Electronic Spectroscopy 1

ELECTRONIC SPECTROSCOPY This can also be called as ultraviolet-visible-near infrared spectroscopy. This mainly deals with the electronic excitation in molecules. The region from 200 to 1100 nm will be useful in the study of complexes. This study helps to determine the structure of a complex in the following ways: 

This spectrum helps to determine the oxidation state of the central metal atom and thus the number of delectrons in the metal.



The geometry of the complex like octahedral, tetrahedral, square planar etc.

In order to arrive at these, the following steps are followed: 

The number and position of absorption maxima are identified which in turn gives the number of transitions in the complex.

The oxidation state of the central metal ion is determined from the number of transitions. 

Then the geometry that is in agreement with these transitions is proposed.

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In order to understand the electronic spectroscopy, the following steps must be followed: Electronic configuration Terms Term Symbol Spectroscopic

Term

Symbol Splitting of the Terms under Various Fields Splitting of d1 – d9 systems Energy Level Diagram Energy Levels and their Dependence on Crystal Field Orgel Diagram Tanabe – Sugano Diagrams Distortions: Jahn – Teller Effects Electronic Configuration This tells how the electrons are arranged in a certain set of orbitals. In other words, the electronic configuration assigns a

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given number of electrons to a certain set of orbitals. For example, if there are six electrons in an atom, they are arranged as 1s22s22p2. Another example is d5 system. This configuration merely tells that there are five electrons in d-orbitals. But it does not tell how these electrons are arranged within the different d-orbitals. Terms This tells us how the given numbers of electrons are arranged within a particular orbital and how many different arrangements are possible. Each arrangement has its own energy and there are different numbers of arrangement possible from a configuration. Thus an electronic configuration gives rise to a number of terms. Example. Let us consider the electronic configuration d4. This gives rise to the following arrangements: t2g3eg1; t2g2eg2; t2g1eg3; t2g4eg0; t2g0eg4 etc. Each one represents one energy level or term. In the above example, we can say that the d4 electronic configuration has given rise to five terms or five energy levels. Term symbol

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The various energy levels are represented by symbols called term symbols. The following steps derive the term symbols: Step 1: Write the electronic configuration. Step 2: Omit all the closed shells and sub shells. Step 3: Determine the different electron arrangements allowed by the Pauli exclusion principle (Pigeonhole diagram). Step 4: Determine all the possible values of ML and MS of the different arrangements. Step 5: Check up with the formula N = Nl! / {x! ( Nl – x)! } where Nl = 2(2l+1) and x = number of electrons. Step 6: Set up a chart of microstate consisting of one combination of ML and MS. Step 7: Resolve the chart of microstates in to appropriate atomic states: (2S+1) columns and (2L+1) rows. Step 8: Determine the J values: L+S , L+S-1, L+S-2, ……….., |L-S|

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Step 9: Determine the ground state by applying the three Hund’s rules: 

2S+1 should be maximum.



If more than one term possesses the same multiplicity, the one with the highest L value is the ground state.

If shells are less than half-full, states with lower J value are lower in energy. Example 1: Carbon Step 1: Write down the electronic configuration: 1s22s22p2 Step 2:Omit all the closed shells and sub-shells omit 1s2 and 2s2 . Step3 & 4: Determine the different electron arrangements allowed by the Pauli exclusion principle. Determine all the possible values ML and MS of the different arrangements by Pigeonhole diagram:     

 

   

 

 

 

 

 

 

5





 

Step 5: Check up the arrangements with the formula, N = Nl / {x!(Nl-x)!}; Nl

=

2(2l+1); x = No. of electrons. l=1 for ‘p’

orbital. Nl = 2(2.1 + 1) = 6; x =2.

N = 6! / 2!(6-2)! =

6.5.4.3.2.1 / 2.1.4.3.2.1 = 15. There are 15 columns in

the

pigeon hole diagram and hence correct. Step 6: Step 7: Resolve the pigeonhole arrangement in to appropriate atomic states. The highest ML = +2. This corresponds to symbol ‘D’. Underline ML = +2, +1, 0, -1, -2. ML = +2 appears only once. Hence, 2S+1 = 1. Therefore, these constitute the term 1D. Next higher value of ML = +1; the symbol is P. ML = +1 occurs thrice. Therefore, the multiplicity is 3. Underline ML = +1, 0, -1. Therefore, the term symbol is 3P. Now, we are left with ML = 0. The symbol is S. ML occurs only once. Therefore, 2S+1 = 1. Hence, the term symbol is 1S. Thus, the pigeonhole diagram is resolved as 1D, 3P, 1S. II Method

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The maximum ‘L’ value is +2. The different L values are +2,+1,0,-1,-2. MS = +1,0,-1. Now the following table is constructed. MS +1

0

-1

2

1+,1-

1

1+,0+

1+,0-; 1-,0+

1-,0-

0

1+,-1+ 1+,-1-;1-,-1+ 1-,-10+,0-

-1 -2

-1+,0+ -1+,0-;-1-,0+ -1-,0-1+,-1-

Resolve the above microstate chart in to appropriate atomic states; that is, (2S+1) columns and (2L+1) rows. (a)

In the above chart, ML = 2 occurs only once. That

is, it is a singlet state. ‘2’ corresponds to the symbol ‘D’. Therefore, the atomic state is 1D. An atomic state forms an array of microstates consisting of 2S+1 columns and 2L+1 rows. For 1D state, 2S+1 = 1; i.e. one column. 2L+1 = 2(2) + 1 = 5 rows. These are marked and removed from the chart. (b)

ML=1 occurs thrice. Hence, it is a triplet. ‘1’

corresponds to the symbol ‘P’. Therefore, the atomic state

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occurs in three columns; 2L+1 = 2(1)+1 = 3 rows. Again these are marked and removed from the chart. (c)

Now, we are left with only one microstate with

ML=0. This occurs in only one column. Hence, the multiplicity is ‘1’. ‘0’ corresponds to the symbol ‘S’. Therefore, the atomic state is 1S. Thus the p2 configuration gives rise to three terms, viz., 1S, 3P and 1D. Step 7: Determination of the ‘J’ values. J=L+S 1

D Term: 2S+1 = 1; i.e. only one ‘J’ value is possible. L = 2; S = 0; J = L+S = 2+0 = 2. Therefore, the term symbol is 1D2. 1

S Term: 2S+1 = 1; i.e. only one J value is possible.

L = 0; S= 0; J=L+S=0. Therefore, the term symbol is 1S0. 3

P Term: 2S+1=3. i.e. three J values are possible.

L=1; S=1; J=L+S= 1+1 =2. The possible J values are: L+S,L+S-1,……|L-S| ; i.e. 2,1,0. The different P terms are; 3

P2,3P1,3P0

Determination of ground state:

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Hund’s Rule 1: 2S+1 should be a maximum. Therefore, of the three values,1D,1S and 3P, 3P is the ground state. Hund’s Rule 2: There are two singlet terms, viz., 1D and 1S . The one with the highest L value has lower energy. Therefore, 1

D has lower energy than 1S. The order of the terms is: 3

P<1D<1S 1

3

D

P

Hund’s Rule 3: If shells are less than half full, states with lower J value are lower in energy.i.e.3P0<3P1<3P2. Thus the complete arrangement of energy levels is given below: 1

S

……………..

1

D

……………..

1

1

3

3

3

P0

9

D2

P2 3

P

S0

P1

Example 2: d2 system: Step 1: The electronic configuration is 1s22s22p63s23p63d2 Step 2: Omit the closed shells and sub shells. Therefore, we are left with 3d2. Step 3. Determine all the possible values of ML and MS. (a) There are five ‘d’ orbitals. +2 +1 0 -1 -2 (b) The maximum value of L is +4:  +2 +1 0 -1 -2 (c) The different values of ML are: 4,3,2,1,0,-1,-2,-3,-4. (d) The possible spin values are: +1 0 -1    Step 4: Determine all the electronic configurations allowed by Pauli exclusion principle.

