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How would you account for the following ...

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.

Text Solution

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(i) `Cr^(2+)` has the configuration `d^(4)` and easily changes to `d^(3)` has filled orbitals and hence stable. Therefore, `Cr^(2+)` is reducing .On the other hand `Mn^(2+)` is more stable due to half filled `d^(5)` configuartion and `Mn^(3+)` easliy changes to `Mn^(2+)` and hence is oxiding .
(ii) This is due to the large no.of unpaired in d-orbitals in the middle of the series.
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