sblock Elements 2 Question 14
14. The oxidation state of the most electronegative element in the products of the reaction, $\mathrm{BaO}_2$ with dil. $\mathrm{H}_2 \mathrm{SO}_4$ are
(1991, 1M)
(a) 0 and -1
(b) -1 and -2
(c) -2 and 0
(d) -2 and -1
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Answer:
Correct Answer: 14. (d)
Solution: The reaction involved is
$ \mathrm{BaO}_2+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{BaSO}_4+\mathrm{H}_2 \mathrm{O}_2 $
The most electronegative atom, oxygen, in $\mathrm{BaSO}_4$ and $\mathrm{H}_2 \mathrm{O}_2$ has -2 and -1 oxidation state respectively.
BETA
