MCQ
The color of $KMnO_4$ is due to :
- ✓$L\to M$ charge transfer transition
- B$\sigma - \sigma ^*$ transition
- C$M\to L$ charge transfer transition
- D$d - d$ transition
✓
Answer
Correct option: A.
$L\to M$ charge transfer transition
a
$K M n O_{4} \rightarrow K^{+}+M n O_{4}^{-}$
$K M n O_{4} \rightarrow K^{+}+M n O_{4}^{-}$
$\therefore \ln M n O_{4}^{-},$ Mn has $+7$ oxidation state having no electron in d-orbitals. It ts considered that higher the oxidation state of metal, greater is the tendency to occur $L \rightarrow M$ charge transfer,
because ligand is able to donate the electron into the vacant ( - orbital of metal.
since, charge transfer is laporate as well as spin allowed, therefore, it shows colour
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