Chapter-24 Lesson-| I— Metal Complexes A) Complex Ions « An ion containing a central metallic cation to which two or more groups (ligands) are attached by coordinate covalent bonds is called a complex ion. e A ligand can be a neutral molecule or anion bonded (coordinate) covalently to a central metallic cation in a complex ion. Complex Ton Complex ton Cation | Cation [Ligand] Ligand lame Formula) Name Formula| Name [Ag(NH3)2]° eo Ag! | silver (1)/ NH3_ {ammonia [Cu(H20),P* copper ID Cu* rid 10 | water joys xan ayn Me co | on [Ni(CsHsN) 62" Pe a Niz* “ny CsHsN| pyridine [Fe(CN)c} fem Fe at CNF | cyanide [Ag(CN)2]" eset Ag'* “o CN" | cyanide [AIF] cemnaeptit Ar ce F* | fluorine [Zn(OH) 4] ea Zn" | zine(It) | OH" {hydroxide e The number of bonds formed from the central metallic cation to the ligands is called the cation's coordination number. e The overall charge of the complex ion is equal to the sum of the charges of the central metal and the ligands.e There are examples of cation-ligand groups that are neutral. These are simply called complexes. C 1 Complex Cation | Cation | Ligand | Ligand omnes Name __|Formula| Name |Formulas| Names diammine . NH3_ Jammonia| PU(NE)2Ch | dichloro | Pe a «| & platinum (IT) Cl'-_| chlorine triammine NH3_ Jammonia| balt ° Co(NH3)3Cl3 | trichloro Co* “am & & cobalt (III) Cl | chlorine e Compounds that include a complex ion are called coordination compounds. Coordination Compound's Coordination Compound's Formula Name [Ni(NH3)6]Clo hexaamminenickel(II) chloride [Co(H20)«6](NO3)2 hexaaquacobalt(II) nitrate Ks[Au(CN)a] potassium tetracyanoaurate(I) Naa[FeBro] sodium hexabromoferrate(II) f Lig: ds (See table 24.2 in textbook for examples ) . Ordinarily, ligands contain at least one unshared pair of electrons. H +k : :N-H ‘0-H ‘0-H scan: H H e These are monodentate ligands. Ifthe ligand contains two or more unshared pairs of electrons on different, nonadjacent atoms, it can act as a chelating agent, forming more than one bond with me central metal atom. 26, 36: * i I HH omen H-N-C-C-N-H 70 “or one oxalate ion (ox”) ethylenediamine (en) ¢ These are bidentate ligands (bidentate chelates)e EDTA* (ethylenediaminetetraacetate ion) is a polydentate ligand. One EDTA* ion can form six bonds with the metal atom. See page 1032 in textbook for structural formula. e EDTA* can be used to treat heavy metal poisoning such as lead. The ion can "scavenge" for Pb** ions and form the stable complex [Pb(EDTA)]* which is harmless. Compl Complex | Cation | Cation | Ligand | Ligand complex Name ___|Formula| Name _|Formulas| Names tris(ethylene- [Co(en)3]** | diamine) | Co** “ti en otnylene cobalt (II) ._ | bis(oxalato) platinum 2 2. 2+ 2. [Pt(ox)2] platinate (II) Pt (I) Ox! oxalate ethylene ethylene| x ».| diamine on . 4.-diaminge PO(EDTA)}” | etraacetato | PO’ | lead (II) |EDTAS ET plumbate (IT) acetate QJ: Determine the coordination number for each of the following complexes: (a) [Cu(en)(NH3)2]*", (b) [Ni(CN)4]*, (c) [Pt(ox)2]*, (d) [Mn(H20)6}”", (e) [CuCl], (f) [Co(en)s]*", (g) [Po(EDTA)]* Al:(a)4 (b()4. ()4 «(6 (e)2,— (6H (g)6 Q2: Determine the overall charge for each of the following complexes. (a) [Au(CN),4]’, where Au is +1; (b) [Pt(en)Cl2]’, where Pt is +2; (c) [Cu(en)2(NH3)2]’, where Cu is +2; (d) [AgCh]’, where Ag is +1 :(a)-3 (b)0 (ce) #2) B II - Geometry of Metal Complexes A) Linear Structure (180° Bond Angles) e Complex ions in which the central metal atom has a coordination number equal to two (2) form a linear structure with the ligands. [cl-Cu-cly" [HsN-Ag-NH3]!