Coordination Chemistry: Spectra PDF

COORDINATION CHEMISTRY: SPECTRA Why are so many coordination compounds colored? Emerald Cr3+ in Be3Al2Si6O18 Amethyst Fe2+/3+ and Ti4+ in Al2O3 CuSO4.5H2O COORDINATION CHEMISTRY: SPECTRA Absorption of Light Complementary colors: if a compound absorbs light of one color, we see the complement...

COORDINATION CHEMISTRY: SPECTRA Why are so many coordination compounds colored? Emerald Cr3+ in Be3Al2Si6O18 Amethyst Fe2+/3+ and Ti4+ in Al2O3 CuSO4.5H2O COORDINATION CHEMISTRY: SPECTRA Absorption of Light Complementary colors: if a compound absorbs light of one color, we see the complement of that color. Ex: A substance that absorbs red light is green in color. tering and reflection of light from cell surfaces) at a given wavelength by an absorbing species in solution: Io log = A = elc I COORDINATION CHEMISTRY: SPECTRA Absorption Spectrum of [Cu(H2O)6]2+ Visible LightLight and Colors TABLE 11.1 Visible andComplementary Complementary Colors Chemistry III: Electronic Spectra Cu2+(d 9) 10 e 5 0 5,000 10,000 15,000 20,000 25,000 30,000 35,000 cm-1 2,000 1,000 667 500 nm 400 333 Wavelength Range (nm) Wave Numbers (cm−1) < 400 > 25,000 400 – 450 22,000 – 25,000 Violet Yellow 450 – 490 20,000 – 22,000 Blue Orange 490 – 550 18,000 – 20,000 Green Red 550 – 580 17,000 – 18,000 Yellow Violet 580 – 650 15,000 – 17,000 Orange Blue 650 – 700 14,000 – 15,000 Red Green > 700 < 14,000 Color Complementary Color Ultraviolet Infrared 286 An example from coordination chemistry is the deep blue color of aqueous solutions • The blue color is a consequence of the absorption of light between ca. 600 and 1000 f copper(II) compounds, containing the ion [Cu(H2O)6]2 + . The blue color is a conse(maximum near approximately 800 nm); 600 in the yellow to IR of uence of the absorption of light between and 1000 nm (maximum nearthe spectrum. 00 nm; Figure ), incolor the yellow to infrared region of theisspectrum. The color observed, • 11.1 The observed, blue, the average complementary color of the light absorbed. lue, is the average complementary color of the light absorbed. It is not always possible to make a simple prediction of color directly from the absorpion spectrum, in large part because many coordination compounds contain two or more bsorption bands of different energies and intensities. The net color observed is the color redominating after the various absorptions are removed from white light. nm gives a spectrum characteristic of the molecule or ion in question, as in Figure y Io at a given wavelength passes through a solution containing a species will see, by thisaspectrum the light emerges with intensity I, which may be measured suitableis a consequence of transitions between states of differ and can provide valuable information about those states and, in turn, about t 1.2). COORDINATION CHEMISTRY: SPECTRA and bonding of the molecule or ion. mbert law may be used to describe the absorption of lightAlthough (ignoring scatthe quantity most commonly used to describe absorbed light is the Beer–Lambert law at a given wavelength ion ofThe light from cell surfaces) by and an frequency absorbing energy are also used. In addition, the wavenumber, the number Relations between energy, frequency, centimeter (a quantity proportional to the energy), is frequently used, especially in n: wavelength and wavenumber ofquantities light. are given by t infrared light. For reference, the relations between these Io log = A = elc hc 1 I E = hv = = hc a b = hcv 11.2 Quantum Numbers of Multielectron Atoms | 405 le Light and Complementary Colors I 0 Wave Numbers (cm−1) > 25,000 Light source 22,000 – 25,000 where Cell h = Planck constant = 6.626 * 10 - 34 Js c = speed of light = 2.998 * 108 m s - 1 Complementary Color Ultraviolet l Violet l EFIGURE = energy 11.2 Absorption of Light by Solution. I Detector Color l v = frequency (s - 1) Yellow 20,000 – 22,000 Blue Orange where A = absorbance e = molar absorptivity (L mol - 1 cm - 1) (also known as molar 18,000 – 20,000 Green Red extinction coefficient) l = length through solution (cm) 17,000 – 18,000 Yellow Violet c = concentration of absorbing species (mol L - 1) l = wavelength (often reported in nm) 1 -1 n wavenumber (cm ) = = l 11.2 Quantum Numbers of Multielectron Atoms 15,000 – 17,000 Orange Blue Absorbance is a dimensionless quantity. An absorbance of 1.0 corresponds to 90% absorp- Absorption of light results in the excitation of electrons from lower to higher e tion at a given wavelength,* an absorbance of 2.0 corresponds to 99% absorption, and so on. 14,000 – 15,000 Red Green Spectrophotometers commonly obtain spectra as plots of absorbance versus such wave- states are quantized, we observe absorption in “bands” (as in because length. The molar absorptivity is a characteristic the light < 14,000 Infraredof the species that is absorbing with the energy of each band corresponding to the difference in energy betwe and is highly dependent on wavelength. A plot of molar absorptivity versus wavelength COORDINATION CHEMISTRY: SPECTRA Quantum Numbers of Multielectron Atoms Recall: As a result of repulsions between electrons (characterized by energy Πc), electrons tend to occupy separate orbitals; Consider: Carbon 1s2 2s2 2p2. As a result of exchange energy (Πe), electrons in separate orbitals tend to have parallel spins. • there are three major energy levels for the p2 electrons; differing in energy by pairing and exchange energies) • the lowest major energy level is split into three slightly different energies, for a total of five energy levels. COORDINATION CHEMISTRY: SPECTRA Each energy level can be described as a combination of the ml and ms values of the 2p electrons. Each of the 2p electrons could have any of six possible ml, ms combinations: n=2 ml = +1 l=1 0 ms = +1/2 (quantum number defining 2p orbital) -1 (three possible values) -1/2 (two possible