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Chemical Bonding
Ionic bond: Type of chemical bond that involves the electrostatic attraction
between oppositely charged ions. These ions represent atoms that have lost one or
more electrons (cations) and atoms that have gained one or more electrons (anions).
Here Sodium molecule is donating its 1 valence electron to the Chlorine molecule.
This creates a Sodium cation and a Chlorine anion. Notice that the net charge of the
compound is 0.
Basic Concept: Chemical Bonding
Some examples of ionic bonds and ionic compounds:
NaBr - sodium bromide NaF - sodium fluorideKI - potassium iodide KCl - potassium chlorideCaCl2 - calcium chloride KBr - potassium bromide
Ionic bonding in sodium chloride
Formation of ionic bond in lithium fluoride
Covalent bond: A chemical bond that involves the sharing of electron
pairs between atoms. The stable balance of attractive and repulsive forces between
atoms when they share electrons is known as covalent bonding.
Here Phosphorous molecule is sharing its 3 unpaired electrons with 3 Chlorine
atoms. In the end product, all four of these molecules have 8 valence electrons and
satisfy the octet rule.
Examples of covalent bonding
In chemistry, sigma bonds (σ bonds) are the strongest type of covalent chemical bond.
They are formed by head-on overlapping between atomic orbitals. Sigma bonding is
most clearly defined for diatomic molecules.
Sigma bond
Pi bonds (π bonds) are covalent chemical bonds where two lobes of one involved atomic
orbital overlap two lobes of the other involved atomic orbital. Each of these atomic
orbitals is zero at a shared nodal plane, passing through the two bonded nuclei.
π bond
Coordinate bond : A dipolar bond, more commonly known as a dative covalent
bond or coordinate bond is a kind of 2-center, 2-electron covalent bond in which the
two electrons derive from the same atom.
Metallic Bond: Metallic bonding constitutes the electrostatic attractive
forces between the delocalized electrons, called conduction electrons, gathered
in an electron cloud and the positively charged metal ions.
Valence Shell Electron Pair Repulsion (VSEPR) Theory Valence shell electron pair repulsion (VSEPR) theory is a model in chemistry,
which is used for predicting the shapes of individual molecules.
The theory was suggested by Sidgwick and Powell in 1940 and was developed
by Gillespie and Nyholm in 1957. It is also called the Gillespie-Nyholm Theory
after the two main developers.
VSEPR theory is based on the idea that the geometry of a molecule or polyatomic
ion is determined primarily by repulsion among the pairs of electrons associated
with a central atom.
The pairs of electrons may be bonding or nonbonding (also called lone pairs).
Only valence electrons of the central atom influence the molecular shape in a
meaningful way.
VSEPR theory may be summarized as:
The shape of the molecule is determined by repulsions between all of the
electron pairs present in the valence shell.
A lone pair of electrons takes up more space around the central atom than a
bond pair. Three types of repulsion take place between the electrons of a
molecule:
The lone pair-lone pair repulsion (lp-lp)
The lone pair-bonding pair repulsion (lp-bp)
The bonding pair-bonding pair repulsion. (bp-bp)
The best spatial arrangement of the bonding pairs of electrons in
the valence orbitals is one in which the repulsions are minimized.
lp-lp> lp-bp> bp-bp
The magnitude of the repulsions between bonding pairs of electrons
depends on the electronegativity difference between central atom
and other atoms.
Double bonds cause more repulsion than single bonds, and triple
bonds cause more repulsion than a double bond.
Predicted molecular shapes from Sidgwick- Powell Theory:No. of electron pairs in outer shell
Arrangement of electron pairs Electron-pair geometry
Bond angles
2
3
4
5
6
Linear
Trigonal Planar
Tetrahedral
Trigonal bipyramid
Octahedral
180 0
120 0
109.50
90 0
120 0
90 0
Some examples using VSEPR Theory
SnCl2
Lewis model:
Shape : bent
lp-bp repulsions cause the Cl-Sn-Cl bond angle close to less than 120 0 (approx 950)
NH3
Lewis model:
Shape : Trigonal Pyramid
lp-bp repulsions cause the H-N-H angles to close to less than 109.5 o (107.3o).
