SGCh22 - Chapter Twenty-Two TRANSITION METAL CHEMISTRY AND...

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Unformatted text preview: Chapter Twenty-Two TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS • • • • Properties of the Transition Metals Coordination Compounds The Structure of Coordination Compounds Bonding in Coordination Compounds PROPERTIES OF THE TRANSITION METALS STUDY OBJECTIVES 1. 2. Describe several general characteristics of the transition metal elements. Explain why there is only a gradual decrease in atomic radius when moving across a series of transition elements. Electron Configurations of Atoms and Ions. In this chapter, we will focus on the first-row transition elements from scandium to copper. Recall that the transition metals have atoms or ions with incompletely filled d subshells. The electron configurations of the first-row transition elements were discussed in Section 7.10 of the text. For the representative elements potassium and calcium, the 4s orbital is lower in energy than the 3d orbital. The 3d orbitals begin to fill only after the 4s orbital is complete. For elements after Ca [Ar]4s2 , electrons are added one at a time to the 3d subshell beginning with scandium, the first transition element. The electron configurations are given in Table 22.1 below. The configurations of chromium [Ar]4s1 3d5 and copper [Ar]4s1 3d10 are exceptions. The basis for these exceptions is that the energies of a 3d orbital and a 4s orbital are close for these elements. For chromium, the 4s and 3d orbitals are almost identical in energy, and so the electrons enter all six of the available orbitals singly in accordance with Hund's rule. In the case of copper, the exception shows the importance of extra stability associated with a completely filled 3d subshell. Table 22.1 Electron Configurations of Atoms and Ions _________________________________________________ Atom +2 ion +3 ion _________________________________________________ Sc 4s2 3d1 3d1 [Ne] 2 3d2 Ti 4s 3d2 3d1 2 3d3 3 V 4s 3d 3d2 1 3d5 4 Cr 4s 3d 3d3 2 3d5 5 Mn 4s 3d 3d4 2 3d6 6 Fe 4s 3d 3d5 2 3d7 7 Co 4s 3d 3d6 2 3d8 8 Ni 4s 3d ___ Cu 4s1 3d10 3d9 ___ _________________________________________________ 4 30 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 31 To understand the electron configurations of the ions shown in Table 22.1, it helps to recall that electrons are removed from the 4s orbital before they are taken out of the 3d. The energies of the 3d and 4s orbitals are not as close together in the ions of the transition metals as they are in the neutral atoms. In fact, for the ions of transition elements the 3d orbitals are lower in energy than the 4s orbitals. Therefore, the electrons most easily lost are those in the outermost principal energy level, the ns. Additional electrons may then be lost from the (n – 1)d orbital. Properties of Transition Elements. For the representative elements, properties such as the atomic radius, ionization energy, and electronegativity vary markedly from element to element as the atomic number increases across any period. In contrast, the chemical and physical properties of the transition metal elements vary only slightly as we read across a period. Table 22.1 in the text gives a number of properties of the transition elements for comparison. The characteristics of transition metals are summarized below. 1. General physical properties. Transition metals have relatively high densities, high melting and boiling points, and high heats of fusion and vaporization. See Table 22.2 in the text. 2. Atomic Radius. Recall that the atomic radii of representative elements decrease markedly as we read across a period of elements. In contrast, the atomic radii of transition metals decrease only slightly as we read across a series of these elements. The decrease in radii is consistent with an increase in effective nuclear charge as atomic number increases. All the first-row transition elements have 4s orbitals as the outermost occupied orbitals. The atomic radius will depend on the strength of the nuclear charge felt by the 4s electrons. The greater the charge, the smaller the atom. The effective nuclear charge experienced by the outermost electrons depends on the shielding provided by inner electrons. For transition metals the nuclear charge increases from scandium to copper, and electrons are being added to an inner 3d subshell. These 3d electrons shield the 4s electrons from the increasing nuclear charge for the most part. Consequently the 4s electrons of the first-row transition elements feel only a slightly increasing effective nuclear charge as the atomic number increases, and the atomic radii makes only a gradual decrease. 