Why Mn2+ ion is almost colourless wheras Mn7+ ion is coloured
The Correct Answer and Explanation is :
The color of transition metal ions is primarily determined by their electronic configurations and the nature of their d-orbitals. Manganese exists in multiple oxidation states, notably +2 (Mn²⁺) and +7 (Mn⁷⁺), each exhibiting distinct electronic structures that influence their color properties.
Mn²⁺ Ion:
In the Mn²⁺ state, manganese has an electron configuration of [Ar] 3d⁵. This configuration results in a high-spin d⁵ system, where all five d-orbitals are singly occupied. In an octahedral field, these d-orbitals split into two energy levels: the lower-energy t₂g and the higher-energy e_g orbitals. However, due to the high-spin nature of Mn²⁺, the energy difference between these levels is minimal, leading to weak d-d transitions. Moreover, the selection rules governing these transitions often render them spin-forbidden, further diminishing their intensity. Consequently, Mn²⁺ complexes typically appear colorless or exhibit very faint colors.
Mn⁷⁺ Ion:
Conversely, Mn⁷⁺ possesses an electron configuration of [Ar] 3d⁰. With no electrons in the d-orbitals, Mn⁷⁺ cannot undergo d-d transitions. Instead, its color arises from charge transfer transitions, where an electron moves from an oxygen ligand’s orbital to the empty d-orbital of the manganese ion. These transitions are typically allowed and result in the absorption of visible light, imparting a characteristic color to Mn⁷⁺ compounds. For instance, potassium permanganate (KMnO₄), where Mn is in the +7 oxidation state, is deep purple due to such charge transfer transitions.
In summary, the colorless nature of Mn²⁺ is attributed to its high-spin d⁵ configuration, leading to weak and often spin-forbidden d-d transitions. In contrast, Mn⁷⁺ exhibits color due to charge transfer transitions, where electrons move between the metal ion and its ligands, resulting in the absorption of visible light and the manifestation of color.