Lesson 2 · 30 min

Oxidation states and colour

Potassium permanganate is deep purple, copper sulfate pale blue, zinc salts colourless; shake a vanadate solution with zinc and it turns from yellow through green and blue to green again and finally violet. Where do the colours come from?

In this lesson

Relate the colour of a solution to the light it absorbs through complementary colours
Distinguish d–d from charge-transfer transitions, and explain why d⁰ and d¹⁰ ions are usually colourless
Describe the four colours of vanadium from +5 to +2 and use electrode potentials to predict how far each reducing agent takes it
Explain why the “green” that appears during the reduction is not a new oxidation state

A reagent shelf looks like a paint box: potassium permanganate deep purple, potassium dichromate orange, copper sulfate sky blue, nickel sulfate emerald, iron(II) sulfate pale green — while zinc sulfate and scandium sulfate are colourless. Shake a yellow acidified vanadate solution with zinc powder and within minutes it turns green, blue, green again and finally violet. All these colours have one cause: the ions absorb part of the visible light.

Predict first

Which of these ions gives a colourless aqueous solution?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

\ce{Cu^2+}

\ce{Ni^2+}

\ce{Zn^2+}

\ce{Cr^3+}

Where the colour comes from

The colour of a solution is what is left of white light after absorption. $\ce{[Cu(H2O)6]^2+}$ absorbs red and orange light (its band peaks near 800 nm and tails into the visible), and the blue-green light that gets through makes the solution sky blue; permanganate absorbs green (about 526 nm), leaving red and blue — purple.

Transition-metal ions absorb in two main ways:

d–d transitions: ligands split the d orbitals into two sets of different energy (see crystal-field theory), and an electron absorbs a photon to jump from the lower set to the upper. Such transitions are restricted by the Laporte rule and absorb weakly (ε about 1–100 L mol⁻¹ cm⁻¹), so these solutions are fairly pale. d⁰ ions ( $\ce{Sc^3+}$ , $\ce{Ti^4+}$ ) and d¹⁰ ions ( $\ce{Zn^2+}$ , $\ce{Cu+}$ ) have no d–d transitions.
Charge transfer: an electron jumps from a ligand (such as $\ce{O^2-}$ ) to the metal. These transitions are allowed and absorb far more strongly (ε up to thousands), which is why d⁰ permanganate (Mn +7), chromate and dichromate (Cr +6) are intensely coloured even at millimolar concentrations.

A = \varepsilon\, c\, l

The Beer–Lambert law: absorbance is proportional to molar absorptivity, concentration and path length — ε decides how intense a colour is.

$A$: absorbance (dimensionless; A = 1 lets 10 % through)
$\varepsilon$: molar absorptivity, L mol⁻¹ cm⁻¹
$c$: concentration, mol/L
$l$: path length, cm

Computed from absorption bands through the CIE observer

Colours of first-row ions

$\ce{MnO4-}$

d electrons: d0 · charge transfer (strong, ε up to thousands)

Named in the literature: purple

Strongest visible absorption at 526 nm (green light); the light that gets through looks purple.

Concentration · 5.0 × 10⁻⁴ M

Absorption spectrum

Click through the ions and compare their spectra: d–d bands are broad and weak, charge-transfer bands far taller. Note that permanganate and dichromate are shown at only a few thousandths of a mole per litre. Drag the concentration slider and watch the colour deepen.

The vanadium rainbow

Vanadium has four common oxidation states in water, each with its own colour:

\underset{\text{+5, yellow}}{\ce{VO2+}} \xrightarrow{+0.991} \underset{\text{+4, blue}}{\ce{VO^2+}} \xrightarrow{+0.337} \underset{\text{+3, green}}{\ce{V^3+}} \xrightarrow{-0.255} \underset{\text{+2, violet}}{\ce{V^2+}} \xrightarrow{-1.175} \ce{V}

(the standard potential of each step, in volts, is over its arrow). Dissolve a vanadate in acid and add zinc powder: $E^\circ(\ce{Zn^2+/Zn}) = -0.762$ V lies below the first three potentials, so all three steps go and the solution runs from yellow to violet — but not below −1.175 V, so zinc cannot reduce $\ce{V^2+}$ to the metal. Tin (−0.138 V) gets only as far as +3, and iron(II) (+0.771 V) only as far as +4.

