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Question

An element A in a compound ABD has oxidation number A n-. It is oxidised by Cr2O27 in acidic medium. In the experiment 1.68×103 x moles ofK2Cr2O7 were used for 3.26×103 moles of ABD. The new oxidation number of A after oxidation is:


A

3

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B

3-n

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C

n-3

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D

+n

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Solution

The correct option is B

3-n


First - let us write down the reaction:
Cr2O27+AnBDCr3++An+x

From stoichiometry, we have:
Equivalents of Cr2O27 = equivalents of ABD
1.68×103=3.26×103×x
Solving, we get x=3.
Hence, the new oxidation number is: 3-n


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