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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. |
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Answer» (i) `Cr^(2+)` has the configuration `d^(4)` and easily changes to `Cr^(3+)` which has half `t_(2g)` configuration and hence more stable. Therefore `Cr^(2+)` is reducing. On the other hand, the change from `Mn^(3+)` to `Mn^(2+)` results in the half filled , `d^(5)` configuration which has extra stability. Therefore `Mn^(3+)` is oxidising. (ii) the actionoids exhibit a larger number of oxidation state than the correponding member in the lanthanoid series. This is due to very small energy gap between 5f, 6d and 7s. Orbital in the actinoid series. (iii) Most of the transmition metal ions exhibit characteristic in colours in aqueous solution. This is due to the particle absorpation of visible light. The absorbed light promotes the electron from one obrbital to another orbital to the same d subshell. Due to presence of unpaired electrons in d-orbitals and d -d- transition. |
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