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CHEM 1B: GENERAL CHEMISTRY Ch. 23 Transition Elements and Their Coordination Compounds Dr. Orlando E. Raola Santa Rosa Junior College.

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Presentation on theme: "CHEM 1B: GENERAL CHEMISTRY Ch. 23 Transition Elements and Their Coordination Compounds Dr. Orlando E. Raola Santa Rosa Junior College."— Presentation transcript:

1 CHEM 1B: GENERAL CHEMISTRY Ch. 23 Transition Elements and Their Coordination Compounds Dr. Orlando E. Raola Santa Rosa Junior College

2 Chapter 23 Transition Elements and Their Coordination Compounds

3 The Transition Elements and Their Coordination Compounds 23.1 Properties of the Transition Elements 23.2 The Inner Transition Elements 23.3 Coordination Compounds 23.4 Theoretical Basis for the Bonding and Properties of Complexes

4 Figure 23.1The transition elements (d block) and inner transition elements (f block) in the periodic table.

5 Properties of the Transition Metals All transition metals are metals, whereas main-group elements in each period change from metal to nonmetal. Many transition metal compounds are colored and paramagnetic, whereas most main-group ionic compounds are colorless and diamagnetic. The properties of transition metal compounds are related to the electron configuration of the metal ion.

6 Figure 23.2The Period 4 transition metals.

7 Electron Configurations of Transition Metals and their Ions The d-block elements have the general condensed ground-state configuration [noble gas]ns 2 (n – 1)d x where n = 4 to 7 and x = 1 to 10. Periods 6 and 7 elements include the f sublevel: [noble gas]ns 2 (n – 2)f 14 (n – 1)d x where n = 6 or 7. Transition metals form ions through the loss of the ns electrons before the (n – 1)d electrons.

8 Table 23.1 Orbital Occupancy of the Period 4 Transition Metals The number of unpaired electrons increases in the first half of the series and decreases in the second half, when pairing begins.

9 Sample Problem 23.1 Writing Electron Configurations of Transition Metal Atoms and Ions PROBLEM:Write condensed electron configurations for the following: (a) Zr; (b) V 3+ ; (c) Mo 3+. (Assume that elements in higher periods behave like those in Period 4.) PLAN:We locate the element in the periodic table and count its position in the respective transition series. These elements are in Periods 4 and 5, so the general electron configuration is [noble gas]ns 2 (n – 1)d x. For the ions, we call that ns electrons are lost first. SOLUTION: (a)Zr is the second element in the 4d series: [Kr]5s 2 4d 2

10 Sample Problem 23.1 (b)V is the third element in the 3d series, so its configuration is [Ar]4s 2 3d 3. When it forms V 3+, it loses the two 4s e - first, then one of the 3d e - : [Ar]3d 2 (c)Mo lies below Cr in group 6B(6), so we expect the same exception as for Cr. The configuration for Mo is therefore [Kr]5s 1 4d 5. Formation of the Mo 3+ ion occurs by loss of the single 5s electron followed by two 4d electrons: [Kr]4d 3

11 Trends in the Properties of Transition Metals Across a period the following trends are observed: Atomic size decreases at first, then remains relatively constant. - The d electrons fill inner orbitals, so they shield outer electrons very efficiently and the 4s electrons are not pulled closer by the increasing nuclear charge. Electronegativity and ionization energies also increase relatively little across the transition metals of a particular period.

12 Figure 23.3Trends in key atomic properties of Period 4 elements.

13 Trends in the Properties of Transition Metals Within a group the trends also differ from those observed for main group elements. Atomic size increases from Period 4 to 5, but not from Period 5 to 6. - The extra shrinkage from the increase in nuclear charge (called the lanthanide contraction) is roughly equal to the normal size increase due to adding an extra energy level. - A Period 6 element has 32 more protons than its preceding Period 5 group member instead of only 18.

