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B
Cr2O72-2+ < MnO4-
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C
MnO4- < Cr2O72-+
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D
VO2+ < MnO4- < Cr2O72-
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Solution
The correct option is A VO2+ < Cr2O72- < MnO4-
This is due to the increasing stability of the lower species to which they are reduced. MnO−4(Mn7+) is reduced to (Mn2+) which is having half-filled stable d5 configuration.