H2O
Lewis model:
Shape : Bent
lp-bp repulsions cause the H-O-H angle to be lesser than 109.5 0 (104.50 )
ClF3
Lewis model:
Shape : T shape
Lone pairs occupy equatorial positions of trigonal bipyramid
lp-bp repulsions cause F-C-F angle to be lesser than 90 0
Limitations of VSEPR Theory
It fails to predict the shapes of isoelectronic species [ CH4 and NH4+] and
transition metal compounds.
The model does not take relative size of substituents.
Atomic orbitals overlap cannot be explained by VSEPR theory.
The theory makes no predictions about the lengths of the bonds, which is
another aspect of the shape of a molecule.
Valence Bond Theory
Developed by Linus Pauling The sharing of electrons between
atoms is viewed as an overlap of atomic orbitals of the bonding atoms.
When H – H distance = 74 pm,Repulsion = Attraction strongest bond optimal overlap lowest energy
74 pmAt H – H distance > 74 pm,
Repulsion < Attraction weaker bond too little overlap atoms come closerAt H – H distance < 74 pm,Repulsion > Attraction weaker bond too much overlap atoms get further apart
22
Bond Energy• Reaction 2 H· H2 releases 436 kJ/mol• Product has 436 kJ/mol less energy than two atoms: H–H has
bond strength of 436 kJ/mol. (1 kJ = 0.2390 kcal; 1 kcal = 4.184 kJ)
23
Bond Length• Distance between
nuclei that leads to maximum stability
• If too close, they repel because both are positively charged
• If too far apart, bonding is weak
Theories of chemical bonding 24
Energy of Interaction Between Two H Atoms
Potential energy
distance
–346 kJ mol –1 H – H bond
+346 kJ mol –1 antibonding
Energies of attraction and repulsion as functions of distance between two H atoms are shown here.
The minimum of the attraction force occur at H–H bond length of 74 pm, at which, the antibonding orbital is +346 kJ mole–1 above 0, energy when H atoms are far apart.
How does energy affect the two-atom system?
25
The Valence-bond MethodValence bond method considers the covalent bond as a result of overlap of atomic orbitals. Electrons stay in regions between the two atoms. Some bond examples
s-s s-p s-d p-p p-d d-d H-H H-C H-Pd C-C Se-F Fe-Fe (?)Li-H H-N in Pd P-P
H-F hydrideBut overlapping of simple atomic orbitals does not explain all the features. Thus, we have to take another look, or do something about atomic orbitals – hybridization.
How does valence-bond approach explain the formation of chemical bonds?
26
Hybridization of Atomic OrbitalsThe solutions of Schrodinger equation led to these atomic orbitals.1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, etc. However, overlap of these orbitals does not give a satisfactory explanation. In order to explain bonding, these orbitals are combined to form new set of orbitals – this method is called hybridization.During the lecture, these hybridized orbitals will be explained:
sp 2 sp hybrid orbitals from mixing of a s and a p orbitalsp 2 3 sp2 hybrid orbitals from mixing of a s and 2 p orbitalsp3 fill in you explanation please
sp3d 5 sp3d hybrid orbitals from mixing of a s and 3 p and a d orbitalsp3d 2 ____________
Provide a description for hybrid orbitals sp, sp2, sp3, sp3d, and sp3d2
Theories of chemical bonding 27
The sp Hybrid Orbitals The sp hybrid orbitals: formation of two sp hybrid orbitals
+ + + - = + -
+ – + - = - +
hybridization of s and p orbitals = 2 sp hybrid orbitals
_ _ __ __
__ __ __ Two sp hybrid orbitlas => Two states of Be
28
Bonds with sp Hybrid OrbitalsFormations of bonds in these molecules are discussed during the lecture. Be prepared to do the same by yourself.
Cl–Be–Cl; H–CC–H; H–CN : ; O=C=O
Double and triple bonds involve pi p bonding, and the the application of valence bond method to p bonds will be discussed.
You are expected to be able to draw pictures to show the p bonding.
O=C=O
31
A p Bond
Overlap of 2 2p orbitals for the formation of p bond
Sigma (s) bond is symmetric about axis.
