Why does acidified potassium manganate(VII) decolorize when react with cyclohexene?

Cyclohexene is reacted with acidified potassium permanganate solution. What would be observed? In alkaline solution the potassium permanganate would turn green followed by the formation of a brown precipitate of MnO2. The reaction is an oxidation (definitely NOT a substitution) and is quite complicated.

Because the color is due to the permanganate anion, MnO4(-), which oxydizes the alkene to the corresponding 1,2-diol.

Permanganate is thus reduced either to slightly pink Mn(2+) or brown MnO2 thus the solution loses its deep purple color.

KMnO4 is an oxidizing agent and an alkene like cyclohexene has an electron-rich pi cloud, which is to say it is a reducing agent. In the resulting redox reaction, the lovely violet KMnO4 is reduced to brown, insoluble MnO2. The cyclohexene will be oxidized to cis-1,2-cyclohexanediol (and sometimes further, depending on reaction conditions.)

‘Acidified potassium tetraoxo manganate (VII) is more commonly known as potassium permanganate (KMnO4). Under acidic or basic conditions, this compound oxidizes primary alcohols (like ethanol, CH3-CH2-OH) to carboxylic acids. So, for the oxidation of ethanol with potassium permanganate, the product is ethanoic acid (aka, acetic acid, CH3-COOH) and the chemical reaction is:

CH3-CH2-OH + KMnO4 → CH3-COOH + MnO2

Cyclohexene burns and produces more soot because of the higher percentage of carbon compared to cyclohexane.

Also, the presence of multiple covalent bonds between carbon atoms require more energy to break down. So, you need more heat to burn them. Plus they generate more soot.

Both these factors combine to cause a shorter combustion compared to a cyclohexane. This can be true when you are comparing combustion between any single bond alkane vs multiple bond alkene or alkyne.

Manganate (VII) ions are reduced to manganese (II) ions while the alkene is oxidised to a diol (two alcohol groups). The electron configuration of manganese in the manganate (VII) ion is [Ar] 3d° 4s°, while the electron configuration for a manganese (II) ion is [Ar] 3d5 4s°, so it seems odd that the form with the empty d orbitals is highly coloured while the one with partially filled d orbitals is almost colourless (very pale pink).

An explanation needs someone better informed than me, but as I understand it both ions are exceptions. In the MnO4- ion the charge is represented as distributed amongst the 4 oxygens, but electrons from the oxygens can also be transferred to empty orbitals in the manganese. The energy change corresponds to the energy of visible light; absorption of light energy produces the intense purple colour.

Why Mn2+ is almost colourless is harder to explain. Colour in transition metal compounds is normally a result of electron transfers between d orbitals of two slightly different energies (split d orbitals). My inadequate understanding is that for this to happen when all five d orbitals have a single electron there must be two changes: a change to the electron spin (quantum number) and a change between the split d orbitals making the transition unlikely.

Could someone else correct me if I have not got this right or confirm if I have?