Transition Metal Chemistry CHEM261HC/SS1/01 Periodic Table 1. - PDF document

6/4/2011 APPLI ED I NORGANI C CHEMI STRY FOR CHEMI CAL ENGI NEERS Transition Metal Chemistry CHEM261HC/SS1/01 Periodic Table 1. Main-group elements 2. Transition metals  Elements are divided into four categories 3. Lanthanides 4. Actinides Main-group elements Main-group elements ( S -Block) Transition metals ( P -Block) ( ) Lanthanides Actinides CHEM261HC/SS1/02 1

6/4/2011 Transition metals vs. Main-group elements  Main ‐ group metals • malleable and ductile Transition Main ‐ group • conduct heat and electricity metals elements • form positive ions  Transition metals • more electronegative than the main group metals • more likely to form covalent compounds • easily form complexes Cisplatin • form stable compounds with neutral molecules  There is some controversy about the classification of the elements i.e. Zinc (Zn), Cadmium (Cd) and Mercury (Hg) e ‐ configuration [ ]n s 2 n ‐ 1 d 10 IUPAC ‐ A transition metal is " an element whose atom has an incomplete d sub ‐ shell, or which can give rise to cations with an incomplete d sub ‐ shell .” CHEM261HC/SS1/03 Electron configuration of Transition-metal ions  The relationship between the electron configurations of transition ‐ metal elements and their ions is complex. Example  Consider the chemistry of cobalt which forms complexes that contain either Co 2+ or Co 3+ ions. Co has 27 electrons 4 s 2 3 d 7 [Ar] has 18 electrons Co : [Ar] Co 2+ : [Ar] 3 d 7 Co 3+ : [Ar] 3 d 6  In general, electrons are removed from the valence shell s orbitals before they are removed from valence d orbitals when transition metals are ionized. CHEM261HC/SS1/04 2

6/4/2011  How do we determine the electronic configuration of the central metal ion in any complex? • Try to recognise all the entities making up the complex • Need knowing whether the ligands are neutral or anionic • Then you can determine the oxidation state of the metal ion. …same as M(+2) or M 2+ A simple procedure exists for the M(II) case 22 23 24 25 26 27 28 29 Ti V Cr Mn Fe Co Ni Cu Cross off the first 2 gives you total No. of valence electrons left 2 3 4 5 6 7 8 9 CHEM261HC/SS1/05 EXAMPLES Oxidized elements Configuration Elements Configuration Sc Sc [Ar]4 s 2 3 d 1 [Ar]4 s 3 d Sc(III) Sc(III) [Ar] [Ar] V [Ar]4 s 2 3 d 3 V(II) [Ar]3 d 3 Cr [Ar]4s 1 3 d 5 Cr(III) [Ar]3 d 3 Fe [Ar]4 s 2 3 d 6 Fe(II) [Ar]3 d 6 [Ar]4 s 2 3 d 8 [Ar]3 d 8 Ni Ni(II) [Ar]4 s 1 3 d 10 [Ar]3 d 10 Cu Cu(I) Zn [Ar]4 s 2 3 d 10 Zn(II) [Ar}3 d 10 …variety of oxidation states !! 3

6/4/2011 Evaluating the oxidation state [CoCl(NO 2 )(NH 3 ) 4 ] + Net charge on complex ion (+1) Neutral zero charge zero charge X + (- 2) + 0 = +1 X - 2 = +1 x = +3 Co 3+ CHEM261HC/SS1/06 Oxidation states and their relative stabilities  Why do these elements exhibit a variety of oxidation states?  Because of the closeness of the 3d and 4s energy states Sc Sc +3 +3 Ti +1 +2 +3 +4 V +1 +2 +3 +4 +5 Cr +1 +2 +3 +4 +5 +6 Mn +1 +2 +3 +4 +5 +6 +7 Fe +1 +2 +3 +4 +5 +6 Co C +1 +1 +2 +2 +3 +3 +4 +4 +5 +5 Ni +1 +2 +3 +4 Cu +1 +2 +3 Zn +2  The most prevalent oxidation numbers are shown in green. CHEM261HC/SS1/07 4

