CH22_VB_CF - VBLEBQE Boob Tuna (Lomum Ewan»; mom) Or-...

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Unformatted text preview: VBLEBQE Boob Tuna (Lomum Ewan»; mom) Or- Comrmcs On. couomnnw communes Ottawa.” TEN-h «mean. 58») use Futon. Guam» Hem 1mm OctfiaEbup COMPLfles menu-no Primaries: men Ito» spun Tmuepeau [swans Ptan announces Octahedral Complexes Among the first complexes studied were those of chromium(III). There are a great number of these, including the violet complex hexaaquachromium(III) chloride, [Cr(H20)6]Cl3, and the yellow complex hexaamminechromium(III) chloride, [Cr(NH3)6]Cl3. Almost all chromium(III) complexes have the coordination number 6 and thus are octahedral complexes. All are paramagnetic, displaying a magnetism equivalent to that of three unpaired electrons. Let us see how valence bond theory explains the bonding and paramagnetism of the complex ions of chromium(HI), such as Cr(NH3)63+. The electron configuration of the chromium atom is [Ar]3d54s1. In the forma— tion of a transition-metal ion, the outer s electrons (here, 451) are lost first, then the 3 outer d electrons. Therefore, Cr3+ has the configuration [Ar]3d3. The orbital dia- d‘ gram is [Ar] ooooo 0 COO @0000 3d 4s 4p 4d Note that the 3:13 electrons are placed in separate orbitals with their spins in. the same direction, following Hund’s rule. Two empty 3d orbitals, in addition to orbitals of the n = 4 shell, may be used for bonding to ligands. If electron pairs from six :NH3 ligands are to bond to Cr“, forming six eQuiv— alent bonds, hybrid orbitals will be required. Hybrid orbitals with an octahedral arrangement can be formed with two d orbitals, the 4s orbital, and the three 4p orbitals. The d orbitals could be either 3d or 40'; the two available 3d orbitals are used because they have lower energy. We will call these dzsp3 hybrid orbitals (rather £593. h'bh J»! 80.“th s 3d} hybri ii to.“ M than sp3d2) tovemphasize that the d orbitals have a principal quantum number 1 less than that of the s and p orbitals. We can now write the orbital diagram for the metal atom in the complex. ' [Ar] ®®®i ® @0000 4p 4d 3d I 4S Cr in Cr(NH3)63+: 11er3 bonds to ligands Electron pairs donated from ligands are shown in color. Note that there are three unpaired electrons in 3d orbitals on the chromium atom, which explains the para- magnetism of this complex ion. The bonding in other octahedral complexes of chromium(lII) is essentially the same. The bonding in complexes of iron(II) is more diverse. Most of the complexes are octahedral and paramagnetic, and the magnitude of this paramagnetism indi- cates four unpaired electrons. An example is the pale green hexaaqua'ironfll) ion, Fe(H20)62+. However, the yellow hexacyanoferratefll) ion, Fe(CN)64“, is an exam— 1 ple of a diamagnetic iron(II) complex ion. (Iron in oxidation state +2 also occurs occasionally in complexes of coordination number 4, with tetrahedral geometry.) Consider the bonding in Fe(H20)62+. The configuration of the iron atom is [Ar]3d64s2, and the configuration of Fe2+ is [Ar] 3d 6. The orbital diagram of the ion is Fe“: [Ar] @0000 0 000 ©0000 3d 4s 4p 4d According to Hund’s rule, the first five of the 3d 6 electrons go into separate orbitals; the sixth electron must then pair up with one of the others. Because the 3d orbitals have electrons in them, they cannot be used for bonding to ligand orbitals unless some of the electrons are moved. Suppose you use two of the empty 4d orbitals instead. We will call the hybrid orbitals formed from them sp3d2 (rather than dzsp3) to emphasize that the d orbitals have the same principal quantum number as that of the s and p orbitals. Then the orbital diagram of the complex ion would be FeinFetHao)62+: [Ar] @0600 ® 000 3d 4d 4s 4p 5173112 bonds to ligands This bonding picture correctly explains the paramagnetism of the complex ion and shows that it should correspond to that of four unpaired electrons, in agreement with experiment. Suppose, however, that in forming a complex ion of Fe“, the 3d electrons air up so that two of the 3d orbitals are unoccupied and can be used to form d sp3 hybrid orbitals. The configuration of this excited state of the Fe2+ ion is 000000 4s 4p Z Fe2+(excited): [AI] ' 3d . 