Question

What are the oxidation numbers of oxygen atom $\left(O\right)$ in $H_2{O}_{2}andK{O}_2$ respectively.

• A. $+1and+\frac{1}{2}$
• B. $-1and-\frac{1}{2}$
• C. $-1and+1$
• D. $+1and-1$

Answer 1

The correct option is B $-1and-\frac{1}{2}$

In hydrogen peroxide $\left(H_2{O}_2\right)$ peroxide linkage $(-O-O-)$ linkage is present. Again electronegativity of oxygen is more than hydrogen so , hydrogen carries positive charge and oxygen carries negative charge.
Let the oxidation number of oxygen in $H_2{O}_2$ be $x.$
So , $2x+2=02x=-2x=-1$
Therefore, the oxidation state of each oxygen is $-1$.

In $K{O}_2$ the $K$ atom carries $+1$ charge as it has only one electron in its valence shell so , $O_2$ should have $-1$ charge i.e. it acts as a superoxide.
In $K{O}_2$​, let oxidation number of oxygen be $x.$
$+1+2x=0x=-\frac{1}{2}$
Therefore, the oxidation number of oxygen in $K{O}_2$​ is $-\frac{1}{2}$