values) Six combinations: (+1)(+1/2), (+1)(-1/2), (0)(+1/2), (0)(-1/2), (-1)(+1/2), (-1)(-1/2) COORDINATION CHEMISTRY: SPECTRA The 2p electrons are not independent of each other. The orbital angular momenta (ml) and spin angular momenta (ms) interact in a manner called Russell-Saunders coupling or LS coupling. The interactions produce atomic states called microstates that can be described by new quantum numbers: ML = Σml (Total orbital angular momentum) MS = Σms (Total spin angular momentum) Once the ml-ms combinations are known, the corresponding ML and MS values are determined. COORDINATION CHEMISTRY: SPECTRA For shorthand, use the designation: superscript + or + for ms = +1/2 ; - for ms = -1/2 Example: An electron having ml =+1 and ms = +1/2 will be denoted as 1+ One possible set of values for two electrons in p are 1st electron: ml = +1 ms = +1/2 1+ 2nd electron: ml = 0 ms = -1/2 0- Notation: 1+ 0- COORDINATION CHEMISTRY: SPECTRA Each set of possible quantum numbers (such as 1+0- ) is called a microstate. The next step is to tabulate the possible microstates. Two precautions: (1) make sure that no two electrons in the same microstate have identical quantum numbers (the Pauli exclusion principle applies); and (2) count only the unique microstates. Examples: The microstates 1+0- and 0-1+; 0+0- and 0-0+ in a p2 configuration are duplicates (only one is listed). COORDINATION CHEMISTRY: SPECTRA The number of microstates = i!/[ j! (i - j)!] 11.2 Q 2), where i = number of ml-mTABLE (6 in p s combinations 11.2 Microstate Table for p2 and j = no. of electrons. MS For p2: -1 number of microstates = 6!/[ 2! (6 - 2)!] = 15 The 15 microstates can be arranged according to their ML and Ms values and listed in a microstate table: ML 1- 0- 0 -1- 1- -1 -1- 0- -2 +1 1+ 1- +2 +1 0 1+ 01- 0+ -1+ 10+ 0-1- 1+ -1+ 0-1- 0+ 1+ 0+ -1+ 1+ -1+ 0+ -1+ -1- quantum numbers (the Pauli exclusion principle applies); and (2) to count microstates. For example, the microstates 1 + 0 - and 0 - 1 + , 0 + 0 - and 0 - COORDINATION Example: Determine the possible microstates for an microstate table. figuration are duplicates and only one of each pair will be listed. If we determine all possible microstates and tabulate them accordi MS values, we obtain a SPECTRA total of 15 microstates.* These microstates can b CHEMISTRY: ing to their ML and MS values and listed conveniently in a microstate Table 11.2. E1X A M P L E 11 .1 1 s p configuration and use them to prepare a Determine the possible microstates for an s1p1 configuration, and use a microstate table. 1 The s electron can have ml = 0 and ms = { 2. The s electron can have ml = 0 and ms = +1/2 , -1/2 1 The p electron can have ml = +1, 0, -1 and ms = { 2. The p electron can have ml = +1, 0, -1 and msThe = +1/2 , -1/2 resulting microstate table is then MS # of microstates = 2!/[1! (2 - 1)!] x 6!/[1! (6 - 1)!] = 12 The microstate table: + - -1 0 +1 +1 0- 1- 0- 1+ 0+ 1- 0 + 1+ 0 0- 0- 0+ 00- 0+ 0+ 0+ -1 0- -1- 0- -1+ 0+ -1- 0+ -1+ ML - + In this case, 0 0 and 0 0 are different microstates, because the fir and the second electron is a p; both must be counted. COORDINATION CHEMISTRY: SPECTRA Seatwork Determine the possible microstates for a d2 configuration and use them to prepare a microstate table. ML and Ms give atomic quantum numbers L, S, and J. L, S, and J - describe the energy and symmetry of an atom or ion - determine the possible transitions between states of different energies. The quantum numbers that describe states of multielectron atoms are defined as follows: L = total orbital angular momentum quantum number S = total spin angular momentum quantum number J = total angular momentum quantum number COORDINATION CHEMISTRY: SPECTRA Quantum numbers L and S describe collections of microstates, whereas ML and Ms describe the microstates themselves. L and S are the largest possible values of ML and Ms. ML is related to L much as ml is related to I , and the values of Ms and ms are similarly related: Atomic States Individual Electrons ML = 0, ±1, ±2,..., ±L ml = 0, ±1, ±2,..., ±l MS = S, S - 1, S - 2,..., -S ms = +1/2, -1/2 The values of L correspond to atomic states described as S, P, D, F, etc. The values of S are used to calculate the spin multiplicity, defined as 2S + 1. States having spin multiplicities of 1, 2, 3, and 4, are described as singlet, doublet, triplet, and quartet states. an electron’s orbit magnetic field ass COORDINATION CHEMISTRY: SPECTRA due to an electron and MS describes The spin multiplicity is designated as a left superscript. spin for a microsta The values of Atomic states characterized by S and L are often called free-ion terms (sometimes Russell-similar t a manner Saunders terms) because they describe individual atoms or ions, free of ligands. ing from MS) are u states having spin Examples of Atomic States (Freeand quartet states ion Terms) and Quantum Numbers atomic states are g Atomic states Term L S S state L = 0 Russell–Saunders 1S 0 0 Their labels are of P state L = 1 2S 0 1/2 ing to the value of D state L = 2 3P 3 1 1 symbol D corresp L = 3 F state 4D 5 2 3/2 F marks a state i 5F 3 2 *Unfortunately, S is us having L = 0. Chemist **Although term and st COORDINATION CHEMISTRY: SPECTRA The labels of free ion terms are often called term symbols. 1 Term symbols are composed of a letter relating to the value of L and a left superscript for the spin multiplicity. TABLE 11.3 Examples of Atomic States (Free-Ion Terms) and Qu Term Examples: 3D 5F 1S corresponds to a state in which L = 2 and the spin multiplicity (2S + 1 ) is 3. marks a state in which L = 3 and multiplicity of 2S + 1 = 5. (singlet S) L S 0 0 2 0 1 2 3 1 1 4 2 3 2 5 3 2 1 S S P D F An S term has L = 0 and must thereforeFree-ion have M L = 0. terms are very important in the interpretation of the sp compounds. The spin multiplicity (the superscript) is 2S + 1. The following examples show how to determine the va MS for a given term and how to prepare microstate tables from them e following examples show how to determine the values of L, ML, S, and erm and how to prepare microstate tables from them. COORDINATION CHEMISTRY: SPECTRA 1. 