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Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 24 Transition Metals and Coordination Compounds Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Copyright 2011 Pearson Education, Inc. Gemstones • The colors of rubies and emeralds are both due to the presence of Cr3+ ions – the difference lies in the crystal hosting the ion Some Al3+ ions in Be3Al2(SiO3)6 are replaced by Cr3+ Some Al3+ ions in Al2O3 are replaced by Cr3+ Tro: Chemistry: A Molecular Approach, 2/e 2 Copyright 2011 Pearson Education, Inc. Properties and Electron Configuration of Transition Metals • The properties of the transition metals are similar to each other and very different to the properties of the main group metals high melting points, high densities, moderate to very hard, and very good electrical conductors • The similarities in properties come from similarities in valence electron configuration – they generally have 2 valence electrons Tro: Chemistry: A Molecular Approach, 2/e 3 Copyright 2011 Pearson Education, Inc. Electron Configuration • For 1st & 2nd transition series = ns2(n−1)dx 1st = [Ar]4s23dx; 2nd = [Kr]5s24dx • For 3rd & 4th transition series = ns2(n−2)f14(n−1)dx • Some individuals deviate from the general pattern by “promoting” one or more s electrons into the underlying d to complete the subshell Group 1B • Form ions by losing the ns electrons first, then the (n – 1)d Tro: Chemistry: A Molecular Approach, 2/e 4 Copyright 2011 Pearson Education, Inc. Example 24.1: Write the ground state electron configuration of Zr identify the noble gas that precedes the [Kr] element and write it in square brackets Count down the periods to determine the outer principal quantum level—this is the quantum level for the s orbital. Subtract one to obtain the quantum level for the d orbital. [Kr]5s4d If the element is in the third or fourth transition series, include (n − 2)f14 electrons in the configuration. count across the row to see how many Zr has 4 more electrons are in the neutral atom and fill the electrons than Kr orbitals accordingly [Kr]5s24d2 Tro: Chemistry: A Molecular Approach, 2/e 5 Copyright 2011 Pearson Education, Inc. Example 24.2: Write the ground state electron configuration of Co3+ identify the noble gas that precedes the element and write it in square brackets Count down the periods to determine the outer principal quantum level—this is the quantum level for the s orbital. Subtract one to obtain the quantum level for the d orbital. [Ar] [Ar]4s3d count across the row to see how many electrons are in the neutral atom and fill the orbitals accordingly Co has 9 more electrons than Ar [Ar]4s23d7 for an ion, remove the required number of electrons, first from the s and then from the d orbitals. lost 3 e− [Ar]3d6 Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright 2011 Pearson Education, Inc. Practice – Use the Periodic Table to write the short electron configuration and short orbital diagram for each of the following [Ar]4s23d8 • Ni (at. no. 28) 5s 4d [Kr]4d5 • Tc2+ (at. no. 43) 5s [Xe]6s24f145d4 • W (at. no. 74) 6s Tro: Chemistry: A Molecular Approach, 2/e 4d 4f 7 5d Copyright 2011 Pearson Education, Inc. Irregular Electron Configurations • We know that because of sublevel splitting, the 4s sublevel is lower in energy than the 3d; and therefore the 4s fills before the 3d • But the difference in energy is not large • Some of the transition metals have irregular electron configurations in which the ns only partially fills before the (n−1)d or doesn’t fill at all • Therefore, their electron configuration must be found experimentally Tro: Chemistry: A Molecular Approach, 2/e 8 Copyright 2011 Pearson Education, Inc. Irregular Electron Configurations • • • • • • • • • • • • Expected Cr = [Ar]4s23d4 Cu = [Ar]4s23d9 Mo = [Kr]5s24d4 Ru = [Kr]5s24d6 Pd = [Kr]5s24d8 Tro: Chemistry: A Molecular Approach, 2/e 9 Found Experimentally Cr = [Ar]4s13d5 Cu = [Ar]4s13d10 Mo = [Kr]5s14d5 Ru = [Kr]5s14d7 Pd = [Kr]5s04d10 Copyright 2011 Pearson Education, Inc. Atomic Size • The atomic radii of all the transition metals are very similar small increase in size down a column • The 3rd transition series atoms are about the