Transcript Chapter 6 Lecture 2 Hard-Soft Acid-Base Concepts I.

Chapter 6 Lecture 2 Hard-Soft Acid-Base Concepts
I.
Hard and Soft Acids and Bases
A.
Factors other than acid/base strength determine acid/base reactivity
1) Silver Halide solubility
a) AgX(s) + H2O
Ag+(aq) + X-(aq)
b) Ksp’s: AgF = 205, AgCl = 1.8 x 10-10, AgBr = 5.2 x 10-13, AgI = 8.3 x 10-17
2) Explanation of Solubility
a) Solvation: F- much better solvated (small, high charge)
b) Degree of Ag—X Interaction must also play a role
3) HSAB Theory can help explain this data
a) Hard acids/bases are small and nonpolarizable
b) Soft acids/bases are large and polarizable
c) Hard/Hard and Soft/Soft interactions are the most favorable
d) Polarizable = easily distorted by other charged ions
4) AgX data and HSAB
a) Ag+ is large and polarizable = Soft
b) Softness of Halides: I- > Br- > Cl- > Fc) AgI has the strongest interaction, thus the lowest solubility
d) Softness is also associated with covalent bonds, not ionic bonds
5)
Coordination of Thiocyanate (SCN-) to metal ions
a) SCN- binds to large, polarizable metals, through S: Hg2+----SCN
b) SCN- binds to smaller, less polarizable metals through N:
Zn2+----NCS
c) Explanation: Hard/Hard and Soft/Soft interactions are favored
6)
Exchange Reactions of [CH3Hg(H2O)]+
a) [CH3Hg(H2O)]+ + HCl
CH3HgCl + H3O+ K = 1.8 x 1012
b) [CH3Hg(H2O)]+ + HF
CH3HgF + H3O+ K = 4.5 x 10-2
c) Explanation: Hard/Hard and Soft/Soft interactions are favored
7)
B.
LiX solubility: LiBr > LiCl > LiI > LiF
a) Li+ is a hard ion
b) LiF would be expected to be very ionic and soluble
c) Very favorable hard-hard LiF interaction even overcomes solubility
d) LiBr, LiCl are more soluble because of less favored interactions
e) LiI is out of order because of poor I- solvation
Pearson’s Hard and Soft Acids and Bases (1963)
1) Most metal ions are hard acids (class a), some borderline depending on charge
2) Large polarizable metal ions are soft (class b)
3) Lewis bases can also be categorized as hard or soft
4) Reactions favor hardness matches
5) Hard/hard more energetically favored than soft/soft interactions
Soft and borderline
soft metal ions
6)
Polarizability = degree to which an atom’s electron cloud is distorted by
interactions with other ions
a) Hard = small, compact charge, nonpolarizable = M3+, O2b) Soft = large, polarizable = M0, S2c) Comparison is easiest within a column of the periodic table
C.
Pearson’s Absolute Hardness = h
1) Quantitative method to measure hardness and softness, predict matches
2) Formula uses Ionization energy (I) and Electron Affinity (A)
IA
η
2
Related to Mulliken’s definition of Electronegativity
Defines Hardness as a large difference between I and A
a) I = HOMO energy
b) A = LUMO energy
5) Softness = s = 1/h
6) Halogens as an example
a) Trend in h parallels HOMO
energy (LUMO’s are about the same)
b) F = most electronegative, smallest,
least polarizable = hardest
c) ClBrI h decreases as
HOMO energy increases
7) Problem: h doesn’t always match
reactivity (hard, but still weak acid)
3)
4)
χ
IA
2
D.
Drago’s Quantitative Approach
1) -DH = EAEB + CACB
a) DH = enthalpy of reaction for A + B
AB
b) E = capacity for ionic interactions (calculated from various reactions)
c) C = capacity for covalent interactions (for various reactions)
I2 used as reference E =
1.00, C = 1.00
Primarily covalent, so
most E values > 1,
most C values < 1
II.
2)
Example: I2 + C6H6
I2 • C6H6
-DH = (1.00 x 0.681 + 1.00 x 0.525) = 1.206 kcal/mol (weak adduct)
3)
Advantages of Drago’s System
a) Emphasis on the 2 factors involved in acid-base strength
i. Electrostatics
ii. Covalency
b) Pearson’s Hardness only considers covalency
c) Good predictability if E and C have been tabulated
d) If no data is available, Pearson’s HSAB method still allows for a
rough prediction of the strength of an acid-base reaction
Frontier Orbitals and Acid-Base Chemistry
A.