2 Because 2S + 1 = 1, S must equal 0 (and Ms = 0). 1 There can be only one microstate, having M = 0 and Ms = 0 for a S term. L L = 0 and must therefore have ML = 0. The spin multiplicity (the 2S + 1. Because 2S + 1 = 1, S must equal 0 (and MS = 0). There can For the maximum configuration of two electrons: 1 microstate having ML = 0 and MS = 0 for a S term. For the minimum of two electrons we have the following: ML 0 MS MS 0 0 + 0 0 or ML 0 x Each microstate is designated by x in this second form of the table. state is designated by x in the second form of the table. P) L = 1; therefore, ML can have three values: +1, 0, and –1. The spin 1 1 1 0 0 COORDINATION CHEMISTRY: SPECTRA or ML 0 0+02P ML 0 x (doublet P) Each microstate is designated by x in the second form of the table. A P term has L = 1 ; therefore, ML can have three values: + 1,0, and -1. 2P (doublet P) The spin multiplicity is 2 = 2S + A 1.PTherefore, , and M MLs can canhave have two values: term has L S= =1;1/2 therefore, three values: +1, 0,+1/2 and –1. The 1 and -1/2. multiplicity is 2 = 2S + 1. Therefore, S = 2 , and MS can have two values: + 2 There are six microstates in a P term (3 rows * 2 columns). For the minimu There are six microstates in a 2one P term (3 rows x 2 columns). electron we have the following: MS MS For the minimum case of one electron: 1 2 ML 1 2 1 + 2 1 x x 0 x x -1 x x 1 + 2 1 1- 1+ 0 0- 0+ -1 -1- -1+ or ML The spin multiplicity is equal to the number of possible values of MS; therefo spin multiplicity is simply the number of columns in the microstate table. COORDINATION CHEMISTRY: SPECTRA The spin multiplicity is equal to the number of possible values of Ms The spin multiplicity is simply the number of columns in the microstate table. Seatwork: For each of the following free-ion terms, determine the values of L, ML, S, and Ms, and diagram the microstate table as in the preceding examples: 2D, 1P, and 2S. Return to the p2 microstate table and reduce it to its constituent atomic states (terms). To reduce the p2 microstate table into its terms, all that is necessary is to find rectangular arrays. ent degrees of electron–electron interactions. For our example of a p configuration for a carbon atom, the 3P, 1D, and 1S terms have three distinct energies—the three major energy levels observed experimentally. COORDINATION CHEMISTRY: SPECTRA TABLE 11.4 The Microstate Table for p2 and Its Reduction to Free-Ion Terms The Microstate Table for p2 and its Reduction to free-ion Terms MS 0 -1 x +2 ML +1 x x x x 0 x x x x x -1 x x x x -2 x MS MS 0 -1 ML +1 +1 0 -1 MS +1 +2 x +2 +1 x +1 x x x 0 x 0 x x x -1 x -1 x x x -2 x -2 1D ML 0 -1 +1 +2 +1 ML x 0 -1 -2 3P 1S NOTE: The 1S and 1D terms have higher energy than the 3P term. The relative energies of higher energy terms like these cannot be determined by simple rules. COORDINATION CHEMISTRY: SPECTRA The final step in this procedure is to determine which term has the lowest energy. This can be done by using two of Hund's rules: 11.2 Quantum Numbers of Multielectron Atoms | 411 1 . The ground term (term of lowest energy) has the highest spin multiplicity. inal step in this procedure is to determine which term has the lowest energy by 2p of Hund’s rules: In our example of p2, the ground term is the 3P. This ground term term (term of lowest energy) has the highest spin multiplicity. can be identified as having the following 2 3 r example of p , the ground term is the P. This term can be identifed as having configuration: onfiguration in the margin. This is sometimes called Hund’s rule of maximum 2s plicity, introduced in Section 2.2.3. Hund’s rule of maximum multiplicity o or more terms share the maximum spin multiplicity, the ground term is the one ng the highest value of L. 4 4 example, P and F terms are both found an electron configuration, 2. Ififtwo or more terms share the for maximum spin multiplicity,the 1s 4 4 as lowerthe energy: F has L =is3the , andone P has L = 1.the highest value of L. ground term having P L E 11 . 3 1 1 the microstate table for the s p configuration to its component free-ion terms, ntify the ground-state term. uration are duplicates and only one of each pair 2. willIfbe listed. two or more terms share the maximum spin multiplicity, the ground term is the If we determine all possible microstates and tabulate them according to their ML and having the highest value of L. * accordS values, we obtain a total of 15 microstates. These microstates can be arranged 4 4 if Pasand F terms are both found for an electron configuration g to their ML and MS values and listed convenientlyFor in a example, microstate table, shown in 4 4 4 F has lower energy: F has L = 3 , and P has L = 1. ble 11.2. COORDINATION CHEMISTRY: SPECTRA E X A MExample: P L E 11 .1 E X A M P L E 11 . 3 Reduce the microstates microstate forguration, the s1and p1 use configuration free-ion terms, and Determine the possible for antable s1p1 confi them to prepare to its1component 1 Reduce the microstate table for the s p confi guration to its component free-ion ter a microstate table.the ground-state term. identify The s electron can have ml = 0 and ms = { 12. and identify the ground-state term. 1 The p electron can have ml = +1, 0, -1 and ms = The { 2. microstate table (prepared in Example 11.1) is the sum of the microstate tab Answer: 3 1 The resulting microstate table is then for the P and P terms: MS MS -1 0 +1 +1 0- 1- 0- 1+ 0+ 1- 0 + 1+ 0 0- 0- 0+ 00- 0+ 0+ 0+ -1 0- -1- 0- -1+ 0+ -1- 0+ -1+ ML ML MS -1 0 +1 +1 x x x 0 x x x -1 x x x -1 ML 3P 0 +1 x 0 x -1 x +1 1P 3 Hund’s rule of maximum multiplicity requires P as the ground state. 