When potassium permanganate reacts in a faintly alkaline or neutral medium, it forms a brown solid product. The number of d-electrons present on the transition metal in this brown solid is the same as that in:

Step-by-step solution

In a faintly alkaline or neutral medium, potassium permanganate ( KMnO ₄ ) is reduced to manganese dioxide ( MnO ₂ ), which precipitates as a brown solid. The oxidation state of Mn in MnO ₂ is +4. The atomic number of Mn is 25, and its ground state electronic configuration is [ Ar ] 4 s ^2 3 d ^5 . Removing four electrons to form Mn ⁴⁺ gives the configuration [ Ar ] 3 d ^3 . Thus, it has 3 d-electrons. Now, let us check the d-electron count for the given ions: Cr ³⁺ (Z=24): [ Ar ] 3 d ^3 (3 d-electrons) Fe ³⁺ (Z=26