3. Ionization Energy. The first ionization energies of the first transition metal series are remarkably similar, increasing very gradually from left to right. There is a slight increase over the first five elements then the ionization energy barely changes from iron to copper. 3. Variable oxidation states. The common oxidation states for the transition elements from Sc to Cu are +2 and +3 (Table 22.1 text). Some of the elements exhibit the +4 state, and Mn even shows +5, +6, and +7. Transition metals usually exhibit their highest oxidation states in compounds with oxygen, fluorine, or chlorine. KMnO4 and K2 Cr2 O7 are examples. The variability of oxidation states for transition metal ions results from the fact that the 4s and 3d subshells are similar in energy. Therefore, an atom can form ions of roughly the same stability by losing different numbers of electrons. The +3 oxidation states are more stable at the beginning of the series, but toward the end the +2 oxidation states are more stable. _______________________________________________________________________________ EXAMPLE 22.1 Transition Metal Elements What distinguishes a transition element from a representative element? •Method of Solution The representative elements are those in which all the inner subshells are filled and the last electron enters an outer s or p subshell. The transition elements are those in which an inner d subshell is partially filled. For example, the outer electron configuration of Sc is 4s 2 3d1 and for Co it is 4s2 3d7 . _______________________________________________________________________________ EXAMPLE 22.2 Properties of Transition Elements Why do the transition metals have higher densities than the Group 1A and 2A metals? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 32 / Transition Metal Chemistry and Coordination Compounds •Method of Solution Densities of solids are related to atomic and ionic radii. As the atoms get smaller, the solids tend to get more dense. The effective nuclear charge experienced by the outermost electrons in transition metal atoms is greater than that experienced in Group 1A and 2A metal atoms. Answer: The atomic radii of the transition elements, in general, are smaller than those of the representative elements of the same period. _______________________________________________________________________________ EXERCISES 1. 2. 3. Write the electron configurations of: a. a Mn atom b. a Mn2+ ion c. a Mn3+ ion. Which atom in each pair has the larger atomic radius? a. V or Fe b. V or Nb Which has the higher ionization energy, Ti or Fe? COORDINATION COMPOUNDS STUDY OBJECTIVES 1. 2. 3. Define the terms: complex ion, ligand, donor atom, coordination number, and coordination compound. Determine the oxidation number and coordination number of metal atoms in coordination compounds. Name coordination compounds when given their formulas, and write their formulas when given their names. Terminology. A c omplex ion consists of a central metal cation to which several anions or molecules are bonded. The complex ion may be positively or negatively charged. The metal cation is called the c entral atom , and the attached anions and molecules are called l igands . The free ligands each have at least one unshared pair of electrons which can be donated to the electron-deficient metal ions. The d onor atom is the atom in the ligand that is directly bonded to the metal. Some typical ligands are Cl– , CN– , NH 3 , H 2 O, and H2 NCH2 CH2 NH2 . A neutral species containing a complex ion is called a c oordination compound . These compounds usually have complicated formulas. Two examples are : [Ag(NH3 )2 ]Cl and K 3 [Fe(CN)6 ] The complex ion is shown enclosed in brackets. In the silver compound, Cl– is a free (meaning uncomplexed) chloride ion, and in the iron compound each K+ is a free potassium ion. K+ and Cl– ions in the above formulas are examples of counter ions. They serve to balance or neutralize the charge of the complex ion. The coordination number is defined as the number of donor atoms surrounding the central metal atom. Thus the coordination number of Pt2+ in [Pt(NH 3 )4 ]2+ is 4, and that of Co2+ in [Co(NH 3 )6 ]2+ i s 6. Some ligands contain more than one donor atom. Ligands that coordinate through two bonds are called bidentate ligands . Those with more than two donor atoms are referred as p olydentate ligands . Ethylenediamine (en) and oxalate ion (ox) are bidentate ligands: O || .. .. H2 N—CH2 —CH2 —NH2 O || C—C / \ :O:– ˙˙ ˙˙ Oxalate ion (ox), C2 O4 2– – :O: Ethylenediamine (en) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 33 Bidentate