In acidic medium
$MnO _{4}^{-}+8 H ^{+}+5 e ^{-} \longrightarrow Mn ^{2+}+4 H _{2} O$
In neutral medium
$MnO _{4}^{-}+2 H _{2} O +3 e ^{-} \longrightarrow MnO _{2}+4 OH ^{-}$
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.