Question

calculate the oxidation states of oxygen in H2O2 and KO2​

Answer 1

Answer:

-1 and -1/2

Explanation:

H2O2 is peroxide.In peroxide oxygen oxidation state is -1 and KO2 is Super oxide . in super oxide Oxygen oxygen oxidation state is -1/2

Answer 2

Answer: $H_2O_2$  : -1

$KO_2$  :  $\frac{-1}{2}$

Explanation:

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2.

The oxidation number of a Group 1 element in a compound is +1.

a)  $H_2O_2$

Let the oxidation state of O be, 'x'

$+2+2(x)=0\\\\x=-1$

Hence, the oxidation state of O is -1

b) $KO_2$

Let the oxidation state of O be, 'x'

$+1+2(x)=0\\\\x=\frac{-1}{2}$

Hence, the oxidation state of O is $\frac{-1}{2}$