Question
The oxidation states of che most electronegative element in the products of the reaccion, $\mathrm{BaO}_{2}$ with dil. $\mathrm{H}_{2} \mathrm{~S} Q_{4}$ are(a) 0 and $-1$(b) $-1$ and $-2$(c) $-2$ and 0(d) $-2$ and $+1$
Step 1
The oxidation state of $\mathrm{Ba}$ is +2 because it is an alkaline earth metal and typically loses two electrons to form a stable ion. The oxidation state of $\mathrm{O}$ is -2 because it is a highly electronegative element and typically gains two electrons to Show more…
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Topic 2 : Oxidation Number
The oxidation states of the most electroregative 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 $+\mathrm{l}$
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