3 In this case, 0 0 and 0 0 are different microstates, because the firequires rst electron isPanas s the g.s. Hund’s rule 2 and the second electron is a p; both must be counted. EXERCISE 11.3 In Exercise 11.1, you obtained a microstate table for the d configur this to and its component free-ion terms, and identify the ground-state term EXERCISE 11.1 Determine the possible microstatestion. for a Reduce d2 configuration use them + - - + to prepare a microstate table. (Your table should contain 45 microstates!) 3 P 1 P COORDINATION CHEMISTRY: SPECTRA 3 Hund’s rule of maximum multiplicity requires P as the ground state. For example, if 4P and 4F terms are both found for an electron configuration, the 4F has 2 4 4 lower energy F has L = 3; 11.1 P has L = obtained 1). EXERCISE 11.3( In Exercise , you a microstate table for the d configura- tion. Reduce this to its component free-ion terms, and identify the ground-state term. SPIN-ORBIT COUPLING 11.2.1 Spin-Orbit Coupling The spin and orbital angular momenta couple with each other, a phenomenon known as Upspin-orbit to this point in the discussion of multielectron atoms, the spin and orbital angular coupling. momenta have been treated separately. An additional factor is important: the spin and orbital angular momenta magnetic fields numbers associatedcombine with them) with each In multielectron atoms,(or thethe S and L quantum intocouple the total angular other, a phenomenon known as spin-orbit coupling. In multielectron atoms, the S and L momentum quantum number J. quantum numbers combine into the total angular momentum quantum number J. The The quantum number J may have the following values: quantum number J may have the following values: J = L + S, L + S - 1, L + S - 2, c , ! L - S ! The value of J is given as a subscript. • The value of J is given as a subscript. E X A M P L E 11 . 4 COORDINATION CHEMISTRY: SPECTRA Example: Determine the possible values of J for the Carbon terms. Answer: The 1D and 1S terms each have only one J value. J can have only the value 0 for the 1S term (0 + 0) and only the value 2 for the 1D term (2 + 0). The 3P term has three slightly different energies, each described by a different J. For the 3P term, J can have the three values 2,1, and 0 ( 1+1, 1 + 1 - 1 , 1 + 1 - 2). Spin-orbit coupling acts to split free-ion terms into states of different energies. 2 Chapter 11 | Coordination Chemistry III: Electronic Spectra COORDINATION CHEMISTRY: SPECTRA 3 Spin-orbit coupling splits free-ion terms into states of different energies. The P term The 3P term therefore splits into states of three different energies, and the total energy level therefore splits into states of three different energies, and the energy level diagram for the diagram for thecarbon carbon atom can be shown as atom is as follows: Energy (cm-1) 1S 1S 0 21648.8 1D 1D 2 10193.7 3P 2 43.5 3P 1 3P 0 16.4 3P 0 Spin-orbit coupling (exaggerated scale for 3P) LS coupling only These are the five energy states for the carbon atom referred to at the beginning of The state of lowest energy (spin-orbit coupling included) can be predicted from Hund's this section. The state of lowest energy (spin-orbit coupling included) can be predicted third rule. from Hund’s third rule: 2 3. For subshells (such as p ) that are less than half filled, the state having the lowest J value has the lowest energy (3P0 for p2); for subshells that are more than half filled, COORDINATION CHEMISTRY: SPECTRA 3. For subshells (such as p2) that are less than half-filled, the state having the lowest J value has the lowest energy (3P0 above); For subshells that are more than half-filled, the state having the highest J value has the lowest energy. Half-filled subshells have only one possible J value. Spin-orbit coupling can have significant effects on the electronic spectra of coordinations compounds, especially those involving fairly heavy metals (atomic number >40). COORDINATION CHEMISTRY: SPECTRA 11.3 Electronic Spectra of Coordination Compounds | 413 Free-Ion Terms for dn Configurations n TABLE 11.5 Free-Ion Terms for d Configurations Configuration d1 Free-Ion Terms 2 D d2 1 1 S D 1G d3 2 d4 5 d5 2 d6 Same as d 4 d7 Same as d3 d8 Same as d2 d9 Same as d1 d10 1 D P 3F 4 2 3 3 1 1 4 2 2 2 P 4F D D 3 1 1 S D 1G P 3F P 4F P 2D 2F 2G 2H P 3D 3F 3G 3H P 2D 2F 2G 2H S D 1F 1G 1I S D 2F 2G 2I 4 D 4G 6 S S NOTE: For any configuration, the free-ion terms are the sum of those listed; for example, for the d 2 configuration, the free-ion terms are 1S + 1D + 1G + 3P + 3F. In the interpretation of spectra of coordination compounds, it is often important to identify the lowest-energy term. A quick and fairly simple way to do this is given here, 10 d 1 S COORDINATION CHEMISTRY: SPECTRA NOTE: For any configuration, the free-ion terms are the sum of those listed; for example, for the d 2 configuration, the free-io Spectra of Coordination Compounds Make the connectionInbetween electron–electron interactions and the absorption spectra of the interpretation of spectra of coordination compounds, it is often important coordination compounds. identify the lowest-energy term. A quick and fairly simple way to do this is given her 3 * symmetry. using an example a d configuration in octahedral In the interpretation of as spectra of coordination compounds, it is often important to identify the lowest-energy term. 1. Sketch the energy levels, showing the Example: d3 configuration in octahedral symmetry d electrons. 1. Sketch the energy levels, showing the d electrons. 2. Spin multiplicity2.ofSpin lowest-energy multiplicity state of lowest-energy state = number of unpaired = number of unpaired electrons +1 electrons ** +1. Spin multiplicity = 3 + 1 = 4 Spin multiplicity = 3 + 1 = 4 3. Determine the maximum possible 3. Determine the maximum possible value of ML value MLconfiguration (sum of ml values) for the con(sum of ml values) forofthe as shown. Maximum possible value of ML for thr 2 + 1 + 0 = 3; therefore, F term electrons as shown: 2 + 1 + 0 = 3; ther figuration as shown. This determines fore, F term 4F 4. Combine results ofthe Steps 2 and 3 to get the ground term. type of free-ion term (e.g., S, P, D). 