| Statement I Cone. \[{{H}_{2}}S{{O}_{4}}\]cannot be used to prepare HBr or HI from KBr or KI. |
| Statement II Cone. \[{{H}_{2}}S{{O}_{4}}\]is a strong oxidizing agent while HBr or HI are reducing agent. |
A) Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I
B) Both Statement I and Statement II are true but the Statement II is not the correct explanation of the Statement I
C) Statement I is false but Statement II is true
D) Both Statement I and Statement II are false
Correct Answer: A
Solution :
Both Statement I and Statement II are correct and Statement II is correct explanation of Statement I. \[{{H}_{2}}S{{O}_{4}}\] oxidizes Kl to \[{{l}_{2}}\]and KBr to \[B{{r}_{2}}\].
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