6/4/2011  An increase in the No. of oxidation states from Sc to Mn.  All seven oxidation states are exhibited by Mn.  There is a decrease in the No. of oxidation states from Mn to Zn. WHY? Sc +3 Ti +1 +2 +3 +4  Because the pairing of d -electrons occurs after Mn (Hund's rule) V +1 +2 +3 +4 +5 which in turn decreases the number of available unpaired electrons Cr +1 +2 +3 +4 +5 +6 and hence, the number of oxidation states. Mn +1 +2 +3 +4 +5 +6 +7 Fe +1 +2 +3 +4 +5 +6  The stability of higher oxidation states decreases in moving from Sc Co +1 +2 +3 +4 +5 to Zn. Ni +1 +2 +3 +4 Cu +1 +2 +3  Mn(VII) and Fe(VI) are powerful oxidizing agents and the higher oxidation states of Co, Ni and Zn are unknown. Zn +2 CHEM261HC/SS1/09  The relative stability of +2 state with respect to higher oxidation states increases in moving from left to right. On the other hand +3 state becomes less stable from left to right. Why? Sc +3 Ti +1 +2 +3 +4  This is justifiable since it will be increasingly difficult to remove the V +1 +2 +3 +4 +5 third electron from the d -orbital. Cr +1 +2 +3 +4 +5 +6 Example Mn +1 +2 +3 +4 +5 +6 +7 Fe +1 +2 +3 +4 +5 +6 22 23 24 25 26 27 28 29 Co +1 +2 +3 +4 +5 Ti Ti V V Cr Cr Mn Mn Fe Fe Co Co Ni Ni Cu Cu Ni +1 +2 +3 +4 Cu +1 +2 +3 M = [Ar]4 s 2 3 d x Zn +2 M +2 = [Ar]3 d x loss of the two s electrons M +3 = [Ar]3 d x-1 more difficult CHEM261HC/SS1/10 5

6/4/2011 Chromium Oxidized by HCl or H 2 SO 4 to form blue Cr 2+ ion • Cr 2+ oxidized by O 2 in air to form green Cr 3+ • A Assignment 1 i t 1  Write down balance equations that show the two reactions 2 − and • Cr also found in +6 state as in CrO 4 2 − are strong oxidizer Cr 2 O 7 Cr 2 O 7 are strong oxidizer Assignment 2  Use balanced equations to show that CrO 4 2 − 2 − are strong oxidizing agents and Cr 2 O 7 Assignment 1 Solution Cr (S) + H 2 SO 4(aq) Cr 2 SO 4(aq) + H 2(g) Cr + H SO Cr SO + H 2 Cr (s) + 4 HCl (aq) 2 CrCl 2(aq) + 2H 2(g) 2CrCl 2(aq) + O 2(g) Cr 2 O 2 Cl 2(aq) + Cl 2(g) 6

6/4/2011 Iron • Fe exists in solution in +2 or +3 state • Elemental Fe reacts with non-oxidizing acids to form Fe 2+ , which oxidizes in air (O 2 ) to Fe 3+ • Brown water running from a faucet is caused by insoluble Fe 2 O 3 Fe 3+ soluble in acidic solution, but forms a • hydrated oxide as red-brown gel in basic solution solution Assignment 3  Complete and balance the following equation Fe 2 O 3 + HCl Coordination Chemistry  A coordination compound (complex), contains a central metal atom (or ion) surrounded by a number of oppositely charged ions or neutral molecules (possessing lone pairs of electrons) which are known as ligands.  If a ligand is capable of forming more than one bond with the central metal atom or ion, then ring structures are produced which are known as metal chelates  the ring forming groups are described as the ring forming groups are described as chelating agents or polydentate ligands.  The coordination number of the central metal atom or ion is the total number of sites occupied by ligands.  Note: a bidentate ligand uses 2 sites, a tridentate 3 sites etc. CHEM261HC/SS1/13 7

