Home
Class 12
CHEMISTRY
Why is Cr^(2+) reducing and Mn^(3+) oxid...

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

Text Solution

Verified by Experts

`E^0` value for `Cr^(3+)//Cr^(2+)` is negative (-0.41V), this means that `Cr^(3+)` ions are more stable than `Cr^(2+)` . Therefore, `Cr^(2+)` can readily lose electrons to undergo oxidation to form `Cr^(3+)` ion and hence Cr(II) is strongly reducing. On the other hand the `E^0` value for `Mn^(3+)//Mn^(2+)` is positive (+1.57V), this means that `Mn^(3+)` ions can be readily reduced to `Mn^(2+)` and hence Mn (III) is strongly oxidising,
Promotional Banner

Topper's Solved these Questions

  • D AND F-BLOCK ELEMENTS

    MODERN PUBLICATION|Exercise PRACTICE PROBLEM|33 Videos

  • D AND F-BLOCK ELEMENTS

    MODERN PUBLICATION|Exercise CONCEPTUAL QUESTION|49 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|20 Videos

Similar Questions

Explore conceptually related problems

(a) What happens when (i) Manganate ions (MnO_(4)^(2-)) undergoes diSQProportionation reaction in acidic medium ? (ii) Lanthanum is heated with Sulphur? (b) Explain the following trends in the properties of the members of the First series of transition elements: (i) E^(@) (M^(2+)//M) value for copper is positive (+ 0.34 V) in contrast to the other members of the series. (ii) Cr^(2+) is reducing while Mn^(3+) is oxidising, though both have d^(4) configuration. (iii) The oxidising power in the series increases in the order VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_94)^(-) .

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.