* [N=C-Au-C=N]"B) Square Planar Structure (90° Bond Angles) © Complex ions in which the central metal atom has a coordination number equal to four (4) may form a square planar structure with the ligands. e Square planar complex ions may exhibit geometric isomerism. HO H.0]* far _a]* [Bra] Cu Ni Ni 0” ~H:0 Bro CL cl7 Br [Cu(H20),]** cis-[Ni(Br)2(C])2]*"_ trans-[Ni(Br)2(Cl)2]* C) Tetrahedral Structure (109.5° Bond Angles) ¢ Complex ions in which the central metal atom has a coordination number equal to four (4) may form a tetrahedral structure with the ligands. MMs 2+ $ 2- [Zn(NH3)s)** | Zn, [Co(Br)x(C)2P*] Co HN | /NH3 cr | Br NH3 Br D) Octahedral Structure (90° & 180° Bond Angles) © Complex ions in which the central metal atom has a coordination number equal to six (6) forms an octahedral structure with the ligands. 1 + NH3 I+ iin C cl c NES fo RT I ink Hy Hi NH3 NHs3 cis [Co(NH3)4(Cl)2]'* trans1 ‘1 c cl c NH ‘Co Co . i, - an, Hs Hs NH cl [Co(NHs)s(C))sI 3+ [Co(en)3}* NH; K+ NH; [+ OLN ° Cl cl [Co(NHs)2(CI):(en)]'"*Q3: Draw all optical isomers for the complex [Pd(en)CI(SCN)]. en NCS A3: Pd mirror Q4: Draw all structural isomers for the complex [Fe(H20)2CIsBr]. Ad: <— This structure has an optical isomer as wellChapter-24 Lesson-2 Il — Electronic Structure of Metals in Complexes e According to Valence Bond Theory, electron pairs contributed by the ligands enter hybrid orbitals of the central metal cation. e A central metal cation with a coordination number equal to... - 2 forms sp orbitals and has a linear geometry. - 4 forms dsp’ orbitals and has a square planar geometry if the metal ion has a d® electron configuration. - 4 forms sp? orbitals and has a tetrahedral geometry if the metal ion has a d'” electron configuration. - 6 forms d?sp* orbitals and has an octahedral geometry. B) Orbital Diagrams (the Valence Bond Theory) ¢ To determine the orbital diagram for the metal ion in a complex... 1) first determine the coordination number of the metal ion. 2) then determine the outer d*s*p* electron configuration of the metal ion before hybridization and draw its orbital diagram. 3) and finally determine the hybridization of the metal ion and draw the hybridized orbital diagram. ¢ (CuONES)2]"" GDENEDEDH aK Oo) sp Ss iP metal electron [Zn(NH3)4}°" AAD) I) nina) sp’ s P » INN NOHONCDRD_ MD) MO) dsp? 3d 4s 4p ligand electron [Co(NEs)6P* GOCHANIMAD ON) (ON) dsp? 3d 48 4pQl: Al: Q2: Q3: A3: (i) Draw the outer orbital diagram of the central metal ion in [Ag(CN)2]" and (ii) state the geometry of the complex. Coordination number =_2_ and Ag'*-__4d!°5s°5p° (i) Ag" GYADADUNG) ©) CIC IC) 4d 5s 5p GOaDEDADED @bed CY) Pp Ssp (ii) [Ag(CN)2]"" has a linear geometry. (i) Draw the outer orbital diagram of the central metal ion in [PtCl* and (ii) state the geometry of the complex. : Coordination number =__4 _ and Pt®*- __5d*6s°6p" (i) Pe GOGDGNL yoy ©) CIO) 6s 6p GAGHANM) Ghahanay ( ) Sd 6dsp? 6p (ii) [PtCl.]* has a square planar geometry (i) Draw the outer orbital diagram of the central metal ion in [Cd(OH)4]* and (ii) state the geometry of the complex. Coordination number =__4__ and Cd?" - ___ 4d!°5s°5p° (i) Cd"! GDUNGNGDGD ©) CCI) 4d 5s 5p UDG {SEREBLC1 sp (ii) [Cd(OH)4]* has a tetrahedral geometry.Q4: (i) Draw the outer orbital diagram of the central metal ion in [Fe(CN)«]* and (ii) state the geometry of the complex. A4: Coordination number =__6 __ and Fe**-__3d°4s°4p° (i) Fe* HADCNCDCNCD €) CIO) 3d 4s 4p GOGDAD MaMa), 3d 4d’sp° (ii) [Fe(CN)s}* has an octahedral geometry. Q5: (i) Draw the outer orbital diagram of the central metal ion in [Cr(NH3)s]** and (ii) state the geometry of the complex. AS: Coordination number =__6 _ and Cr**- __3d°4s°4p° (i) cre CHCHNCNC IC) ©) CIC IO) 3d 4s 4p COCA) Maannhananny 3d 4d?sp (ii) [Cr(NHs3)o]** has an octahedral geometry. Q6: (i) Draw the outer orbital diagram of the central metal ion in [Cu(H20)4}** and (ii) state the geometry of the complex. : Coordination number =__4 and Cu**-__3d°4s°4p° (i) Cu GOADGNANC ©) CIC IC) 3d 4s 4p & BOONE) (MINIM) t) ) ( dsp? 4p (ii) [Cu(H20)s]** has a square planar geometry.Q7: (i) Draw the outer orbital diagram of the central metal ion in [Fe(H20)<]°* which is found to be paramagnetic with four unpaired electrons and (ii) state the geometry of the complex. AZ: Coordination number =__6 _ and Fe**-___3d°4s°4p° (i) Fe ACNCDCNCD C) CII) 3d 4s 4p The valence bond model does not always predict the correct magnetic behavior of complex ions. The orbital diagram below predicts that [Fe(H20)«]** is diamagnetic. GYGDU) MbiMhaananinyan) 3d 4d?sp* The “explanation” below showing sp*d? hybridization will predict the paramagnetic behavior of [Fe(H20)s]** but it is awkward. GNC NC HNC ACH GY YON) (MMC CIC) 3d 4s 4p 4d Gi) [Fe(H20).}** has an octahedral geometry. Q8: Which of the five d-orbitals shown below would be affected the most by six monodentate ligands in an octahedral complex? A8: Since the lobes of the d,-,’ and d,’ orbitals point directly toward the corners of an octahedral structure, it is reasonable to assume that these orbitals would be affected more than the other three.Chapter-24 Lesson-3 IV — Crystal-Field Theory e The valance bond model cannot explain the colorful nature of coordination compounds. ¢ The valance bond model cannot always correctly predict the magnetic properties of coordination compounds. e The crystal-field theory is able to explain both the color and the magnetism of coordination compounds. © The crystal-field theory interprets the metal-ligand bonding as an electrostatic field interaction around the d-orbitals of the metal ion. For octahedral complexes, this electrostatic field splits the energies of the five d-orbitals into two levels called the low spin orbitals and the high spin orbitals. ¢ The energy difference between the two levels is called the crystal- field splitting energy (A). The value of A depends on the ligand.e Ligands that cause the value of A to be large are called strong- field ligands and they produce low spin complexes. e Ligands that cause the value of A to be small are called weak- field ligands and they produce high spin complexes. ( a 9 )(8,10) d? di? (4,9 GS, 10) d? de? iF Low ‘ High smal spin pliting spin apliting (1,6 (2,7 (3,8 ) (1,4 (2,5 (3,6 ) dy due dye dy du dy, ¢ The field strength of the ligands are given in the following spectrochemical series: (weakest to strongest) Cl B->V ->R->O-yellow because the energy jump increases between the low and high d-orbitals. A complex ion of metal M, ML¢*, is violet. The same metal M forms a complex ion with another ligand, MJ", and is green. Is the ligand L a stronger or weaker field ligand than J? Explain. Since ML¢** is violet then it is absorbing yellow light and since MJ." is green then it is absorbing red light. Thus L is a stronger field ligand than J since L is absorbing light of higher energy. If a complex absorbs light at 610 nm, what is the energy of this absorption in kJ/mol? E=he/A E = (6.63x10™4J-s)(3.00x108m/s) x (1.00kJ ) x (6.02x10?5mol"!) (610x10° m) (1000 J) 196 kJ/mo!Q4: The [Cr(NH3)sCI}* ion is purple, whereas the [Cr(NH3)o]** ion is yellow-orange. (i) Predict the color of light absorbed by each ion. (ii) Which ion absorbs light with the shorter wavelength? (iii) Do your conclusions agree with the spectrochemical series? : (i) Since [Cr(NH3)sCIP* is purple, it must be absorbing light in the yellow-orange region of the spectrum and since [Cr(NH3)6]** is yellow-orange, it must be absorbing light in the blue-violet region of the spectrum. (ii) [Cr(NH3)6]** is absorbing light with the shorter wavelength. (iii) YES! Ammonia is a stronger field ligand than chlorine causing [Cr(NHs3)«]** to have a higher crystal-field splitting energy (A) than [Cr(NH3)sCl}*". Blue-violet light has higher energy than yellow-orange light. Q5: [Ti(NH3)o**] has a d-orbital transition energy equal to 300kJ/mol. Determine the color of this complex (hint: find the wavelength). Br 2h=he/E 2 = (6.63x103F-s)(3.00%10%m/s) x (6.0210°2mol") x (1.00 ) (300 kJ/mol) (1000 J) 2.