14 Trends in the Properties of Transition Metals Electronegativity increases within a group from Period 4 to 5, then generally remains unchanged from Period 5 to 6. The heavier elements often have high EN values. Although atomic size increases slightly down the group, nuclear charge increases much more, leading to higher EN values. Ionization energy values generally increase down a transition group, also running counter to the main group trend. Density increases dramatically down a group since atomic volumes change little while atomic masses increase significiantly.

15 Figure 23.4Vertical trends in key properties within the transition elements.

16 Oxidation States of Transition Metals Most transition metals have multiple oxidation states. Elements in Groups 8B(8), 8B(9) and 8B(10) exhibit fewer oxidation states. The higher oxidation state is less common and never equal to the group number. - The +2 oxidation state is common because the ns 2 electrons are readily lost. The highest oxidation state for elements in Groups 3B(3) through 7B(7) equals the group number. - These states are seen when the elements combine with the highly electronegative oxygen or fluorine.

17 Figure 23.5Aqueous oxoanions of transition elements. Mn 2+ MnO 4 2− MnO 4 − The highest oxidation state for Mn equals its group number. VO 4 3− Cr 2 O 7 2− MnO 4 − Transition metal ions are often highly colored.

18 Table 23.2 Oxidation States and d-Orbital Occupancy of the Period 4 Transition Metals *

19 Metallic Behavior of Transition Metals The lower the oxidation state of the transition metal, the more metallic its behavior. Ionic bonding is more prevalent for the lower oxidation states, whereas covalent bonding occurs more frequently for higher oxidation states. Metal oxides become less basic (more acidic) as the oxidation state increases. A metal atom in a positive oxidation state has a greater attraction for bonded electrons, and therefore a greater effective electronegativity, or valence-state electronegativity, than in the zero oxidation state. This effect increases as its oxidation state increases.

20 Table 23.3 Standard Electrode Potentials of Period 4 M 2+ Ions Half-ReactionE°(V) Ti 2+ (aq) + 2e − Ti(s) V 2+ (aq) + 2e − V(s) Cr 2+ (aq) + 2e − Cr(s) Co 2+ (aq) + 2e − Co(s) Fe 2+ (aq) + 2e − Fe(s) Mn 2+ (aq) + 2e − Mn(s) Ni 2+ (aq) + 2e − Ni(s) Cu 2+ (aq) + 2e − Cu(s) Zn 2+ (aq) + 2e − Zn(s) In general, reducing strength decreases across the series.

21 Color and Magnetic Behavior Most main-group ionic compounds are colorless and diamagnetic because the metal ion has no unpaired electrons. Many transition metal ionic compounds are highly colored and paramagnetic because the metal ion has one or more unpaired electrons. Transition metal ions with a d 0 or d 10 configuration are also colorless and diamagnetic.

22 Figure 23.6Colors of representative compounds of the Period 4 transition metals. titanium(IV) oxidesodium chromate potassium ferricyanide nickel(II) nitrate hexahydrate zinc sulfate heptahydrate scandium oxidevanadyl sulfate dihydrate manganese(II) chloride tetrahydrate cobalt(II) chloride hexahydrate copper(II) sulfate pentahydrate

23 Table 23.4 Some Properties of Group 6B(6) Elements Element Atomic Radius (pm)IE 1 (kJ/mol) E° (V) for M 3+ (aq)/M(s) Cr Mo W IE 1 increases down the group, so reactivity decreases. This trend is opposite to that seen in main-group elements.

24 Lanthanides and Actinides The lanthanides are also called the rare earth elements. The atomic properties of the lanthanides vary little across the period, and their chemical properties are also very similar. Most lanthanides have the ground-state electron configuation [Xe]6s 2 4f x 5d 0. All actinides are radioactive, and have very similar physical and chemical properties. The +3 oxidation state is common for both lanthanides and actinides.

