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Why is Cr^(2+) reducing and Mn^(3+) oxid...

Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^(4)` configuration ?

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`Cr^(2+)` has the configuration `3d^(4)` . It can lose electron to form`Cr^(3+)` which has the stable `3d^(3)` configuration ( as it has half- filled `t_(2g)` level - explained in unit). Hence, it is reducing . On the other hand, `Mn^(3+)` also has `3d^(4)` configuration but it can gain electron to form `Mn^(2+)` which has stable `3d^(5)` configuration ( as it is exactly half- filled ). Hence, it is oxidizing.
Alternatively. `E^(@)` value for `Cr^(3+) // Cr^(2+) ` is negative `( - 0.41V)` whereas `E^(@) `value for `Mn^(3+) // Mn^(3+)` is positive `( + 1.57V)` . Hence, `Cr^(2+)` ion can easily undergo oxidation to give `Cr^(3+)` ion and , therefore , acts as strong reducing agent where as `Mn^(3+)` can easily undergo reduction to give`Mn^(2+)` and hence acts as oxidizing agent.
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Assertion : Cr^(2+) is reducing and Mn^(3+) is oxidising. Reason : Cr^(2+) and Mn^(3+) have d^4 configuration.

The ion Cr^(2+) is redusing agent while that of Mn^(3+) is an oxidising agent though both have 3d^(4) configuration. This is because

Assertion :- Cr^(+2) is a reducing agent and Mn^(+3) is oxidising agent. Reason :- Mn^(+3) has d^(5) configuration.

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.

How would you account for the following? (i) Cr^(2+) is reducing in nature while with the same d-orbital configuration (d^14) Mn^(3+) is an oxidising agent. (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (iii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than for the 3d series.

Which of the following statement are correct about Cr^(2+) (Z = 24) and Mn^(3+) (Z = 25) ? (i) Cr^(2+) is a reducing agent (ii) Mn^(3+) is an oxidizing agent (iii) Both Cr^(2+) and Mn^(3+) exhibit d^(4) configuration (iv) When Cr^(2+) is used as a reducing agent, the chromium ion attains d^(5) electronic configuration