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How would you account for the following: (i) `Cr^(2+)` is reducing in nature while with the same d-orbital configuration `(d^(4)) Mn^(3+)` is an oxidising agent (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation state occurs in the middle of the series. |
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Answer» Correct Answer - (i) It is because `Cr^(3+)` loses electron to become `Cr^(3+)` which is more stable due to half filled `t_(2g)` orbitals. Whereas `Mn^(3+)` will gain electrons to become `Mn^(2+)` which is more stable due to half filled d-orbitals. (ii) It is due to large of unpaired electrons in d-orbitals in middle of the series. |
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