and polydentate ligands are often called c helating ligands (pronounced key-lateing). The name derives from the Greek word chele meaning "claw." Chelate complexes are extra stable because two bonds must be broken to separate a metal from a ligand. EDTA4– is an excellent chelating ligand because it has six donar atoms. Oxidation Number. Each complex ion carries a net charge that is the sum of the charges on the central 3– atom (or ion) and on each of the ligands. In Fe(CN)6 , for example, each cyanide ion (CN– ) contributes a –1 charge, and the iron ion a +3 charge. The charge of the complex ion then is; (+3) + 6(–1) = –3. The oxidation number of iron is +3 which is the same as its ionic charge. See Example 22.3 for a calculation. Naming Coordination Compounds. Thousands of coordination compounds are known. The rules that have been developed for naming then are summarized below: 1. 2. The cation is named before the anion. Within the complex ion, the ligands are named first, followed by the metal ion. Ligands are listed in alphabetical order. 3. The names of anionic ligands end with the letter o, whereas neutral ligands names are usually the same as the names of the molecules. The exceptions are H2 O (aquo), CO (carbonyl), and NH3 (ammine). Table 22.3 in the textbook lists the names of some common ligands. 4. When several ligands of a particular kind are present, use the Greek prefixes di-, tri-, tetra-, penta-, and hexa. Thus the ligands in [Co(NH 3 )4 Cl 2 ]+ are "tetraamminedichloro." If the ligand itself contains a Greek prefix, use the prefixes bis (2), tris (3), and tetrakis (4) to indicate the number of ligands present. The ligand ethylenediamine already contains the term di; therefore bis(ethylenediamine) is used to indicate two ethylenediamine ligands. 5. The oxidation number of the metal is written in Roman numerals following the name of the metal. 3– 6. If the complex is an anion, attach the ending -ate to the name of the metal. Fe(CN)6 is named hexacyanoferrate (III) ion. See Examples 22.3, 22.4, and 22.5 for the application of terminology and nomenclature rules to coordination compounds. _______________________________________________________________________________ EXAMPLE 22.3 Terminology of Coordination Compounds A certain coordination compound has the formula [Co(NH3 )4 Cl 2 ]Cl. a. Which atom is the central atom? b. Name the ligands and point out the donor atoms. c. What is the charge on the complex ion? d. What is the oxidation number (O.N.) of the central atom? •Method of Solution a. b. c. d. Inside the brackets the metal atom is written first and this is the central atom to which all ligands are bonded. Cobalt is the central atom. The ligands are written next. These are ammonia and chloride ion. The donor atoms are N and Cl, respectively. Since only one chloride ion is needed to balance the charge of the complex ion, the complex ion must have a charge of +1. The charge of the central ion plus the sum of charges of ligands = the charge of the complex ion. Ammonia is a neutral ligand and will not affect the complex ion charge. Since two chloride ions are –2, in order for the complex ion to have a +1 charge, then Co must be a +3 ion. O.N.(Co) + 2 O.N.(Cl) = complex ion charge O.N.(Co) + 2(–1) = + 1 O.N.(Co) = +3 _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 34 / Transition Metal Chemistry and Coordination Compounds _______________________________________________________________________________ EXAMPLE 22.4 Naming Coordination Compounds Give a systematic name for each of the following compounds: a. [Ag(NH3 )2 ]Br b. Ni(CO)4 c. K2 [Cd(CN)4 ] •Method of Solution a. The two ammonia ligands are represented by the term "diammine." The oxidation state of the Ag ion is +1. The bromide ion is not complexed with silver. It is a counter ion. The systematic name is diamminesilver(I) bromide. b. The term tetracarbonyl signifies that there are four carbon monoxide ligands. The oxidation state of the nickel atom is zero. Systematic name: tetracarbonylnickel(0). c. In this case the complex ion is an anion, thus cadmium will be referred to as cadmate. Since K is +1 and each cyanide ligand is –1, Cd must have a +2 charge. The name is potassium tetracyanocadmate(II). _______________________________________________________________________________ EXAMPLE 22.5 Writing Formulas from the Names Write formulas for the following coordination compounds: a. diaquodicyanocopper(II) b. potassium hexachloropalladate(IV) c. dioxalatocuprate(II) ion •Method of Solution a. diaquo refers to two water molecules, and dicyano to two cyanide ions. Since the copper ion has a +2 charge, this coordination compound is neutral. Answer: [Cu(H 2 O)2 (CN)2 ] b. The ending -ate indicates that the complex ion must be an anion. Answer: K 2 [Pd(Cl)6 ]. c. The complex ion is an anion. Answer: [Cu(C 2 O4 )2 ]2– _______________________________________________________________________________ EXERCISES . 