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How would you account for the following ...

How would you account for the following :
(i) Out 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 oxidized.
(iii) The `d^(1)` configuration is very unstable in ions.

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(i). `E^(ɵ)` value for `((Cr^(3+))/(Cr^(2+)))` is negative `(-0.41V)` whereas `E^(ɵ)` value for `((Mn^(3+))/(Mn^(2+)))` is positive `(+1.57V)`. Hence `Cr^(2+)` ions can easily undergo oxidation to give `Cr^(3+)` ions and therefore, act as strong reducing agent whereas `Mn^(2+)` can easily undergo reduction to give `Mn^(2+)` and hence act as oxidising agent.
(ii). `Co^(3+)` has greater tendecny to form coordination complexes than `Co^(2+)`. Hence in the presence of ligands `Co^(2+)` changes to `Co^(3+)` i.e., is easily oxidised.
(iii) The ions with `d^1` configuration have the tendency to lose the only electron present in d-subshell to acquire stable `d^0` configuration. Hence, they are unstable and undergo oxidation or disproportionation.
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How would you account for the following: A) Of the d^(4) species Cr^(2+) is strongly reducing while manganese(III) is strongly oxidizing. B) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidized. C) The D^(1) configuration is very unstable in ions.

How and you account for the following : (i) Cabalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (ii) The transition elements exhibit high enthalpy of atomization. (iii) of the d^4 species, Cr^(2+) is strongly reducing while Mn (III) is strongly oxidsing .

How would you account for the following ? (i) With the same d-orbital configuration (d^(4)) Cr^(2+) is reducting agent while Mn^(3+) is an oxidizing agent. (ii) The actionoids exhibits a larger numbe of oxidation states than the corresponding members in the lanthanoid series. (iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

How would you account for the following ? (i) With the same d-orbital configuration (d^(4)) Cr^(2+) is reducting agent while Mn^(3+) is an oxidizing agent. (ii) The actionoids exhibits a larger numbe of oxidation states than the corresponding members in the lanthanoid series. (iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

Give reasons for the following: (i) Transition metals exhibit a wide range of oxidation states. (ii) Cobalt (II) is very stable in aqueous solutions but gets easily oxidised in the presence of strong ligands. (iii) Actinoids exhibit a greater range of oxidation states than lanthanoids.

How would you account for the following : (i) Cr^(2+) is reducing in nature while with the same d-orbital configuration (d^(4)),Mn^(2+) is oxidising in nature. (ii) In the transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series.

Account for the following statements : (i) Co (ii) is stable in aqueous solution but in the presence of strong ligands and air it can get oxidised to Co (III) (at. No Co=27) (ii) of the d^4 species Cr (II) is strongly reducing but Mn (III) is strongly oxidising (at. no Cr= 24 , Mn = 25, Co = 27)

Account for the following (i) Co(II) is stable in aqueous solution but in the presence of strong ligand and air , it can get oxidized to Co(III) . (ii) [Ni(CN)_(4)]^(2-) is square planar and diamagnetic whereas [NiCl_(4)]^(2-) is tetrahedral and paramagnetic .

How would you account for the following : (i) H_(2)S is acidic than H_(2)O . (ii) The N-O bond in NO_(2)^(-) is shorter than the N-O bond in NO_(3)^(-) . (iii) Both O_(2) and F_(2) stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of flurine.

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