4d available for dzsp3 bonds The 3d orbitals are lower in energy than the 4d orbitals, so they will be preferred for bonding if available. However, pairing of the electrons to make the two 3d orbitals available for 0125p3 bonding requires energy (because two electrons in the same orbital strongly repel one another). You would expect this type of bonding to occur only if the bonds formed are sufficiently strong to provide the energy needed for electron pairing. This is apparently the case for the hexacyanoferrate(ll) ion, Fe(CN)64_. (The CN‘ ion tends to give strong bonding to metal ions, in contrast to- the weaker bonding of H20.) The orbital diagram for the iron atom in this complex is it 3 -- A sf Hemmer rm] on o noel ooooo 3d 4s 4p 4d dzsp3 bonds to ligands This diagram correctly describes the ion as diamagnetic (no unpaired electrons). If you similarly examine any transition-metal ion that has configurations d4, d5, d6, or d7, you find two bonding possibilities. In one case, the bonding is sp3d2. In the other, some additional electrons are paired and the bonding is dzsp3 . There are fewer unpaired electrons in the latter type of complex than that with sp3d2 bonding. A high-spin complex ion is a complex ion in which there is minimum pairing of electrons in the orbitals of the metal atom. A low-spin complex ion is a complex ion in which there is more pairing of electrons in the orbitals of the metal atom than in a corresponding high-spin complex ion. Because low-spin complexes require energy for pairing of electrons, they are expected to occur with ligands that bond relatively strongly. Weakly bonding ligands form high-spin complexes. EXAMPLE 23.5 I Describing the‘Bonaiiig ” ' at? ‘ (Valence BondTheqny- (CobaltaI) has ihighfspm; i such. the pinkion; 7 BSCfibe the bondingifiablo'th ‘ I 'sonufrroN'} -' Theajelectron configuratio . ' [mart “A high-spin complex, r, hybrid orbitals. The comple we 59‘“ “Wall St») .,, :3 3d orbital, bu! the, othe‘fiwo‘ Md)- The orbital diagram r 1 4‘ “6?; Low .89») l . frelwwwwi' at not EXPLQNOED blew Bu VMw boot) TREND! *6 Exercise 23.8 i Cobalt(III) forms many stable complex ions, including Co(NH3)63+. Most of these are 3* octahedral and diamagnetic. The complex C01363 ", however, is paramagnetic. to Describe the bonding in Co(NH3)63+ and C0F63_, using valence bond theory. How many unpaired electrons are there in each complex ion? » r If you examine metal ions with configurations d8 and d9, you find only one spin type of octahedral complex possible. (Recall that the d3 configuration, as in i Cr“, gives only one type of octahedral complex; d1 and d2 are similar.) An exam- ple of a d3 complex is the green ion Ni(H20)62+. The nickel atom has the configu- ration [Ar]3d84s2, and the Ni2+ ion has the configuration [Ar]3d8. The orbital dia- d‘g gram for this ion is . Ni“: no «new» 0 000 ©0000 3‘1 4S 4 4d P ‘ ' Because three of the 3d orbitals are doubly occupied, two empty orbitals cannot be created by pairing of electrons. However, you could obtain octahedral hybrid orbitals using the 4d orbitals. Thus, NiinNi(H20)oz+i [Ar] «961969630 ® 000 3d 4 4d 4s p m" sp3d2 bonds to ligands This would predict a paramagnetic complex ion with two unpaired electrons, which is what is observed experimentally. Tetrahedral and Square Planar Com lexes Care must be taken when pre— Coordination number 6 (octahedral complexes) is very common, and coordination dieting geometry solely from number 4 is nearly as common. In the latter case, the geometries are either tetrahe- magnetic Characteristics- A dral or square planar. Table 23.7 lists the hybrid orbitals used to describe various “3ka complex With the geometries. The tetrahedral geometry is described by sp3 hybrid orbitals. The bidentate ligand acetylacetonate Square planar geometry uses dspz orbitals. (.2 2mm has the empmcffl Nickel(H) complex ions with four ligands are common. Most of these, such as parafiggggifi‘f‘iffjjggeag‘fufi Ni(CN)42‘, are diamagnetic. That _Ni(CN)42‘ is diamagnetic is evidence for a coordinaga mu would predict a square planar geometry. 4 To see this, recall that the electron configuration of the ’ Ni2+ ion is [Ar]3d8, with three doubly occupied 3d orbitals and two singly occupied tetrahedral geometry about Ni. . 2 I _ . The actual fommlajs 3d orbitals. To form dSp hybr1d orbitals, the two unpaired 3d electrons must first be [Nit‘acac)3]3. and the geometry about Ni is or T 0.5, A alt _..___——- f," ‘ Coordination Number Geometry Hybrid Orbital rants 23.7 ———————— Hybrid Orbitals for Various 2 Linear SP3 Coordination Numbers and 4 Tetrahedral sp 2 Geometries Square planar dsp ‘ 6 Octahedral dzsp3 or sp3d2 ___—______.