Pi (p) electron distribution above and below axis with a nodal plane, on which probability of finding electron is zero; p bond is not as strong as sigma - less overlap.
Nodal plane
Bonding of C2H4
C2s 2p 2p 2p
sp2 sp2 sp2 2p
How are pi bonds formed?
Theories of chemical bonding 32
Triple Bonds in H-CC-HH-C-C-H: three s bonds due to overlapping of 1sH – spC; spC – spC; and spC – 1sH.
Two p bonds in HCCH and HCN triple bonds are due to overlapping of p orbitals results.
Draw and describe how atomic orbitals overlap to form all bonds in acetylene, H–CC–H
py over lap
px over lap H H
sp hybrid orbitals
Two nodal planes of p bonds are perpendicular to each other.
in p bond
in p bond
C2s 2p 2p 2psp sp 2p 2p
Theories of chemical bonding 33
Two p Bonds in H–CC–H
A triple bond consists of a sigma and two pi bonds. Overlaps of two sets of p orbitals form of two p bonds.
Theories of chemical bonding 34
Bonding of CO2For CO2, the C atom forms a s bond and a p bond with each of two O atoms. The two nodal planes of the two p bonds are also perpendicular.
During the lecture, I draw diagrams and explain the two s two p bonds in CO2. You are expected to be able to do the same, in a test.
O=C=O or H2C=C=CH2
py over lap in p bond
Overlap p–p in s bonds
px over lap in p bond
Discuss the bonding of allene H2C=C=CH2
See extra problems B17 in the handout
Resonance structures
: O – C O :
: O C – O :
. .
Theories of chemical bonding 35
09_174
O C O
sigma bond(1 pair of electrons) pi bond
(1 pair ofelectrons)
pi bond(1 pair ofelectrons)
(a)
(b)
O C O
Bonding in CO2 – another view
Compare with H2C=C=CH2
36
The sp2 Hybrid OrbitalsGround state and excited state electronic configuration of B
_ _ _ __
_ __ __
The hybridization of a s and two p orbitals led to 3 sp2 hybrid orbitals for bonding.
Compounds involving sp2 hybrid orbitals: BF3, CO3
2–, H2CO, H2C=CH2, NO3
–, etc
37
An example of using sp2 hybrid orbitals
__ orbitals for bonding?
Dipole moment = ____?
Theories of chemical bonding 38
Bonding of H2C=CH2 moleculesUtilizing the sp2 hybrid orbitals, each C atom form two H–C s bonds for a total of 4 s H–C bonds. The C–C s bond is common to both C atoms.
A C–C p bond is formed due to overlap of p orbitals from each of the C atoms.
Hybrid orbitals (sp2) for H–C and C–C s bond
Overlap of p orbital for C–C p bond
C2s 2p 2p 2p
sp2 sp2 sp2 2p
39
The sp3 Hybridized OrbitalsGround state and excited state electronic configuration of C
_ _ _ _
_ _ __
The hybridization of a s and three p orbitals led to 4 sp3 hybrid orbitals for bonding.
Compounds involving sp3 hybrid orbitals: CF4, CH4, : NH3, H2O::, SiO4
4–, SO42–, ClO4
–, etc
Theories of chemical bonding 40
C
2s 2p 2p 2psp3 sp3 sp3 sp3
41
The sp3d Hybrid OrbitalsHybridization of one s, three p, and a d orbitals results in 5
sp3d hybrid orbitals. The arrangement of these orbitals is a
trigonal pyramid. Some structures due to these type of
orbitals are PClF4, TeCl4E, and BrF3E2.
How many unshared electron pairs are present in TeCl4 and BrF3?
What are their shapes?
42
The sp3d2 Hybrid Orbitals
Hybridization of one s, three p, and two d orbitals results in 6 sp3d2 hybrid orbitals. The arrangement of these orbitals is an octahedron. Compounds using these type of orbitals are shown here.
AX6, AX5E, AX4E2 AX3E3 and AX2E4
IOF5, IF5E, XeF4E2
No known compounds of AX3E3 and AX2E4 are known or recognized, because they are predicted to have a T shape and linear shape respectively when the lone pairs of electrons are ignored.
