Question

In which of the following compounds oxygen has highest oxidation state and in which it has lowest oxidation state? $OF _{2}, H _{2} O _{2}, KO _{2}, O _{2} F _{2}$

• A. Highest $= KO _{2}$, lowest $= H _{2} O _{2}$
• B. Highest $= OF _{2}$, lowest $= K _{2} O _{2}$
• C. Highest $= OF _{2}$, lowest $= KO _{2}$
• D. Highest $= KO _{2}$, lowest $= H _{2} O $

Answer 1

Answer: (C)

Oxidation number of oxygen in $OF _{2}=+2$
In $KO _{2}=\frac{-1}{2}$