25 Sm is the eighth element after Xe. Two electrons go into the 6s sublevel. In general, the 4f sublevel fills before the 5d, so the remaining six electrons go into the 4f sublevel. Sample Problem 23.2 SOLUTION: Finding the Number of Unpaired Electrons PROBLEM:The alloy SmCo 5 forms a permanent magnet because both samarium and cobalt have unpaired electrons. How many unpaired electrons are in Sm (Z = 62)? PLAN:We write the condensed electron configuration of Sm and then, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram and count the unpaired electrons. The condensed configuration of Sm is [Xe]6s 2 4f 6.

26 Sample Problem d5d6p6p ↑ 4f4f ↑↑↑↑↑ 6s6s ↑↓ Sm has six unpaired electrons. The partial orbital diagram is:

27 Coordination Compounds A coordination compound contains at least one complex ion, which consists of a central metal cation bonded to molecules and/or anions called ligands. The complex ion is associated with counter ions of opposite charge. The complex ion [Cr(NH 3 ) 6 ] 3+ has a central Cr 3+ ion bonded to six NH 3 ligands. The complex ion behaves like a polyatomic ion in solution.

28 Coordination Number The coordination number is the number of ligand atoms bonded directly to the central metal ion. Coordination number is specific for a given metal ion in a particular oxidation state and compound. The most common coordination number in complex ions is 6, but 2 and 4 are often seen. - [Cr(NH 3 ) 6 ] 3+ has a coordination number of 6.

29 Figure 23.7Components of a coordination compound. [Co(NH 3 ) 6 ]Cl 3 dissolves in water. The six ligands remain bound to the complex ion. [Pt(NH 3 ) 4 ]Br 2 has four NH 3 ligands and two Br - counter ions.

30 Table 23.5 Coordination Numbers and Shapes of Some Complex Ions Coordination NumberShapeExamples 2Linear[CuCl 2 ] -, [Ag(NH 3 ) 2 ] +, [AuCl 2 ] - 4Square planar [Ni(CN) 4 ] 2-, [PdCl 4 ] 2-, [Pt(NH 3 ) 4 ] 2+, [Cu(NH 3 ) 4 ] 2+ 4Tetrahedral [Cu(CN) 4 ] 3-, [Zn(NH 3 ) 4 ] 2+, [CdCl 4 ] 2-. [MnCl 4 ] 2- 6Octahedral[Ti(H 2 O) 6 ] 3+, [V(CN) 6 ] 4-, [Cr(NH 3 ) 4 Cl 2 ] +, [Mn((H 2 O 6 ] 2+, [FeCl 6 ] 3-, [Co(en) 3 ] 3+ The geometry of a given complex ion depends both on the coordination number and the metal ion. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

31 Ligands The ligands of a complex ion are molecules or anions with one or more donor atoms. Each donor atom donates a lone pair of electrons to the metal ion to form a covalent bond. Ligands are classified in terms of their number of donor atoms, or “teeth”: - Monodentate ligands bond through a single donor atom. - Bidentate ligands have two donor atoms, each of which bonds to the metal ion. - Polydentate ligands have more than two donor atoms.

32 Table 23.6 Some Common Ligands in Coordination Compounds

33 Chelates Bidentate and polydentate ligands give rise to rings in the complex ion. A complex ion containing this type of structure is called a chelate because the ligand seems to grab the metal ion like claws. EDTA has six donor atoms and forms very stable complexes with metal ions.

34 Formulas of Coordination Compounds A coordination compound may consist of –a complex cation with simple anionic counterions, –a complex anion with simple cationic counterions, or –a complex cation with complex anion as counterion. When writing the formula for a coordination compound –the cation is written before the anion, –the charge of the cation(s) is/are balanced by the charge of the anion(s), and –neutral ligands are written before anionic ligands, and the formula of the whole complex ion is placed in square brackets.

