(i) A powered of (A) on treatment with fusion mixture gives a green coloured compound(B)
(ii) The solution of (B) in boiling water on acidification with dilute `H_(2)SO_(4)` gives a pink coloured compound (C) and brown colour compounds(D)
(iii) The aqueous solution of (A) on treatment with `NaOH` and `Br_(2)`-water give the compound(D).
(iv) A solution (D) in concentrated `HNO_(3)` on treatment with lead peroxide at boiling temperature produce a compound (E) which was the same colour at that of (C)
(v) A solution of (A) on treatment with a solution of barium chloride gave a white percipitate of compound(F) which was insoluble in concentrated `HNO_(3)` and concentrated `HCl`
Consider the following statement,
anions of both (B) and (C) are diamagnetic and have tetrahedral geometry. (II)
anion of both (B) and (C) are paramagnetic and have tetrahedral geometry.(III)
anion of (B) is paramagnetic and that (C) is dimagnetic but both have tetrahedral geometry. (Iv)
Green coloured compound (B) in a neutral or acidic medium disproportionates to give (C) and (D). Of these select the correct one from the codes given:

Options

A. I and III only
B. II and III only
C. II and IV only
D. III and IV only

Answer

D

Explanation

(III)(B)`MnO_4^(2-)=[Ar]^18 3d^1` paramagnetic and tetrahedral
(C )`MnO_4^(-)=[Ar]^18 3d^0` diamagnetic and tetrahedral
(IV)`3MnO_4^(2-)+4H^(+)to2MnO_4^(-)+MnO_2 + 2H_2O`
Solution of passage :
`MnSO_4(A)+2KNO_3 + K_2CO_3 to K_2MnO_4(B)+2KNO_2 + 2CO_2 + K_2SO_4`
`MnO_4^(2-)+e^(-)overset(H^+)toMnO_4^(-)`(C )
`Mn^(2+)+2OH^(-)toMn(OH)_2 darr , Mn(OH)_2 +Br_2 +2NaOH to MnO_2(D)+2NaBr+2H_2O`
`MnO_2 + 4HNO_3 to 2Mn(NO_3)_2 + 2H_2O +O_2`
`Mn(NO_3)_2+5PbO_2 + 6HNO_3 to 2HMnO_4(E)+5Pb(NO_3)_2 + 2H_2O`
`SO_4^(2-)+Ba^(2+)toBaSO_4darr`(white)(F)