Q. 155.0( 1 Vote )

How would you acc

Answer :

(i) We know that the stability of a half-filled orbital (like d5) or a full filled orbital (like d10) is always higher than the other configurations. So, an element or an ion would always prefer the half-filled or full-filled configuration to gain stability, thus they always try to go somehow to that configuration.

Now, the electronic configuration of Cr is [Ar]3d54s1

When Cr loses 2 electrons to become Cr2+, the electronic configuration becomes [Ar]3d4. But it always tries to gain the more stable Cr3+ configuration state by losing another electron and get oxidised to form d3 due to the stable t2g configuration.

Whereas the electronic configuration of Mn is [Ar]3d54s2

Now, when Mn loses 3 electrons to form Mn3+, d4 configuration is achieved. So, it would try to gain an electron and get reduced to the more stable d5 configuration (Mn2+).

So, Cr2+ is reducing in nature while Mn3+ is an oxidizing agent.

(ii) It is obvious that in the middle of a transition series, due to the presence of more unpaired electrons and more number of partially filled orbitals the metals exhibit the greatest number of oxidation states.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

When a chromite oChemistry - Exemplar

Give reasonChemistry - Board Papers