35 Determining the Charge of the Metal Ion The charge of the cation(s) is/are balanced by the charge of the anion(s). K 2 [Co(NH 3 ) 2 Cl 4 ] contains a complex anion. The charge of the anion is balanced by the two K + counter ions, so the anion must be [Co(NH 3 ) 2 Cl 4 ] 2-. There are two neutral NH 3 ligands and four Cl - ligands. To have an overall charge of 2-, the metal ion must have a charge of 2+. Charge of complex ion = charge of metal ion + total charge of ligands 2- = charge of metal ion + [(2 x 0) + (4 x -1)] Charge of metal ion = (-2) – (-4) = +2 or 2+ The metal ion in this complex anion is Co 2+.

36 [Co(NH 3 ) 4 Cl 2 ]Cl contains a complex cation. The charge of the cation is balanced by the Cl - counter ion, so the cation must be [Co(NH 3 ) 4 Cl 2 ] +. There are four neutral NH 3 ligands and two Cl - ligands. To have an overall charge of 1+, the metal ion must have a charge of 3+. Charge of complex ion = charge of metal ion + total charge of ligands 1+ = charge of metal ion + [(4 x 0) + (2 x 1-)] Charge of metal ion = (+1) – (2-) = +3 or 3+ The metal ion in this complex cation is Co 3+.

37 Naming Coordination Compounds The cation is named before the anion. Within the complex ion, the ligands are named in alphabetical order before the metal ion. –Anionic ligands drop the –ide and add –o after the root name. A numerical prefix is used to indicate the number of ligands of a particular type. –Prefixes do not affect the alphabetical order of ligand names. –Ligands that include a numerical prefix in the name use the prefixes bis (2), tris (3), or tetrakis (4) to indicate their number. A Roman numeral is used to indicate the oxidation state for a metal that can have more than one state. If the complex ion is an anion, we drop the ending of the metal name and add –ate.

38 Table 23.7 Names of Some Neutral and Anionic Ligands Neutral Anionic NameFormulaNameFormula AquaH2OH2OFluoroF-F- AmmineNH 3 ChloroCl - CarbonylCOBromoBr - NitrosylNOIodo I-I- HydroxoOH - CyanoCN -

39 Table 23.8 Names of Some Metal Ions in Complex Anions MetalName in Anion IronFerrate CopperCuprate LeadPlumbate SilverArgentate GoldAurate TinStannate

40 Sample Problem 23.3 PLAN:We use the rules for writing formulas and names of coordination compounds. Writing Names and Formulas of Coordination Compounds PROBLEM: (a) What is the systematic name of Na 3 [AlF 6 ]? (b) What is the sytematic name of [Co(en) 2 Cl 2 ]NO 3 ? (c) What is the formula of tetraamminebromochloroplatinum(IV) chloride? (d) What is the formula of hexaamminecobalt(III) tetrachloroferrate(III)? SOLUTION: (a) The complex ion is [AlF 6 ] 3-. There are six (hexa-) F - ions (fluoro) as ligands. The complex ion is an anion, so the ending of the metal name must be changed to –ate. Since Al has only one oxidation state, no Roman numerals are used. sodium hexafluoroaluminate

41 (b) There are two ligands, Cl - (chloro) and en (ethylenediamine). The ethylenediamine ligand already has a numerical prefix in its name, so we indicate the two en ligands by the prefix bis instead of di. Sample Problem 23.3 The complex ion is a cation, so the metal name is unchanged, but we need to specify the oxidation state of Co. The counter ion is NO 3 -, so the complex ion is [Co(en) 2 Cl 2 ] +. Charge of complex ion = charge of metal ion + total charge of ligands 1+ = charge of metal ion + [(2 x 0) + (2 x 1-)] Charge of metal ion = (+1) – (-2) = +3 or 3+ The ligands must be named in alphabetical order: dichlorobis(ethylenediamine)cobalt(III) nitrate