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MS ML

+1  ++

4 3 2

2+1+ 2+0+

1

2+-1+ 1+0+

0

2+-2+ 1+-1+

0  +2+22+1- ; 2-1+ 2+0- ; 2-0+ 1+12+-1-, 2--1+ 1+0- , 1-0+

-1  --

2+-2- ,2- -2+ 1+-1- ,1- -1+ 0+0-

2- -21- -1+

2-12-02- -11-0-

Step 5: Check up with the formula

N = Nl! / x!(Nl – x)! Where Nl = 2(2l + 1); x = No. of electrons. Nl = 2(2l + 1) = 2( 2 x 2 + 1) = 10; x = 2 N = 10! / 2!(10 –2 )! = 10.9.8.7.6.5.4.3.2.1. / 2.1 .8.7.6.5.4.3.2.1. = 10 x 9 / 1 x 2 = 45 There are 45 microstates in the above chart. Hence correct. Step 6: Resolve the above microstate chart in to appropriate atomic states; that is, 2S+ 1 columns and 2L + 1 rows. In the above chart, ML = 4 occurs only once. Hence, it is a singlet state. ‘4’ corresponds to the symbol ‘G’. Hence, the atomic state is 1G. An atomic state forms an array of microstates consisting of 2S+1 columns and 2L+1 rows. These are marked and removed from the chart. ML = 3 occurs thrice, i.e. in three columns. Hence, it is a triplet state. ‘3’ corresponds to the symbol ‘F’. Therefore, the 11

atomic state is 3F. For a 3F state, there are 2S+1 = 3 columns and 2L+1 = 7 columns (2 x 3 + 1 = 7). These are marked and removed from the chart. ML = 2 occurs only once, i.e. in only one column. Hence, it is a singlet state. ‘2’ corresponds to the symbol ‘D’. Hence, the atomic state is 1D. For a 1D state, there are one column and 2L+1, i.e., 2 x 1 + 1 = 5 rows. These are marked and removed from the chart. ML = 1 occurs thrice, i.e., in three columns. Hence, it is a triplet state. ‘1’ corresponds to the symbol P and hence, the atomic state is 3P. For a 3P state , there are 2S+1 = 3 columns and 2L+1 = 3 rows. These are marked and removed from the chart. ML = 0 occurs only once, i.e., in only one column and hence it is a singlet. ‘0’ corresponds to the symbol S and hence the atomic state is 1S. For a 1S state, there are 2S+1 = 1 column and 2L+1 = 1 row. Thus the different terms are: 1G, 3P, 1D, 3P and 1S. Step 7: Determine the J values. J = L+S. For a 1G term, S = 0 and L = 4. L + S = 4. Since it is a singlet state, only one ‘J’ value is possible. Therefore, the term symbol is 1G4 . For a 3F term, S = 1 and L = 3. L+S=4;L-S=2. The possible ‘J’ values are 4,3 and 2. Hence the different terms are 3F4,3F3 and 3F2. J = 2 has the lowest energy since the ‘d’ level is less than half full. Therefore, the order of the energy levels is 3F2<3F3<3F4. For a 1D term, S = 0 and L = 2. L = S = 2. Since it is a singlet state, there is only one J value. The term is 1D2. In the case of a 3P term, S = 1 and L = 1. L+S=2 and L-S=0. The possible ‘J’ values are 2,1 and 0. Since the ‘d’ level is less

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than half full, J=0 has the lowest energy. Hence, the order of the energy levels is 3P0<3P1<3P2. Step 8: Determination of the ground state. Rule 1. That state with the highest multiplicity has the lowest energy. Therefore, 3F and 3P have lower energy than others. Of these two states, 3F has the lowest energy because its ‘L’ value is greater. Hence the order of the terms is: 3 F<3P<1G<1D<1S The complete arrangement of the energy levels is given below: Degeneracy (2S+1)(2L+1) 1 S ……. 1S0 1 x1 = 1 1

D

……. 1D2

1x5=5

1

G

…….. 1G4

1x9=9

3 3

P2 P1 3 P0 3

P

3

3

F4 3

F 3

3x3=9

F3

3 x 7 = 21

F2 Total

= 45

Degeneracy of a term: The degeneracy of each term is given by the formula (2S+1)(2L+1):

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3

F term: 3 x 7 = 21

3

P term: 3 x 3 = 09

1

G term: 1 x 9 = 09

1

D term: 1 x 5 = 05

1

S term: 1 x 1 = 01 Total

= 45

Spectroscopic Terms or

Molecular Terms

When atomic terms are surrounded by ligands, these terms are split and molecular or spectroscopic terms result. In an octahedral field, the ‘S’ and ‘P’ terms are not split. The ‘D’ and ‘F’ terms are split as shown in Figure (a) and Figure (b). Eg D

A2g T2g F

T2g T1g Figure (a)

Figure (b)

Reason for splitting of the atomic terms

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The atomic term split due to electron interaction. In an atom, there is no interaction between electrons in an unfilled shell. However, in an ion, the electrons present in an unfilled shell suffer two interactions: (i) Inter-electronic repulsion. (ii) Spin-orbit coupling. Inter-electronic repulsion The electrons in a shell repel each other. Therefore the energy depends up on how the electrons occupy the orbitals of the shell relative to one another. Spin-orbit coupling The magnetic field produced by the spin of an electron interacts with that produced by the orbital motion of the electron. This is called spin-orbit coupling. Effect of inter-electronic repulsion: The inter-electronic repulsion splits the highly degenerate energy level of the unperturbed configuration into a number of terms.

Effect of spin-orbit coupling: The spin-orbit coupling splits the terms into states. These effects are schematically represented in the Figure 1. 1 S 1 1

D G

d2

3 3

P2

P 3

3

P1

F

terms states Figure 1. Effect of Spin-Orbit Coupling

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Number of states arising from terms: The number of states arising from a term is (2S+1) or (2L+1) whichever is smaller. These effects are schematically represented as follows: Electronic configuration Electronic repulsion Terms

Spin-orbit coupling States Magnetic field Micro states

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Specification of a State When a particular state arises from a term, the value of ‘J’ is added as a subscript to the term symbol. The full symbol of a state is: 2S+1 LJ; e.g. 3P1 Racah Parameters The inter-electronic repulsion is responsible for the differences in energy between the terms. It is expressed in terms of Racah parameters B and C. Significance of ‘B’ and ‘C’ When the terms have the same multiplicity, the Racah parameter ‘B’ is used. When the terms have different multiplicities, the inter-electronic repulsion is expressed by a combination of ‘B’ and ‘C’. Kinds of Coupling There are two types of coupling, viz., Russell-Saunders coupling and jj coupling. They are used depending upon the circumstances. RS coupling RS coupling is used when the electronic repulsion is greater than the spin-orbit coupling. This coupling splits each of the terms into states. RS coupling is valid for most of the lighter elements because the effects of spin-orbit coupling are small in these cases. In particular, this coupling applies well to the first transition series elements. RS coupling does not apply the heavy elements like the second and third transition series elements. In these cases, the spin-orbit coupling is not small but greater than the interelectronic repulsion. In particular, this fails in the case of third

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transition series elements. The degeneracy of each state obtained by the splitting of terms is given by the formula, (2J+1). jj coupling It is exactly opposite to that of RS coupling. Here, the spinorbit coupling is more important than the electronic repulsion. At first, the configuration is split into levels and not terms. The ions considered under ligand-field theory do not conform to the jj coupling scheme.

