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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 states of the underlined elements in them are, respectively:
A
+1, -1, and +6
B
+2, -2, and +6
C
+1, -2, and +4
D
+4, -4, and +6
Explanation
In $\text{KO}_2$, $\text{K} = +1$ (superoxide). In $\text{H}_2\text{O}_2$, $\text{O} = -1$ (peroxide). In $\text{H}_2\text{SO}_4$, $\text{S} = +6$.
Detailed Solution
1. In potassium superoxide ($\text{KO}_2$), potassium is an alkali metal and always has an oxidation state of $+1$ (the superoxide ion is $\text{O}_2^-$ where each oxygen is $-1/2$). 2. In hydrogen peroxide ($\text{H}_2\text{O}_2$), the peroxide linkage gives oxygen an oxidation state of $-1$. 3. In sulfuric acid ($\text{H}_2\text{SO}_4$), $2(+1) + x + 4(-2) = 0 \implies x = +6$. Hence, the oxidation states are $+1, -1$, and $+6$.