Hard-Soft Acids and Bases: Altering the Cu + /Cu 2+ Equilibrium Objectives: (1) Calculate/predict stability of copper oxidation states (2) Use ligands to change stabilities of oxidation states HSAB theory: qualitative predictions Redox potentials: quantitative results

Oxidation States Sum of oxidation states = overall charge on species Assumes unequal sharing of electrons –more electronegative atom gets all electrons, preferred oxidation state Examples: –MnO, MnO 2, [K + MnO 4 - ] What differences are found between metals in different oxidation states? Atomic radius, reactivity Hard/soft, redox potential

Hard vs. Soft Ligands and Metals Bonding trends of Lewis acids / Lewis bases --electron acceptors / electron donors Polarizable (soft) vs non-polarizable (hard):

Thermodynamics of Hard/Soft Ligand/Metal Binding --Hard-hard / soft-soft thermodynamically stronger binding / interaction --Hard-soft / soft-hard thermodynamically weaker binding / interaction HSAB theory: Preferential selection of oxidation states by hard or soft ligand set

K stability = [AB] / [A][B] softerharder most stable complexes least stable complexes

Lewis acids and bases Hard acids H +, Li +, Na +, K +, Rb +, Cs + Be 2+, Mg 2+, Ca 2+, Sr 2+, Ba 2+ BF 3, Al 3+, Si 4+, BCl 3, AlCl 3 Ti 4+, Cr 3+, Cr 2+, Mn 2+ Sc 3+, La 3+, Ce 4+, Gd 3+, Lu 3+, Th 4+, U 4+, Ti 4+, Zr 4+, Hf 4+, VO 4+, Cr 6+, Si 4+, Sn 4+ Borderline acids Fe 2+, Co 2+, Ni 2+, Cu 2+, Zn 2+ Rh 3+, Ir 3+, Ru 3+, Os 2+ R 3 C +, Sn 2+, Pb 2+ NO +, Sb 3+, Bi 3+ SO 2 Soft acids Tl +, Cu +, Ag +, Au +, Cd 2+ Hg 2+, Pd 2+, Pt 2+, M 0, RHg +, Hg 2 2+ BH 3 CH 2 HO +, RO + Hard bases F - H 2 O, OH -, O 2- CH 3 COO -, ROH, RO -, R 2 O NO 3-, ClO 4- CO 3 2-, SO 4 2-, PO 4 3- RNH 2 N 2 H 4 Borderline bases Cl -, Br - NH 3, NO 2-, N 3- SO 3 2- C 6 H 5 NH 2, pyridine N 2 Soft bases H -, I - H 2 S, HS -, S 2-, RSH, RS-, R 2 S SCN - (bound through S), CN -, RNC, CO R 3 P, C 2 H 4, C 6 H 6 (RO) 3 P

 G 0 = -nFE 0 n = mol e - F = 96,500 Coulombs / mol e - E 0 = standard reduction potential in volts  G 0 = free energy in joules Electrochemical potentials E 0 --Related to thermodynamic stability:

(1) Cu 2+ + Iˉ + eˉ  CuI0.86V (2)Cu 2+ + Clˉ + eˉ  CuCl0.54V (3)I 2 + 2eˉ  2Iˉ0.54V (4)Cu + (aq) + eˉ  Cu(s)0.52V (5)Cu 2+ (aq) + 2eˉ  Cu(s)0.37V (6)CuCl + eˉ  Cu(s) + Clˉ0.14V (7)Cu(NH 3 ) eˉ  Cu(s) + 4NH V (8) Cu 2+ (aq) + eˉ  Cu + (aq)-0.15V (9)CuI + eˉ  Cu(s) + Iˉ-0.19V (10)Cu(en) eˉ  Cu + 2en-0.50V Electrochemical Potentials Used in Experiment 1: E 0

Redox Potential Calculation Cu(aq) NH 3  Cu(NH 3 ) 4 +2 (5)Cu 2+ (aq) + 2eˉ  Cu(s)0.37V (7)Cu(NH 3 ) eˉ  Cu(s) + 4NH V Reduction: Cu 2+ (aq) + 2eˉ  Cu(s) E 0 = +0.37V Oxidation: Cu(s) + 4NH 3  Cu(NH 3 ) eˉ E 0 = +0.12V Net: Cu 2+ (aq) + 4NH 3  Cu(NH 3 ) 4 2+ E 0 = +0.49V (5) + (7*)

Disproportionation 2 Fe 4+ →Fe 3+ + Fe 5+ 2 H 2 O 2 → 2 H 2 O + O 2 --Two identical atoms in same oxidation state exchange one electron --Take on two different oxidation states 2 Cu + → Cu 0 + Cu 2+

Summary --Make Cu compounds, force into desired oxidation state with correct H/S ligands --Determine if products agree with HSAB and electrochemical potentials --Determine whether two theories useful in applications