4. 5. 6. 7. 8. 9. 10. Back What is the charge on the complex ion in K2 [PtCl 4 ]? What is the coordination number of Hg in [Hg(en)2 ]2– ? What is the oxidation number of Fe in [Fe(CN)6 ]4– ? What is the coordination number and oxidation number of the central atom in K[Cr(NH3 )3 Cl 3 ]? Give the systematic name for [CoCl 3 Br3 ]4– . Give the systematic name for [Ni(en)2 (NH3 )2 ]2+ . Write formulas for the following coordination compounds: a. diammineoxolatocopper(II) b. potassium amminetrichloroplatinate(II) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 35 THE STRUCTURE OF COORDINATION COMPOUNDS STUDY OBJECTIVES 1. 2. Describe geometric and optical isomers. Define the terms chiral and enantiomers. Common Structures. The geometry of a coordination complex is defined by the arrangement of the donar atoms of the ligands around the central metal atom. Complexes with coordination numbers of two, four, and six n+ are the most common. Complexes with a general formula ML2 , where M is the metal atom or ion and L is a donar atom, are linear with ligands on opposite sides of the metal atom. Other geometries are as follows: ___________________________________________________________ General formula Coodination Structure number ___________________________________________________________ n+ ML2 2 Linear n+ ML4 4 Tetrahedral or square planar n+ ML6 6 Octahedral ___________________________________________________________ Geometric Isomers. When two or more compounds have the same composition but a different arrangement of atoms, the compounds are called i somers . Isomerism is a characteristic feature of coordination compounds. S tereoisomers are compounds that have the same types and numbers of atoms bonded together in the same sequence, but with different spatial arrangements. There are two types of stereoisomers: geometric isomers and optical isomers. Geometric, or c is-trans , isomers are distinguished by the position of like ligands or groups. The isomer with the like-groups in adjacent positions is called the the cis isomer, and the one with like-groups across from each other is called the trans isomer. Cis and trans isomers of coordination compounds generally have different physical and chemical properties. Two forms of the complex Pt(NH 3 )2 Cl 2 have been prepared. Both have square planar structures. The trans form has no dipole moment, whereas the cis form has an appreciable dipole moment. H3N Cl Cl Pt H3N N H3 Pt Cl H3N cis Cl trans The simplest type of geometric isomerism of octahedral complexes occurs in cases where four of the six ligands of the complex are the same. Then the cis and trans forms correspond to those shown below. The two X ligands are closest to each other in the cis isomer. In the trans isomer they are across the complex ion from each other. L L L L X M M X L Forward Main Menu TOC L X L cis Back X L trans Study Guide TOC Textbook Website MHHE Website 4 36 / Transition Metal Chemistry and Coordination Compounds Optical Isomers. O ptical isomers are nonsuperimposable mirror images of one another. They are also called enantiomers. Enantiomers have the same relationship to one another as do your right and left hands. If you place your left hand parallel to your right hand facing each other, you get the same effect that you would get if you placed one hand in front of a mirror. Your left hand is the mirror image of your right hand. However, your left hand is not superimposable upon your right hand. Your hands are enantiomers, that is, nonsuperimposable mirror images of one another. The enantiomers of cis-dichlorobis(ethylenediamine)cobalt(III) ion are shown in Figure 22.13 in the text. The mirror image of the trans isomer can be superimposed on the original after rotating it by 90°. Therefore, there is only one form of the trans isomer. The mirror image of the cis isomer cannot be superimposed on the original no matter how it is rotated. The cis isomer exists as enantiomers. Not all objects are enantiomers. An ordinary chair, for example, looks the same in the mirror as when viewed directly. A chair is superimposable on its mirror image. An object which is not identical with its mirror image is said to be chiral, from the Greek word for hand (see Figure 22.12 in the text). Example 23. 