Consider the following compounds:underline{K} | vedclass

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(A) $1$. For $\underline{K}O_2$: Potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds.$2$. For $H_2\unde

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$+1, -1$ and $+6$

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(A) $1$. For $\underline{K}O_2$: Potassium $(K)$ is an alkali metal and always exhibits an oxidation state of $+1$ in its compounds.$2$. For $H_2\underline{O}_2$: Let the oxidation state of oxygen be $x$. Since hydrogen is $+1$,we have $2(+1) + 2x = 0$,which gives $2x = -2$,so $x = -1$.$3$. For $H_2\underline{S}O_4$: Let the oxidation state of sulfur be $x$. Hydrogen is $+1$ and oxygen is $-2$. Thus,$2(+1) + x + 4(-2) = 0$,which simplifies to $2 + x - 8 = 0$,so $x = +6$.Therefore,the oxidation states are $+1, -1$ and $+6$.

Consider the following compounds:
$\underline{K}O_2$,$H_2\underline{O}_2$ and $H_2\underline{S}O_4$.
The oxidation states of the underlined elements in them are,respectively,

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Consider the following compounds:$\underline{K}O_2$,$H_2\underline{O}_2$ and $H_2\underline{S}O_4$.The oxidation states of the underlined elements in them are,respectively,