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The oxidation states of the most electro...

The oxidation states of the most electronegative elements in the products of the reaction between `BaO_(2)` and `H_(2)SO_(4)` are

A

`-1`

B

`+1`

C

`-2`

D

`0`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`BaO_(2)+underset(dil)(H_(2)SO_(4))toBaSO_(4)+H_(2)O_(2)`
The most electronegative elements in the products formed is oxygen. The oxidation state of oxygen on `BaSO_(4)` is `-2` whereas in `H_(2)O_(2)` is `-1`.
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Oxidation number is the charge which an atom of an element has in its ion or appears to have when present in the combined state. It is also called oxidation state. Oxidation number of any atom in the elementary state is zero. Oxidation number of a monoatomic ion is equal to the charge on it. In compounds of metals with non metals, metals have positive oxidation number while non metals have negative oxidation numbers. In compounds of two difference elements, the more electronegative element has negative oxidation number whereas the other has positive oxidation number. In complex ions, the sum of the oxidation number of all the atoms is equal to the charge on the ion. If a compound contains two or more atoms of the same element, they may have same or different oxidation states according as their chemical bonding is same or different. The oxidation state of the most electronegative element in the products of the reaction between BaO_(2) and H_(2)SO_(4) are

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