same size as the 2nd the lanthanide contraction is the decrease in expected atomic size for the 3rd transition series atoms that come after the lanthanides Tro: Chemistry: A Molecular Approach, 2/e 10 Copyright 2011 Pearson Education, Inc. Why Aren’t the 3rd Transition Series Atoms Bigger? • 14 of the added 32 electrons between the 2nd • • and 3rd series go into 4f orbitals Electrons in f orbitals are not as good at shielding the valence electrons from the pull of the nucleus The result is a greater effective nuclear charge increase and therefore stronger pull on the valence electrons – the lanthanide contraction Tro: Chemistry: A Molecular Approach, 2/e 11 Copyright 2011 Pearson Education, Inc. Ionization Energy • The 1st IE of the transition • metals slowly increases across a series The 1st IE of the 3rd transition series is generally higher than the 1st and 2nd series indicating the valence electrons are held more tightly – why? trend opposite to main group elements Tro: Chemistry: A Molecular Approach, 2/e 12 Copyright 2011 Pearson Education, Inc. Electronegativity • The electronegativity of the transition metals slowly increases across a series except for last element in the series • Electronegativity slightly increases between 1st and 2nd series, but the 3rd transition series atoms are about the same as the 2nd trend opposite to main group elements Tro: Chemistry: A Molecular Approach, 2/e 13 Copyright 2011 Pearson Education, Inc. Oxidation States • Unlike main group metals, transition metals often • • exhibit multiple oxidation states Vary by 1 Highest oxidation state is the same as the group number for Groups 3B to 7B Tro: Chemistry: A Molecular Approach, 2/e 14 Copyright 2011 Pearson Education, Inc. Complex Ions • When a monatomic cation combines with • • multiple monatomic anions or neutral molecules it makes a complex ion The attached anions or neutral molecules are called ligands The charge on the complex ion can then be positive or negative, depending on the numbers and types of ligands attached Tro: Chemistry: A Molecular Approach, 2/e 15 Copyright 2011 Pearson Education, Inc. Coordination Compounds • When a complex ion combines with counter-ions • • to make a neutral compound it is called a coordination compound The primary valence is the oxidation number of the metal The secondary valence is the number of ligands bonded to the metal coordination number • Coordination numbers range from 2 to 12, with the most common being 6 and 4 CoCl36H2O = [Co(H2O)6]Cl3 Tro: Chemistry: A Molecular Approach, 2/e 16 Copyright 2011 Pearson Education, Inc. Coordination Compound Tro: Chemistry: A Molecular Approach, 2/e 17 Copyright 2011 Pearson Education, Inc. Complex Ion Formation • Complex ion formation is a type of Lewis acid– base reaction • A bond that forms when the pair of electrons is donated by one atom is called a coordinate covalent bond Tro: Chemistry: A Molecular Approach, 2/e 18 Copyright 2011 Pearson Education, Inc. Ligands with Extra Teeth • Some ligands can form more than one coordinate covalent bond with the metal atom lone pairs on different atoms that are separated enough so that both can bond to the metal • A chelate is a complex ion containing a multidentate ligand the ligand is called the chelating agent Tro: Chemistry: A Molecular Approach, 2/e 19 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 20 Copyright 2011 Pearson Education, Inc. EDTA a Polydentate Ligand Tro: Chemistry: A Molecular Approach, 2/e 21 Copyright 2011 Pearson Education, Inc. Complex Ions with Polydentate Ligands Tro: Chemistry: A Molecular Approach, 2/e 22 Copyright 2011 Pearson Education, Inc. Geometries in Complex Ions Tro: Chemistry: A Molecular Approach, 2/e 23 Copyright 2011 Pearson Education, Inc. Naming Coordination Compounds 1. Determine the name of the noncomplex ion 2. Determine the ligand names and list them in alphabetical order 3. Determine the name of the metal cation 4. Name the complex ion b: a) naming each ligand alphabetically, adding a prefix in front of each ligand to indicate the number found in the complex ion b) following with the name of the metal cation 5. Write the name of the cation followed by the name of the anion Tro: Chemistry: A Molecular Approach, 2/e 24 Copyright 2011 Pearson Education, Inc. Common Ligands Tro: Chemistry: A Molecular Approach, 2/e 25 Copyright 2011 Pearson Education, Inc. Common Metals found in Anionic Complex Ions Tro: Chemistry: A Molecular Approach, 2/e 26 Copyright 2011 Pearson Education, Inc. Examples of Naming Coordination Compounds Name [Cr(H2O)5Cl]Cl2 Name K3[Fe(CN)6] Identify the cation and anion, and the name of the simple ion [Cr(H2O)5Cl]2+ is a complex cation, Cl− is chloride K+ is potassium, [Fe(CN)6]3− is a complex ion Give each ligand a name and list them in alphabetical order H2O is aqua Cl− is chloro CN− is cyano Cr3+ is chromium(III) Fe3+ is ferrate(III) because the complex ion is anionic [Cr(H2O)5Cl]2+ is pentaquochlorochromium(III) [Fe(CN)6]3− is hexacyanoferrate(III) [Cr(H2O)5Cl]Cl2 is pentaquochlorochromium(III) chloride K3[Fe(CN)6] is potassium hexacyanoferrate(III) Name the metal ion Name the complex ion by adding prefixes to indicate the number of each ligand followed by the name of each ligand followed by the name of the metal ion Name the compound by writing the name of the cation before the anion. The only space is between ion names. Tro: Chemistry: A Molecular Approach, 2/e 27 Copyright 2011 Pearson Education, Inc. Practice – Name each of the following K2CuI4 potassium tetraiodocuprate(II) [Co(NH3)2(CO)4](NO3)3 diamminotetracarbonylcobalt(III) nitrate Tro: Chemistry: A Molecular Approach, 2/e 28 Copyright 2011 Pearson Education, Inc. Isomers • Structural isomers are molecules that have • the same number and type of atoms, but they are attached in a different order Stereoisomers are molecules that have the same number and type of atoms, and that are attached in the same order, but the atoms or groups of atoms point in a different spatial direction Tro: Chemistry: A Molecular Approach, 2/e 29 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 30 Copyright 2011 Pearson Education, Inc. Linkage Isomers • Linkage isomers are structural isomers that have ligands attached to the central cation through different ends of the ligand structure yellow complex = pentamminonitrocobalt(III) red complex = pentamminonitritocobalt(III) Tro: Chemistry: A Molecular Approach, 2/e 31 Copyright 2011 Pearson Education, Inc. Geometric Isomers • In fac–mer isomerism three identical ligands • •In cis–trans isomerism identical ligands inGeometric an octahedral complex either are adjacent isomers aretwo stereoisomers that are either adjacent toorientation eachface other to each making one (fac) oror form an differ inother the spatial of(cis) ligands arc around the other center(trans) (mer) in in the the structure structure opposite each • cis–trans isomerismininoctahedral octahedral complexes MA34B B32 2B2 fac–mer isomerism square-planar complexes complexes MA MA Tro: Chemistry: A Molecular Approach, 2/e 32 Copyright 2011 Pearson Education, Inc. Example 24.5: Draw the structures and label the type for all isomers of [Co(en)2Cl2]+ the ethylenediamine ligand (en = H2NCH2CH2NH2) is bidentate each Cl ligand is monodentate octahedral MA4B2 Tro: Chemistry: A Molecular Approach, 2/e 33 Copyright 2011 Pearson Education, Inc. Practice – Identify each of the following stereoisomers as cis, trans, fac, or mer trans Tro: Chemistry: A Molecular Approach, 2/e mer 34 Copyright 2011 Pearson Education, Inc. [Co(en)3]3+ Optical Isomers • Optical isomers are stereoisomers that are nonsuperimposable mirror images of each other Tro: Chemistry: A Molecular Approach, 2/e 35 Copyright 2011 Pearson Education, Inc. Ex 24.7 – Determine if the cis-trans isomers of [Co(en)2Cl2]+ are optically active • Draw the mirror • image of the given isomer Check to see if the two are superimposable cis isomer trans isomer mirror image identical is to nonsuperimposable its mirror image nooptical opticalisomers isomerism Tro: Chemistry: A Molecular Approach, 2/e 36 Copyright 2011 Pearson Education, Inc. Practice – Determine if the following is optically active Tro: Chemistry: A Molecular Approach, 2/e 37 Copyright 2011 Pearson Education, Inc. Practice – Determine if the following is optically active mirror images are superimposable, therefore not optically active Tro: Chemistry: A