Theories of chemical bonding 43
Molecules with more than one central atomDescribe the structure of CH3NCO.
Draw the skeleton and add all valence electrons
H3C – N – C – O
Which Lewis dot structure is the most important (stable)?
N = C = O
H–C
H H
120o
109o
180o
What hybridized orbitals are used for bonding in N and C? Why are the bond angles as indicated? No of s and p bonds = __, __? Give formal charges to all atoms in all structures.
Take a new look at slide 22 in Bonding Basics
Which structure is more stable, and why?
NC–O
H–C
H H
N–CO
H–C
H H
Because of orbital overlap, the bonding electrons localize in the region between the bonding nuclei such that
There is a high probability of finding the electrons in the region between the bonding nuclei.
Overlap of two half-filled orbitals leads to the formation of a covalent bond.
1s
1s
1s-1s overlap gives a H – H single bond
F
2s 2p
1sH
The 1s-2p overlap gives a H – F single bond
Non-bonding electrons
F
2s 2p
1sH
F
2s 2p
The 2p-2p overlap gives a F – F single bond
F
2s 2p
F
2s 2p
Non-bonding electrons
F
2s 2p
Each F atom has three pairs of non-bonding electrons.
F F
By VB Theory,
Three 2p-1s(half-filled) overlaps lead to the formation of three N – H single bonds.
N
2s 2p 3H H+
1s 1s
N
H
H
H
By VB Theory,
One 2s(fully-filled)-1s(vacant) overlap leads to the formation of one N H dative bond.
N
2s 2p 3H H+
1s 1s
N
H
H
H
H++ N
H
H
H
H
(b) HCN
By Lewis model, the structure is H-CN one H-C single bond and
one CN triple bond.
By VB Theory,
C
Only 2 single bonds can be formed. Promotion of a 2s electron to a 2p
orbital.
2s 2p
C*
2s 2p
The overlap of one orbital (?) of C* with an 1s orbital of H gives the C-H single bond.
Overlaps of three orbitals (???) of C* with three 2p orbitals of N give the CN triple bond.
C*
2s 2p
N
2s 2p
H
1s
The 2s electrons on N are non-bonding electrons.
The energy released by forming a stronger triple bond outweighs the energy required for promoting an electron from a 2s orbital to a 2p orbital.
C*
2s 2p
N
2s 2p
H
1s
H C N
(c) SO2
By Lewis model, the three possible structures are
OS=O, O=SO, O=S=O
Most stable no separation of opposite formal
charges.
By VB Theory,
Only two single bonds can be formed. One 3p electron has to be promoted to a 3d orbital. Expansion of Octet.
S
3s 3p
By VB Theory,
S
3s 3p
S*
3s 3p
3d
octet expansion
Overlaps of two half-filled orbitals (??) of S* with two half-filled 2p orbitals of an oxygen atom give a S=O double bond.
A total of two S=O bonds are formed with two O atoms
2O
2s 2p
S*
3s 3p
3d
O
S
O
Non-bonding electrons : S* 3s2 ; O 2s2 and 2p2
2O
2s 2p
S*
3s 3p
3d
O
S
O
The energy released by forming of two stronger double bonds outweighs the energy required for promoting an electron from a 3p orbital to a 3d orbital.
S
3s 3p
S*
3s 3p
3doctet expansion
Bent’s Rule
In a molecule, smaller bond angles are formed between electronegative ligands
since the central atom, to which the ligands are attached, tends to direct
bonding hybrid orbitals of greater p character towards its more electronegative
substituents.
Structure of water illustrating how the bond angle deviates from
the tetrahedral angle of 109.5°.
The carbon atoms are directing sp3, sp2, and sp orbitals towards the
hydrogen substituents. This simple system demonstrates that hybridised
atomic orbitals with higher p character will have a smaller angle
between them.
Why Molecular Orbital (MO) TheoryLewis dot and valence bond theories do not always give satisfactory account for various properties of molecules.
For example, the dot and VB theory does not explain the fact that O2 is paramagnetic and has a double bond.
Dot and VB structures : O O : • O O •are unsatisfactory.