Spin-orbit coupling parameters Two parameters are used to describe the spin-orbit coupling. They are: (i) Single electron spin-orbit coupling parameter, (ii) Parameter, (i) Parameter, This parameter is a property of the configuration. This measures the strength of the interaction between the spin and orbital angular momentum of a single electron of the configuration. This is equal to  times the scalar product between the spin and orbital angular momentum vectors of the electron, viz., l.s.  is a positive quantity. (ii) Parameter,  This property is a property of the term. This is equal to  times the scalar product between the spin and orbital angular momentum of the term, viz., L.S Relation between the two parameters The two parameters are related as follows:

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 =   / 2S The ‘+’ sign holds good for a shell less than half full. The ‘-’ sign holds good for a shell more than half full. Sign of   is positive for a shell less than half full and the lowest value of ‘J’, viz., L-S lies lowest in energy. Normal multiplet When a term is split into a set of states of increasing value of ‘J’, it is said to be a normal multiplet. Inverted multiplet When the shells are more than half full,  is negative. This inverts the diagram. In the resulting multiplet, the lowest energy will have the highest value of ‘J’. These are summarized in Table 1. To find out  The spin-orbit contribution to the energy of any level is found out by the following expression: ½ [J (J+1)-L (L+1)-S (S+1)] For a 3p state, when J=0;S=1;L=1. Substituting the values we get, ½ [0-1(1+1)-1(1+1)]=1/2(-4)=-2 Slater-Condon and Racah parameters Energies of the terms above the ground term are determined by two electron repulsion parameters, viz., the Condon-Shortley or Racah parameters. The two set of repulsion parameters are related as follows: B = F2 –5F4 C = 35F4 B and C are called Racah parameters and F’s are called SlaterCondon parameters.

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Advantages of Racah parameters The separations between the terms of same multiplicity within a configuration involve only the parameters ‘B’ and that between different multiplicities involve both ‘B’ and ‘C’. However, in the case of Condon-Shortley parameter, separations between the terms of even the same multiplicity are functions of F2 and F4. Characteristics of d-d transitions d-d transitions are also known as ligand field spectra. These occur in the near i.r., visible and u.v. regions, viz., 10000 – 30000 cm-1. The lower regions are not accessible experimentally. The higher frequencies are accessible but are overshadowed by the charge transfer and the inter-ligand transitions. Hence, the study of the d-d transitions is limited to the visible regions of the spectrum. Selection rules 1. Spin selection rule: This rule may be stated as follows: If the spin multiplicity changes, those transitions are not allowed. (e.g.) s2 s1p1 is not allowed because it is a singlet to triplet transitions (the spin of the two electrons is +1/2.) . But if the spin of the two electrons in the s1p1 state is +1/2 and –1/2 , it becomes a singlet to singlet transition and is allowed. 2. Laporte selection rule: This rule is to be applied if a molecule has a center of symmetry. This rule can be stated as follows: “ If the molecule has center of symmetry, transitions within a given set of ‘p’ or ‘d’ orbitals are Laporte forbidden. That is, those involve only a redistribution of electrons in the given sub shell, are Laporte forbidden.” In other words, d-d transitions in octahedral complexes are forbidden. Any electron transition must have l = 1. Thus only the transitions between an even state ‘g’ and an uneven

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state ‘u’ are allowed. That is, g to g and u to u transitions are not allowed. But g u and u g transitions are allowed. Breakdown of the selection rules Spin selection rule: A d5 ion has high pairing energy. Even then spin forbidden transitions are observed. But the intensities are very low. The reason for the breakdown of the rule here is that the spin-orbit angular momentum coupling mixes the singlet and triplet states. Even 1% mixing of the two states makes the ground state 99% singlet and 1% triplet. The excited state becomes 99% triplet and 1% singlet. Now the intensity is derived from singlet-singlet and triplet-triplet transitions. The extent of mixing depends up on the differences in energies between the states and spin-orbit coupling constants. Thus, the octahedral spin-free complexes of d5 ions (Mn2+and Fe3+) must gain whatever intensity they can through the breakdown of the spin multiplicity rule because all the higher excited states have different and lower multiplicity than the ground states, viz., 6S. Laporte selection rule: The Laporte selection rule can be broken down by mixing of the orbitals on the metal ion, viz., d-f or d-p or by vibronic coupling. Orbital mixing. The ligand vibrations may produce a temporary distorted field and the orbitals can mix. Now the metal atom M will not be at the center of the symmetric field all the time during which electron transition takes place. In tetrahedral complexes, there is no center of symmetry and therefore d-p orbital mixing leads to more intense absorption bands than those for the octahedral complexes. Vibronic coupling. This is nothing but the vibrational and electronic coupling. This removes the center of symmetry. If a

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forbidden band lies near an allowed band, the two will mix due to vibronic coupling and intense absorptions will be obtained. This is called intensity stealing and depends up on the energy differences between the two states. Band widths and band shapes The width of absorption bands for the d-d transitions at room temperature is about 1000 cm-1, which is broad. The reasons are: (i) Molecular vibrations (ii) Spin-orbit coupling (iii) Jahn-Teller distortions The multiplicity forbidden bands are sharp while the multiplicity allowed bands are broad because the excited state has longer bond lengths and the electronic transitions are faster than the vibrational transitions. d-d transition spectra The following points must be understood in order to understand the d-d transitions in metal complexes. (i) Russell-Saunders states for a dn ion in the order of increasing energies. (ii) Group theory and quantum mechanics to find out the effect of an external field on these states. All the energy levels of the system must be known to explain the spectra. Splitting of ‘d’ orbitals in an octahedral field The ‘d’ orbitals are split into t2g and eg sets. The ‘s’ orbital is not split because it is spherically symmetrical. The ‘p’ orbitals are not split because all the three orbitals interact to the same extent. The ‘f’ orbitals are split into t1g, t2g and a2g sets. The splitting of the ‘f’ orbitals is shown below:

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a2g 10Dq t2g f

2Dq

--------------------------------------------------6Dq t1g The atomic states transform in an octahedral field as follows:

S ---------------------------------------- A1g P --------------------------------------- T1g D ---------------------------------------- T2g+Eg F ---------------------------------------- T1g+T2g+A2g G --------------------------------------- T1g+T2g+Eg+A1g H ---------------------------------------- T1g+T1g+T2g+A1g I ----------------------------------------- T1g+T2g+Eg+A1g+A2g The low energy states for all dn configurations are S,P,D or F and arise as shown below: d1 and d9 ------------- 2D d2 and d8 ------------- 3F and 3P

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d3 and d7 ---------------- 4F and 4P d4 and d6 ---------------

5

D

d5 ----------------------- 6S Hole formalism The dn system may be viewed by means of the ‘hole formalism’. According to this, a d9 system is described as spherically symmetrical d10 system with a hole or missing electron. From this point of view, the hole behaves exactly the way the electron does. But the difference is that, instead of finding the lowest available orbital, the hole tends to ‘float’ to the top. An analogous scheme for viewing such systems is to consider a dn system as an inverted d10-n system as shown in Figure 2. hole floats d9configninverted d1 x2-y2 eg eg z2

b2g

xy

a1g

t2g xz,yz b1g hole Figure 2 Hole formalism

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In other words, a d10-n configuration will behave in the same way as the corresponding dn configuration except that all energies of interaction with the environment will have the opposite sign. The following figures show the splitting for D and F terms in an octahedral environment. Eg A2g 6Dq 10Dq D T2g -4Dq F 2Dq T2g -6Dq T1g Figure(a)

Figure(b)

