxidation state of O)=0
Oxidation state of O=+2
Thus, the oxidation state of oxygen is F2O 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 1s22s22p5.
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 2p shell.
Fluorine requires three more electrons to complete its octet. But only one electron can be filled in the valence 2p orbital. The octet cannot be expanded as there is no d orbital. Thus, fluorine cannot have a positive oxidation state.