MO theory, different from VB in that MO theory considers the orbitals of the whole molecules. However the approach of linear-combination-of-atomic-orbitals (LCAO) is usually used.
There are other reasons, but it’s human nature to theorize. The theory is beautiful, and worth learning or teaching.
Limitations of Valence Bond Theory:
(i) It involves a number of assumptions.
(ii) It does not give quantitative interpretation of magnetic data.
(iii) It does not explain the color exhibited by coordination compounds.
(iv) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.(v) It does not make exact predictions regarding the tetrahedral and square planar structures of 4-coordinate complexes.(vi) It does not distinguish between weak and strong ligands.
Molecular Orbital Theory• Molecular orbitals result from the combination of atomic orbitals.
Since orbitals are wave functions, they can combine either constructively (forming a bonding molecular orbital), or destructively (forming an antibonding molecular orbital).
• Consider the H2 molecule, for example. One of the molecular orbitals
in this molecule, it is constructed by adding the mathematical functions for the two 1s atomic orbitals that come together to form this molecule. Another orbital is formed by subtracting one of these functions from the other
Bonding Molecular Orbital Theory
The bonding orbital results in increased electron density between the two nuclei, and is of lower energy than the two separate atomic orbitals.
Antibonding Molecular Orbital Theory
The antibonding orbital results in a node between the two nuclei, and is of greater energy than the two separate atomic orbitals.
Overlap of s & p Orbitals
+ -
+
+
+
+- -
- -
++
+
--
-
Sigma bonding orbitals
• From s orbitals on separate atoms
+ +
s orbital s orbital
+ ++ +
Sigma bondingmolecular orbital
Molecular Orbitals of the Second Energy Level
If we arbitrarily define the Z axis of the coordinate system for the O2 molecule as the axis along which the bond forms, the 2pz orbitals on the adjacent atoms will meet head-on to form a 2p bonding and a 2p* antibonding molecular orbital
Sigma bonding orbitals
• From p orbitals on separate atoms
p orbital p orbital
Sigma bondingmolecular orbital
The 2px orbitals on one atom interact with the 2px orbitals on the other to form molecular orbitals that have a different shape. These molecular orbitals are called pi (π ) orbitals because they look like p orbitals when viewed along the bond.
Pi bonding orbitals
• P orbitals on separate atoms
Pi bondingmolecular orbital
s-p mixing
Molecular Orbital Diagram
Molecular Orbital Diagram of H2
Molecular Orbital Diagram of N₂
Molecular Orbital Diagram of O2
Coordination Chemistry
Transition elements: partly filed d or f shells
What is a transition metal?
• Transition metals [TM’s] have characteristic properties – e.g. coloured compounds, variable oxidation states
• These are due to presence of an inner incomplete d or f-sub-shell
• Electrons from both inner d or f-sub-shell and outer s sub-shell can be involved in compound formation
What is a transition metal?
• Not all d block elements have incomplete d sub-shells – e.g. Zn has e.c. of [Ar]3d104s2, the Zn2+ ion ([Ar]
3d10) is not a typical TM ion– Similarly Sc forms Sc3+ which has the stable e.c of
Ar. Sc3+ has no 3d electrons
• For this reason, a transition metal is defined as being an element which forms at least one ion with a partially filled sub-shell of d-electrons.
Sc +3
Ti +1 +2 +3 +4
V +1 +2 +3 +4 +5Cr +1 +2 +3 +4 +5 +6Mn +1 +2 +3 +4 +5 +6 +7
Fe +1 +2 +3 +4 +5 +6
Co +1 +2 +3 +4 +5
Ni +1 +2 +3 +4
Cu +1 +2 +3
Zn +2
Cu is the only element which affords CuI compounds without p-acceptor ligands
TM complex: Variable valence
Complexes: Have metal ion (can be zero oxidation state) bonded to number of ligands.
Lewis acid = metal = center of coordination
Transition metals can act as Lewis acidLewis base = ligand = molecules/ions covalently bonded to metal in
complex
Coordination compoundCompound that contains 1 or more complexesExample
[Co(NH3)6]Cl3
[Cu(NH3)4][PtCl4]
[Pt(NH3)2Cl2]
• Ligands– classified according to the number of donor
atoms– Examples
• monodentate = 1• bidentate = 2• tetradentate = 4• hexadentate = 6• polydentate = 2 or more donor atoms
chelating agents
monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced, hence the idea of bite angle etc.
