Which of the following is/are correct about the redox reaction ? $Mn{O}_4^{-}+S_2{O}_3^{2-}+H^{\oplus }\to M{n}^{2+}+S_4{O}_6^{2-}$

A) $S{e}^{-}+Mn{O}_4^{-}\to M{n}^2(n=5)$

$2S_2{O}_3^{-}\to S_4{O}_6^{2-}+2e^{-}[n=\frac{2}{2}=1]$

Eq. of  $Mn{O}_4^{-}\equiv Eq.of{S}_2{O}_3^{-}$

5× moles of  $Mn{O}_4^{-}\equiv 1\times molesof{S}_2{O}_3^{2-}$

$\therefore 1molof{S}_2{O}_3^{2-=5}=5molofMn{O}_4^{-}$

B) pH changes from 4 to 10 (acidic to strongly basic)

$e^{-}+Mn{O}_4^{-}\to M{n}_4^{2-}(n=1)$

$S_2{O}_3^{2-}\to 2S{O}_4^{2-}+8e^{-}(n=8)$

Eq. of $Mn{O}_4^{-}=Eqof{S}_3{O}_3^{2-}$

$\therefore 1molof{S}_2{O}_3^{2-}=\frac{1}{8}molofMn{O}_4^{-}$

Hence with change of pH from 4 to 10, will change the stoichiometry of reaction and also changes the product.

C) pH changes from 4 to 7 (acidic to neutral medium)

$3e^{+}+Mn{O}_4^{-}\to Mn{O}_2(n=3)$

$S_2{O}_3^{2-}\to 2HS{O}_4^{-}+8e^0(n=8)$

Hence it will also effect the stoichiometry of reaction and nature of product.

D) At  $pH=7,S_2{O}_3^{2-}$  is oxidized to  $HS{O}_4^{-}$ ion