Colours of inorganic ions and complexes

In a nutshell

Electrons can absorb energy from visible light. Frequencies of light that are not absorbed, are transmitted and can be seen. The concentration of a solution can be determined using spectroscopy.

Split 3d sub-shell

The $$3d$$​ orbitals of transition metals usually have the same energy. However when ligands bind to metal ions, the $$3d$$ orbitals split into different energy levels.

In the ground state, electrons occupy orbitals at lower energies. Electrons can jump up to orbitals of higher energy, and exist in the excited state, if they take in energy from visible light. The energy absorbed by the electrons must be equal to the energy gap $$(\Delta E)$$.

The size of the energy gap $$(\Delta E)$$, and therefore the frequency of light absorbed, depends on the central metal ion involved, its oxidation number, the type of ligands and the coordination number. As the energy gap between the ground state and excited state increases, light of higher frequency needs to be absorbed.

Complementary colours 

Complexes absorb some frequencies of light when the $$d$$​ sub-shell splits; the frequencies of light that are not absorbed are transmitted and can be seen. The colour of light transmitted is complementary to the colour of light absorbed. Complementary colours are opposite each other on the colour wheel. 

Examples

The complex ion $$[Mn(H_2O)_6]^{2+}$$ absorbs frequencies of light in the green region of the spectrum, therefore it appears pink.

The complex ion $$[Cu(H_2O)_6]^{2+}$$ has a light blue appearance whereas the complex ion $$[Cu(NH_3)_4 (H_2O)_2]^{2+}$$ has a dark blue appearance; the ligands surrounding the copper ion affect the energy gap and therefore the complexes colour.

Complexes will not absorb energy if there are no $$3d$$​ electrons to excite or if the $$3d$$ sub-shell is full, therefore they will appear white or colourless.

Aqueous complexes 

When a transitional metal ion is dissolved in water, water ligands will surround it and an aqueous complex forms in the solution. Transition metal ions can be identified by the colour of aqueous complexes. 

You should learn the colours of the vanadium, chromium, iron, cobalt and copper complexes.

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Variable oxidation numbers  

​When a redox reaction takes place, metals are oxidised or reduced and their oxidation number changes. 

Example

Vanadium can have four different oxidation numbers in solution. The four vanadium ions have different colours.

Vanadium(V) is reduced using zinc metal in acidic solution. Each reaction has a reduction potential $$(E^{\circ})$$. Reduction potentials can be used to determine if a reaction involving transition metals is likely to happen.

When vanadium(V) is reduced to vanadium(IV), the solution goes from yellow to blue.    

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$$2VO_2 \ ^+ {(aq)} \, + \, Zn {(s)} \, + \, 4H^+{(aq)} \rightarrow 2VO \ ^{2+} {(aq)} \, + \, Zn^{2+} {(aq)} \, + \, 2H_2O {(l)} \\ VO_2 \ ^+ {(aq)}\, + \, 2H^+ {(aq)} \, + e^- \rightleftharpoons VO \ ^{2+} {(aq)} \, + \, H_2O {(l)} \newline ​E^{\circ}= +1.00 \, V$$

VO_2 \ ^+ \, _{(aq)}\, + \, 2H^+ \, _{(aq)} \, + e^- \rightleftharpoons VO \ ^{2+} \, _{(aq)} \, + \, H_2O \, _{(l)}​

When vanadium(IV) is reduced to vanadium(III), the solution goes from blue to green. ​

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$$2VO^{2+} {(aq)} \, + \, Zn {(s)} \, + \, 4H^+ {(aq)} \rightarrow 2V \ ^{3+} {(aq)} \, + \, Zn^{2+} {(aq)} \, + \, 2H_2O {(l)} \\ VO \ ^{2+} {(aq)}\, + \, 2H^+ {(aq)} \, + e^- \rightleftharpoons V \ ^{3+} {(aq)} \, + \, H_2O {(l)} \newline E^{\circ}= +0.34 \, V$$

VO \ ^{2+} \, _{(aq)}\, + \, 2H^+ \, _{(aq)} \, + e^- \rightleftharpoons V \ ^{3+} \, _{(aq)} \, + \, H_2O \, _{(l)}VO \ ^{2+} \, _{(aq)}\, + \, 2H^+ \, _{(aq)} \, + e^- \rightleftharpoons V \ ^{3+} \, _{(aq)} \, + \, H_2O \, _{(l)}VO \ ^{2+} \, _{(aq)}\, + \, 2H^+ \, _{(aq)} \, + e^- \rightleftharpoons V \ ^{3+} \, _{(aq)} \, + \, H_2O \, _{(l)}

When vanadium(III) is reduced to vanadium(II), the solution goes from green to violet. 

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​$$2V^{3+} {(aq)} \, + \, Zn {(s)} \rightarrow 2V \ ^{2+} {(aq)} \, + \, Zn^{2+} {(aq)} \\ V \ ^{3+} {(aq)} \, + e^- \rightleftharpoons V \ ^{2+} {(aq)} \\ E^{\circ}= -0.26 \, V$$

Vanadium(II) can be reduced to vanadium.

​$$V^{2+} {(aq)} \, + \, Zn {(s)} \rightarrow V {(s)} \, + \, Zn^{2+} {(aq)} \\ V \ ^{2+} {(aq)} \, + 2e^- \rightleftharpoons V {(s)} \\ E^{\circ}= -1.18 \, V$$

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​​Tollens' reagent   

The transition metal silver is usually has an oxidation number of +1. $$Ag^+$$ is easily reduced to silver metal because of the large electrode potential.

$$Ag^+(aq) \space + \space e^- \rightarrow Ag (s) \newline Standard \space electrode \space potential= +0.80 \, V$$​​

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Tollens' reagent uses the reduction of $$Ag^+$$ to distinguish between aldehydes and ketones.  

Tollens' reagent is prepared by adding a small amount to ammonia to silver nitrate solution, which produces a colourless solution containing $$[Ag(NH_3)_2]^+$$complex ions. 

When Tollens' reagent is added to an aldehyde, a silver mirror forms on the inside of the test tube. The $$Ag^+$$ ions are reduced to silver atoms and the aldehyde is oxidised to a carboxylic acid.

When Tollens' reagent is added to ketones a silver mirror does not form as Tollens' reagent cannot oxidise ketones.

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Concentration and spectroscopy 

The concentration of a solution can be determined using spectroscopy which measures how much light is absorbed.

White light Colorimeter 1. 2. Filter 3. Solution 4.

Procedure 

1.	White light is shone through a specific filter. This filters only allows light of a given colour to pass through depending on what the sample absorbs.
2.	Light passes through the sample and then a colorimeter. The colorimeter calculates how much light the sample absorbed. More concentrated coloured solutions will absorb more light.
3.	Measure the absorbance of multiple solutions of known concentrations. Use the results to plot a calibration curve.
4.	Measure the absorbance of the sample using the colorimeter and determine its concentration using the calibration curve.