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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 - Prussian Blue:
Fe4[Fe(CN)6],
                      Turnbull Blue:
Fe4[Fe(CN)6],

Related Question:



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.

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:



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 (↑) (↑)(↑)   →Δoct     (↑)(↑)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 (↑↓) (↑)(↑)     →Δoct      (↑)(↑) 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. 

Why [Mn(H2O)6]+2 is colourless although in which Mn+2 ion had five unpaired electrons ?

There are 5 unpaired electrons in Mn+2 ion and we know that.
                                 Spin multiplicity is = (2S+1) 
                     Where S is some of spin thus  S = 5×1/2=5/2
                                            Hence multiplicity = (2S+1)= 6 

we know multiplicity of excited state cannot be six, thus electronic transition in Mn+2  are 
spin forbidden, hence [Mn(H2O)6]+2 salt appear colourless.

Also [Mn(H2O)6]+2 has centre of symmetry and in such cases electronic transition are expected to be laporate forbidden.

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

The intense colour of the latter complex is strongly suggestive of a charge transfer transition and since the metal is already fully reduced (zero oxidation state), it is highly likely that this involves a MLCT transition. The π* levels of the bipy or CO ligands are possible acceptors but the fact that Fe(CO)5 doesn’t show this colour suggests that it is the bipy π* levels that are involved in Fe(bipy)(CO)3. Since there should be MLCT transitions to the CO π* levels as well, we assume that the lack of colour for Fe(CO)5 means that these transitions fall in the UV rather than the visible.

Why all the tetrahedral Complexes are high spin Complexes ?

The magnitude of crystal field splitting energy (CFSE) in tetrahedral Complexes is quite small and it is always less than the pairing energy. Due to this reason pairing of electron is energetically unfavorable. Thus all the tetrahedral Complexes are high spin Complexes. In fact no tetrahedral Complex with low spin has been found to exist.


FACTOR'S AFFECTING STABILITY OF COMPLEXES:

 (1) Charge on central metal cation:
In general stability of Complexes is directly proportional to the magnitude of charge on central metal atom. Thus Complexes of Fe +3 are more stable than Fe+2
(2) Size of central metal cation:
As size of the central atom ion decreases the stability of the Complex increases. This is applicable when Oxidation state of central metal ion is same in all the cases. For example Zn+2 < Cu+2 < Ni+2 < Co+2 < Fe+2 < Mn+2
(3) Nature of the ligands:
The size and charge of ligands is also an important factor of in deciding the stability of complexes.
(i) If the Ligand is smaller, it can approach the central metal ion more closely forming a stable bond.
(ii) High charge ligand will form a strong bond. Thus high charge and small size of ligands leads to Formation of stable Complexes.
(iii) Charge density of the ligand= charge/size. More the charge density more is the stability of the Complex. For example fluoride (F- ) will form more stable Complexes and Iodide (I-) forms least stable Complexes.
(4) Chelating effects:
Formation of five and six membered chelate from polydentate ligands enhance the stability of Complexes in comparison to monodentate ligands. This is called chelate effect of chelation. In general, the more number of chelate rings in the Complex, the more will be stability of the Complex.
(5) Formation constant of Complexes:
Stronger is the metal- ligand bond, less is the dissociation of Complex ion in the Solution and hence greater is the stability of complex. Thus the larger the numerical value of formation constant, the thermodynamically more stable is the complex.

What is d-d transition in complexes and explain colour of complex by d-d transition ?

Most of the transition metal compounds are coloured both in the solid state and in aqueous solution. This is because of the presence of incompletely filled d-orbitals.  When a transition metal compound is formed the degenerate d-orbitals of the metal  split into two sets, one having three orbitals dxy, dyz and dxz called t2g orbitals with lower energy and the other having two orbitals dx2 –y2 and dz2 called eg orbitals with slightly higher energy in an octahedral field. This is called crystal field splitting When white light falls on these compounds, some wavelength is absorbed for promotion of electrons from one set of lower energy orbitals to another set of 
slightly higher energy within the same d-subshell. This is called d-d transition. The remainder light is reflected which has a particular colour.