Example 1. Cu(II) – d9 ion is same as inverted d1 Eg T2g D T2g

Eg

d1

d9 inverted d1

Example 2. Ni(II) – d8 ion is same as inverted d2 d2 P inverted d2 T1g(P) A2g 10Dq T1g(F) T2g F F T2g 10Dq T1g A2g

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Inversion applies only to the ‘F’ states. 3F is always lower in energy than 3P (whether from d2 or d8). d5+n is identical to that of dn. i.e. d7 = d5+2 That is, the diagram for d7 is identical to that of d2. d3 = d5-2. i.e. the diagram for d3 is the inverted diagram of d2. Note: + sign indicates no inversion and ‘-’sign indicates inversion. Always express in terms of 1 or 2 for easy prediction. The results are shown in Table 2. Example: Cr(III) ion. It is a d3 ion.  d5-2 i.e. inverted d2 diagram as given below: -- 4P

4 4

T1g(P) T1g(F)

6Dq 4

F 2Dq 4

T2g

10Dq 4

The transitions are: 4 A2g

4

A2g

T2g(1)

4

4 A2g T1g(2) The first transition seems to be equal to 10Dq from the diagram for the inverted ‘F’ state. The second transition would be expected to be equal to 18Dq. But it is less than 18Dq. The reason is that the ground state4F mixes with the excited 4P level as shown in Figure 3.

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Table 1 Relation between the sign of  and the multiplets Free ion term (2S+1)(2L+1) degenerate 1

S (1) S=0 L=0 1 G (1)(9) S=0 L=4 3 P (3)(3) =9 S=1 L=1 L+S=2 1 D (1)(5)=5 S=0 L=2 J=2 3 F (3)(7)=21 S=1 L=3 J=4

d2( +ve) (2J+1)degenerate (L-S) lowest (1)

J=0

2(4)+1=9

J=4

d8( -ve) (L+S) lowest

(1)J=0

(9) J=4 (1) J=0

(5)  J=2 (3) J=1 (3) - J =1 (1) -2 J=0

(5)

J=2

(9) 3 J=4

(7) - 3 (5) -4 2

27

(5) J=2

(5) J=2 -4 2 -

3

3

4

Table 2 Energy level diagram of dn states

Confign.

G.S of Free ion

Confign. of complexed ion

Energy level diagram

d1

2

D

t2g1

Fig.(a)

d2

3

F

t2g2

Fig.(b)

d3

4

F

t2g3

Inverted Fig. (b)

d4

5

D

t2g3eg1

Inverted Fig. (a)

d5

6

S

t2g3eg2

No splitting

d6

5

D

t2g4eg2

Fig. (a)

d7

4

F

t2g5eg2

Fig. (b)

d8

3

F

t2g6eg2

Inverted Fig. (b)

d9

2

D

t2g6eg3

Inverted Fig. (a)

d10

1

S

t2g6eg4

No Splitting

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4

P

mixing 4

T1g(P)

15B 4

6Dq

4

2Dq 4

T1g(F)

<18Dq T2g

F 10Dq 4

Figure 3

A2g

Orbital mixing

The wave functions for 4T1g(F) and 4T1g(P) states have identical symmetry. Hence, they will mix. The amount of mixing is inversely proportional to the energy difference between the 4F and 4P levels. This is similar to the LCAO of two atomic orbitals giving rise to two molecular orbitals, one having higher energy and another having a lower energy then the contributing orbitals. Due to this mixing, the 4T1g(F) will lie somewhat lower in energy and the 4T1g(P) will lie somewhat higher in energy than the expected values in the absence of mixing.

Comparison of Experimental and Theoretical data The experimental data do not agree with the theoretical data. Two corrections must be made in order to improve the interpretation and correction of the spectra.

29

The extent of mixing of the F and P terms must be considered while calculating the energies.  The nephelauxetic effect must also be taken into account. Due to nephelauxetic effect, the interionic repulsion is reduced. Hence, the apparent value of ‘B’ is always smaller in complexes when compared to the free ion. If B represents the repulsion in the complex, then 

 = B/B ------- (1)  is always less than ‘1’ and decreases when the delocalisation increases. It can be obtained from the following relation: (1 - ) = hx.kM---------- (2) where hx is the parameter for the ligand and kM is for the metal. In the case of Cr(III), if all the three transitions are obtained, B can be found out easily by applying the following equation: 15B = 3 + 2 - 31 -------- (3) where 1 is the absorption occurring at the lowest frequency. If only two transitions are observed and the third one is obscured by a charge transfer band, B can still be calculated but more complicated.

ORGEL DIAGRAM The magnitude of splitting of the energy levels for an ion is plotted against the increasing ligand field strength for a dn system taking in to consideration the spin – orbit coupling and mixing of the different energy states. The resulting diagram is called the Orgel diagram for the dn ion. In other words, Orgel diagram is the pictorial representation of the energy level diagrams obtained by 30

incorporating the adjusted B and C values including the effect of variable field strength. The diagram is shown in Figure 4.

Figure 4.

Orgel Diagram

The Orgel diagram for the P and F states in octahedral as well as in the tetrahedral field is shown in the Figure 4. Characteristic features of Orgel diagram 1. Orgel diagram treats only the weak field or high – spin case. 2. States with identical designations never cross. 3. The crystal field states have the same spin multiplicity as the free ion states from which they originate.

31

4. States that are the only ones of their type have energies that depend linearly on the crystal field strength. 5. When there are two or more states of identical designations, their lines in general will show curvature because such states interact with one another as well as with the crystal field. Correlation diagrams and analysis of the spectra If a spectrum is to be completely analyzed, a correlation diagram showing all energy terms (ground as well as the excited state) should be shown and analyzed. That is, the following points must be taken into account:  Free ion energy terms  Terms in a weak field  Variations in the field of intermediate strength  Terms in a strong field configuration d1 ion: The ground term is 2D. This splits in to high energy Eg level and low energy T2g level. The electron transition from the lower to the upper level is a direct measure of the ligand field strength. 2

Eg

E 2

D

10Dq 2

Free ion

o Field Strength

32

T2g

d2 ion: When more than one electron is present, the energy levels depend upon the following factors:  Ligand field strength, 10Dq  Interelctronic interaction In the ground state, the configuration is t2g2. The two electrons can have any one of the following arrangements: dxy1dyz1dzx0; dxy0dyz1dzx1; dxy1dyz0dzx1 Thus, the ground state is triply degenerate. The spin multiplicity is 2S+1 = 2(1) + 1 = 3.Thus, the state is represented as 3T1g(F) state. When one electron is excited, the configuration becomes t2g1eg1. There are two possibilities of excitation. When dxz or dyz electron is excited, it can go to either dz2 or dx2-y2 orbital. The resulting configuration will be dxy1dyz0dzx0 (dz2)1 (dx2-y2)0 or dxy1dyz0dzx0 (dz2)0(dx2-y2)1 In the first case, the repulsion will be less and in the second case, the repulsion will be more. Similarly, when an electron is promoted from the dxy orbital, it will go to the dx2-y2 orbital leaving the other electron in the dyz or dxz orbital. Thus, three degenerate excited states exist as shown below: dxy1(dz2)1 ; dyz1(dx2-y2)1 ; dzx1(dx2-y2)1 Put together, these three form the 3T2g(F) state. The other degenerate arrangement is: dxy1(dx2-y2)1 ; dyz1(dz2)1 ; dzx1(dz2)1 These three put together constitute the 3T1g(P) state.

