Coordination Compounds 2 Question 27
29. The colour of $KMnO _4$ is due to
(2015 Main)
(a) $M \rightarrow L$ charge transfer transition
(b) $d \rightarrow d$ transition
(c) $L \rightarrow M$ charge transfer transition
(d) $\sigma \rightarrow \stackrel{*}{\sigma}$ transition
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Answer:
Correct Answer: 29. (c)
Solution:
- $KMnO _4 \longrightarrow K^{+}+MnO _4^{-}$
$\therefore$ In $MnO _4^{-}$, Mn has $+7$ oxidation state having no electron in $d$-orbitals.
It is 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 $d$-orbital of metal.
Since, charge transfer is laporate as well as spin allowed, therefore, it shows colour.
Time saving Technique There is no need to check all the four options. Just find out the oxidation state of metal ion. If oxidation state is highest and ligand present there is of electron donating nature, gives LMCT, which shows more intense colour.
BETA
