Question
The oxidation states of the most electronegative element in the products of the reaction, $\mathrm{BaO}_{2}$ with dil. $\mathrm{H}_{2} \mathrm{SO}_{4}$ are(a) 0 and $-1$(b) $-1$ and $-2$(c) $-2$ and 0(d) $-2$ and $+1$
Step 1
$\mathrm{H}_{2} \mathrm{SO}_{4}$. The reaction is as follows: \[\mathrm{BaO}_{2} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{BaSO}_{4} + \mathrm{H}_{2} \mathrm{O}_{2}\] Show more…
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The oxidation states of the most electronegative element in the products of the reaction, $\mathrm{BaO}_{2}$ with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ are (a) 0 and $-1$ (b) $-1$ and $-2$ (c) $-2$ and 0 (d) $-2$ and $+1$
The oxidation states of the most electronegative element in the products of the reaction of $\mathrm{BaO}_{2}$ with dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ are (a) 0 and $-1$ (b) $-1$ and $-2$ (c) $-2$ and 0 (d) $-2$ and $+1$
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