The correct order of increasing oxidising power in the series is

Options

A. `VO_2^(+) lt Cr_2O_7^(2-) lt MnO_4^(-)`
B. `Cr_2O_7^(2-) lt VO_2^(+) lt MnO_4^(-)`
C. `Cr_2O_7^(2-) lt MnO_4^(-) lt VO_2^(+)`
D. `MnO_4^(-) lt Cr_2O_7^(2-) lt VO_2^(+)`

Answer

A

Explanation

`overset((x))MnO_4^(-) : x + 4 (-2) = -1 rArr x = + 7`
`overset((x))Cr_2O_7^(2-) : 2x + 7(-2) = -2 rArr x = + 6`
` overset((x)) VO_2^(+) : x + 2 (-2) = 1rArr x = + 5`
All the metals are in their highest oxidation state, so all are good oxidizing agent. But `MnO_4^(-)` is the best oxidizing agent in acidic medium due to its highest SRP value :
`MnO_4^(-) to Mn^(2+) , E^@ = 1.51 V`
`Cr_2 O_7^(2-) to Cr^(3+) , E^@ = 1.31 V`.