___—————"‘ paired, giving an unoccupied 3d orbital. The orbital diagram for the complex ionis Ni in Ni(CN)42‘: [Ar] @6196!) 3d ®®OOOOOO 4s 4p 4d dsp2 bonds to ligands EXAMPLE 23.6 Describing the Bonding in a Four-Coordinate Complex Ion (Valence Bond Theory) Nickelal) forms some tetrahedral complex ions such as Ni(NH3)42+. Discuss the bonding in this complex ion and describe its magnetic characteristics. SOLUTION The orbital diagram for Ni“, whose electron configuration is [Ar]3d8, is as follows: Ni“: [Ar] 0 COO @0000 34 4s 4p 4d A tetrahedral geometry uses sp3 hybrid orbitals. Therefore, none of the 3d electrons in Ni2+ needs to be paired up or promoted. The orbital diagram of Ni(NH'3)‘42-+ follows; . NinNrNHsiftz“ [Ar] one _' ,_ I 3d} -. sp’bonds to ligahdsi ' Becauséithere are two‘unpairedi electrons, ther'teh'ahedral‘féomplgx ions, of nickelal) are paramagnetic»? in contrast to the square planar complex ions, which are diamag.‘ "netic.*"-,..::: I V " ‘ A ’ ‘ 1 v. d r“ .. ,1 . _ ' ipThe.’ — is pararnagnetic,;with a magnetism corresponding to three unpairede ectrons‘. If the complex ion indeed has, four ligands, as suggested by the for-' a? hat com tryis mdrcated?‘ ' “ CRYSTAL FIELD THEORY Although valence bond theory explains the bonding and magnetic properties of , complexes in straightforward fashion, it is limited in two important ways. First, the {theory cannot simply explain the color of complexes. Second, the theory is difficult s'tt‘riextend- quantitatively. Consequently, another theory—crystal field theory—has temerged as the prevailing view of transition-metal complexes. " Crystal field theory is a model of the electronic structure of transition-metal complexes that considers how the energies of the d orbitals of a metal ion are aflected by the electric field of the ligands. According to this theory, the ligands in a transition-metal complex are treated as point charges. So a ligand anion becomes simply a point of negative charge. A neutral molecule. with its electron pair that it donates to the metal atom, is replaced by a partial negative charge, representing the negative end of the molecular dipole. In the electric field of these negative charges, the five d orbitals of the metal atom no longer have exactly the same energy. The result, as you will see, explains both the paramagnetism and the color observed in certain complexes. V The simplifications used in crystal field theory are drastic. Treating the ligands as point charges is essentially the same as treating the bonding as ionic. However, it turns out that the theory can be extended to include covalent character in the bond- ing. This simple extension is usually referred to as ligand field theory, but after including several levels of refinements, the theory becomes equivalent to molecular orbital theory. Effect of an Octahedral Field on the d Orbitals All five (I orbitals of an isolated metal atom have the same energy. But if the atom is brought into the electric field of several point charges, these d orbitals may be affected in different ways and therefore may have different energies. To understand how this can happen, you must first see what these d orbitals look like. You will then be able to picture what happens to them in the crystal field theory of an octahedral complex. Figure 23.21 shows the shapes of the five d orbitals. The orbital labeled dz: has a dumbbell shape along the z-axis, with a, collar in the x—y plane surrounding this dumbbell. Remember that this shape represents the volume most likely to be occu- pied by an electron in this orbital. The other four d orbitals have “Cloverleaf” shapes, each differing from one another only in the orientation of the orbitals in space. The “Cloverleaf” orbital d12_.y2 has its lobes along the x—axis and the y—axis. Orbitals dxy, dyz, and dxz have their lobes directed between the two sets of axes designated in the orbital label. Orbital dxy, for example, has its lobes lying between the x- and y—axes. A complex ion with six ligands will have the ligands arranged octahedrally about the metal atom to reduce mutual repulsion. Imagine that the ligands are replaced by negative charges. If the ligands are anions, they are replaced by the anion charge. If the ligands are neutral molecules, they are replaced by the partial negative charge from the molecular dipole. The six charges are placed at equal dis- tances from the metal atom, one charge on each of the positive and negative sides of the x-, y—, and z—axes. (See