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The chemistry of the transition metals Chapter 22 1 Electron configurations • Let’s write the 3d metals’ and their ions’ econfigurations • Sc, V, Cr, Fe, Ni, Cu, Zn 2 Atomic size • Atomic size decreases across • • period & increases down a column Not a big change to the radii across the row – Due to e-’s being stable in outermost orbital (4s) while adding to 3d But 3rd row roughly same as 2nd, not larger – Due to e-’s going into 4f – The f-subshell is ineffective at shielding outer e-’s from nuclear charge • Outer e-’s held more tightly by nucleus – Called lanthanide contraction 3 Ionization energy • Increase across row • But increase smaller than • for main-group elements Also, 3rd transition row has higher ionization E (generally) than first 2 rows – Runs counter to maingroup elements • Due to outer e-’s being held more tightly 4 Electronegativity • Increase across row, same as main-group • However, increase going down a group – Conversely, main-group decrease going down a group • 1st & 2nd rows different, but 2nd & 3rd same – Due to small change in atomic size going down group w/large increase in nuclear charge • Au = 2.4 ! • Au- found to exist! 5 Oxidation states • Variable – Up to +8 in Os & Ru • Re has widest range: -3 +7 ! 6 Coordination compounds • Can form complex ions – Ion containing central metal ion bound to one or more ligands • Lewis base (or e- donor) that forms bond w/metal • When complex ion combines w/counterions (non-ligands) they yield a neutral compound – Coordination compound 7 Its history • Alfred Werner • Central metal ion has 2 types of interactions: • 1. primary valence: oxidation state on central • • metal atom 2. secondary valence: # of molecules/ions bound to metal atom, called coordination #. Ex: CoCl3 6NH3 has primary valence of 3 and coordination # of 6 8 More • Can think of metalligand complex as Lewis acid/base adduct – Since ligand donates epair to empty orbital on metal • Called coordinate covalent bond 9 More on ligands • Those that donate one e- pair = monodentate • • (why is it CO and not OC?) Those that donate 2 = bidentate More than 2 = polydentate • Latter two called chelates • Coordinating ligand = chelating agent 10 Naming coordination compounds 1. Name ligands – Neutral ligands named as molecules except for H2O (aqua), NH3 (ammine), and CO (carbonyl) – Anionic ligands have suffix changes: • -ide -o fluoride fluoro • -ate -ato sulfate sulfato • -ite -ito nitrite nitrito 11 Naming coordination compounds 2. List names of ligands in alphabetical order before name of metal cation – For example: ammine comes before bromo 3. Use prefix to indicate # of ligands – – – – – 2 3 4 5 6 di tri tetra penta hexa diammine triiodo tetranitrito 12 Naming coordination compounds – If the ligand name already has a number prefix, put () around ligand – 2 bis bis(ethylenediamine) – 3 tris tris(edta) – 4 tetrakis – Prefixes don’t affect order in which ligands listed 13 Naming coordination compounds 4. Name metal – When complex ion is cation, use name of metal followed by oxidiation state w/Roman numeral • Pt2+ = platinum (II) – If complex ion anionic, drop ending of metal • Add –ate followed by oxid. state w/Roman numeral – Pt2+ = platinate(II) 14 Naming coordination compounds 5. Write entire name by listing ligands first then metal – Chemical formula: • Symbol of metal first, then neutral molecules and anions last (all one word!) – – – Hexaamminecobalt(III) ion= [Co(NH3)6]3+ Diamminetetrachloroplatinate(II) ion = [Pt(NH3)2Cl4]2- If more than one anion or neutral molecule in ligand, list in alphabetical order based on chemical symbol • • [Mn(CO)(NH3)5]SO4 is correct [Mn(NH3)5(CO)]SO4 is incorrect – SO4 is called a “counter-ion” There should be a space between the complex ion and the counter-ion 15 Practice naming • [Cr(H2O)5Cl]Cl2 • K3[Fe(CN)6] • Na2[PtCl4] • [Mn(CO)(NH3)5]SO4 16 Practice writing chemical formulae • Tetraaquaplatinum(II) hexachloroplatinate(IV) • Ammonium diaquatetrabromovanadate(III) • Tris(ethylenediamine)cobalt(III) trioxalatoferrate(III) 17 Structure and isomerization • 3 categories 1. Structural isomers: atoms connected in different ways 1. Coordination isomers 2. Linkage isomers 2. Geometric isomers: ligands have different spatial arrangement 1. Cis-trans isomers 2. Octahedral complex isomers 3. Optical isomers: nonsuperimposable mirror-images (enantiomers) 18 Structural