33

There is another arrangement, viz., t2g1 eg1 in which, the spin of one of the electrons is reversed:

t2g eg Now, the spin multiplicity becomes 2S + 1 = 2(0) + 1 = 1. That is, the ground state is a triplet while the excited state is a singlet. Therefore, the transition will be triplet to singlet. This transition will be multiplicity forbidden. Lastly, both the electrons may be excited to the eg level giving the configuration eg2. One electron will be present in each of the two eg orbitals. Thus, there is only one arrangement possible and hence it is a singly degenerate state giving 3A2g(F) state. The splitting of a d2 configuration (2D term) gives rise to the following states: 3

F, 1D, 3P, 1G and 1S. Considering the ground state,3F and the next higher energy state,3P, the following energy level diagram is obtained: 3

3

P

A2g(F) 3

E 1

T1g(P)

D 3

3

T2g(F)

F 1

T2g 3



34

T1g(F)

The spectra of the d2 ion complexes show two transitions, 3 3 viz., T1g(F) T2g(F) and 3T1g(F) T1g(P). The transition from 3T1g(F) to 3A2g will involve the excitation of both the electrons to the higher level and hence is not likely to be observed. For example, the spectrum of [V(OX)3]3- shows two absorptions at 17000 and 24000 cm-1 corresponding to the transitions shown in the above figure. 3

d3 ion: The ground state for the d3 ion is 4F. It splits as follows: 4

A2g, 4T2g and 4T1g Figure 3 is drawn again to show the splitting of the F and P terms and mixing of the T1g terms from F and P levels.

4

P

4

T1g

4

T1g

---------

15B 4

F 4

T2g 1

4

A2g

1 = 10Dq; 2 < 18Dq

35

2

2 should be equal to 18Dq but it is always less than 18Dq due to mixing of the similar 4T1g(F) and 4T1g(P) states. In general, the transitions formulated above do not give the observed spectral bands. This is due to  incorrect mixing of the F and P states  nephelauxetic effect These values decrease the value of C and B in the complexes when compared with the free ion values. d4 ion: The energy level diagram is same as inverted d1. The ground state is 5d. This splits as follows: 5 T2g 5

D

---------------------5

Eg

Here there is no mixing from the higher energy levels. The 5 transition is 5Eg T2g d5 ion: High spin d5 ions like [Mn(H2O)6]2+ or [Fe(H2O)6]3+ are almost colorless. The features of the spectrum are:  The extinction coefficients are very low. Here, all the ‘d’ orbitals are singly occupied. Hence, transition of the electron must cause spin pairing in one of the orbitals. Thus, d-d transitions will be spin-forbidden.  The absorption spectrum has many bands. The reason is that, in the energy level diagram many states lie close together. As a result, many closely spaced multiple absorption bands appear.  In the spectrum, two bands are sharper than the others are. They correspond to the transitions, 36

6

4 4 S Eg(G) + 4A1g and 6S Eg(D). These are not affected by ligand field strength. Hence, they are sharp. However, in the case of other transitions, the energy depends on the strength of the ligand field, so that they are broadened.

d6 ion: The ground state is 5D for the weak field case. It is similar to d1 case. The low spin state for d6 ion is 1I. This is a highenergy term. The energy level diagram is given below: 5

1

I 1

5

Eg

T2g 1

T1g

1

A1g T2g

D 5

It is clear from the diagram that the 1A1g and 1T1g states are stabilized (lowered in energy) more than the 5T2g state as the strength of the ligand field increases. Hence, at higher field strength the ground state of the ion becomes 1A1g. (e.g.) cobalt(III) – d6 ion. The spectra can be predicted easily for this ion. The high spin complex, [CoF6]3- shows only one absorption band at 5 13000 cm-1 due to 5T2g Eg transition, where as the low spin cobalt(III) complex shows two peaks corresponding to

37

1

1 1 A1g T1g and 1A1g T2g. For example, the 3+ complex, [Co(en)3] shows transitions at 21400 and 29500cm-1 and the complex, [Co(OX)3]3- at 16500 and 23800 cm-1.

d7ion: The ground state term is 4F. When the 4P state is not too high, the energy level diagram becomes similar to the diagram of d2 ion. d8 ion: The energy level diagram of the d8 resembles that of the d2 system, but inverted. Three transitions are observed: 3

A2g

3

T2g ; 3A2g

3

T1g(F) ; 3A2g

3

T1g(F)

d9ion: This is same as a d1 system but inverted. Visible spectrum of a d9 system in an octahedral field is rare due to Jahn – Teller effect. In actual practice, a non-symmetric absorption band results due to Jahn – Teller effect. The resulting distortion is more in d9 than in the d1 system. Further, the ground state in Cu2+ (d9) is doubly degenerate and the excited state is triply degenerate. Hence, a broad spectrum results. The Tanabe – Sugano Diagram The energies of the states in terms of E/B are plotted against Dq/B taking the ground state of the system on the x-axis. The resulting diagram is called Tanabe – Sugano diagram. Merits These diagrams are applicable to both strong and weak field cases unlike the Orgel diagram, which is applicable only to weak field cases. Demerits

38



 

The diagrams are drawn to a specific C/B ratio and hence not valid for all similar systems, which generally differ in C/B ratio. There is no accurate way to determine C and B values for the metal ions in complexes. The C and B values for the metal ions in complexes are lower than the values for the free metal ions. However, the Orgel and the Tanabe – Sugano diagrams do not consider this point at all.

The transitions for the Co3+ complexes can be predicted from the Tanabe – Sugano diagram for d6 system. Let us consider only the transitions occurring without change in spin multiplicity (selection rule). For high - spin complexes, like CoF63-, only the quintet state, 5 D, will be important. Therefore, only one transition, viz., 5 5 T2g Eg should be observed. The blue color of this complex is due to a single peak at 13000 cm-1. For low spin 1 Co(III) complexes we expect two transitions, viz., 1A1g T1g 1 1 and A1g A2g.

Splitting of orbitals under various symmetries Trigonal (ligands lie in xy plane) x2-y2 , xy

5.46

z2

-3.21

xz, yz

-3.86

39

Tetrahedral

xz,yz,zx 1.78 x2-y2,z2

-2.67

Square planar (ligands lie in xy plane) x2-y2

12.28

xy

2.28

z2

-4.28

xz,yz

-5.14

Trigonal bipyramid (pyramid base in xy plane) z2

7.07

x2-y2, xy

-0.82

xz,yz

-2.72

40

Square pyramid (pyramid base in xy plane) x2-y2

9.14

z2

0.86

xy

-0.86

xz, yz

-4.57

Octahedron x2-y2, xy

6.00

xy,yz,zx

-4.00

41

Trigonal prism xz,yz

5.36

z2

0.96

x2-y2,xy

-5.84

Pentagonal bipyramid z2

4.93

x2-y2,xy

2.82

xz,yz

-5.28

Cube xz,yz,zx

3.56

z2,x2-y2

-5.34

42

Square antiprism xz,yz

3.56

x2-y2,xy

-0.89

z2

-5.34

Icosahedron All the five ‘d’ orbitals are degenerate. Selection rules The different colors of the transition metal ions arise from the absorption of part of the visible spectrum by ions and their dissociated ligand groups. The absorption of light by complex ions takes place by the electric dipole mechanism. In other words, complexes absorb light by this only important mechanism. If the radiation of frequency applied is equivalent to the energy separation between the two energy levels, the light may be absorbed. If the absorption is to take place by the electric dipole mechanism, the transition moment integral, Q, must be nonzero. Q = < 1| r| 2> ‘r’ is the radius vector. This has the symmetry of of an electric dipole. Components of wave functions A wave function may be considered a product of the different wave functions as follows:  = orbital . spin . vibrational . rotational . translational 43