42 Sample Problem 23.3 (c) The central metal ion is written first, followed by the neutral ligands and then (in alphabetical order) by the negative ligands. Charge of complex ion = charge of metal ion + total charge of ligands = (4+) + [(4 x 0) + (1 x 1-) + (1 X 1-)] = +4 + (-2) = +2 or 2+ We will therefore need two Cl - counter ions to balance the charge on the complex ion. [Pt(NH 3 ) 4 BrCl]Cl 2

43 Sample Problem 23.3 (d) This compound consists of two different complex ions. In the cation, there are six NH 3 ligands and the metal ion is Co 3+, so the cation is [Co(NH 3 ) 6 ] 3+. The anion has four Cl - ligands and the central metal ion is Fe 3+, so the ion is [FeCl 4 ] -. The charge on the cation must be balanced by the charge on the anion, so we need three anions for every one cation: [Co(NH 3 ) 6 ][FeCl 4 ] 3

44 Constitutional Isomers of Coordination Compounds Compounds with the same formula, but with the atoms connected differently, are constitutional isomers. Coordination isomers occur when the composition of the complex ion, but not the compound, is different. - This can occur by the exchange of a ligand and a counter ion, or be by the exchange of ligands. Linkage isomers occur when the composition of the complex ion is the same but the ligand donor atom is different. - Some ligands can bind to the metal through either of two donor atoms.

45 Figure 23.8A pair of linkage (constitutional) isomers The nitrite ion can bind either through the N atom or either one of the O atoms.

46 Ligands that have more than one donor atom

47 Stereoisomers of Coordination Compounds Stereoisomers are compounds that have the same atomic connections but different spatial arrangements of their atoms. Geometric or cis-trans isomers occur when atoms or groups can either be arranged on the same side or on opposite sides of the compound relative to the central metal ion. Optical isomers (enantiomers) are non-superimposable mirror images of each other.

48 Figure 23.9AGeometric (cis-trans) isomerism. In the cis isomer, identical ligands are adjacent to each other, while in the trans isomer they are across from each other. The cis and trans isomers of [Pt(NH 3 ) 2 Cl 2 ]. The cis isomer (cisplatin) is an antitumor agent while the trans isomer has no antitumor effect.

49 Figure 23.9BGeometric (cis-trans) isomerism. The cis and trans isomers of [Co(NH 3 ) 4 Cl 2 ] +. Note the placement of the Cl - ligands (green spheres).

50 Figure 23.10AOptical isomerism in an octahedral complex ion. Structure I and its mirror image, structure II, are optical isomers of cis-[Co(en) 2 Cl 2 ] +.

51 Figure 23.10BOptical isomerism in an octahedral complex ion. The trans isomer of [Co(en) 2 Cl 2 ] + does not have optical isomers. Structure I can be superimposed on its mirror image, structure II.

52 Figure 23.11Important types of isomerism in coordination compounds.

53 Sample Problem 23.4 Determining the Type of Stereoisomerism PROBLEM:Draw stereoisomers for each of the following and state the type of isomerism: (a) [Pt(NH 3 ) 2 Br 2 ] (square planar)(b) [Cr(en) 3 ] 3+ (en = H 2 NCH 2 CH 2 NH 2 ) PLAN:We determine the geometry around each metal ion and the nature of the ligands. If there are different ligands that can be placed in different positions relative to each other, geometric (cis-trans) isomerism occurs. Then we see whether the mirror image of an isomer is superimposable on the original. If it is not, optical isomerism also occurs.

54 Sample Problem 23.4 SOLUTION: (a) The square planar Pt(II) complex has two different types of monodentate ligands. Each pair of ligands can be next to each other or across from each other. Thus geometric isomerism occurs. These are geometric isomers; they do not have optical isomers since each compound is superimposable on its mirror image.

55 Sample Problem 23.4 (b)Ethylenediamine (en) is a bidentate ligand. The Cr 3+ ion has a coordination number of 6 and an octahedral geometry, like Co 3+. The three bidentate ligands are identical, so there is no geometric isomerism. However, the complex ion has a nonsuperimposable mirror image. Thus optical isomerism occurs.