=0.000000399 m — 399 nm (= purple) Purple’s complementary color i:Q6: Using Crystal-Field Theory, explain why Mn** forms both high spin and low spin octahedral complexes but Mn** does not. A6; In order to be able to make both a low spin and a high spin complex, the metallic ion would need between four and seven d-electrons. Mn** has four d-electrons. Mn** has only three d-electrons and thus cannot make low and high spin complexes. Cy a2 dey? (tf ) C Mw) d2 des? a2 des? (t4yt ot) Ht ED dy dye dy dy dz dy dy dw dy low spin Mn** high spin Mn** Mn* Q7: [Co(NH3)6}**, [Co(NH3)sH20}**, [Co(NH3)sCl ?* are complex ions whose aqueous solutions are, in random order, red, green, and yellow. Match the complex ion to its color. Explain. AZT: In the spectrochemical series, Cl! is a weaker field ligand than H20 and H20 is a weaker field ligand than NH3._ Thus Cl" will produce a smaller crystal field splitting energy than H2O which in turn will produce a smaller crystal field splitting energy than NH3. Thus [Co(NH3)sCl }* will absorb the lowest energy light, [Co(NH3)sH2O]** will absorb the next lowest energy light, and [Co(NH3)o]** will absorb the highest energy light of the three. Since the color seen is the complementary color to the color absorbed, then the color of the three ions is as follows: [Co(NH3)o]" — yellow 230m [Co(NH3)sH2O}* — red [Co(NH3)sCl?* — greenQ8: From the following empirical formulas, write the proper formulas for the following palladium complexes: (a) Pd(IT)K(NH3)Cl3 (b) Pd(IV)K2Cle (c) Pd(IV)(NH3)sCla (d) Pd(ID)(NH3)2Clo A8: (a) K[Pd(NH3)Cl3] Q9: AM: (b) Ko[PdClo] (c) [Pd(NH3).CL Cle (a) Pd(NH3)2Ch Explain why complexes of Cu’ are brightly colored while those of Zn** are colorless. In order for a complex to have color it must be able to absorb visible light and excite a d-electron from an tz set d-orbital to an e set d-orbital. Since Cu’ has eight d-orbital electrons, the two e set d-orbitals are half full and can receive an excited electron from a tz set d-orbital. In this way Cu** can form brightly colored complexes. Zn** has ten d-orbital electrons — a filled d-orbital. There are no empty or half filled e set d-orbitals to receive an excited electron from a tz set d-orbital. Thus complexes of Zn** will not absorb visible light and will be white. (t \Ct ) mm eset d-orbitas ——> (+ +)(t +) a? dey? d2 dea? CH H)CT HCE 4H) — te set orbitals — (tH )(t Ht 4) dy dy dye dy du dy Cu* Zn**Chapter-24 Lesson-4 V—Nomenclature of Complex Ions 1) The ligand is named before the metal. If there is more than one type of ligand, name them in alphabetical order. A Roman num- eral is put directly after the metal’s name to indicate the oxidation number. Example: [Cr(H20)«Ch]'" is tetraaquadichlorochromium(II1). 2) Anionic ligands whose names end with the "ide" suffix are changed to the "o" suffix. Example: F! - fluoro, Cl’ - chloro, Br" - bromo, I" - iodo, CN - cyano, SCN'* - thiocyano, & OH" - hydroxo 3) Anionic ligands whose names end with the "ate" suffix are changed to the "ato" suffix. Example: C20,” - oxalato, (CH2)2N2(CH2COO"), — ethylenediaminetetracetato (EDTA*), & CO3* - carbonato, SO, - sulfato 4) Anionic ligands whose names end with the “ite" suffix are changed to the "ito" suffix. Example: NO»! - nitrito 5) Molecular ligands keep their names. Example: CsHsN — pyridine, CH3NH2 — methylamine, & NH2(CH2)2NH2 — ethylenediamine (en). Exceptions: H2O - aqua, NH3 - ammine, CO - carbonyl, & NO - nitrosyl. 