4. Combine results of Steps 2 and 3 to get 4 F 11 | Coordination Chemistry III: Electronic Spectra COORDINATION CHEMISTRY: SPECTRA E X A M P L E 11 . 5 Example: 4 (low What is the ground d 4 (low spin)? What is the ground term term for dfor spin)? 1. 2. 3. 1. 2. Spin multiplicity = 2 + 1 = 3 Spin multiplicity = 2 + 1 = 3 3. Highest possible value of ML = 2 + 2 + 1 + 0 = 5; therefore, H term. Note that here, m = 2 for the first two electrons does not violate the exclusion principle l Highest possible value of ML = 2 + 2 + 1 + 0 = 5, therefore H term. because the electrons have opposite spins. Note that here, ml = 2 for the first 3two electrons does not violate the exclusion principle 4. Therefore, the ground term is H . because the electrons have opposite spins. 6 EXERCISE 11.5 Determine the ground terms for high-spin and low-spin d 3H . 4. Therefore, the ground term is configurations in O symmetry. h With this review of atomic states, may nowand consider the electronic states of in EXERCISE: Determine the ground terms forwe high-spin low-spin d6 configurations coordination compounds and how transitions between these states give rise to the observed Oh symmetry. spectra. Before considering specific spectra, however, we must also consider which types of transitions are most probable and, therefore, give rise to the most intense absorptions. COORDINATION CHEMISTRY: SPECTRA Selection Rules The relative intensities of absorption bands are governed by a series of selection rules. 1. Transitions between states of the same parity (symmetry with respect to a center of inversion) are forbidden (Laporte rule). e.g. transitions between d orbitals (g ⎯ g transitions) are forbidden transitions between d and p orbitals (g ⎯ u) transitions are allowed 2. Transitions between states of different multiplicities are forbidden (Spin rule). e.g. transitions between 4A2 and 4T1 are spin-allowed transitions between 4A2 and 2A2 are spin-forbidden These rules seem to rule out transitions for TM complexes. COORDINATION CHEMISTRY: SPECTRA Mechanisms by which the Selection Rules can be relaxed 1. The bonds in TM complexes are not rigid but undergo vibrations that temporarily change the symmetry (i.e. center of symmetry is lost). This phenomenon is called vibronic coupling. • This provides a way to relax the Laporte rule. • Thus, d ⎯ d transitions (ε = 10 - 50 M-1.cm-1) commonly occur and responsible for the bright color of TM complexes. 2. The Td complexes absorb more strongly than Oh complexes the same M and oxidation state. • M ⎯ L σ bonding in Td involves combination of sp3 and sd3 hybridization of M. • Mixing p with d also relaxes the Laporte rule. COORDINATION CHEMISTRY: SPECTRA 3. Spin-orbit coupling in some cases provides a mechanism for relaxing the spin rule. • For 1st-row TM, such transitions are weak (ε < 1 M-1.cm-1). • Spin-orbit coupling becomes more important in complexes of 2nd and 3rd row TM. CORRELATION DIAGRAMS • Show two extremes Example: d2 and octahedral [V(H2O)6]3+ 1. Free ion (no ligand field) • For d2 the term symbols are 3F (lowest E), 3P, 1G, 1D, and 1S. These describe the energy of d2 V3+ ion. • In CD the free ion terms are shown on the far left. COORDINATION CHEMISTRY: SPECTRA 2. Strong ligand field) • Three possible configurations for d2 in an octahedral field • These states are shown in the far right of the CD as the strong field limit • Here, the effects of the ligands are so strong that it completely overrides the effects of the LS coupling. COORDINATION CHEMISTRY: SPECTRA 416 Chapter 11 | • But in an actual complex the situation is intermediate between the two extremes. Correlation Diagram for d2 Coordination Chemistry III: Electronic Spectra FIGURE 11.3 Correlation Diagram for d2 in Octahedral Ligand Field. 1 in Octahedral Ligand Field. 1S 3 1 Eg 1g Eg eg2 A2g 1 1G T1g 1 T2g 1 A1g Energy 3 P 3T 1g 1 1D •The correlation diagram shows the full range of possibilities where the ligand field is not sufficiently strong to quench the coupling between the states generated by the metal valence electrons. 1A 1 •At zero field, the ml and ms values of the individual electrons couple to form, for d2, the five terms 3F, 3P, 1G, 1D, and 1S, representing five atomic states with different energies. •At a very high ligand field strength, the t2g2, t2g eg, and eg2 configurations predominate. A1g Eg 1T 1g 1T 2g 3T 1g 3 T2g t2geg 1T 2g 3 A2g 3F 3T 2g 3T 1g 1A 1g 1E •Transitions to states with the same multiplicity as the ground state are more likely (red arrows). g 1T 2g 3T 1g Free Weak Interaction Ion Strong Interaction t2g2 q Strong Interaction octahedral ligand field, the free-ion terms will be split into states corresponding to the irreducible representations, as shown in Table 11.6. Similarly, irreducible representations may be obtained for the strong-field limit configurations (in our example, t2g2, t2g eg, and eg2 ). Note especially the following characteristics of this correlation diagram: COORDINATION CHEMISTRY: SPECTRA 1. The free-ion states (terms arising from LS coupling) are shown on the far left. In an octahedral ligand field, free-ion terms willonbe into states corresponding to the 2. The extremely strongthe field states are shown thesplit far right. irreducible representations: TABLE 11.6 Splitting of Free-Ion Terms in Octahedral Symmetry Term Irreducible Representations S A1g P T1g D Eg + T2g F A2g + T1g + T2g G A1g + Eg + T1g + T2g H Eg + 2T1g + T2g I A1g + A2g + Eg + T1g + 2T2g NOTE: Although representations based on atomic orbitals may have either g or u symmetry, the terms given here are for d orbitals and as a result have only g symmetry. See F. A. Cotton, Chemical Applications of Group Theory (3rd ed., Wiley InterScience, New York, 1990, pp. 263–264) for a discussion of these labels. in the correlation diagram (Figure 11.3) joining the T1g state arising from the F free-ion term with the 3T1g state arising from the strong-field term, t2g 2. In the Tanabe–Sugano diagram (Figure 11.4), this line is made horizontal*; it is labeled 3T1g (F) and is shown to arise from the 3F term in the free-ion limit (left side of diagram).