56 Bonding in Complex Ions In terms of valence bond theory, the filled orbital of the ligand overlaps with an empty orbital of the metal ion. When a complex ion is formed, each ligand donates an electron pair to the metal ion. The ligand acts as a Lewis base, while the metal ion acts as a Lewis acid. This type of bond is called a coordinate covalent bond since both shared e - originate from one atom in the pair. The VB model proposes that the geometry of the complex ion depends on the hybridization of the metal ion.

57 Figure 23.12Hybrid orbitals and bonding in the octahedral [Cr(NH 3 ) 6 ] 3+ ion.

58 Figure 23.13Hybrid orbitals and bonding in the square planar [Ni(CN) 4 ] 2- ion.

59 Figure 23.14Hybrid orbitals and bonding in the tetrahedral [Zn(OH) 4 ] 2- ion.

60 Figure 23.15An artist’s wheel. Each color has a complementary color; the one opposite it on the artist’s wheel. The color an object exhibits depends on the wavelengths of light that it absorbs. An object will have a particular color because it reflects light of that color, or it absorbs light of the complementary color.

61 Table 23.9 Relation Between Absorbed and Observed Colors Absorbed Colorλ (nm)Observed Colorλ (nm) Violet400Green-yellow560 Blue450Yellow600 Blue-green490Red620 Yellow-green570Violet410 Yellow580Dark blue430 Orange600Blue450 Red650Green520

62 Crystal Field Theory Crystal field theory explains color and magnetism in terms of the effect of the ligands on the energies of the d-orbitals of the metal ion. The bonding of the ligands to the metal ion cause the energies of the metal ion d-orbitals to split. Although the d-orbitals of the unbonded metal ion are equal in energy, they have different shapes, and therefore different interactions with the ligands. The splitting of the d-orbitals depends on the relative orientation of the ligands.

63 Figure 23.16The five d-orbitals in an octahedral field of ligands. The ligands approach along the x, y and z axes. Two of the orbitals point directly at the ligands, while the other three point between them.

64 Figure 23.17Splitting of d-orbital energies in an octahedral field of ligands. The d orbitals split into two groups. The difference in energy between these groups is called the crystal field splitting energy, symbol Δ.

65 Figure 23.18The effect of ligands and splitting energy on orbital occupancy. Weak field ligands lead to a smaller splitting energy. Strong field ligands lead to a larger splitting energy.

66 Figure 23.19The color of [Ti(H 2 O) 6 ] 3+. The hydrated Ti 3+ ion is purple. Green and yellow light are absorbed while other wavelengths are transmitted. This gives a purple color.

67 Figure 23.19The color of [Ti(H 2 O) 6 ] 3+. When the ion absorbs light, electrons can move from the lower t 2g energy level to the higher e g level. The difference in energy between the levels (Δ) determines the wavelengths of light absorbed. The visible color is given by the combination of the wavelengths transmitted.

68 The Colors of Transition Metal Complexes The color of a coordination compound is determined by the Δ of its complex ion. For a given ligand, the color depends on the oxidation state of the metal ion. For a given metal ion, the color depends on the ligand.

69 Figure 23.20Effects of oxidation state and ligand on color. [V(H 2 O) 6 ] 2+ [V(H 2 O) 6 ] 3+ [Cr(NH 3 ) 6 ] 3+ [Cr(NH 3 ) 5 Cl ] 2+ A change in oxidation state causes a change in color. Substitution of an NH 3 ligand with a Cl - ligand affects the color of the complex ion.

70 Figure 23.21The spectrochemical series. I - < Cl - < F - < OH - < H 2 O < SCN - < NH 3 < en < NO 2 - < CN - < CO WEAKER FIELD STRONGER FIELD LARGER  SMALLER  LONGER SHORTER As Δ increases, shorter wavelengths (higher energies) of light must be absorbed to excite electrons. For reference H 2 O is considered a weak-field ligand.