Molecular Approach, 2/e 38 Copyright 2011 Pearson Education, Inc. Bonding in Coordination Compounds: Valence Bond Theory • Bonding takes place when the filled atomic • orbital on the ligand overlaps an empty atomic orbital on the metal ion Explains geometries well, but doesn’t explain color or magnetic properties Tro: Chemistry: A Molecular Approach, 2/e 39 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 40 Copyright 2011 Pearson Education, Inc. Bonding in Coordination Compounds: Crystal Field Theory • Bonds form due to the attraction of the electrons • • • on the ligand for the charge on the metal cation Electrons on the ligands repel electrons in the unhybridized d orbitals of the metal ion The result is the energies of the d orbitals are split The difference in energy depends on the complex formed and the kinds of ligands crystal field splitting energy strong field splitting and weak field splitting Tro: Chemistry: A Molecular Approach, 2/e 41 Copyright 2011 Pearson Education, Inc. Crystal Field Splitting The repulsions other ligands d orbitals in an lie between electron octahedral thecomplex axes pairs and in the ligands and have are located nodes in directly the any on potential electrons in the axes, same space which as the results the d orbitals in less lobes ofrepulsion the result and andin an increase in the lower energies for orbitals energies of these these three orbitals orbitals Tro: Chemistry: A Molecular Approach, 2/e 42 Copyright 2011 Pearson Education, Inc. Splitting of d Orbital Energies Due to Ligands in an Octahedral Complex The size of the crystal field splitting energy, D, depends on the kinds of ligands and their relative positions on the complex ion, as well as the kind of metal ion and its oxidation state Tro: Chemistry: A Molecular Approach, 2/e 43 Copyright 2011 Pearson Education, Inc. Color and Complex Ions • Transition metal ions show many intense colors in host crystals or solution • The color of light absorbed by the complexed ion is related to electronic energy changes in the structure of the complex the electron “leaping” from a lower energy state to a higher energy state Tro: Chemistry: A Molecular Approach, 2/e 44 Copyright 2011 Pearson Education, Inc. Complex Ion Color • The observed color is the complementary color of the one that is absorbed Tro: Chemistry: A Molecular Approach, 2/e 45 Copyright 2011 Pearson Education, Inc. Complex Ion Color and Crystal Field Strength • The colors of complex ions are due to • electronic transitions between the split d sublevel orbitals The wavelength of maximum absorbance can be used to determine the size of the energy gap between the split d sublevel orbitals Ephoton = hn = hc/l = D Tro: Chemistry: A Molecular Approach, 2/e 46 Copyright 2011 Pearson Education, Inc. Example 24.8: Estimate the Crystal Field Splitting Energy for a blue solution of [Cu(NH3)6]2+ use the color wheel to determine approximately what wavelength is being absorbed Because the solution is blue it is absorbing orange. The midpoint of orange is 615 nm Calculate the energy of a photon with this wavelength. It is the same as the energy gap between the d orbitals, D. Tro: Chemistry: A Molecular Approach, 2/e 47 Copyright 2011 Pearson Education, Inc. Practice – Estimate the crystal field splitting energy for a red solution of [Fe(CN)6]3− Tro: Chemistry: A Molecular Approach, 2/e 48 Copyright 2011 Pearson Education, Inc. Practice – Estimate the crystal field splitting energy for a red solution of [Fe(CN)6]3− use the color wheel to determine approximately what wavelength is being absorbed Because the solution is red it is absorbing green. The midpoint of green is 525 nm Calculate the energy of a photon with this wavelength. It is the same as the energy gap between the d orbitals, D. Tro: Chemistry: A Molecular Approach, 2/e 49 Copyright 2011 Pearson Education, Inc. Ligands and Crystal Field Strength • The size of the energy gap depends on what kind of ligands are attached strong field ligands include CN─ > NO2─ > en > NH3 weak field ligands include H2O > OH─ > F─ > Cl─ > Br─ > I─ • The size of the energy gap also depends on the type of cation increases as the charge on the metal cation