7 illiustrates chiral complexes. Enantiomers are called optical isomers because they differ with respect to the direction in which they rotate the plane of polarization of plane-polarized light when light is passed through the substance. The isomer that rotates the plane of polarized light to the right is said to be dextrorotatory. Its mirror image will rotate the plane to the left, levorotatory. A racemic mixture is an equimolar mixture of the two optical isomers. Such a mixture produces no net rotation of plane polarized light. _______________________________________________________________________________ EXAMPLE 22.6 Geometric Isomers Sketch the geometric isomers of [Cr(en)2 Br2 ]+. •Method of Solution First identify the ligands. Recall that "en" stands for ethylenediamine, a neutral bidentate ligand. Bromide ion is the other ligand. The, the central ion is Cr3+ and its coordination number is 6 because it bonds to six donar atoms. The complex will be octahedral. N — N represents ethylenediamine.. For an octahedral complex, ligands at adjacent corners are referred to as cis to each other. The ligands at opposite corners are trans to each other. Two geometric isomers are possible. One has two Br– ions in cis positions, and in the other isomer the Br– ligands are trans. Br Br N N Br Cr Cr N N N cis N N N Br trans _______________________________________________________________________________ EXAMPLE 22.7 Optical Isomers For the complex ion in the preceding example, indicate if either of the isomers exhibit optical isomerism. •Method of Solution Sketch the mirror images of the cis and trans isomers, and look for nonsuperimposable mirror images, the enantiomers. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 37 Br Br N Br N Br Cr Cr N N N N cis N N cis mirror mirror image nonsuperimposable Br Br N N N N Cr Cr N N N N Br Br trans trans mirror mirror image superimposable The mirror image of the trans isomer can be superimposed on the original. Therefore, there is only one form of the trans isomer. The mirror image of the cis isomer cannot be superimposed on the original no matter how it is rotated. The cis isomer exists in two forms (enantiomers). The cis isomer is chiral. The trans isomer is not. _______________________________________________________________________________ EXERCISES 11. Distinguish between geometric isomers and optical isomers. 12. Descirbe the term chiral. BONDING IN COORDINATION COMPOUNDS STUDY OBJECTIVES 1. 2. 3. Back Explain how the interaction between the ligands and the d orbitals of the central metal atom in a coordination compound produces crystal-field splitting. Explain the relationship of the color of a complex ion to the wavelength of light absorbed. Predict the number of unpaired electrons in a given complex ion. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 38 / Transition Metal Chemistry and Coordination Compounds Crystal Field Theory. In the case of coordination compounds a satisfactory theory must account for properties such as color, magnetism, and stereochemistry. Although the crystal field theory is not the only theory of coordination compounds, it explains satisfactorily their colors and magnetic properties. The crystal field model considers the ligand to metal bonding in a complex ion to be primarily electrostatic, rather than covalent. In an isolated transition metal ion all five 3d orbitals have the same energy. The effect of the ligands is to change the relative energies of these orbitals through electrostatic interactions. There are two types of electrostatic interactions. One is the force that holds the ligands to the metal ion. This is the attraction between the positive charge of the metal ion and the lone electron pairs of the ligands. Second, there is the repulsion between the lone pairs on the ligand donor atoms and the electrons in the 3d orbitals of the metal. It is this latter interaction that gives rise to crystal field splitting, and its effect on the color and magnetic properties of the complex ion. 3– Consider the Fe(CN)6 complex ion, for example. An isolated Fe2+ ion has five 3d electrons, all with the same energy. Fe3+ [↑ ]↑ ]↑ ]↑ ]↑ ] When the six cyanide ligands become positioned in an octahedral arrangement, all the 3d orbitals are raised in energy due to the repulsion just mentioned. The repulsion energy is not the same for all 3d orbitals. Rather, those orbitals that are directed straight toward a ligand experience greater repulsion than those directed between ligands. Thus, in an octahedral complex the five 3d orbitals are split into two groups in terms of energy. One group, comprising the d xy , dyz and dxz orbitals, is not raised in energy as much as the other group consisting of the dz2 and dx 2–y 2 orbitals. The situation is shown