`{:("Column I","Column II"),((A)underset(3.5"mole")(Sn^(+2))+underset(1.2"mole")(MnO_4^(-)("acidic")),(p)"Amount of oxidant available decides the number of electrons transfer"),((B)underset(8.4"mole")(H_2C_2O_4)+underset(3.6"mole")(MnO_4^(-)("acidic")),(q)"Amount of reductant available decides the number of electrons transfer"),((C )underset(7.2"mole")(S_2O_3^(-2))+underset(3.6"mole")(l_2),(r)"Number of electrons involved per mole of oxidant gt Number of electrons involved per mole of reductant"),((D )underset(9.2"mole")(Fe^(+2))+underset(1.6"mole")(Cr_2O_7^(-2)"(acidic)"),(s)"Number of electrons involved per mole of oxidant lt Number of electrons involved per mole of reductant"):}`

Answer

a-p,r ; B-q,r ; C-p,q,r ; D-q,r

Explanation

(A)Eq of `Sn^(2+)`=Moles x v.f. =3.5x2=7
Eq of `MnO_4^(-)` =Moles xv.f=1.2x5=6
Since `MnO_4^(-)`(O.A) is the LR, so the amount of oxidant available decides the number of electron transfer.
Also, electron involved per mole of OA `(5)gt` electron involved per mole of RA(2).
(B)Eq of `H_2C_2O_4`=Moles x v.f. =8.4x2=16.8
Eq of `MnO_4^(-)` =Moles x v.f=3.6x5=18
Since `H_2C_2O_4`(RA) is the LR, so the amount of reductant available decides the number of electron transfer.
Also, electron involved per mole of `OA (5)gt` electron involved per mole of RA(2).
(C )Eq of `S_2O_3^(2-)`=Moles x v.f. =7.2x1=7.2
Eq of `I_2` =Moles x v.f=3.6x2=7.2
Since `S_2O_3^(2-)`(RA) and `I_2` (OA) both completely get consumed, so both the amount of reductant and oxidant decides the number of electron transfer.
Also, electron involved per mole of `OA (2)gt` electron involved per mole of RA(1).
(D)Eq of `Fe^(2+)`=Moles x v.f. =9.2x1=9.2
Eq of `Cr_2O_7^(2-)` =Moles x v.f=1.6x6=9.6
Since `Fe^(2+)`(RA) is the LR, so the amount of reductant available decides the number of electron transfer.
Also, electron involved per mole of `OA (6)gt` electron involved per mole of RA(1).