The colours of some 3d metal ions:

SN
d-configuration
Examples with colour
1
d0 (No d-d transition)

2
d1
Ti3+ (3d1) Purple,  V+4(3d1) Blue
3
d2
V+3(3d1) Green
4
d3
Cr3+ (3d3) Violet green
5
d4
Mn+3(3d4) Violet , Cr2+ (3d4) Blue
6
d5
Mn+2(3d5) Pink, Fe+3(3d5) Yellow
7
d6
Fe+2(3d6) Brown , Co+2(3d6) Green ,
8
d7

9
d8
Ni+2(3d8) Green
10
d9
Cu+2(3d9) Blue
10
d10
Sc+3(3d0)  colourless

ILLUSTRATION (1): The mechanism of light absorption in coordination compounds is that photons of appropriate energy can excite the coordination entity from its ground state to an excited state. Consider [Ti(H2O)6]3+  In which  Ti(+3) ion has one electron in d sub shell ( in lower energy t2g d- orbital) . In aqueous solution, [Ti(H2O)6]3+. Appear as purple due to the absorption of light from visible range ( green and yellow portion) resulting  d-d transition ( electron jump from t2g level to eg level) as result  complex has complementary ie purple .
The variety of color among transition metal complexes has long fascinated the chemists.

FACTOR'S AFFECTING MAGNITUDE OF CFSE:

CRYSTAL FIELD STABILISATION ENERGY(CFSE)

Magnitude of CFSE depends upon the following factors.
(1) Nature of central metal cation: the the value of CFSE depends other following factors of central metal cation as given as
(a) For the Complex having same geometry and same ligands but having different numbers of d-electrons then CFSE decrease on increasing number of d-electrons in the central metal cation.
(b) When numbers of d-electrons are same then CFSE increasing on increasing Oxidation number.
(c) For same Ligand, Oxidation state, same d-electrons, CFSE increasing on increasing principle quantum Number of d- orbitals like 3d < 4d < 5d etc.
Thus the elements of the second and third transition series have greater tendency to form low spin complexes than the first transition series. It is possible to arrange the metals according to a spectrochemical series as well. The approximate order is
(2) Nature of ligands: the magnitude of CFSE varies from stronger ligand to weaker ligands it meant CFSE increasing on increasing of splitting power of ligands and decreasing on decreasing of splitting power of ligands.

Splitting power of ligands decide according to spectrochemical series.

(3) Geometry of the Complex: the value of CFSE will change with geometry of Complexes. It is estimated that CFSE of tetrahedral Complexes is approximately 50% as large as that of octahedral Complexes.
ILLUSTRATIVE EXAMPLE (1): Which of the Complex of the following pairs has the largest value of CFSE?
(1) [Co(CN)6]3-  and [Co(NH3)6]3+
(2) [Co(NH3)6]3+ and [CoF6]3-
(3) [Co(H2O)6]3+  and [Rh(H2O)6]3+
(4) [Co(H2O)6]2+ and [Co(H2O)6]3+
SOLUTION:
(1)  CN is the stronger ligand than NH3 therefore CFSE of [Co(CN)6]3-  will be more than  [Co(NH3)6]3+
(2) NH3 is stronger ligand than F therefore CFSE of [Co(NH3)6]3+ will be more than [CoF6]3- .
(3) Co belong to 3d series whereas The Rh belong to 4d series. More the value of n more is CFSE therefore CFSE of  [Rh(H2O)6]3+  is more than [Co(H2O)6]3+ .
(4) Oxidation number of Co in [Co(H2O)6]3+ is more than the Oxidation number of [Co(H2O)6]2+  therefore, CFSE of [Co(H2O)6]3+ is more than  [Co(H2O)6]2+ .

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