** COORDINATION CHEMISTRY: SPECTRA TS Diagram for d2 in Octahedral Ligand Field. TANABE-SUGANO DIAGRAMS 3A • modified correlation diagrams that are useful in the interpretation of electronic spectra of coordination compounds. • for the d2 configuration, the lowest-energy state is 3 described by the line joining the T1g state arising from the 3F free-ion term with the T1g state arising from the strong-field term, t2g. • the excited states of the same spin multiplicity as the ground state are the 3T2g, 3T1g(P), and the 3A2g. *This 60 3 T (P) 1g 1S 50 3T Increasing energy • the lowest-energy state is plotted along the horizontal axis; consequently, the vertical distance above this axis is a measure of the energy of the excited state above the ground state. 2g 2g 40 E B 30 1G 20 3P 1D 10 3F 3T 1g(F) 0 0 10 20 30 40 ¢o B Increasing field strength does not imply that there is no dependence between the absolute energy of the ground state term and the ligand field strength, but defining the horizontal axis in this way is useful from a spectroscopic perspective, where differences in energy between the ground and excited states are examined. 2 The reader should verify that these are the same triplet excited states shown in the d correlation diagram. Excited states of other spin multiplicities are also shown; but, as we COORDINATION CHEMISTRY: SPECTRA will see, they are generally not as important for spectral interpretation. Theplotted quantities plotted in a Tanabe–Sugano as follows: The quantities in a Tanabe–Sugano diagramdiagram are as are follows: Horizontal axis: !o B where ! o is the octahedral ligand field splitting, described in Chapter 10. B= Racah parameter, a measure of the repulsion between 2 terms of the same multiplicity. For d , for example, 3 3 the energy difference between F and P is 15B. 3P 15 B 3F Vertical axis: E B where E is the energy (of excited states) above the ground state. Racah aboution coordination complexes. B for a • B for a free ionparameters is typicallyprovide greateruseful than information B for the same in a coordination complex. free ion is typically greater than B for the same ion in a coordination complex. This com• theparison decrease in B related between a free ionmore and volume a coordination complex can usedintothe is often to how much the valence electrons canbe access assess the degree of covalency in the metal–ligand bonds. complex relative to the free ion. Electrons will experience less repulsion (or their magnetic fields will interact less intensely) if they occupy more volume. In this regard, the decrease in B between a free ion and a coordination complex can be used to assess the degree of COORDINATION CHEMISTRY: SPECTRA 3+ 2 [V(H2O)6] (d ) • The ground state is 3T1g(F); this is the only electronic state that is appreciably occupied under normal conditions. Spin-allowed transitions for d2 configuration • Absorption of light should occur primarily to excited states also having a spin multiplicity of 3. 3 • There are three of these: T2g, 3 3 T1g(P), and A2g. • Two bands are readily observed at 17,800 and 25,700 -1 cm , a third band at 38,000 cm-1 is observed in solid state. ν1 ν2 COORDINATION CHEMISTRY: SPECTRA Tanabe-Sugano Diagrams 2 d 3 d 4 d low-spin high-spin COORDINATION CHEMISTRY: SPECTRA Tanabe-Sugano Diagrams low-spin d5 high-spin d6 low-spin high-spin COORDINATION CHEMISTRY: SPECTRA 7 d 8 d COORDINATION CHEMISTRY: SPECTRA • These are all high-spin complexes because water is a relatively weak field ligand. Spectra of 1st row TM complexes of the formula: [M(H2O)6]n+ •The absorptivities for most bands are similar (1 to 20 L mol-1 cm-1) except for [Mn(H2O)6]2+,which has much weaker bands. •Solutions of [Mn(H2O)6]2+ are extremely pale pink, much more weakly colored than solutions of the other ions shown. •That [Mn(H2O)6]2+ is colored at all is a consequence of very weak forbidden transitions to excited states of spin multiplicity other than 6 (there are many such excited states, hence the rather complicated spectrum). COORDINATION CHEMISTRY: SPECTRA Determining Δo from Spectra Each of these cases corresponds to a simple excitation of an electron from a t2g to an eg orbital with the final (excited) electron configuration having the same spin multiplicity as the initial configuration. COORDINATION CHEMISTRY: SPECTRA Example: -1. Using Electronic Spectra of Coordination Compounds [V(H2O)6]3+ has absorption bands at 17,800 and 25,700 11.3 cm the Tanabe-Sugano diagram for d2, estimate values of Δo, and B for this complex. FIGURE 11.13 Spin-Allow Transitions for d 2 Configura 3 A2g(F) d2 70 60 ¢o /B 3 T1g(P) 50 3 n1: T1g (F) n2: 3T1g (F) n3: 3T1g (F) n3 3 T2g(F) 3 T1g(P) 3 A2g(F) E 40 B n2 3T (F) 2g 30 n1 20 3 P 10 3 T1g(F) 3 F 10 20 ¢o B 30 40 3 impossible. However, we can indirectly measure the difference between T2g (for the t2g eg 3 2 | E X A M P L E 11 . 