Colour of Complexes due to charge transfer:

Colour originated by charge transfer when electronic transition occurs from one part of the Complex to other part i.e.  Such type is also called internal Redox reaction.
Intensity of colour in such type transition is very high as they do not require following any selection rule.
(A) Charge transfer from ligands to metal:
Examples , MnO₄ ^- , MnO₄ ^-2 , CrO₄ ^-2, Cr₂O₇ ^-2 , [Fe(H₂O)₅(NO)]⁺² , Na₄[Fe(CN)₅(NOS)].

(B) Charge transfer from metal to ligands:
Examples, [Fe(CO)₅] , [Fe (pi-C₂H₅)₂],  [Cr (pi-C₆H₆)₂],

(C) Charge transfer from metal to metal:
Examples:

(2) Prussian Blue: Fe₄[Fe(CN)₆]3 

IUPAC Name: Iron(III)hexacyanoferrate(II) 

Common Name:

(Ferric ferrocyanide)

(2) Turnbull Blue: Fe3[Fe(CN)₆]2

IUPAC Name: Iron(II)hexacyanoferrate(III) 

Common Name:

(Ferrous ferricyanide)

Related Question:

(1) Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

(2) Which of the Complex of the following pairs has the highest value of CFSE?

(3) Colour of Complexes due to charge transfer:

(4) Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 

(5) Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?

(6) Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ?

 
(7) Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?

(8) Why all the tetrahedral Complexes are high spin Complexes ?

Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ?

[Ti(H₂O)₆]Cl₃ is an octahedral violet colour complex; the violet colour of this complex is due to d-d transition of a single (d¹) electron from t₂g lower level to eg higher level by absorbing corresponding energy of yellow green region of light and emitted energy corresponding to violet region and hence appears violet in colour. But on heating it is dehydrated and water molecules (ligand) removed so in absence of ligand splitting of D orbitals does not occur hence it becomes colourless.

Similarly  anhydrous CuSO₄ is white , but CuSO₄ .5H₂O is Blue in Colour.

Related Question:

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ? 
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

COLOURS OF COMPLEXES AND SELECTION RULE:

Electronic transitions in a complex are governed by Selection rules A selection rule is a quantum mechanical rule that describes the types of quantum mechanical transitions that are permitted.they reflect the restrictions imposed on the state changes for an atom or molecule during an electronic transition. Transitions not permitted by selection rules are said forbidden, which means that theoretically they must not occur (but in practice may occur with very low probabilities).

(1) Laporte Selection Rule:

Laporte Selection Rule is given by Otto Laporte a German American Physicist

According to Laporte selection rule only allowed transitions are those occurring with a change in parity (flip in the sign of one spatial coordinate.) OR During an electronic transition the azimuthal quantum number can change only by ± 1 (Δ l = ±1) The Laporte selection rule reflects the fact that for light to interact with a molecule and be absorbed, there should be a change in dipole moment.

Practical meaning of the Laporte rule:

Laporte allowed transitions: are those which occur between gerade to ungerade or ungerade to gerade orbitals.

Laporte forbidden transitions: are those which occur between gerade to gerade or ungerade to ungerade orbitals.

Gerade = symmetric with respect to centre of inversion i.e. atomic or molecular orbital with center of symmetry or number of nodal plane = 0, 2, 4 (even number)

Ungerade = anti symmetric with respect to centre of inversion i.e. atomic or molecular orbital without center of symmetry or number of nodal plane = 1, 3, 5, (odd numbers)

Important Note: 

This rule affects Octahedral and Square planar complexes as they have center of symmetry. Tetrahedral complexes do not have center of symmetry therefore this rule does not apply

(2) Spin Selection Rule:

Spin selection rule states that transitions that involve a change in spin multiplicity as compare to ground state  are forbidden. 

(1) According to this rule, any transition for which Δ S = 0 (it means no change in spin multiplicity after d-d transition) is allowed. 

(2) If Δ S ≠ 0 ( change in spin multiplicity after transition) then it is  forbidden (transition not allowed)

ILLUSTRATIVE EXAMPLE: [Mn(H₂O)₆]²⁺ and [FeF₆]^3-  both have a d⁵ configuration and high-spin complexes. Electronic transitions are not only Laporte-forbidden, but also spin-forbidden. The dilute solutions of Mn²⁺ and Fe ⁺³ complexes are therefore colorless

Important Note:

For first transition series d⁵ system, weak ligand field, and coordination number six (6) Complexes are found to be colourless due to violation of selection rule.

Related Question:

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ? 
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?

In case of [Ni(CN)₄]^-2 Ni is in +2  and  3d⁸ configuration but presence of strong ligand (CN) , the two unpaired electron in the 3d orbital pair up thus there is no unpaired electrons present hence it is colourless.

In case of [Ni(H₂O)₄]⁺² Ni is in +2  and  3d⁸ configuration and two unpaired electrons which  I do not pair up in the presence of weak ligand (H₂O) , hence it is colour  due to  d-d transition , red colour light is absorbed and give it complementary  colour green.

Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ?

[FeF₆]^3- is a Fe (III) complex hence [Ar] 3d⁵. F^- is a weak field ligand and the complex is high spin: 

                                     t₂g (↑) (↑)(↑)   →Δ_oct→     (↑)(↑)eg.

And there is 5 unpaired electrons hence it's spin multiplicity = (2S+1) = 6  , and multiplicity of  a excited is cannot be six thus The transitions in Fe ⁺³ ion are spin forbidden and are extremely weak so as to make [FeF₆]^3- almost colorless.

[CoF₆] ^3- this is a Co (III) complex [Ar] 3d⁶. Again F^- is a weak field ligand and [CoF₆]^3- is one of only two common cmplxs that is high spin:

                       t₂g (↑↓) (↑)(↑)     →Δ_oct→      (↑)(↑) eg

Only one band is expected namely the t₂g→eg transition (d¹). The complex is blue consistent with this transition being in the low energy red.