Except for the gas phase, the last two terms in the above expression are not likely to change for a molecule in a time comparable to the life times of excited electronic states. Hence, they integrate out to unity in the expression for Q. From the remaining part of the expression, it is clear that ‘Q’ will be equal to zero only if 1,spin  2,spin In other words, ‘Q’ will not be equal to zero if 1,spin = 2,spin -------------- (1) and , < 1,orbital . 1,vibrational| r | 2,orbital . 2,vibrational >  0.------(2) The first relation will be satisfied, when both the wave functions have the same spin quantum number,S. The radius vector,r, is antisymmetric to inversion through the center of the system. Hence, integration of the function, 1,orbital |r | 2,orbital from - to + yields zero if both 1 and 2 are ‘g’ or both ‘u’ in symmetry. Since the wave functions of ‘d’ orbitals have ‘g’ symmetry, ‘Q’ is zero for d-d transitions in an octahedral complex (the vibrations are not considered). The group theory tests whether the second relation is equal or not equal to zero. The basic principle is that if 1 and 2 transform as the irreducible representations 1 and 2 of the group to which the molecule belongs, and r transforms as r of that group, then, Q = 0, if the reducible representation 1r2 does not contain the totally symmetric representation of that group,A1g. Q  0, if the irreducible representation contains A1g. The radius vector, r = x + y + z and therefore, it transforms in the same manner as the set of Cartesian coordinates in the groups Oh and Td. ‘r’ transforms as T1u for Oh and T2 for Td. When the direct product does not contain A1g, the transitions are orbitally forbidden. When the inversion properties are not

44

obeyed, the transitions are Laporte forbidden. Singlet-triplet, quartet-doublet etc., transitions are spin – forbidden. d1 – octahedral complex: The M.O. diagram is shown below: a1ga t1ua ega t2gn

T2g+Eg

a1g+eg+t1u T1u

egb xx xx

A1g t1ub xx xx xx

a1g xx (  - bonding only) x – denotes ligand paired electrons -

denotes metal ‘d’ electrons

45

Excitation of the ‘d’ electron from the t2gn molecular orbital to the ega or the a1ga molecular orbitals is forbidden because each has subscript ‘g’. When the transition to the t1u orbital is considered, it is governed by the direct product, T2g x T1u x T1u Ground state

nature of ‘r’ state to which e- is excited

The character table for ‘O’ group is given below: O

E

6C4

3C2(=C42)

8C3

6C2

A1

1

1

1

1

1

A2

1

-1

1

1

-1

E

2

0

2

-1

0

T1

3

1

-1

0

-1

T2

3

-1

-1

0

1

T2 =

3

-1

-1

0

1

T1 =

3

1

-1

0

-1

T1 =

3

1

-1

0

-1

-1

0

1

T2xT1xT1 = 27 -1

46

The above reducible representation is reduced as follows: A1 = 1/24[(27.1.1)+(-1.6.1)+(-1.3.1)+(0.8.1)+(1.6.1)] =1 A2 = 1/24[(27.1.1)+(-1.6.-1)+(-1.-3.-1)+(0.8.0)+(1.6.-1)] =1 E = 1/24[(27.1.2)+(-1.6.0)+(-1.3.2)+(0.8.-1)+(1.6.0)] =2 T1 = 1/24[(27.1.3)+(-1.6.1)+(-1.3.-1)+(0.8.0)+(1.6.-1)] = 3 T2 = 1/24[(27.1.3)+(-1.6.-1)+(-1.3.-1)+(0.8.0)+(1.6.1)] = 4 Therefore, T2g.T1u.T1u = A1g+A2g+2Eg+3T1g+4T2g Thus, the reducible representation contains A1g and therefore, the transition is allowed. This transition is not purely d-d transition and contains some charge transfer character. Molecular Orbital Diagrams for -bonding in Tetrahedral Complexes a1a t2

t2 a

a1 10Dq

t2 +e

en t2 b xxxxxx a1b xx

47

t2+a1

The symmetry of the ‘p’ orbitals is t1u. It transforms as t2 in a tetrahedral field. ‘x’ denotes ligand paired electrons denotes metal ‘d’ electrons The transition is en to t2a. These transitions are largely of the ligand field type. The integral is <en|r|t2a> r = x+y+z transforms as t2 in tetrahedral field. Therefore, the integral is <E.T2.T2>. The direct product, E x T2 x T2 = 2T1+2T2+2E+A2+A1 This contains A1. However, the inversion properties(g or u) have not been investigated. Hence, the presence of A1 alone is not sufficient to establish the existence of the matrix element <en|r|t2a> Although the molecular orbitals in a tetrahedral complex do not possess g or u properties, but the atomic orbitals out of which they are constructed do possess. The en molecular orbital is composed of metal d orbitals only, while the t2a molecular orbital may contain both d and p atomic orbitals of the metal ion. The d orbitals have g character, while the p orbitals have u character. That part of the integral, which involves the d orbital contribution to the t2a molecular orbital must vanish, but that part, which involves the p orbitals may be non – zero. Hence, the ligand field transitions, which are forbidden in octahedral complexes, may be partially allowed in tetrahedral complexes because of d-p mixing.

48

Break down of selection rules If the spin and Laporte selection rules were rigorously obeyed, only a few absorption bands would be observed for transition metal complexes. However, neither of the rules holds very strictly because some mechanisms offer relaxation. Spin selection rule break down This rule is not fully valid in the presence of spin – orbit coupling. The coupling between spin and orbital angular momenta within an ion factorizes the total wave function accurately in to spin and orbital wave function products. However, the factorization can only be approximate. Hence, the spin selection rule cannot be strict. Therefore, the spin – forbidden bands appear in the transition metal complexes. However, the intensity is one or two orders of magnitude less than that of the allowed transitions. The intensity of spin – forbidden bands relative to spin – allowed bands increases as spin – orbit coupling constants increase. The spin – orbit coupling increases as one moves to the right hand side of a transition series, or from the first to the third transition series within a group of the periodic table. Laporte selection rule break down This rule can be broken down in two ways: 1. Vibronic coupling 2. Reduction of symmetry Vibronic coupling The Laporte selection rule assumes that vibrations are absent. However, the ligand atoms of a complex do vibrate. Therefore, a suitable combination of vibrations in the ground

49

and the excited electronic levels relaxes the rules. The coupling between electronic and vibrational wave functions is called ‘vibronic’. The vibronic wave function has symmetric representations determined by the direct product of the representations of electronic and vibronic. In general, there are a number of possible modes of vibration to the ground and excited terms of a complex. It is almost certain that for some vibrations, 1,vibronic.r.2,vibronic contains A1g and hence the integral, <1,vibronic.|r|2,vibronic>  0 Therefore, the transition between 1 and 2 is allowed to at least some extent. For example, in an octahedral complex, and considering only d-d transitions, if the ground term is mixed with a g – type vibration, the excited term must be mixed with a u – type vibration and vice-versa. Laporte – forbidden bands are usually three orders of magnitude or more weaker than allowed transitions. Reduction of symmetry This also causes relaxation in selection rule. For example, consider reduction of cubic symmetry. Consider the 3T1g to 3 A2g transition of the d2 configuration. The transition moment integral is: The direct product in the group, Oh is: T1g x T1u x A2g T1 x T1 = A1+E+T1+T2 gxu=u Therefore, T1g x T1u = A1u+Eu+T1u+T2u T1g x T1u x A2g = [A1u+Eu+T1u+T2u]A2g 50

