Step 1: Understanding the Concept:
In coordination chemistry, the intense colors of transition metal complexes typically arise from electronic transitions within the visible light spectrum. While many transition metal compounds owe their colors to $d-d$ transitions, certain complexes exhibit vivid colors despite having empty $d$-orbitals. This requires an alternative electronic mechanism known as charge transfer.
Step 2: Key Formula or Approach:
Let's determine the oxidation state and electronic configuration of the central metal atom in potassium permanganate ($\text{KMnO}_4$):
- In the permanganate ion ($\text{MnO}_4^-$), oxygen has an oxidation state of $-2$.
- Let $x$ be the oxidation state of Manganese ($\text{Mn}$):
$$ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 $$
- The ground-state electronic configuration of atomic Manganese ($Z=25$) is $[\text{Ar}] 3d^5 4s^2$.
- For the $\text{Mn}^{7+}$ ion, all 7 valence electrons are removed:
$$ \text{Mn}^{7+} = [\text{Ar}] 3d^0 4s^0 $$
Step 3: Detailed Explanation:
Let's analyze why $d-d$ transition cannot explain the color, and why charge transfer is responsible:
1. Rule out $d-d$ transitions: Because the $\text{Mn}^{7+}$ central ion possesses a completely empty $d$-orbital configuration ($3d^0$), there are no electrons available in lower $d$-orbitals to absorb light and jump to higher $d$-orbitals. This rules out options (A) and (C).
2. Mechanism of Charge Transfer: The intense deep-violet color is caused by a Ligand-to-Metal Charge Transfer (LMCT) transition.
3. An electron absorbs a photon of visible green-yellow light and momentarily shifts from a filled $p$-orbital of an oxide ligand ($\text{O}^{2-}$) into an empty, low-lying $3d$-orbital of the central $\text{Mn}^{7+}$ ion.
4. Because this transition is Laporte-allowed and spin-allowed, it is highly probable, creating an exceptionally intense color absorption compared to weak, forbidden $d-d$ transitions.
The absorbed light falls in the green-yellow region, causing our eyes to perceive the complementary deep violet color. This corresponds to option (B).
Step 4: Final Answer:
The violet color of $\text{KMnO}_4$ is due to a charge transfer transition.