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(a) Why are Mn^(2+) compounds more stabl...

(a) Why are `Mn^(2+)` compounds more stable than `Fe^(2+)` towards oxidation to their +3 state?
(b) What are interstitial compounds ? Why are such compounds well-known for transition metals?
(c) Write electronic configuration of `Pm^(3+)`.

Text Solution

Verified by Experts

(a)`Mn^(2+) (3d^(5))` - exactly half filled and extra stable and has high 3rd ionisation enthalpy `Fe^(2+) (3d^(6))` can lose electron easily to get `Fe^(3+) (3d^(5))` i.& extra stable configuration.
(b) In such compounds, the small atoms such as H, C, B, N etc. occupy interstitial sites in their lattices eg. Tic, `TiH_(2), Fe_(3)H` Steel etc. Such compounds are called interstitial compounds. Transition metals form such compounds because they can take up small atoms of C, H, B, Netc. in their interstitial sites in their lattices.
(c) `Pm^(3-1)(61-3=58e)`
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)4p^(6)4d^(10)5s^(2)5p^(6)4f^(4)` in `[Xe]_(54)4f^(4)`
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Fe^(3+) compounds are more stable than Fe^(2+) compounds because

S_(1) : Mn^(2+) compounds more stable than Fe^(2+) towards oxidation to their +3 state. S_(2) : Titanium and copper both in the first series of transition metals exhibits +1 oxidation state most frequently. S_(3) : Cu^(+) ions is stable in aqueous solutions. S_(4) : The E^(0) value for the Mn^(3+)//Mn^(2+) couple much more positive than that for Cr^(3+)//Cr^(2+) or Fe^(3+)//Fe^(2+) ,