Figure 23.21.) ' Fundamentally, the bonding in this model of a complex is due to the attraction of the positive metal ion for the negative charges of the ligands. However, an elec- tron in a d orbital of the metal atom is repelled by the negative charge of the ligands. This repulsion alters the energy of the d orbital depending on whether it is directed toward or between ligands. For example, consider the difference in the repulsive effect of ligands on metal—ion electrons in the dz: and the dxy orbitals. Because the dz: orbital is directed at the two ligands on the z—axis (one on the -z side and the other on the +2 side), an electron in the orbital is rather strongly repelled by them. The . energy of the dz: orbital becomes greater. Similarly, an electron in the d,y orbital is - repelled by the negative charge of the ligands, but because the orbital is not pointed directly at the ligands, the repulsive effect is smaller. The energy is raised, but less than the energy of the d,: orbital is raised. If you look at the five d orbitals in an octahedral field (electric field of octahe- drally arranged charges), you see that you can divide them into two sets. Orbitals dz: and d,._,. are directed toward ligands, and orbitals dxy, dyz, and d,“ are directed between ligands. The orbitals in the first set (dz: and dxz_yz) have higher energy than FIGURE 23.21 The five (I orbitals. The dz: orbital has a dumbell shape with a collar; the other orbitals have cloverleaf shapes. In an isolated atom, these orbitals have the same energy. However, in an octahedral complex, the orbitals split into two sets, with the dz: and dx2_yz orbitals having higher energy than the other three. Note 'High-Spin and Low-Spin ComElexes k that the lobes of the dz: and d;.,,: , , orbitals point toward. mggfigaflds . Once you have the energy levels for the d orbitals 1n octahedral complex, you can to}, decrde how the d electrons of the metal ton are distributed in them. Knowing this charges), whereas '_ distribution, you can predict the magnetic characteristics of the complex. the other orbitals point; Consider the complex ion Cr(NH3)63+. According to crystal field theory, it ligands. The repulsionlrsgreatai consists of the Cr3+ ion surrounded by NH3 molecules treated as partial negative in the case Of the dz’ anddae—y charges. The effect of these charges is to split the d orbitals of Cr3+ into two sets as orbitals- shown in Figure 23.22. The guestion now is how the d electrons are distributed among the d orbitals of the Cr + ion. Because the electron configuration of Cr“ is [Ar]3d3, there are three at electrons to distribute. They are placed in the d orbitals of those in the second set (an, dyz, and dxz). Figure 23.22 shows the energy levels of the d orbitals in an octahedral field. The difi‘erence in energy between the two sets of d orbitals on a central metal ion that arises from the interaction of the orbitals with the electric field of the ligands is called the crystal field splitting, A0. (represented hereb'; FIGURE 23.22 Energy levels of d orbitals in an octahedral field. The posi- tive metal ion is attracted to the negative charges (ligands). but electrons in the d orbitals are repelled by them. Thus, although there is an overall attraction, the a’ orbitals no longer have the same energy. FIGURE 23.23 Occupation of the 3d orbitals in an octahedral complex of Cr“. The electrons occupy different lower-energy orbitals but have the same spin (Hund’s rule). n 9.“, a | Energy of attraction of E5 point charges 6122 dxz _ yz d.) n.- 5 es l A0 dx)’ dyz xz i " . 1.43 lower energy, following Hund’s rule (Figure 23.23). You see that the complex ion Cr(NH3)63+ has three unpaired electrons and is therefore paramagnetic. As anOther example, consider the complex ion Fe(H20)62+. What are its mag- netic characteristics? Remember, you need look at only the d electrons of the metal ion, Fe”. The electron configuration of the ion is [Ar]3d6. You distribute six elec- trons among the d orbitals of the complex in such a way as to get the lowest total energy. If you place all six electrons into the loWer three d orbitals, you get the dis- tribution shown by the energy-level diagram in Figure 23.24A. All of the electrons are paired, so you would predict that this distribution gives a diamagnetic complex. The Fe(H20)62+ ion, however, is paramagnetic. Therefore, the distribution given in Figure 23.24A is not correct. What was the mistake? The mistake was to ignore