Teeth of a ligand ( teeth dent)
Chelate effect
Typical Ligands
Chelate Complex
• EDTA occupies 6 coordination sites, for example [CoEDTA]- is an octahedral Co3+ complex.
• Both N atoms (blue) and O atoms (red) coordinate to the metal.
Equilibrium log β ΔG ΔH /kJ mol−1 −TΔS /kJ mol−1
Cd2+ + 4 MeNH2 = Cd(MeNH2)42+ 6.55 -37.4 - 57.3 19.9
Cd2+ + 2 en = Cd(en)22+ 10.62 -60.67 - 56.48 - 4.19
The enthalpy term should be approximately the same for the two reactions. The difference between the two stability constants is due to the entropy term. In equation (1) there are two particles on the left and one on the right, whereas in
equation (2) there are three particles on the left and one on the right. This means that less entropy of disorder is lost when the chelate complex is
formed than when the complex with monodentate ligands is formed.
These data show that the standard enthalpy changes are indeed approximately equal for the two reactions and that the main reason for the greater stability of the chelate complex is the entropy term.It is clear that the chelate effect is predominantly an effect of entropy.
Cu2+ + en [Cu(en)]2+ (1)
Cu2+ + 2 MeNH2 [Cu(MeNH2)2]2+ (2)
ΔG = −RT ln K = ΔH – TΔS
The Chelate Effect• How to Make a Strong Complex
– Factors Effecting M—L Binding Strength = Molecular Organization
• Complementarity = sum of size, geometry, and electronic matching between the metal ion and the ligand(s)
– The individual components are simple and can be predicted or found experimentally
– Example: HSAB Theory predicts Fe3+/O2- is more complementary than Fe3+/S2- – Example: d8 Ni2+ should have good complementarity with cyclam– Complementarity is only the first step towards complex stability
• Constraint = the number and flexibility between ligand donor atoms– Topology = interconnectedness of donor atoms– Rigidity = how fixed in space donor atoms of the ligand are with respect to each
other– These constraint factors are more difficult to grasp than complementarity– Maximizing these factors can lead to huge increases in complex stability
Constraint and Binding Affinity
BindingAffinity
Constraint
Size Geometry
Electronics
Topology
Rigidity
Complementarity
CoCl3 . 6 NH3 ORANGE-YELLOWCoCl3.5NH3.H2O REDCoCl3.5NH3 PURPLECoCl3.4NH3 GREEN
Werner’s coordination theory
Measurements of the conductivity of aqueous solutions of the above complexes suggest:
CoCl3.6NH3 and CoCl3.5NH3.H2O complexes dissociate in water to give a total of four ions.
CoCl3.5NH3 dissociates to give three ions
CoCl3.4NH3 dissociates to give only two ions.
Werner explained these observations by suggesting that transition-metal ions such as the Co3+ ion have a primary valence and a secondary valence.
The primary valence is the number of negative ions needed to satisfy the charge on the metal ion. In each of the cobalt(III) complexes previously described, three Cl- ions are needed to satisfy the primary valence of the Co3+ ion.
The secondary valence is the number of ions of molecules that are coordinated to the metal ion. Werner assumed that the secondary valence of the transition metal in these cobalt(III) complexes is six.
The formulas of these compounds can therefore be written as follows.
[Co(NH3)63+][Cl-]3 orange-yellow
[Co(NH3)5(H2O)3+][Cl-]3 red
[Co(NH3)5Cl2+][Cl-]2 purple
[Co(NH3)4Cl2+][Cl-] green
The cobalt ion is coordinated to a total of six ligands in each complex, which satisfies the secondary valence of this ion.
Each complex also has a total of three chloride ions that satisfy the primary valence.
Some of the Cl- ions are free to dissociate when the complex dissolves in water. Others are bound to the Co3+ ion and does not dissociate.
Chelating agents:
(1) Used to remove unwanted metal ions in water.