here schematically in Figure 22.2. The energy difference between these two sets of d orbitals is called the crystal field splitting energy and is given the symbol ∆ . d z 2 d x 2– y 2 ∆ d xy d yz d xz d x 2 y 2 d z 2 d xy d yz d xz – Figure 22.2 Crystal field splitting. On the left are the energy levels of the five 3d orbitals in a free Fe3+ ion. On the right are the same d orbitals in the same ion when it is a part of the octahedral complex ion [Fe(CN)6 ]3– . Color. The color of a coordination compound results from electron transitions from the lower-energy set of orbitals to the higher-energy orbitals. Absorption of light occurs when the energy of an incoming photon is equal to the difference in energy ∆ , and an electron transition occurs. The requirement for light absorption is ∆ = Ephoton Recall from Chapter 7 that the energy of a photon is Ephoton = h ν , where h is Planck's constant, and ν is the frequency of the radiation. Therefore, ∆ = hν . A substance has a color because it absorbs light at certain wavelengths in the visible part of the electromagnetic spectrum (from 400 to 700 nm) and it reflects the other wavelengths. Light that is a combination of all visible wavelengths appears white. When white light impinges on a coordination compound, and light of a certain wavelength is absorbed, the reflected light is missing that component and no longer appears white to the eye. If the complex absorbs all colors except orange, the complex appears orange. The complex will also appear orange if it absorbs only light of the color blue. In a complementary manner, if the complex ion absorbed only orange, it would appear blue. Blue and orange are referred to as c omplementary colors . Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 39 The following "color wheel" allows you to estimate the color of reflected light when light of a given color (or wavelength) is absorbed. The colors that appear across from each other are complementary colors. For example, the hexacyanoferrate(II) ion absorbs light in the visible region of the spectrum at about 410 nm. The absorbed light is violet. The color of the complex is the color directly across the wheel from violet. The complex will appear yellow, the complementary color of violet. 750 400 nanometers RE D 620 VIOLET ORANG E 580 430 BLU E YELLO W GREE N 490 560 The extent of crystal field splitting, that is, the magnitude of ∆ , determines the wavelength of light absorbed, and thus the color of the complex. Each ligand has a different strength of interaction with the d orbitals of the metal ion. Thus each splits the orbital energies by a different amount. The ligands can be arranged according to increasing values of ∆ . This establishes a spectrochemical series. I– < Br – < C l – < OH – < F – < H2 O < NH 3 < en < CN– < CO CO and CN– are called strong-field ligands because they produce a relatively large splitting, whereas I– and Br– are weak-fields ligands because they split d orbital energies to a lesser extent. Magnetic Properties. In Chapter 7 you learned that paramagnetic substances have at least one unpaired electron and diamagnetic substances have all their electrons paired. The Fe3+ ion has a d 5 configuration. Two of its octahedral complexes, [FeF6 ]3– and [Fe(CN)6 ]3– are both paramagnetic, and yet they have different magnetic properties. [FeF6 ]3– has five unpaired electrons, but [Fe(CN)6 ]3– has only one. The former is called a high-spin complex , and the latter a low-spin complex. The reason for this difference lies in the spectrochemical series and the value of ∆ . Remember that according to Hund's rule, electrons in orbitals with similar energies prefer to be unpaired. Energy is required to make electrons pair up. Therefore, if ∆ is small, as in the case of the weak-field ligand F– , the electrons enter all five orbitals one at a time without pairing in accord with Hund's rule. For small ∆ values, pairing of electrons would not occur until the sixth electron is added. However, when ∆ is large enough, Hund's rule no longer applies to the entire set of d orbitals. The lowest energy configuration corresponds to placing all of the electrons in the three lowest-energy 3d orbitals first, before any enter the two higher-energy orbitals. This is the case for [Fe(CN)6 ]3– because CN– is a strong-field ligand and produces large ∆ values. Figure 22.2 shows the energy level diagrams for hexafluoroferrate (III) ion and hexacyanoferrate (III) ion. So far we have focused on octahedral complexes. Crystal field splitting occurs for tetrahedral and square planar complexes as well. See Figures 22.23 and 22.24 in the text for the splitting diagrams of these complexes. