The oxide that gives $H_{2} O_{2}$ on treatment with dilute $H_{2} S O_{4}$ is —

M n O_{2}

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B a O_{2} .8 H_{2} O + O_{2}

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P b O_{2}

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T i O_{2}

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Solution

The correct option is B $B a O_{2} .8 H_{2} O + O_{2}$
Oxides such as $B a O_{2}$ or $N a_{2} O_{2}$ contain peroxy linkage (O−O) and thus when they are reacted with dilute $H_{2} S O_{4}$ , they give $H_{2} O_{2}$ . Thus, when $B a O_{2} .8 H_{2} O + O_{2}$ is reacted with dilute $H_{2} S O_{4}$ , following reaction takes place:
$B a O_{2} . 8 H_{2} O + O_{2} + H_{2} S O_{4} ↓ B a S O_{4} + H_{2} O_{2} + 8 H_{2} O$
Here, $H_{2} O_{2}$ is produced in this reaction.
Hence, option (B) is correct.

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Question 7

The oxide that gives $H_{2} O_{2}$ on treatment with dilute $H_{2} S O_{4}$ is

(a) $P b O_{2}$ (b) $B a O_{2} .8 H_{2} O + P_{2}$

The oxide that gives hydrogen peroxide ( $H_{2} O_{2}$ ) on the treatment with a dilute acid ( $H_{2} S O_{4}$ ) is

H_{2} O_{2}

(H_{2} O_{2})

(H_{2} {S O}_{4})

H_{2} O_{2}

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