Q104 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
Consider the following compounds :
$\underline{\text{K}}\text{O}_2, \text{H}_2\underline{\text{O}}_2$ and $\text{H}_2\underline{\text{S}}\text{O}_4$
The oxidation state of the underlined elements in them are respectively,
$\underline{\text{K}}\text{O}_2, \text{H}_2\underline{\text{O}}_2$ and $\text{H}_2\underline{\text{S}}\text{O}_4$
The oxidation state of the underlined elements in them are respectively,
- A$+1, -2$ and $+4$
- B$+4, -4$ and $+6$
- Correct answerC$+1, -1$ and $+6$
- D$+2, -2$ and $+6$
Solution
In $\text{KO}_2$, potassium is an alkali metal, so its oxidation state is always $+1$.
In $\text{H}_2\text{O}_2$, oxygen is present as a peroxide ion ($\text{O}_2^{2-}$), so the oxidation state of oxygen is $-1$.
In $\text{H}_2\text{SO}_4$, let the oxidation state of sulfur be $x$.
$2(+1) + x + 4(-2) = 0$
$x - 6 = 0$
$x = +6$
The oxidation states of the underlined elements are $+1, -1$ and $+6$ respectively.
Answer: $+1, -1$ and $+6$
In $\text{H}_2\text{O}_2$, oxygen is present as a peroxide ion ($\text{O}_2^{2-}$), so the oxidation state of oxygen is $-1$.
In $\text{H}_2\text{SO}_4$, let the oxidation state of sulfur be $x$.
$2(+1) + x + 4(-2) = 0$
$x - 6 = 0$
$x = +6$
The oxidation states of the underlined elements are $+1, -1$ and $+6$ respectively.
Answer: $+1, -1$ and $+6$
Hence, the correct answer is option: 'C'.