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 40 / Transition Metal Chemistry and Coordination Compounds ∆ ∆ [FeF6 ]3– 5 unpaired electrons (high spin) [Fe(CN)6 ]3– 1 unpaired electrons (low spin) Figure 22.2 Effect of crystal field splitting on pairing of d electrons in two Fe(III) complex ions. When ∆ is small, Hund's rule applies to all five orbitals. When ∆ is large, the three lower-energy orbitals fill before the two higher orbitals. _______________________________________________________________________________ EXAMPLE 22.8 Crystal Field Theory [Ti(H2 O)6 ]3+ ion absorbs light at a wavelength of 498 nm. a. Calculate the crystal field splitting energy ∆ . b. Determine the color of the complex. •Method of Solution for (a) a. Absorption of light occurs when the energy of an incoming photon is equal to the difference in energy ∆ between the lower energy 3d orbitals, and the higher energy 3d orbitals of the titanium ion. The requirement for light absorption is ∆ = Ephoton ∆ = hν = hc/λ where h is Planck's constant, c is the velocity of light, and λ is the wavelength of light, 498 nm. •Calculation for (a) ( 6.63 × 1 0 –34 J s )(3.00 × 1 0 8 m /s) 10–9 m 498 nm × 1 nm ∆ = 3.99 × 10 –19 J ∆= •Method of Solution for (b) The complex absorbs light with a wavelength of 498 nm. On the color wheel this is green light. A solution that absorbs green light will reflect all the colors except green. According to the color wheel the solution will be red. _______________________________________________________________________________ EXAMPLE 22.9 Crystal Field Theory [CoF 6 ]4– is a high-spin paramagnetic complex, and [Co(en)3 ]2+ is a low-spin paramagnetic complex. Draw the crystal field splitting diagrams for these two complex ions and show the proper relationship of the two ∆ values. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 41 •Method of Solution The crystal field splitting diagram for a Co2+ ion (3d7 ) must show 7 electrons in the 3d orbitals. In a high-spin complex ∆ is small enough so that electrons can enter 3d orbitals one at a time according to Hund's rule. Pairing of electrons will not occur until after the fifth electron has been added. In the low-spin complex, ∆ is much larger and the electrons enter the lower-energy orbitals first. Only after 6 electrons are positioned in the lowerenergy set of orbitals, can electrons then be placed in the upper two orbitals. ∆ ∆ [CoF 6 ]4– [Co(en)3 ]2+ 3 unpaired electrons 1 unpaired electrons (high-spin) (low-spin) _______________________________________________________________________________ EXERCISES 13. 14. 15. 16. If a complex ion absorbs blue light, what color is the complex ion? Which ligand is the stronger-field ligand, ethylenediamine or bromide ion? Distinguish between a high-spin complex and a low-spin complex. n+ In the complex ion ML6 , M n+ has six d electrons and L is a weak field ligand. How many unpaired electrons are there in this complex? _____________________________________________________________________________ CONCEPTUAL QUESTION 1. The color of oxyhemoglobin is due to complex ions involving Fe(II) ions. Oxyhemoglobin (with O2 bound to iron) is a low-spin, Fe(II) complex. Five of the donor atoms are N atoms and do not change. The sixth donor atom is O in the O2 molecule. Deoxyhemoglobin (without the O2 ) has a water ligand in place of O2 and is a high-spin complex. Both complexes are octahedral Fe2+ complexes. a. How many unpaired electrons occupy the 3d orbitals of the iron(II) ion in each case? b. Oxyhemoglobin is red and deoxyhemoglobin is blue. Explain in terms of the crystal field theory why the two forms of hemoglobin have different colors. PRACTICE TEST 1. 2. 3. 4. Back Write the electron configuration of an Fe atom; an Fe 2+ ion; an Fe3+ ion. Write the formula of the coordination compound containing a Pt4+ central atom, three chlorides, three ammonia ligands, and one uncomplexed nitrate ion. Find the oxidation number of the central atom in Na 2 [Co(H2 O)2 I4 ]. For the coordination compound [Co(NH3 )6 ]Cl3 give the oxidation state and coordination number of the central atom. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 42 / Transition Metal Chemistry and Coordination Compounds 5. For the coordination compound [Ni(en)2 (NH3 )2 ]Cl2 give the oxidation state and coordination number of the central atom. 6. Name the following coordination compounds and complex ions. a. [Pt(NH3 )4 Cl 2 ]Cl2 b. [Ni(NH3 )6 ]2+ c. [Cr(OH)4 ]– 3– d. Co(CN)6 e. K 2 [Cu(CN)4 ] 7. Write the formula for each of the following coordination compounds and ions. a. tetrahydroxoaluminate(III) ion b. tetraiodomercurate(II) ion c. potassium dichlorobis(oxalato)cobaltate(III) d. tris(ethylenediamine)nickel(II) sulfate 8. Sketch two geometrical isomers for [Ni(NH3 )2 (CN)4 ]2– . 