(i)

2MnO 4 – + 2NH 3 ⎯→ 2MnO 2 + N 2 + 2OH – + 2H 2 O

It is due to charge transfer from O 2– to empty d-orbital of Mn(VII).

MnO 4 – is in highest oxidation state i.e. +VII and thus can not be oxidised further.

MnO 2 + OH ⎯→ MnO 2 2– MnO 4 –

(ii) In MnSO 4 , = +II ; MnO 4 2– = +VI ; MnO 4 – = +VII

(iii)

(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 + 2MnO 4 – + MnO 2 + 2H 2 O

Solution : (i) to (iii)

A = MnSO 4 , B = K 2 MnO 4 , C = KMnO 4 , D = MnO 2 , E = HMnO 4 , F = BaSO 4 .

MnSO 4 + 2KNO 3 + K 2 CO 3 K 2 MnO 4 + 2KNO 2 + 2CO 2 + K 2 SO 4

2MnO 4 2– + 4H + MnO 4 – + MnO 2 + 2H 2 O.

Mn 2+ + 2OH – Mn(OH) 2 ↓ ; Mn(OH) 2 + Br 2 + 2NaOH MnO 2 + 2NaBr + 2H 2 O

MnO 2 + 4HNO 3 2Mn(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+ BaSO 4 ↓ (white) (F)