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In the reaction A^(+x) + MnO(4)^() to AO...

In the reaction `A^(+x) + MnO_(4)^() to AO_(3)^() + Mn^(++) +(1)/(2)O, ` if one mole fo `MnO_(4)^()` oxidises 1.67 moles of `A^(+x) ` to `AO_(3)^()` , then what will be the value of x ?

A

5

B

3

C

2

D

0

Text Solution

Verified by Experts

The correct Answer is:
C
`overset(+7)(MnO_(4)^(-))+5e^(-)rarroverset(+2)(Mn)`
Since 1 mole of `MnO_(4)^(-)` accepts 5 moles of electrons, therefore 5 moles of electron are lost by 1.67 moles of `M^(x+)`
1 mole of `M^(x+)` will lose electrons
`= 5//1.67 =3` mol (approx)
Since `M^(x+)` changes to `MO_(3)^(-)` (where O.N of M = 5) by accepting 3 electrons
`:.` oxidising state of `M^(x+)` i.e., `x = +5 - 3 = +2`
`overset(+2)(M^(2+))rarroverset(+5)(M)O_(3)^(-)+3e^(-)`.
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