9. Sketch the structures of the square planar complex trans-[Pt(Cl)2 (NH3 )2 ], and its mirror image. Are the two structures enantiomers? 10. Why are chiral substances said to be optically active? 11. Illustrate each of the following by giving formulas, geometry, or energy diagrams of: 4– a. octahedral complex of NiCl6 b. cis and trans isomers of Ni(CO) 2 (CN)2 . c. [Fe(H2 O)6 ]3+ is a high-spin complex d. a bidentate ligand e. a coordination number of 2 12. If green light is absorbed by a complex in solution, what is the color of the solution? 13. Sketch the crystal field splitting diagram for the low-spin [Fe(CN)6 ]4– ion. 14. Sketch the crystal field splitting diagram for the tetrahedral complex [FeCl4 ]– . 15. Distinguish between inert and labile complexes. 16. Complexes of zinc are never paramagnetic. Explain. 17. In the complex ion [ML 6 ]n+ , M n+ has four d electrons, and L is a weak-field ligand. How many unpaired electrons are there in this complex? 18. [Ni(H2 O)6 ]2+ is green and [Ni(en)3 ]2+ is violet. Which has the larger value of ∆ ? ANSWERS Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Back a. 3d5 4s2 b. 3d5 c. 3d4 a. V b. Nb Fe –2 4 +2 coordination number = 6, oxidation number = +2 tribromotrichlorocobalt(II) ion. diamminebis(ethylenediamine)nickel(II) ion a. [Cu(NH3 )2 (C2 O4 )] b. K[PtNH3 Cl 3 ] Geometric isomers have the same numbers and kinds of atoms but the arrangement of atoms is different even though the same bonds are present. Optical isomers (enantiomers) are mirror images that cannot be superimposed on each other. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Transition Metal Chemistry and Coordination Compounds / 4 43 12. Chiral objects are mirror images that cannot be superimposed on one another no matter how they are rotated. 13. orange 14. ethylenediamine 15. The terms low-spin and high-spin refer to magnetic properties of complex ions. In a high-spin complex the electrons are arranged so that they are unpaired as much as possible. In a low-spin complex there are more electrons that are paired. High-spin complexes are more paramagnetic than low-spin complexes. 16. 4 Conceptual Question 1. a. Six d-electrons total. In the low-spin O2 containing complex there are no unpaired electrons. In the highspin complex with H2 O there are four unpaired electrons. b. The two forms of hemoglobin have different colors due to the differing amount of crystal field splitting (∆ ) in the iron complexes. O2 must be a stronger-field ligand than H2 O. The splitting is much greater in the low-spin O 2 -containing complex than in the high-spin H2 O-containing complex. Therefore the color of light absorbed will be different for the two complexes, leading to different observed colors. From the complementary colors you can tell that the O 2 complex absorbs green light and the H2 O complex absorbs orange light. Practice Test 1. 2. 3. 4. 5. 6. 7. Fe [Ar]4s2 3d6 Fe [Ar]3d6 Fe [Ar]3d5 [Pt(NH3 )3 Cl 3 ]NO3 +2 The coordination number is 6; the oxidation state is +3. The coordination number is 6; the oxidation state is +2. a. tetraamminedichloroplatinum(IV) chloride b. hexaaminenickel(II) ion c. tetrahydroxochromate(III) ion d. hexacyanocobaltate(III) ion e. potassium tetracyanocuprate(II) a. [Al(OH)4 ]– b. [HgI4 ]2– c. K 3 [Co(C2 O4 )2 Cl 2 ] d. [Ni(en)3 ]SO4 8. NH 3 NH 3 NH 3 NC NC Ni NC CN NC Back CN NH 3 CN cis 9. CN Ni trans The two structures are not enantiomers. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 44 / Transition Metal Chemistry and Coordination Compounds 10. Because when plane-polarized light passes through a solution of a chiral substance, the plane of the polarized light is rotated. 11. a. Cl Cl Cl Ni Cl Cl Cl b. NC OC CO Ni NC CN Ni CO NC cis CO trans c. d. OO || || – O—C—C—O– e. [H3 N—Ag—NH3 ]+ 12. Red 13. 14. 15. The terms inert and labile are kinetic terms, that is, they refer to relative rates of ligand substitution reaction. Complexes that undergo rapid ligand exchange are said to be labile complexes. In inert complexes the bonds between the central atom and the ligands are broken and re-formed relatively infrequently. 16. Zinc has ten d electrons that completely fill the five d orbitals. 17. 4 18. [Ni(en)3 ]2+ _______________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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This note was uploaded on 09/15/2009 for the course CHEM 102 taught by Professor Bastos during the Spring '08 term at Adelphi.

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