A1u x A2g = A2u Eu x A2g = Eu T1u x A2g = T2u T2u x A2g = T1u Hence, T1g x T1u x A2g = A2u + Eu + T2u + T1u Thus, the direct product does not contain A1g or even A1 if the inversion properties are ignored. Consequently, the transition is doubly forbidden by the symmetry of the electronic wave function. Suppose, now the octahedron is distorted by compression or elongation along the z-axis. The resultant figure belongs to the Group D4h. The irreducible representation T1g of the group Oh must correspond to other irreducible representations of the Group D4h, since there is no triply degenerate irreducible representation in that group. T1g in Oh corresponds to A2g+Eg in D4h, and T1u corresponds to A2u+Eu. A2g in Oh corresponds to B1g in D4h. The correlation is given in Table 3. Now, the direct product T1g x T1u x A2g = (A2g + Eg)(A2u + Eu)(B1g) The second term in the product is concerned with the selection rule for Qx,y and is A1u+A2u+B1u+B2u+Eu The first term is concerned with Qz and is Eu+B1u From these results, it is deduced that the reduction in symmetry from Oh to D4h partially relaxes the orbital selection rule for radiation incident in the plane of the molecule (xyplane) because the direct product of the representation of the ground and excited terms with that of the transition moment in the plane, Qx,y, contains the A1 irreducible representation. However, the rule is not relaxed for the radiation down the C4 (z) axis of the molecule. In other words, the transition is no longer as forbidden as it was in cubic symmetry, and is anisotropic. An absorption band corresponding to the transition is polarized. The reduction in

51

symmetry has no effect on the inversion properties of the wave functions, and so the transition is still Laporte forbidden. The Laporte selection rule must be overcome by some other mechanism like vibronic coupling. Band Intensities Intensity of a band is measured in terms of oscillator strength, f. ‘f’ is the area under the band in a plot of extinction coefficient,, versus frequency, . Very often, the intensities are compared by means of their extinction coefficients. The intensities of various types of transition in metal complexes are given below: Type of transition App.f App.  Spin-forbidden, Laporte forbidden 10-7 10-1 Spin – allowed, Laporte forbidden 10-5 10 -3 Spin-allowed, Laporte forbidden, but with 10 102 d-p mixing (Td symmetry) Spin-allowed, Laporte forbidden, but with Intensity stealing

10-2

103

Spin-allowed, Laporte allowed (CT)

10-1

104

Band Widths If the terms of the transition metal complexes correspond to a single energy level, then the absorption bands in the spectra should be exceedingly narrow. However, actually the bands at room temperature and in solution are mostly rather wide (1000cm-1). This means that the terms themselves are spread over a comparable range of energies (that is, non-degenerate). Several factors are responsible for lifting the degeneracies of

52

terms. Some or all of them may operate together in any particular situation. The important factors are:  Vibration  Spin-orbit coupling  Jahn-Teller effect Vibration Actually, 10Dq depends upon the metal – ligand separation. Vibrations modulate the ligand field magnitude. Hence, they spread energy of each term over a range of values. If the transition is between two terms, whose energies depend upon 10Dq in different manners, then the energy of transition is spread over a range, and the corresponding band is of considerable width. Bandwidths of up to about 1000cm-1 may be accounted for in this way. In certain instances, the energies of some excited states run parallel to the ground term, at least over a part of the diagram. In such regions, transitions to these excited terms are independent of 10Dq, and hence the magnitude of the ligand field. These are not broadened by the vibrational mechanism. Absence of vibrational fine structure If vibrational interaction were the only band broadening mechanism, the vibrational fine structure should have been observed instead of a single broad band. Nevertheless, in solution and at room temperature, interaction with the solvent occurs and many vibrational levels are thermally accessible. Hence, it is mostly difficult to resolve the fine structure. All that is observed is the envelope of the fine structure line, leading to a singlewide absorption peak. In solid state and at low temperatures, it is possible to resolve the fine structure in a great many bands. The reason is that as the temperature is lowered, fewer vibrational levels are occupied. Consequently, the chance that there are suitable 53

vibrations that couple the ground and excited terms are reduced. Generally, for the Laporte forbidden transitions, band intensities depend on vibrational coupling. Hence, as the temperature is lowered, intensities decrease.

Spin- Orbit Coupling T terms are split by spin – orbit coupling. The energy difference between the highest and lowest states of a term is the ‘overall multiplet width’. For the first transition series, this will be in the range 100 to 1000cm-1. The exact value depends up on the ion. The spin – orbit coupling could be observed as a fine structure to bands under ideal conditions. However, in solution and at room temperature, it is not possible to observe the fine structure with the first transition elements because the spin – orbit coupling is small. In the case of second and third transition series, the spin – orbit coupling is large. Hence, the fine structure has been observed in solutions of their complexes. The spin – orbit coupling fine structure may be resolved even in the complexes of the ions of the first transition series in the solid state and particularly at low temperatures. Jahn – Teller Effect Due to distortion from cubic symmetry (Oh and Td), additional energy levels are seen in the spectrum. If these levels are close, broad bands appear. That is, bandwidth increases. These three mechanisms are sufficient to account for the widths of the ligand field bands in the spectra of almost all transition metal complexes.

54

Charge transfer bands tend to be much broader than 1000 cm-1. The reason is that transitions from the ground state to a set of excited terms occur rather than to just a single term. Band Shapes The shapes of the bands are determined by three factors:  Vibrational Interaction  Spin – orbit Coupling  Low symmetry ligand fields Vibrational Interaction This leads to asymmetrically shaped bands. The intensities of transitions between vibrational levels increase as the temperature increases. Therefore, if the intensity is largely due to vibronic coupling, the symmetry is expected only at low temperatures. At higher temperatures, a trail towards low frequency is expected because higher vibrational levels are occupied according to Boltzmann distribution. The position of the band maximum may change with temperature. The band maximum need not represent exactly the electronic energy difference between the ground and excited terms. Spin – Orbit Coupling In most of the cases, spin – orbit coupling does not split a term symmetrically. Hence, bands are expected to be asymmetric in shape. Low – Symmetry ligand fields Low – symmetry ligand fields split E and T terms. The T term may be split into two components. One component will be of two-fold degeneracy and the other component non – degenerate. The two components will have different energy. Therefore, E and T terms may lead to asymmetric bands if the orbital degeneracy is lifted by any factor.

55

In spite of these factors, in solution, bands in the spectra of transition metal complexes are most often symmetrical in shape. The reason may be that all the three factors operate simultaneously to give an average result, which is almost asymmetric. Diffuse Reflectance Spectrum When the complex is insoluble in any of the solvents, the spectrum is recorded in the solid state using diffuse reflectance accessory. The sample is ground intimately with a suitable inert white material, like MgCO3. The light reflected from it is examined. The minima in the reflected light correspond to the maximum absorptions in the complex. If the grinding is not thorough, bands will be broader. Colors and Range The colors of the visible spectrum correspond to energy in roughly the following manner: Red 14000 – 16000 cm-1 Yellow 18000 cm-1 Green 20000 cm-1 Blue 21000 – 25000 cm-1 Violet 25000 cm-1 An absorption band in one part of the spectrum gives the solution its complementary color. For example, a band in the red region leads to a blue solution. Absorbed Color

Complementary Color

Violet Blue Green – Blue Green Yellow – Green

Yellow – green Yellow Orange – Red Purple Violet

56

Yellow Orange Red

Blue Green – Blue Blue – Green

Table 3 – Correlation Table for Oh Group Oh A1g A2g Eg T1g T2g A1u A2u Eu T1u T2u

