sblock Elements 2 Question 14
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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
======= ####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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) 0 and -1
(b) -1 and -2 (1991, 1M)
(c) -2 and 0
(d) -2 and -1
$(1980,1 \mathrm{M})$
(a) electrolysis of molten $\mathrm{CaCl}_{2}$
(b) electrolysis of solution of $\mathrm{CaCl}_{2}$ in water
(c) reduction of $\mathrm{CaCl}_{2}$ with carbon
(d) roasting of limestone
Objective Questions II
(One or more than one correct option)
Show Answer
Answer:
Correct Answer: 14. (b)
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.