isomerism: coordination isomers • Coordination isomers – Coordination ligand exchanges places w/uncoordinated counter-ion • Ex: [Co(NH3)5Br]Cl vs. [Co(NH3)5Cl]Br 19 Structural isomerism: linkage isomers • Either one of atoms in NO2- can bond to metal – When O, nitrito: ONO– When N, nitro: NO2- • Different color compounds (page 1089) 20 Geometric isomerism: cis-trans isomers • Occurs in sq-planar: MA2B2 • And octahedral complexes: MA4B2 21 Geometric isomerism: octahedral complex isomers • MX3Y3 • Fac (facial) isomer – Three identical ligands at corners of a triangular face of octahedron • Mer (meridian) isomer – Three identical ligands at corners of a triangular meridian (inside octahedron) 22 Optical isomerism • Nonsuperimposable mirror-images – Enantiomers • Have chirality/chiral center • Rotation of plane-polarized light in opposite directions 23 Geometries • Depend in part on coordination # • Coordination # Shape Example •2 linear [Ag(NH3)2]+ • 4 d8 sq planar [PdCl4]2• 4 d10 tetrahedral [Zn(NH3)4]2+ •6 octahedral [Fe(H2O)6]3+ 24 Bonding in coordination compounds • Valence Bond Theory: hybridization of orbitals (CHEM&141) • Filled orbital on ligand • Empty orbital on metal cation – Lead to hybridized orbitals 25 Hybridizations of complex ions Geometry • Linear • Tetrahedral • Square planar • Octahedral Hybridization sp sp3 dsp2 d2sp3 26 Crystal Field Theory • VB theory can’t explain color and magnetism of • coordination compounds Basically, complex ions form due to attractions between e- on ligands and positive charge on metal ion – Yet, e- on ligands repel e- in unhybridized metal dorbitals • Crystal Field Theory (CFT) focuses on these repulsions 27 Ligands attack on x, y, z axes • Notice how ligands interact most strongly with top two orbitals’ lobes – Greater interactions • Higher repulsions • In below three, orbitals lie between axes & have nodes directly on axes – Less interactions • Lower repulsions 28 More… • D-orbitals end up being split into low- and highE orbitals – Based on spatial arrangement of ligands • High-E orbitals are top two (see previous figure) • Low-E orbitals are bottomw three (see previous figure) • Diff in E between split d-orbitals – Crystal field splitting E – =∆ • Magnitude of ∆ depends on particular complex (mostly its ligands) 29 Octahedral geometry 30 Tetrahedral • Based on how ligands interact with orbital lobes 31 Square planar • Again, based on how ligands overlap with orbital lobes 32 Comparing all 3 ligand geometries 33 Calculate ∆ (crystal field splitting E) • Using Ephoton = h = hc/ = ∆ • [Ti(H2O)6]3+ has max abs @ 498 nm • What is the CFSE in kJ/mol? • Calculate it! 34 Solution E photon = h = hc = 34 8m (6.626 10 J s)(3.00 10 ) hc s = 9 498 10 m -19 -19 -22 3.99 10 J = 3.99 10 J/ion = 3.99 10 kJ/ion -22 kJ 23 ions 3.99 10 6.022 10 240 kJ ion mol mol 35 Magnitude of CFSE • Depends largely on ligands involved – Spectroscopic studies of various ligands attached to same metal • Yield Spectrochemical series – Arranged from ligands that result in largest ∆ to smallest 36 Spectrochemical series • CN- > NO2- > en > NH3 > (typically strong-field ligands) • H2O > OH- > F- > Cl-> Br-> I- (typically weak-field ligands) • Large ∆ = strong-field ligands • Small ∆ = weak-field ligands 37 More • Also, as oxidation state increases, so does ∆ – Draws ligands closer to metal, increases nodal repulsions 38 Magnetic properties • Magnitude of ∆ • determines d-orbital occupancy and # of unpaired e-’s Large ∆ = strong-field ligands: – Low-spin complexes • Small ∆ = weak-field ligands: – High-spin complexes 39 Magnetic properties • Low-spin complexes tend to be diamagnetic • High-spin complexes tend to be paramagnetic • d1, d2, d3, d8, d9, and d10 do not rely on ligand splitting strength • Why? • Practice: • How many unpaired e-’s would one expect for [FeCl6]3-? • How many unpaired e-’s would one expect for [Co(CN)6]4-? 40 Color of complex ions • The color wheel: absorption of color appears as complementary color 41 Color of complex ions • Color in causes lower d-orbital e- to go up to higher d-orbital state – Specific wavelength of light kicked out • The complement of color absorbed • Colorless complexes are either d0 or d10 – Don’t have d-orbital e-’s to move up 42 Problem • Two ligands form complexes with same metal ion. The first ligand, A, complexes a red soln, while the second ligand, B, complexes a yellow soln. • Which ligand produces the larger ∆? 43