a
The permanganate ion has an intense purple colour. \(Mn\)  \((+ VII)\) has a \(d^0\) configuration. So the colour arises from charge transfer and not from \(d - d\) spectra.

In \(\mathrm{MnO}_{4}^{-}\) an electron is momentarily changing \(\mathrm{O}^{--}\) to \(\mathrm{O}^{-}\) and reducing the oxidation state of the metal from \(Mn\,(VII)\) to \(Mn\, (VI)\). Charge transfer requires that the energy levels on the two different atoms are fairly close.

\(\,\begin{array}{*{20}{c}}
  {O\, = \,(8)\, = 2,} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K} 
\end{array}\,\begin{array}{*{20}{c}}
  {6\,;} \\ 
  L 
\end{array}\)           \(\,\begin{array}{*{20}{c}}
  {Mn\,\,(25)\, = 2,} \\ 
  {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K} 
\end{array}\,\begin{array}{*{20}{c}}
  {8,} \\ 
  L 
\end{array}\begin{array}{*{20}{c}}
  {15} \\ 
  M 
\end{array}\)

Hence the charge transfer occurs from

\(L\to M\)