6) Use di, tri, tetra, penta, or hexa if monodentate ligand appears more than once in the complex. Use bis (2), tris (3), or tetrakis (4) if a polydentate ligand appears more than once in the complex. 7) Ina complex cation, the metal keeps its own name. 8 In a complex anion, "ate" is added to the metal's name. Exceptions: Before adding "ate" to the metal’s name, drop the “ium”, “um”, or “ese” endings. Some metal's use Latin names: Ag - argentate, Fe - ferrate, Cu - cuprate, Pb - plumbate, Sn - stannate, Au - aurate. 9) For coordination compounds, name cations before anions.Ql: Al: Q2: A2: ke IB fe Name the following compounds. (a) Nao[CdCl4]__ (b) [Ni(NH3)s]Cls_— (c) [Zn(en)2](NOs)2 (a) sodium tetrachlorocadmiumate(II) (b) pentaamminenickel(III) chloride (c) bis(ethylenediamine)zinc(II) nitrate Write the formulas of the following compounds. (a) bromopentacarbonylmanganese(I) (b) sodium hexanitritocobaltate(II]) (c) dichlorobis(ethylenediamine)cobalt(III) nitrate (a) [Mn(CO)sBr] (b) Nas[Co(NO>)6] (c) [Co(NH2(CH2)2NH2)2ChJNO3 or [Co(en)xCh]NOs : Name the following compounds. (a) Ka[Fe(CN)«] _ (b) [Ru(NH3)sCIJ]SOs —(c) Na[Ag(CN)2] : (a) potassium hexacyanoferrate(II) (b) pentaamminechlororuthenium(IIT)sulfate (c) sodium dicyanoargentate(I) : Write the formulas of the following compounds. (a) pentaammineiodochromium(III) bromide (b) ethylenediaminemolybdenum(II]) chlorate (c) potassium tetrachloroplatinate(II) Ada: (a) [Cr(NH3)sIJBro (b) [MoNH2(CH2)2NH2](C1O3)3 or [Mo(en)](C103)3 (c) Ko[PtChi]VI -— Coordination Compounds in Our Lives e Except for scandium and titanium, all period 4 transition metals are essential for life. In trace amounts, these metals form various complexes in proteins (enzymes) or vitamins. e Aniron (II) ion is at the center of four heme sites in hemoglobin. aN X is the bonding c Cc site for oxygen. lx | 2 =N 1 N— Cc. 4 ed \ c > Fe c YX “ ' N a C—N 1 N—C | | Y is the bondi The center of X 1g ine bonding’ aheme group @ site for the globin : N J molecule. ce [Hb(H20)s]ag) + 4 Ore) [Hb(O2)sJag + 4 120 Deoxyhemoglobin oxyhemoglobin (blue) (red) At high altitudes, where there is a lower concentration of Ox), the position of the equilibrium will shift to the left. With less oxygenated blood, fatigue and dizziness occur. carbon monoxide (CO) is a stronger field ligand than oxygen gas and thus can easily replace the oxygen gas in the hemoglobin causing death. [Hb(O2)sJag + 4 COw @ [Hb(CO)sJog + 4 Org) The strong field ligand cyanide, CN', coordinates strongly to cytochrome oxidase, an iron containing cytochrome enzyme involved in cellular respiration, and causes rapid death.© Lead (II) ions cause lead poisoning, which is very harmful in young children. If EDTA* is administered soon after the Pb?* ions (mostly from paint chips or paint dust) are ingested, then the harmful effects can be limited as the [Pb(EDTA)]* complex is formed and is eliminated in the urine. EDTA‘ is also used to remove poisonous mercury (II) ions, Hg”*, from the body. © The neutral complex cis-[Pt(NH3)2Cl], known commonly as cisplatin, is a chemotherapy agent used to treat cancer. The cisplatin molecule attaches itself to the cancer DNA by ex- changing the chloride ions with nitrogen atoms in the DNA. The resulting cis-[Pt(NH3)2N-2] complex interferes with DNA replication and kills the cancer. Cl. NH NH Pt. aN cl NH DNA cis-[Pt(NH3)2Ch] eae cis-[Pt(NH3)2N-2] e A magnesium (II) ion is at the center of a porphine ring in chlorophyll. LN Cc Cc Cc | | G ——N N= C 4 NU \ c Mg, c YAN ¢— N—C I | Cc Cc