Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^(4)` configuration ?

Account for the following: (i) Transition elements exhibit higher enthalpies of atomisation. (ii) Cr^(2+) is reducing and Mn^(3+) is oxidising when both have d^4 configuration.

(A) Cr^(2+) is reducing while Mn^(3+) is oxidizing when both have d^4 configuration. (R) Configuration of Cr^(2+) changes from d^3 to d^4

(a) What happens when (i) Manganate ions (MnO_(4)^(2-)) undergoes disproportionation reaction in acidic medium ? (ii) Lanthanum is heated with Sulphur? (b) Explain the following trends in the properties of the members of the First series of transition elements: (i) E^(@) (M^(2+)//M) value for copper is positive (+ 0.34 V) in contrast to the other members of the series. (ii) Cr^(2+) is reducing while Mn^(3+) is oxidising, though both have d^(4) configuration. (iii) The oxidising power in the series increases in the order VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_94)^(-) .

(a) What happens when (i) Manganate ions (MnO_(4)^(2-)) undergoes diSQProportionation reaction in acidic medium ? (ii) Lanthanum is heated with Sulphur? (b) Explain the following trends in the properties of the members of the First series of transition elements: (i) E^(@) (M^(2+)//M) value for copper is positive (+ 0.34 V) in contrast to the other members of the series. (ii) Cr^(2+) is reducing while Mn^(3+) is oxidising, though both have d^(4) configuration. (iii) The oxidising power in the series increases in the order VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_94)^(-) .

Why is Cr^(2+) reducing and Mn^(3+) oxidizing even though both have the same d^(4) electronic configuration.

Why is Cr^(2+) acts as reductant and Mn^(3+) as oxidant eventhough both have d^4 configuration ?

Explain why Cr^(2+) behaves reducing while Mn^(3+) oxidising through both are havign d^9 configuration. What is interstetial compound?