`(i)` A powdered substance (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 )`.
`(iii)` The aqueous solution of (A) on treatment with NaOH and `Br_(2)-` water gives a compound (D).
`(iv)` A solution of (D) in conc. `HNO_(3)` on treatment with lead peroxide at boiling temperature produced a compound (E) which was of the same color at that of (C).
`(v)` A solution of (A) on treatment with a solution of barium chloride gave a white precipitate of compound (F) Which was insoluble in conc. `HNO_(3)` and conc. HCl.
Consider the following statement `:`
`(I) ` Anions of both (B) and (C ) are diamagnetic and have tetradhedral geometray.
`(II)` Anions of both (B) and ( C) are paramagnetic and have tetrahedral geometry.
`(III)` Anions of (B) is paramagnetic and that of (C ) is diamagnetic but both have same tetrahedral geometry.
`(IV)` Green coloured compound (B) in a neutral of acidic medium disproportionates to give (C ) and (D).
Of these select the correct one from the code 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) Electron configuration of Mn(VI) in `MnO_(4)^(2-)` is `[Ar]^(18)3d^(1)`. So it is paramagnetic and tetrahedral. Electron configuration of Mn(VII) in `MnO_(4)^(-)` is `[Ar]^(18)3d^(0)`. So it is diamagnetic and tetrahedral.
(IV) `3MnO_(4)^(2-)+4H^(+)to2MnO_(4)^(-)+MnO_(2)+2H_(2)O`
A=`MnSO_(4),B=K_(2)MnO_(4),C=KMnO_(4),D=MnO_(2),E=HMnO_(4),F=BaSO_(4)`
`MnSO_(4)(A)+2KNO_(3)+K_(2)CO_(3)toK_(2)MnO_(4)(B)+2KNO_(2)+2CO_(2)+K_(2)SO_(4)`
`2MnO_(4)^(2-)+4H^(+)toMnO_(4)^(-)(C)+MnO_(2)+2H_(2)O`
`Mn^(2+)+2OH^(-)toMn(OH)_(2)darr,Mn(OH)_(2)+Br_(2)+2NaOHtoMnO_(2)(D)+2NaBr+2H_(2)O`
`MnO_(2)+4HNO_(3)to2Mn(NO_(3))_(2)+2H_(2)O+O_(2)`
`Mn(NO_(3))_(2)+5PbO_(2)+6HNO_(3)" "2HMnO_(4)(E)+5Pb(NO_(3))_(2)+2H_(2)O`
`SO_(4)^(2-)+Ba^(2+)toBaSO_(4)darr("White")(F)`