(2) Selective removal of Hg2+ and Pb2+ from body when poisoned.
(3) Prevent blood clots.
(4) Solubilizing iron in plant fertilizer.
Crystal Field theory•Compounds of Transition metal complexes solution.
[Fe(H2O)6]3+
[Co(H2O)6]2+
[Ni(H2O)6]2+
[Cu(H2O)6]2+
[Zn(H2O)6]2+
The d-orbitals: the t2gset
the egset
dyz dxy dxz
dz2 dx2-y2
x x x
x x
zzz
zz
y y y
y y
Splitting of the d sub-shell in octahedral coordination
dyz dz2 dx2-y2
the three orbitals ofthe t2g set lie betweenthe ligand donor-atoms(only dyz shown)
the two orbitals of the eg set lie along theCartesian coordinates, and so are adjacentto the donor atoms of the ligands, whichraises the eg set in energy
z z z
blue = ligand donor atom orbitals the egsetthe t2g set
y y y
x x x
energyeg
t2gCo3+ ion
in gas-phase(d6)
Δ
Co(III) in octahedral
complex
3d sub-shell
d-shellsplit bypresenceof liganddonor-atoms
Splitting of the d sub-shell in an octahedral complex
The crystal field splitting parameter (Δ)
Different ligands produce different extents of splitting between the eg and the t2g levels. This energy difference is the crystal field splitting parameter Δ, also known as 10Dq, and has units of cm-1. Typically, CN- produces very large values of Δ, while F- produces very small values.
[Cr(CN)6]3- [CrF6]3-
eg eg
t2g
t2g
energy
Δ = 26,600 cm-1 Δ = 15,000 cm-1
• Color of the Complex depends on magnitude of • 1. Metal: Larger metal larger • Higher Oxidation State larger • 2. Ligand: Spectrochemical series• Cl- < F- < H2O < NH3 < en < NO2
- < (N-bonded) < CN-
• Weak field Ligand: Low electrostatic interaction: small CF splitting.• High field Ligand: High electrostatic interaction: large CF splitting.
Spectrochemical series: Increasing
High and low-spin complexes:
energy
eg eg
t2gt2g
low-spin d6
electrons fill the t2g level first. In this case the complex is diamagnetic
high-spin d6
electrons fill the whole d sub-shell according to Hund’s rule
The d-electrons in d4 to d8 configurations can be high-spin, where theyspread out and occupy the whole d sub-shell, or low-spin, where the t2glevel is filled first. This is controlled by whether Δ is larger than the spin-pairing energy, P, which is the energy required to take pairs of electrons with the same spin orientation, and pair them up with the opposite spin.
Δ > P Δ < P
Paramagnetic4 unpaired e’s
diamagneticno unpaired e’s
Electron Configuration in Octahedral Field• Electron configuration of metal ion:• s-electrons are lost first. • Ti3+ is a d1, V3+ is d2 , and Cr3+ is
d3 • Hund's rule:• First three electrons are in
separate d orbitals with their spins parallel.
• Fourth e- has choice:• Higher orbital if is small; High
spin• Lower orbital if is large: Low
spin.• Weak field ligands• Small , High spin complex• Strong field Ligands• Large , Low spin complex
High Spin Vs. Low Spin (d1 to d10)•Electron Configuration for Octahedral complexes of metal ion having d1 to d10 configuration [M(H2O)6]+n. •Only the d4 through d7 cases have both high-spin and low spin configuration.
Electron configurations for octahedral complexes of metal ions having from d1 to d10 configurations. Only the d4 through d7 cases have both high-spin and low-spin configurations.