Reducing VO₂⁺ step by step: colours computed from absorption bands

The four colours of vanadium

Reducing agent

0.0 min (illustrative)

Time · 0.0

Share of each oxidation state

VO₂⁺ (+5, yellow) 100 %
VO²⁺ (+4, blue) 0 %
V³⁺ (+3, green) 0 %
V²⁺ (+2, violet) 0 %

The solution now holds mainly VO₂⁺ (+5, yellow); computed colour rgb(255 253 32).

Thermodynamic end point: V²⁺ (+2, violet)

Potential ladder (V)

$\ce{VO2+ + 2H+ + e- -> VO^2+ + H2O}$ Goes: E°cell = +1.753 V
$\ce{VO^2+ + 2H+ + e- -> V^3+ + H2O}$ Goes: E°cell = +1.099 V
$\ce{V^3+ + e- -> V^2+}$ Goes: E°cell = +0.507 V

Press “Add and swirl” and watch the colour and the share of each oxidation state (the rates are only illustrative). Early on the solution turns green — look at the bar: it is yellow VO₂⁺ mixed with blue VO²⁺. Then try tin, iron(II) sulfate or potassium iodide and see how the potential ladder decides where the reaction stops.

Worked example

How far can tin reduce vanadium?

$E^\circ(\ce{Sn^2+/Sn}) = -0.138$ V. Decide whether tin can carry out each of VO₂⁺ → VO²⁺, VO²⁺ → V³⁺ and V³⁺ → V²⁺.

Concept check

Which reducing agent can reduce $\ce{VO2+}$ to $\ce{VO^2+}$ but no further?

Zinc (−0.762 V) Tin (−0.138 V) Iron(II) (Fe³⁺/Fe²⁺ +0.771 V) Hydrogen (0 V)

Try the calculation

Find $E^\circ_\mathrm{cell}$ for zinc reducing $\ce{V^3+}$ to $\ce{V^2+}$ : $\ce{2V^3+} + \ce{Zn} \mathrel{\ce{->}} \ce{2V^2+} + \ce{Zn^2+}$ ( $E^\circ(\ce{V^3+/V^2+}) = -0.255$ V, $E^\circ(\ce{Zn^2+/Zn}) = -0.7618$ V).

Model limits

The colours in this lesson are computed from teaching bands: positions are literature maxima, while ε and widths are fitted to reproduce the observed colours, not measured spectra. An ion’s colour also depends on its ligands: hexaaquavanadium(III) in perchlorate solution absorbs slightly higher (17 800 and 25 700 cm⁻¹) than in the chloride or sulfate solution of the demonstration, where anions partly coordinate and the solution is green; hexaaquachromium(III) is blue-violet, whereas the familiar “green” chromium salt solutions contain $\ce{[CrCl2(H2O)4]+}$ . The time course of the vanadium reduction is only illustrative: real rates depend on the zinc’s surface area, the acidity and the stirring, and $\ce{V^2+}$ is soon reoxidised by air.

Key takeaways

A solution shows the complement of the light it absorbs; the larger ε, the concentration and the path, the deeper the colour.
d–d transitions are weak (ε 1–100) and charge transfer strong (ε up to thousands); d⁰ and d¹⁰ ions have no d–d transitions and are often colourless.
Vanadium runs VO₂⁺ yellow → VO²⁺ blue → V³⁺ green → V²⁺ violet; the reductant’s potential decides where the reduction stops.
The early green is a mixture of the yellow and blue species, not a new oxidation state.

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Data and references

P. Vanýsek, “Electrochemical series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010) (vanadium, zinc, tin and iron couples).
A. B. P. Lever, Inorganic Electronic Spectroscopy, 2nd ed., Elsevier (1984) (band positions of aqua ions).
C. J. Ballhausen, H. B. Gray, “The electronic structure of the vanadyl ion”, Inorg. Chem. 1, 111 (1962).
C. E. Housecroft, A. G. Sharpe, Inorganic Chemistry, 5th ed. (2018), ch. 20–21.
N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed. (1997), ch. 22 (vanadium).