Question

The oxidation number of Oxygen in \[O{F_2}\] is :
A. +2
B. -2
C. +1
D. -1

Answer 1

Hint:
1. Total Oxidation number of any atom is always equal to zero, i.e. the sum of the oxidation state of all the elements present in any atom is equal to zero.
2. In \[O{F_2}\] there is one Oxygen molecule and two fluorine molecules.
3. Oxidation state of Fluorine is always -1.
4. In spite of being Oxygen as an electronegative element, here, the oxidation number of Oxygen in \[O{F_2}\] will be positive, as Fluorine is more electronegative than Oxygen.

Complete answer:
As we have,\[O{F_2}\] means, Oxygen difluoride,
Here,
As we know that,
Fluorine is more electronegative than Oxygen,
As, one Fluorine atom have always -1 Oxidation state,
So, due to 2 molecules of Fluorine,
Total oxidation number of Difluoride will be \[ = (2 \times - 1) = - 2\]
We Know that,
Total oxidation state of the element is always zero.
Let the oxidation number of Oxygen in \[O{F_2}\]is “x”.
So,
As per question,
Total Oxidation state of \[O{F_2}\]will be equal to zero.
Hence,
\[
  (x + (2 \times - 1)) = 0 \\
\Rightarrow x - 2 = 0 \\
\Rightarrow x = 0 - ( - 2) \\
\Rightarrow x = 2
 \]

Hence, in \[O{F_2}\] the oxidation number of Oxygen will be equal to +2.

 Additional Information:
1. The value of electronegativity of Oxygen on Pauling scale is 3.5.
2. The value of electronegativity of Fluorine on Pauling scale is 4.0
3. Fluorine always behaves as an electronegative element, except when it is used in the form of fluorine gas. Hence, fluorine in its gaseous form has a zero oxidation state.

Note:
1. Oxidation state of Fluorine is -1,
2. As we know that, Oxygen is electronegative element, but in \[O{F_2}\] the other element i.e. Fluorine is more electronegative than the Oxygen, So, in this case the oxidation number of Oxygen will be positive.