8 [V(H2O)6]3 + has absorption bands at 17,800 and 25,700 cm - 1 . Using the 2 COORDINATION CHEMISTRY: SPECTRA Tanabe–Sugano diagram for d , estimate values of ! o and B for this complex. From the Tanabe–Sugano diagram there are three possible spin-allowed transitions From the Tanabe-Sugano diagram there are three possible spin-allowed transitions (Figure 11.13 ): 3 T1g (F) h 3 T2g (F) n1 (lowest energy) 3 T1g (F) h 3 T1g (P) n2 T1g (F) h 3 A2g (F) n3 (one of the these must be the higher-energy band) 3 ¶ It is often useful to determine the ratio of energies of the absorption bands. In this When working with spectra, it is often useful to determine the ratio of energies of the example, absorption bands. In this example, -1 25,700 cm = 1.44 17,800 cm - 1 the ratio of energy of the higher-energy transition (n2 or n3) to the lowest-energy transition (n1) must therefore be approximately 1.44. From the Tanabe–Sugano diagram, The ratio of energy of the higher energy transition (ν2 or ν3) to the lowest-energy transition (ν1 ) must therefore be approximately 1.44. COORDINATION CHEMISTRY: SPECTRA From the Tanabe-Sugano diagram, the ratio of ν3 to ν1 is approximately 2, regardless of the strength of the ligand field. Thus, ν3 can be eliminated as the possible transition occurring at 25,700 cm-1. This means that the 25,700 cm-1 band must be ν2, corresponding to and ν2/ν1 = 1.44 The ratio ν2/ν1 varies as a function of the strength of the ligand field. By plotting the ratio ν2/ν1 versus Δo/B, we find that ν2/ν1 = 1.44 at approximately Δo/B = 31. e considered to have the 4T1g h 4A2g transition in the small shoulder near 6,000 cm–1 and the 4T1g (F) h 4T1g (P) transition at the peak.** we can see that the ratio of n3 to n1 is approximately 2, regardless of the strength ligand field; the slope of line associated with the 3A2g(F) state is approximately tw that of the line associated with the 3T2g(F) state. We can therefore eliminate n3 as her Configurations: d5 (High Spin), d4 to d7 (Low Spin) possible transition occurring at 25,700 cm - 1 . This means that the 25,700 cm - 1 b h-spin d5 complexes have no excited states of the same spin multiplicity (6) as must the be n , corresponding to 3T (F) h 3T (P), and 1.44 = n2 . 2 1g 1g n1 und state. The bands that are observed are therefore the consequence of spin-forbidden nsitions and are typically very weak as, for example, in [Mn(H2O)6]2 + . The interested The ratio n2 / n1 varies as a function of the strength of the ligand field. By plottin der is referred to the literature11 for an analysis of such spectra. In the case of low-spin ratio n2 / n1 versus ! o / B after extracting these values from the Tanabe–Sugano di (Figure 11.14), we find that n2 / n1 = 1.44 at approximately ! o / B = 31.10,* COORDINATION CHEMISTRY: SPECTRA 2.6 E/B ¢o /B n1 n2 n2/n1 0 10 20 30 40 50 0 8.74 18.2 27.9 37.7 47.6 15 21.5 31.4 40.8 50.4 60.2 — 2.46 1.73 1.46 1.34 1.26 2.2 n2 n1 1.8 1.4 1.0 0 10 20 30 ¢o B 40 50 !o At = 31, B E E 25,700 cm - 1 n2 : = 42 (approximately); B = = = 610 cm - 1 B 42 42 E E 17,800 cm - 1 n1 : = 29 (approximately); B = = = 610 cm - 1 B 29 29 !o Because = 31, B ! o = 31 * B = 31 * 610 cm - 1 = 19,000 cm - 1 This procedure can be followed for d2 and d7 complexes of octahedral geometry estimate values for ! o (and B). EXERCISE 11.8 Use the Co(II) spectrum in Figure 11.8 and the Tanabe–Sugano ferent references report slightly different positions for the absorption bands of [V(H2O)6]3 + and hence htly different values of B and ! o. diagrams of Figure 11.7 to find ! o and B. The broad band near 20,000 cm–1 can he 4T1g S 4A2g transition is generally weak in octahedral complexes of Co2 + because such a transition be considered to have the 4T1g h 4A2g transition in the small shoulder near esponds to simultaneous excitation of two electrons and is less probable than the other spin-allowed 16,000 cm–1 and the 4T1g (F) h 4T1g (P) transition at the peak.** sitions, which are for excitations of single electrons. Other Configurations: d5 (High Spin), d4 to d7 (Low Spin) High-spin d5 complexes have no excited states of the same spin multiplicity (6) ground state. The bands that are observed are therefore the consequence of spin-fo Tetrahedral complexes generally have more intense absorptions than octahedral complexes. This is a consequence of the first (Laporte) selection rule (Section 11.3.1): transitions COORDINATION CHEMISTRY: SPECTRA between d orbitals in a complex having a center of symmetry are forbidden. As a result, absorption bands for octahedral complexes are weak (small molar absorptivities); that they TETRAHEDRAL COMPLEXES absorb at all is partly the result of vibrational motions that spontaneously distort complexes In general, tetrahedral complexes (no inversion center) have more intense absorptions slightly from pure O symmetry. h than octahedral complexes. In tetrahedral complexes, the situation is different. The lack of a center of symmetry This is means a consequence of the first (Laporte) selection rule: transitions between d orbitals in that the Laporte selection rule does not apply. The consequence is that tetrahedral a complex having a center of symmetry are forbidden. * complexes often have much more intense absorption bands than octahedral complexes. As we have seen, the d orbitals for tetrahedral complexes are split in the opposite The d orbitals for tetrahedral complexes are split in the opposite fashion to octahedral fashion to octahedral complexes: complexes: eg t2 e t2g Octahedral Tetrahedral A useful comparison can be drawn between these by using the hole formalism. This is best 1 illustrated by example. Consider a d configuration in an octahedral complex. The one elec9 complexes often have much more intense absorption bands than octahedral complexes.