O A1 A2 E T1 T2 A1 A2 E T1 T2

Td A1 A2 E T1 T2 A2 A1 E T2 T1

D4h A1g B1g A1g+B1g A2g+Eg B2g+Eg A1u B1u A1u+B1u A2u+Eu B2u+Eu

D2d A1 B1 A1+B1 A2+E B2+E B1 A1 A1+B1 B2+E A2+E

C4v A1 B1 A1+B1 A2+E B2+E A2 B2 A2+B2 A1+E B1+E

C2v A1 A2 A1+A2 A2+B1+B2 A1+B1+B2 A2 A1 A1+A2 A1+B1+B2 A2+B1+B2

D3d A1g A2g Eg A2g+Eg A1g+Eg A1u A2u Eu A2u+Eu A1u+Eu

D3 A1 A2 E A2+E A1+E A1 A2 E A2+E A1+E

C2h Ag Bg Ag+Bg Ag+2Bg 2Ag+Bg Au Bu Au+Bu Au+2Bu 2Au+Bu

Calculation of 10Dq, B and Assignment of Bands d2 ion For example, consider an aqueous solution of trivalent vanadium salt. The color of these salts will be green. The spectrum consists of wide band at 17200 and 25600 cm-1. V3+ is a d2 system. Based on the Tanabe – Sugano diagram for d2 configuration, the bands are assigned as follows: 3 T1g (F)  3T2g (F) ------- 17200 cm-1 3 T1g (F)  3T1g(P) -------- 25000 cm-1 The ratio between these two transitions is: 3 T1g (F)  3T1g(P) 25000 = = 1.49 3 3 T1g (F)  T2g(F) 17200

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From the Tanabe – Sugano diagram, Dq/B (x –axis) corresponding to this ratio of 1.49 is found to be 2.8. Perpendiculars are drawn from this point on the x – axis. The perpendicular cuts the lines for the 3T2g (F) and 3T1g(P) terms corresponding to the values of E/B = 25.9 and 38.7 respectively. For the transition, 3T1g(F)  3T2g(F), E/B = 25.9 and E = 17200 cm-1. Therefore, B = 17200/25.9 = 664 cm-1. Substituting this value of ‘B’ in Dq/B = 2.8, we get, Dq = 2.8B = 2.8 x 664 cm-1 = 1859 cm-1 Therefore, 10Dq = 18590 cm-1 The ‘B’ value for V3+ ion is 860 cm-1. However, in the complex, it is found to be 664 cm-1. Thus there is reduction in the value of ‘B’ in the complex. Hence, the assignment of bands is correct. d8, Ni2+ ion Aqueous solutions of bivalent nickel salts have a light gren color. This is due to the presence of weak absorptions in the red and blue portions of the visible spectrum. The bands are found at 8700, 14500 and 25300 cm-1. These bands are assigned as follows with the help of Tanabe – Sugano diagram for d8 system. 3 A2g(F)  3T2g(F) ( = 10Dq) ----- 8700 cm-1 3 A2g(F)  3T1g(F) ----------------14500 cm-1 3 A2g(F)  3T1g(P) ---------------- 25300 cm-1 The ratio, 3 A2g(F)  3T1g(P) 25300 = = 1.74 3 3 A2g(F)  T1g(F) 14500 Dq/B corresponding to this ratio is obtained at Dq/B = 0.98 from the Tanabe – Sugano diagram. The vertical line drawn from this point cuts the 3T1g(F) term at 16. 14500/B = 16 58

Therefore, B = 14500/16 = 906 cm-1 Dq/B = 0.98 Dq = 0.98 x B = 0.98 x 906 = 888 cm-1 10Dq = 8880 cm-1 Observed 10Dq = 8879 cm-1 The ‘B’ value for Ni2+ ion = 1040 cm-1 The ‘B’ value in the complex = 906 cm-1 % reduction = 906/1040 x 100 = 87 That is, there is a reduction of ‘B’ to about 90% in the complex. d3, Chromium(III) ion The aqueous solution of trivalent chromium gives a light green color. The color is due to the absorption in the yellow and blue parts of the spectrum. The absorption spectrum shows bands at 17000, 24000 and 37000 cm-1 (weak band). Based on the Tanabe – Sugano diagram ford3, the assignments are made as follows: 4 A2g(F)  4T2g(F) = 10Dq ----- 17000 cm-1 4 A2g(F)  4T1g(F) --------------- 24000 cm-1 4 A2g(F)  4T1g(F) -------------- 37000 cm-1 The ratio of any of these two bands is fitted to the appropriate figure. The fit occurs for the lower two bands, which are the more obvious in the spectrum, for Dq/B = 2.45. 4 A2g(F)  4T2g(F) = 24.5 = 17000/B B B = 695 cm-1 10Dq = 17000 cm-1 ‘B’ is reduced to about 70% of the free – ion value for Cr3+ (1030 cm-1) d7, Co2+

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Aqueous solutions of the Co2+ are pink. There is a weak absorption at 8000 cm-1 and in the region of 16000 to 22000 cm-1. Magnetic susceptibility studies show that the ground term of cobalt(II) ion solutions in water is a quartet. Hence, the left hand side of the Tanabe – Sugano diagram should be consulted. The bands are assigned as follows: 4 T1g(F)  4T2g(F) -------- 8000 cm-1 4 T1g(F)  4A2g(F) -------- 19600 cm-1 4 T1g(F)  4T1g(P) --------- 21600 cm-1 The ratio of the energies of the upper and lower bands is the most sensitive one to fit to the appropriate diagram. The ratio = 2.7. It corresponds to Dq/B = 0.96. The vertical line drawn from this point cuts the 4T2g(F) term at 8.2. Therefore, 4 T1g(F)  4T2g(F) = 8.2 B 8000/B = 8.2 B = 980 cm-1 10Dq = 9800 cm-1 ‘B’ in the complex is reduced to about 80% of the free ion value for Co2+ (1120 cm-1). d4, Cr2+ Aqueous solutions of Cr2+ is present are pale blue. There is a weak band at 14100 cm-1. From the magnetic studies, it is deduced that Cr2+ in these cases has the quintet ground term (d4 high spin). The band is assigned to 5Eg  5T2g. d6, Fe2+ The aqueous solution gives a pale green color, which is due to the result of a weak absorption band in the red region, at 10000 cm-1. The ground term is known to be quintet from magnetic studies so that left hand side of the Tanabe – Sugano diagram is considered. The band is assigned to the spin –

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allowed transition 5T2g  5Eg. This corresponds to 10Dq. Therefore, 10Dq = 10000 cm-1. d5, Mn2+, Fe3+ The color of aqueous solutions, which contain the manganous ion, is very pale pink. The absorption spectrum of the solutions shows a series of very weak narrower bands stretching through the green and blue portions of the visible spectrum and into the ultraviolet. Magnetic studies show that the ground term of the manganous ion in aqueous solution is sextet. The only sextet term of the d5 configuration in octahedral symmetry is the 6A1g. Hence, there can be no spin – allowed transitions. The bands are assigned as follows: 6 A1g  4T1g(G) -------- 18000 cm-1 6

6

A1g  4T2g(G) -------- 22900 cm-1

A1g  4Eg(G) --------- 24900 cm-1

6

A1g  4A1g(G) --------25150 cm-1

6

A1g  4T2g(D) ------- 27900 cm-1

6

A1g  4Eg(D) -------- 29700 cm-1

6

A1g  4T1g(P) --------32400 cm-1 The set of bands fit such an assignment for Dq/B = 1.1 From the Tanabe – Sugano diagram, 6 A1g  4T1g(G) = 24.0 = 18600/B B B = 770 cm-1; 10Dq = 8500 cm-1

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‘B’ is reduced to about 80% of the free ion value for Mn2+ (960 cm-1). The narrowness of some of the bands can be explained as follows: Energies of the 4Eg(G) and 4A1g(G) terms do not change much relative to the 6A1g ground term as Dq changes. Consequently, bands corresponding to transitions between the ground term and these two terms are not appreciably broadened by vibronic coupling. Spin – Orbit coupling also do not cause any broadening since it does not raise the degeneracy of any of the terms concerned. d1, Ti3+ Aqueous solutions have purple color. Hence, green color is absorbed. Therefore, absorption band near 20000 cm-1 should appear. Actually, the band is located at 20300 cm-1. This is assigned to 2T2g  2Eg = 10Dq.

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