Home
Class 11
CHEMISTRY
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

`0` and `-1`

B

`-1` and `-2`

C

`-2` and `0`

D

`-2` and `+1`

Text Solution

Verified by Experts

The correct Answer is:
B

`BaO_(2)+H_(2)SO_(4)(dil)rarrBaSO_(4)+H_(2)O_(2)`
Oxidation state of `O` in `H_(2)O_(2)= -1`
Oxidation state of `O` in `BaSO_(4)= -2`
Promotional Banner

Topper's Solved these Questions

  • REDOX REACTIONS

    CENGAGE CHEMISTRY|Exercise Archives (Integers)|1 Videos
  • REDOX REACTIONS

    CENGAGE CHEMISTRY|Exercise Exercises (Integers)|11 Videos
  • PURIFICATION OF ORGANIC COMPOUNDS AND QUALITATIVE AND QUANTITATIVE ANALYSIS

    CENGAGE CHEMISTRY|Exercise Assertion Reasoning Type|5 Videos
  • S-BLOCK GROUP 1 - ALKALI METALS

    CENGAGE CHEMISTRY|Exercise Archives Subjective|8 Videos

Similar Questions

Explore conceptually related problems

The oxidation states of the most electronegative element in the products of the reaction between BaO_2 with dilute H_2SO_4 are

The oxidation state of the most electronegative atom in each of the product is H_(2) +O_(2) rarr H_(2)O_(2) +H_(2)O

The major product (A) formed in the reaction- overset(H_(2)SO_(4))rarrA is:

Among C, S and P the element (s) that produces (s) SO_(2) on reaction with hot conc. H_(2)SO_(4) is /are