The d-And-f-Block Elements

Question

# (a) Give reasons for the following:(i) Mn3+ is a good oxidising agent.(ii)   Values are not regular for first row transition metals (3d series).(iii) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn2O7.(b) Complete the following equations:

(a) Outer electronic configuration f Mn is 3d5 4s2.

Outer electronic configuration of Mn3+ is 3d4 4s0.

Now Mn3+ is a strong oxidising agent. A good oxidizing agent reduces itself . i.e. gains electrons from other. Its tends to gain one more electron to acquire stable electronic configuration. If it gains one electron, its configuration will be 3d5, which is stable .this is the reason, it acts as a good reducing agent.

(ii)  Values are not regular which can be explained by the irregular variation of ionisation enthalpies i.e. iH1 + iH2 and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

(iii) The ability of oxygen to stabilise the higher oxidation state exceeds that of fluorine. Also, the ability of oxygen to form multiple bonds with metals favours Mn2O7. Therefore, the highest Mn fluoride is MnF4 whereas highest oxide is Mn2O7. In Mn2O7, each Mn is tetrahedrally surrounded by O's including a Mn—O—Mn bridge.

b)