6/4/2011 Ligands molecular Lewis Lewis donor coordination formula formula base/ligand base/ligand acid acid atom atom number number [Zn(CN) 4 ] 2- CN - Zn 2+ C 4 [PtCl 6 ] 2- Cl - Pt 4+ Cl 6 [Ni(NH 3 ) 6 ] 2+ : NH 3 Ni 2+ N 6 CHEM261HC/SS1/14 Mono-dentate Multidentate ligands Abbreviation Name Formula en Ethylenediamine ox 2- Oxalato EDTA 4- Ethylenediamine- tetraacetanato CHEM261HC/SS1/15 8

6/4/2011  Chelating ligands bond to metal  forms rings – chelate rings  Five or six atoms rings are common (i.e. including metal)  Coordination numbers and geometries Li Linear Square planar Tetrahedral Octahedral CHEM261HC/SS1/16 Nomenclature of Coordination Compounds • The basic protocol in coordination nomenclature is to name the ligands attached to the metal as prefixes before the metal name. • Some common ligands and their names are listed above. 9

6/4/2011  As is the case with ionic compounds, the name of the catio n appears first; the anion is named last.  Ligands are listed alphabetically before the metal. Prefixes denoting the number of a particular ligand are ignored when alphabetizing. Example cation anion [Co(NH 3 ) 5 Cl]Cl 2 Pentaamminechorocobalt(III) chloride 5 NH 3 cobalt in +3 Cl ‐ ligands oxidation states ligands  The names of anionic ligands end in “o”; the endings of the names of neutral ligands are not changed.  Prefixes tell the number of a type of ligand in the complex. If th If the name of the ligand itself has such a prefix, f th li d it lf h h fi alternatives like bis -, tris -, etc., are used. cation Example [Co(NH 2 CH 2 CH 2 NH 2 ) 2 Cl 2 ] + dichlorobis(ethylenediammine)cobalt(III) 2 Cl ‐ 2 en ligands with cobalt in +3 ligands 2 NH 2 groups oxidation states en = ethylenediammine 10

6/4/2011  If the complex is an anion, its ending is changed to - ate .  The oxidation number of the metal is listed as a roman numeral in parentheses immediately after the name of the metal . Example Na 2 [MoOCl 4 ] Exercise 1 Name the following coordination complexes: (i) Cr(NH 3 )Cl 3 (ii) Pt(en)Cl 2 (ii) Pt(en)Cl 2 (iii) [Pt(ox) 2 ] 2- Exercise 2 Give the chemical formular for the following coordination complexes: (i) Tris(acetylacetanato)iron(III) (ii) Hexabromoplatinate(2-) (iii) Potassium diamminetetrabromocobaltate(III) 11

6/4/2011 Solutions (i) Cr(NH 3 )Cl 3 tri chloro chromium (III) ammine Ammine Ammine trichlorochromium(III) trichlorochromium(III) (ii) Pt(en)Cl 2 Platinum (II) ethylenediammine di chloro Dichloro ethylenediammineplatinum(II) y p ( ) (iii) [Pt(ox) 2 ] 2- di Platinate (II) oxalato Dioxalato platinate(II) Solutions (i) Tris(acetylacetanato)iron(III) 3+ ( ) 3 Fe acac Fe(acac) 3 Fe(acac) 3 (ii) Hexabromoplatinate(2-) [ ] 2- Pt Br 6 [PtBr 6 ] 2- (ii) P t (ii) Potassium diamminetetrabromocobaltate(III) i di i t t b b lt t (III) 3+ K ( ) 2 NH 3 Br Co 4 K[Co(NH 3 ) 2 Br 4 ] 12