* As we have seen, the d orbitals for tetrahedral complexes are split in the opposite fashion to octahedral complexes: COORDINATION CHEMISTRY: SPECTRA eg Hole formalism t2 Consider a d1 configuration in an octahedral complex. e t2g The single electron occupies an orbital in a triply degenerate set (t2g) Octahedral Tetrahedral A useful comparison can be drawn between these by using the hole formalism . This is best 9 Now, consider a d configuration in a1tetrahedral complex. illustrated by example. Consider a d configuration in an octahedral complex. The one electron occupies an orbital in a triply degenerate set (t2g). Now, consider a d9 configuration in This configuration has a "hole" in a triply degenerate set of orbitals (t2). a tetrahedral complex. This configuration has a “hole” in a triply degenerate set of orbitals 1 (t2). It can be shown that, in terms of symmetry, the 1d Oh configuration is analogous to the 9 It can be shown that, in terms of symmetry, the d O configuration is analogous to the d h d9 Td configuration; the “hole” in d9 results in the same symmetry as the single electron in d1 . Td configuration eg Hole t2 e Octahedral t2g Tetrahedral In practical terms, this means that, for tetrahedral geometry, we can use the correlation 10 - n n diagram for the d configuration in octahedral geometry to describe the d configuration COORDINATION CHEMISTRY: SPECTRA The "hole" in d9 Td, results in the same symmetry as the single electron in d1. Thus, the correlation diagram for d10-n configuration in octahedral geometry can be used to describe the dn configuration in tetrahedral geometry. Examples: d2 in Oh for d8 in Td d3 in Oh for d7 in Td CHARGE-TRANSFER SPECTRA Coordination compounds exhibit strong charge-transfer absorptions, typically in the UV-VIS spectrum. - much more intense than d-d transitions (20 L mol-1 cm-1 or less) - molar absorptivities of 50,000 L mol-1 cm-1 or greater are common Examples of charge-transfer absorptions are numerous. The octahedral complexes 6 2- 5 32IrBr6 (d ) and IrBr6 (d ) both show charge-transfer bands. For IrBr6 , two bands COORDINATION SPECTRA appear, near 600 nm and near 270 nm; CHEMISTRY: the former is attributed to transitions to the t2g 3levels and the latter to the eg . In IrBr6 , the t2g levels are filled, and the only possible ChargeLMCT transfer to metalis(CTTM) or ligand metal charge transfer (LMCT) absorption therefore to the eto . Consequently, no low-energy absorptions in the g 600-nm range are observed, but strong absorption is seen near 250 nm, corresponding to • electrons are excited from orbitals originating from the ligands to metal-based orbitals. charge transfer to eg . A common example of tetrahedral geometry is the permanganate ion, -transition results in formal reduction of the metal. • this type of MnO4 , which is intensely purple because of a strong absorption involving charge transfer Charge Transfer to Metal d to Metal eg d t2g LMCT Uncoordinated metal Octahedral complex Ligand sigma orbitals absorption in the visible region, and possess significantly long-lived excited states to permit photo-promoted electron and energy transfer processes. It is not surprising that metal CHEMISTRY: SPECTRA complexes withCOORDINATION MLCT bands have received significant attention for these applications on the basis of their typically high MLCT molar absorptivities. Coordination complexes con6 2Metal + 2to + ligand 1 + charge 3 +transfer (MLCT) Chargetaining transfer to ligand (CTTL) or the d metal ions Ru , Os , Re , and Ir with strong field N-heterocyclic organic ligands (for example, pyridine, 2,2'-bipyridine, 1,10-phenantholine) are ubiquitous • involves electron promotion from metal-based orbitals orbitals originating in this area; these “light-harvesting” complexes offerto tunable transitions, andfrom the the redox ligands (typically π* orbital) properties of these ions facilitate electron transfer steps. These low-spin complexes, such as 2+ 2+ [Ru(bpy) and • results in oxidation of [Os(phen) the metal. 3] 3] , afford MLCT excited states in which the metal is formally * oxidized (via excitation from metal-based t2g into orbitals with extensive p ligand character), Charge Transfer to Ligand p* eg d MLCT t2g Uncoordinated metal Octahedral complex Ligand p* orbitals FIGUR Charg COORDINATION CHEMISTRY: SPECTRA MLCT RuII (t2g-based) ➝ PrCN (π*) RuIII ➝ RuII RuII LMCT Cl (π) ➝ RuIII (t2g-based) RuIII Billones, J.B., Ratilla, E.M.A., Heath, G.A. Inorg. Chem. Commun. 2011, 14, 1788-1793. Scheme 1. Synthetic scheme showing the electrochemical generation of accessible Ru(IV) species from their Ru(III) analogue. J.B. Billones et al. / Inorganica Chimica Acta 429 (2015) 209–215 211 COORDINATION CHEMISTRY: SPECTRA COORDINATION CHEMISTRY: SPECTRA 212 J.B. Billones et al. / Inorganica Chimica Acta 429 (2015) 209–215 Fig. 1. Spectral progression accompanying the electrochemical oxidation of (Bun4N)[RuCl4(PrCN)2] into neutral [RuCl4(PrCN)2] in a CH2Cl2 solution at !50 !C and +1.7 Arrows indicate the direction of change. COORDINATION CHEMISTRY: SPECTRA 214 J.B. Billones et al. / Inorganica Chimica Acta 429 (2015) 209–215 Fig. 5. Variation of chloride-based donor orbital and ruthenium(IV)-based acceptor orbital levels involved in ligand-to-metal charge transfer (LMCT) transitions in [RuCl(6!n)Ln]n!2 series of complexes; computed at DFT B3LYP/6-31G⁄ level. Fig. 4. Optical spectra of [RuCl6!n(PrCN)n]n!2 (n = 0–3) in CH2Cl2 showing the chloride-to-RuIV charge-transfer manifold. in transition energies. In the case of Ru(III) series, [RuCl6!n(PrCN)n]n!3 [10], the Cl donor level descended at slower rate compared to the Ru acceptor level. Consequently, the stepwise halide removal shifts the principal XMCT band to lower energy. However, the sharp des- When comparing LMCT bands in spectra of series of complexes with the same oxidation state, the observed uneven movement of the main absorption manifold largely contradicts with essentially linear variations in the DFT calculated halide donor and metal acceptor levels as a function of n (r = !0.9997 in both sets) (Fig. 5). The calculated donor–acceptor gap in fact widened across the exhaustive series (n = 0–5) indicating that the LMCT band should have regularly drifted to the high-energy side of the spectrum as n increases. In particular, the calculated donor–acceptor gap increases from 1.88 eV (in [RuCl6]2!) to 2.50 eV (in [RuCl(PrCN)5]3+). Thus, these dichotomous observations should elicit more expansive