the pairing energy, P, the energy required to put two electrons into the same orbital. When an orbital is already occupied by an elec- tron, it requires energy to put another electron into that orbital because of their mutual repulsion. Suppose that this pairing energy is greater than the crystal field splitting; that is, suppose P > A. In that case, once the first three electrons have singly occupied the three lower-energy d orbitals, the fourth electron will go into one of the higher-energy d orbitals. It will take less energy to do that than to pair up with an electron in one of the lower-energy orbitals. Similarly, the fifth electron will go into the last empty d orbital. The sixth electron must pair up, so it goes into one of the lower-energy orbitals. Figure 23.24B shows this electron distribution. In this case, there are four unpaired electrons and the complex is paramagnetic. We see that crystal field theory predicts two possibilities: a low-spin complex when P < Aoand a high-spin complex when P > AaThe value of A0 as We Will explain later, can be obtained from the spectrum of a complex, and the value of P can be calculated theoretically. Even in the absence of these numbers, however, the theory predicts that a paramagnetic octahedral complex of Fe“ should have a mag“ 45 Fe” Nulle 5m: Energy ‘ 9 Ice Energy —————> FIGURE 23.24 Occupation of the 3d orbitals in complexes of Fe“. (A) Low spin. (B) High spin. d? to 1“ Realm» 5P!” netism equal to that of four unpaired electrons. This is what you find for the 1=e(H20)62+ ion. You would expect low-spin diamagnetic Fe“ complexes to occur for ligands that bond strongly to the metal ion—that is, for those giving large As Ligands that _. might give a low—spin complex are suggested by a look at the specuochemical series. The spectrochcmical series is an arrangement of ligands according to the relative magnitudes of the crystal‘field splittings they induce in the d orbitals ofa metal ion. The following is a short version of the spectrochemical series: weak»bonding ligands strong-bonding ligands I“ <Br‘ <Cl_ <F‘ <0H‘ <H20<NH3<en<N02_ <CN‘ <CO increasing A —> From this series, you see that the CN— ion bonds more strongly than H20, which explains why Fe(CN)64— is a low-spin complex ion and Fe(I-i20)62+ a high-spin complex ion. You can also see why carbon monoxide might be expected to be poi- sonous. You know that 0; bonds reversibly to the Fe(II) atom of hemoglobin, so the. bonding is only moderately strong. According to the spectrochernical series, how- ever, carbon monoxide, CO, forms a strong bond. The bonding in this case is irre- versible (or practically so). It results in a very stable complex of CO and hemoglo- bin, which cannot function then as a transporter of 02. EXAMPLE 23.7 ' Theelectron configuration of Coal" isi[Ar~]3d The high-spin; " ' ‘ H trons dorbitalsiarer I 9 ,_ tributi’on of d electrons in Ni(H20 ’1 retrons are there in this ion? )6?f‘,'iisingc'ryscaté field theory; Howl Tetrahedral' and Square Planar Complexes When a metal ion bonds with tetrahedrally arranged ligands, the d orbitals of the ion split to give two d orbitals at lower energy and three d orbitals at higher energy (just FIGURE 23.25 Energy splittings of the d or- bitals in a complex with four ligands. (A) A tetrahedral field. The crystal field splitting, Aiis smaller than in a comparable octahedral complex. (B) A square planar field. Hts“ 59m Tenancuam. 3‘33 IN m“) amen Lou 5P») Snow PM; 659“ M Minamoto ate in. l >~n 3:” .dx_y £25. in. £5 E l A t 2. ‘12: i fa & A B the opposite of what is found for an octahedral field). See Figure 23.25A. In the field of ligands in a square planar arrangement, the d orbitals split as shown in Figure 23.253. The observed splittings, Atin a tetrahedral field are approximately one—half the size of those in comparable octahedral complexes. Only high-spin complexes are observed, because the pairing energy is always greater than Atln the square planar case, only low-spin complexes have been found. owispin-. complexesthere paramagneticalis pro blyx'te, onfigagauoa (fits: Ni(CN)42" (square planar); ...
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This note was uploaded on 04/13/2010 for the course CHEM 114 taught by Professor Jursich during the Spring '08 term at Ill. Chicago.

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CH22_VB_CF - VBLEBQE Boob Tuna (Lomum Ewan»; mom) Or-...

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