increases Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+ Tro: Chemistry: A Molecular Approach, 2/e 50 Copyright 2011 Pearson Education, Inc. Magnetic Properties and Crystal Field Strength • The electron configuration of the metal ion with • • split d orbitals depends on the strength of the crystal field The 4th and 5th electrons will go into the higher energy if the field is weak and the energy gap is small – leading to unpaired electrons and a paramagnetic complex The 4th thru 6th electrons will pair the electrons in the dxy, dyz, and dxz if the field is strong and the energy gap is large – leading to paired electrons and a diamagnetic complex Tro: Chemistry: A Molecular Approach, 2/e 51 Copyright 2011 Pearson Education, Inc. Low Spin & High Spin Complexes diamagnetic paramagnetic low-spin complex high-spin complex only electron configurations d4, d5, d6, or d7 can have low or high spin Tro: Chemistry: A Molecular Approach, 2/e 52 Copyright 2011 Pearson Education, Inc. Example 24.9: How many unpaired electrons are in [CoF6]3− determine the charge and number of d electrons on the metal cation because F is −1, Co is 3+ Co is atomic number 27, so Co3+ is [Ar]4s03d6 determine if the ligand is strong field or weak field F− is weak field decide if the complex ion is low spin or high spin weak field ligands have small D and yield high spin configurations draw the energy diagram and count the unpaired electrons there are four unpaired electrons Tro: Chemistry: A Molecular Approach, 2/e 53 Copyright 2011 Pearson Education, Inc. Practice – How many unpaired electrons are in [Fe(CN)6]3− Tro: Chemistry: A Molecular Approach, 2/e 54 Copyright 2011 Pearson Education, Inc. Practice – How many unpaired electrons are in [Fe(CN)6]3− because CN is −1, Fe is 3+ Fe is atomic number 26, so Fe3+ is [Ar]4s03d5 determine if the ligand is strong field or weak field CN− is strong field decide if the complex ion is low spin or high spin strong field ligands have large D and yield low spin configurations draw the energy diagram and count the unpaired electrons shere is one unpaired electron Energy determine the charge and number of d electrons on the metal cation Tro: Chemistry: A Molecular Approach, 2/e 55 Copyright 2011 Pearson Education, Inc. Tetrahedral Geometry and Crystal Field Splitting • Because the ligands interact more strongly • • with the planar orbitals in the tetrahedral geometry, their energies are raised This reverses the order of energies compared to the octahedral geometry Almost all tetrahedral complexes are high spin because of reduced metal orbital – ligand interaction Tro: Chemistry: A Molecular Approach, 2/e 56 Copyright 2011 Pearson Education, Inc. Crystal Field Splitting in the Tetrahedral Geometry Tro: Chemistry: A Molecular Approach, 2/e 57 Copyright 2011 Pearson Education, Inc. Square Planar Geometry and Crystal Field Splitting • d8 metals Pt2+, Pd2+, Ir+, Au3+ • The most complex splitting pattern • Almost all are low-spin complexes Tro: Chemistry: A Molecular Approach, 2/e 58 Copyright 2011 Pearson Education, Inc. Applications of Coordination Compounds • Extraction of metals from ores silver and gold as cyanide complexes nickel as Ni(CO)4(g) • Use of chelating agents in heavy metal poisoning EDTA for Pb poisoning • Chemical analysis qualitative analysis for metal ions blue = CoSCN+ red = FeSCN2+ Ni2+ and Pd2+ form insoluble colored precipitates with dimethylglyoxime Tro: Chemistry: A Molecular Approach, 2/e 59 Copyright 2011 Pearson Education, Inc. Applications of Coordination Compounds • Commercial coloring agents Prussian blue = mixture of hexacyanoFe(II) and Fe(III) inks, blueprinting, cosmetics, paints • Biomolecules porphyrin ring cytochrome C hemoglobin chlorophyll Tro: Chemistry: A Molecular Approach, 2/e chlorophyll 60 Copyright 2011 Pearson Education, Inc. Applications of Coordination Compounds • Carbonic anhydrase catalyzes the reaction between water and CO2 contains tetrahedrally complexed Zn2+ Tro: Chemistry: A Molecular Approach, 2/e 61 Copyright 2011 Pearson Education, Inc. Applications of Coordination Compounds • Drugs and therapeutic agents cisplatin anticancer drug Tro: Chemistry: A Molecular Approach, 2/e 62 Copyright 2011 Pearson Education, Inc.