71 Sample Problem 23.5 SOLUTION: Ranking Crystal Field Splitting Energies (Δ) for Complex Ions of a Metal PROBLEM:Rank the ions [Ti(H 2 O) 6 ] 3+, [Ti(NH 3 ) 6 ] 3+, and [Ti(CN) 6 ] 3- in terms of Δ and of the energy of visible light absorbed. PLAN:The formulas show that Ti has an oxidation state of +3 in all three ions. From Figure 23.21, we tank the ligands by crystal field strength: the stronger the ligand, the greater the splitting, and the higher the energy of light absorbed. The ligand field strength is CN - > NH 3 > H 2 O, so the relative size of Δ and energy of light absorbed will be [Ti(CN) 6 ] 3- > [Ti(NH 3 ) 6 ] 3+ > [Ti(H 2 O) 6 ] 3+

72 The Magnetic Properties of Transition Metal Complexes Magnetic properties are determined by the number of unpaired electrons in the d orbitals of the metal ion. Hund’s rule states that e - occupy orbitals of equal energy one at a time. When all lower energy orbitals are half- filled: The number of unpaired e - will depend on the relative sizes of E pairing and Δ. - The next e - can enter a half-filled orbital and pair up by overcoming a repulsive pairing energy, (E pairing ). - The next e - can enter an empty, higher, energy orbital by overcoming Δ.

73 Figure 23.22High-spin and low-spin octahedral complex ions of Mn 2+.

74 Figure 23.23Orbital occupancy for high-spin and low-spin octahedral complexes of d 4 through d 7 metal ions. high spin: weak-field ligand low spin: strong-field ligand high spin: weak-field ligand low spin: strong-field ligand

75 Sample Problem 23.6 Identifying High-Spin or Low-Spin Complex Ions PROBLEM:Iron (II) forms a complex in hemoglobin. For each of the two octahedral complex ions [Fe(H 2 O) 6 ] 2+ and [Fe(CN) 6 ] 4, draw an energy diagram showing orbital splitting, predict the number of unpaired electrons, and identify the ion as low spin or high spin. PLAN:Fe 2+ electron configuration shows the number of d electrons, and the spectrochemical series shows the relative ligand strengths. We draw energy diagrams and separate the t 2g and e g orbital sets more for the strong-field ligand. Then we add electrons, noting that a weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex, whereas the strong-field ligand will give the minimum number of unpaired electrons and a low-spin complex.

76 Sample Problem 23.6 SOLUTION: t 2g egeg Potential energy ↑↑↓↑ ↑↑ [Fe(H 2 O) 6 ] 2+ high-spin t 2g egeg Potential energy ↑↓ [Fe(CN) 6 ] 4- low-spin

77 Figure 23.24ASplitting of d-orbital energies by a tetrahedral field of ligands. The splitting of d-orbital energies is less in a tetrahedral than an octahedral complex, and the relative d-orbital energies are reversed. Only high-spin tetrahedral complexes are known because Δ is small.

78 Figure 23.24BSplitting of d-orbital energies by a square planar field of ligands. Square planar complexes are low-spin and usually diamagnetic because the four pairs of d electrons fill the four lowest-energy orbitals.

79 Figure B23.1Hemoglobin and the octahedral complex in heme. Chemical Connections Hemoglobin consists of four protein chains, each with a bound heme. In oxyhemoglobin (B), the octahedral complex in heme has an O 2 molecule as the sixth ligand for iron(II). (Illustration by Irving Geis. Rights owned by Howard Hughes Medical Institute. Not to be used without permission.)

80 Table B23.1 Some Transition Metal Trace Elements in Humans

81 Figure B23.2The tetrahedral Zn 2+ complex in carbonic anhydrase. Chemical Connections

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