Consider the following compounds:

KO₂, H₂O₂ and H₂SO₄

The oxidation states of the underlined elements in them are, respectively,

The correct option is: +1, −1, and +6

Let us find the oxidation states of the underlined elements in each of the given compounds step-by-step:

1. Oxidation state of K in KO₂ (Potassium superoxide):
Potassium (K) is an alkali metal (Group 1 element). In all its stable compounds, alkali metals always exhibit an oxidation state of +1.
In KO₂, which is potassium superoxide, the superoxide ion is O₂^-, where the two oxygen atoms share a overall charge of -1 (giving an average oxidation state of -0.5 for each oxygen).
Thus, the oxidation state of K in KO₂ is +1.

2. Oxidation state of O in H₂O₂ (Hydrogen peroxide):
Hydrogen peroxide is a peroxide compound. Hydrogen (H) has an oxidation state of +1 when bonded to non-metals.
Let the oxidation state of O be $x$.
Since the overall molecule is neutral, the sum of the oxidation states of all atoms must be zero:
$2(+1)+2\left(x\right)=0$
$2+2x=0$
$2x=-2$
$x=-1$
Therefore, the oxidation state of O in H₂O₂ is −1.

3. Oxidation state of S in H₂SO₄ (Sulfuric acid):
In sulfuric acid, hydrogen (H) has an oxidation state of +1 and oxygen (O) has an oxidation state of -2.
Let the oxidation state of S be $y$.
Since the molecule is neutral, the sum of the oxidation states of all atoms is zero:
$2(+1)+y+4(-2)=0$
$2+y-8=0$
$y-6=0$
$y=+6$
Therefore, the oxidation state of S in H₂SO₄ is +6.

Comparing these values with the options, we get:
Oxidation state of K = +1
Oxidation state of O = −1
Oxidation state of S = +6
Hence, the oxidation states are, respectively, +1, −1, and +6.