Consider the following compounds: The oxidation states of the underlined elements in them are, respectively, [2025]

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Solution

Q.
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, [2025]

1 +1, −2 and +4

2 +4, −4 and +6

3 +1, −1 and +6

4 +2, −2 and +6

Ans.
(3)

${KO}_{2}$ ionises as $K^{+}$ and $O_{2}^{-}$ ions as it is a superoxide.

Thus, oxidation state of K is +1.

$H_{2} O_{2}$ ionises as $H^{+}$ and $O_{2}^{2 -}$ as it is a peroxide.

Oxidation state of each O in $O_{2}^{2 -}$ is -1.

In $H_{2} {SO}_{4}$ , let the oxidation state of S be $x$ ; hence

$2 (+ 1) + x + 4 (- 2) = 0$

$2 + x - 8 = 0$

$x = + 6$

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