How would you account for the following: (i) Of the d^(4) species, Cr^(2+) is strongly reducing while manganese (III) is strongly oxidising. (ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d^(1) configuration is very unstable in ions

How would you account for the following: (i) Of the d^(4) species, Cr

1.	How would you account for the following: (i) Of the d^(4) species, Cr^(2+) is strongly reducing while manganese (III) is strongly oxidising. (ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d^(1) configuration is very unstable in ions
Answer» Solution :(i) Since `Cr^(2+)` is reducing its configuration CHANGES to `d^(3)" from " d^(4).d^(3)` configuration is half filled `t_(2g)` which is stable in aqueous solution. `Mn^(3+)` is oxidizing, its configuration changes to `d^(5)` from `d^(4).d^(5)` is EXTRA stable as it is half-filled. (ii) Co(II) gets oxidized to Co(III) easily because in presence of strong field ligands, teh electrons get paired up forming diamagnetic octahedral complexes. These complexes are highly stable due to high CFSE. (iii) `d^(1)` configuration is highly unstable because after losing one `E^(-)`, the stable configuration `(d^(0))` is attained or such species will UNDERGO disproportionation reaction. For example. `{:(3MnO_(4)^(2-),+,4H^(+),rarr,2MnO_(4)^(-),+,MnO_(2),+,2H_(2)O),((+6),,,,(+7),,(+4),,),((d^(1)),,,,(d^(0)),,,,):}`