HOMO-LUMO Combination
1) Acid-Base Reactions result in new product frontier orbitals
2)
Example: NH3 + H+
NH4+
a) NH3 lone pair reside in a1 MO = HOMO (base)
b) H+ has only an empty 1s AO = LUMO (acid)
c)
Combine these two orbitals
to form a Bonding/
Antibonding pair
d)
Lone pair is stabilized in the
new bonding orbital
e)
NH4+ is more stable than the
separate NH3 and H+
f)
Pick orbitals of similar
symmetry and energy to
combine
g)
If there is no match, there is
no acid-base adduct formed
3)
a1 b1 b2
Symmetry, Energy, and Occupation of Frontier Orbitals allow us to
predict the result of a given reaction
a)
2 H2O + Ca
Ca2+ + H2(g) + 2 OHi. Ca HOMO >> H2O LUMO
ii. Not matched well for acid-base adduct to form
iii. Electron transfer (oxidation of Ca by H2O) is the predicted reaction
b)
n H2O + Cl[Cl(H2O)n]i. Correct energy match for acid-base adduct to form and be stable
ii. Water is the acid; its LUMO is used along with the base (Cl-) HOMO
c)
6 H2O + Mg2+
[Mg(H2O)6]2+
i. Correct energy match for acid-base adduct to form and be stable
ii. Water is the base; its HOMO is used along with the LUMO of acid Mg2+
d)
2 H2O + 2 F2
4 F- + 4 H+ + O2
i. Water HOMO >> F2 LUMO
ii. Not matched well for an acid-base adduct to form
iii. Electron transfer (reduction of F2 by H2O) is the predicted reaction
4)
Restating the Lewis Acid-Base
Definition:
a) Base: has e- pair in
HOMO of correct energy
and symmetry
b) Acid: has LUMO of
correct energy and
symmetry
5)
HSAB using HOMO-LUMO
a) Hard/Hard = simple
electrostatics, little change
in HOMO-LUMO energies,
which remain far apart
b) Soft/Soft = HOMO and
LUMO close in energy
combine to form
energetically favorable new
MO’s
c) Hard/Hard is usually more
favored than it appears due
to +/-charge attraction
B.
Carbon Monoxide as a Lewis Base
1) Electronegativity suggests O is the
e- pair donor
C O
2)
C O
M+
In fact, C is always the donor
a) Formal Charge
-1
+1
C O
b)
MO Frontier Orbitals
i. HOMO that is involved in
bonding is mostly on C
ii. C-like HOMO donated to
the M Lewis acid
O C
M+
C.
Hydrogen Bonding
1) FHF- (symmetric, equivalent bonds to H)
a) In chapter 5, we described the MO’s as H + F•F group orbitals
b) Using a HOMO-LUMO Acid-Base description, we can use HF + Fc) F- HOMO (base) + HF LUMO and HOMO (acid) give 3 new MO’s
i. Filled bonding MO of lowest energy
ii. Filled nonbonding (node through H) orbital
iii. Empty antibonding (nodes between all atoms) orbital of highest energy
d) 3-center 2 electron bond, with ½ bond order per H—F bond
2)
Unsymmetric BHA molecules are simplest example of Hydrogen Bonding
a) Similar MO picture to that of FHFb) Overall lower energy for the 2 e- pairs when bonded than unbonded
c) 3 different possibilities for the energy match
(a) Poor match: H2O + CH4 = overall higher energy; no H-bonding
(b) Good match: H2O + HOAc = overall lower energy; strong H-bond
(c) Very poor match: H2O + HCl = H atom transfer
d)
Why are electronegative atoms good at H-bonding?
Low energy H—A HOMO similar in energy to H+ 1s orbital
D.
Electronic Spectra of I2 (acid) + Donor (base) adducts
1)
I2 gas: violet due to pg*su* transition (blue and red
transmitted = purple)
2)
Nondonor hexane solvent causes no change = purple
3)
Donor solvents benzene and and methanol
a) Donor HOMO interacts with su* LUMO of I2
b) New HOMO/LUMO energies of adduct shift spectrum
c) Color (light transmitted) is changed
i. Red-violet in benzene
ii. Yellow-brown in methanol
4)
KI/H2O solution with I2 forms I2—I- adduct I3a) pg*su* energy increases, LUMO higher in energy
toward blue
b) ss* transition appears = Charge Transfer appears
whenever a good donor interacts with I2
c) Charge transfer is so named because the transition
occurs between an orbital mostly composed of one of
the partners to an orbital mostly composed of the other
partner
hννCT


I 2  Donor [I 2 ] [Donor]
5)
Charge Transfer in [Fe(H2O)5X]2+ complexes
a) HOMO of X- increases in energy from F- to Ib) F- complex has closest HOMO-LUMO energy match forming a
strong complex with stabilized new HOMO = colorless (all UV)
c) Cl- complex has less interaction, new HOMO not stabilized as much,
yellow color
d) Br- even less interaction, less stabilization, yellow-brown color
e) I- has no interaction, charge transfer only
2 Fe3+ + 2 I2 Fe2+ + I2 (purple color of I2 only)