energy
eg eg
t2gt2g
low-spin d5 ([Fe(CN)6]3-)electrons fill the t2g level first. In this
case the complex is paramagnetic
high-spin d5 ([Fe(H2O)6]3+)electrons fill the whole d sub-shell
according to Hund’s rule
For d5 ions pairing is usually very large, so these are mostly high-spin. Thus, Fe(III) complexes are usually high-spin, although with CN- Δ is large enough that [Fe(CN)6]3- is low spin: (CN- always produces the largest Δ values)
Δ > P Δ < P
Paramagnetic5 unpaired e’s
paramagneticone unpaired e
High and low-spin complexes of d5 ions:
[Fe(CN)6]3- Δ = 35,000 cm-1
P = 19,000 cm-1
[Fe(H2O)6]3+ Δ = 13,700 cm-1
P = 22,000 cm-1
energy
eg eg
t2gt2g
low-spin d6 ([Co(CN)6]4-)electrons fill the t2g level first. In this
case the complex is diamagnetic
high-spin d5 ([CoF6]3-)electrons fill the whole d sub-shell
according to Hund’s rule
For d6 ions Δ is very large for an M(III) ion such as Co(III), so all Co(III) complexes are low-spin except for [CoF6]3-.high-spin. Thus, Fe(III) complexes are usually high-spin, although with CN- Δ is large enoughthat [Fe(CN)6]3- is low spin: (CN- always produces the largest Δ values)
Δ >> P Δ < P
Paramagnetic4 unpaired e’s
diamagneticno unpaired e’s
High and low-spin complexes of some d6 ions:
[Co(CN)6]3- Δ = 34,800 cm-1
P = 19,000 cm-1
[CoF6]3- Δ = 13,100 cm-1
P = 22,000 cm-1
energy
eg eg
t2gt2g
low-spin d7 ([Ni(bipy)3]3+)The d-electrons fill the t2g level first,
and only then does an electronoccupy the eg level.
high-spin d7 ([Co(H2O)6]3+)electrons fill the whole d sub-shell
according to Hund’s rule
The d7 metal ion that one commonly encounters is the Co(II) ion. For metalions of the same electronic configuration, Δ tends to increase M(II) < M(III) < M(IV), so that Co(II) complexes have a small Δ and are usually high spin. The (III) ion Ni(III) has higher values of Δ, and is usually low-spin.
Δ > P Δ < P
Paramagnetic3 unpaired e’s
paramagneticone unpaired e
High and low-spin complexes of d7 ions:
Tetrahedral ArrangementExample:[Ni (CO)4]
Ni (0) {d10 system}
Square Planar Field
Square Planar ArrangementExample: [Ni(CN)4]2-
Ni+2 (d8 system)
Octahedral, Tetrahedral & Square Planar
•CF Splitting pattern for various molecular geometry
M
dz2dx2-y2
dxzdxy dyz
M
dx2-y2 dz2
dxzdxy dyz
M
dxz
dz2
dx2-y2
dxy
dyz
OctahedralTetrahedral Square planar
Pairing energy Vs. Weak field < PeStrong field > Pe
Small High SpinMostly d8
(Majority Low spin)Strong field ligandsi.e., Pd2+, Pt2+, Ir+, Au3+
When splitting of the d sub-shell occurs, the occupation of the lower energy t2g level by electrons causes a stabilization of the complex, whereas occupation of the eg level causes a rise in energy. Calculations show that the t2g level drops by 0.4Δ, whereas the eg level is raised by 0.6Δ. This means that the overall change in energy, the CFSE, will be given by:
CFSE = Δ(0.4n(t2g) - 0.6n(eg))
where n(t2g) and n(eg) are the numbers of electrons in
the t2g and eg levels respectively.
Crystal Field Stabilization Energy (CFSE):
The CFSE for some complexes is calculated to be:
[Co(NH3)6]3+: [Cr(en)3]3+
egeg
t2gt2g
Δ = 22,900 cm-1 Δ = 21,900 cm-1
CFSE = 22,900(0.4 x 6 – 0.6 x 0) CFSE = 21,900(0.4 x 3 – 0.6 x 0)
= 54,960 cm-1 = 26,280 cm-1
Calculation of Crystal Field Stabilization Energy (CFSE):
energy
The CFSE for high-spin d5 and for d10 complexes is calculated to be zero:
[Mn(NH3)6]2+: [Zn(en)3]3+
egeg
t2gt2g
Δ = 22,900 cm-1 Δ = not known
CFSE = 10,000(0.4 x 3 – 0.6 x 2) CFSE = Δ(0.4 x 6 – 0.6 x 4)
= 0 cm-1 = 0 cm-1
Crystal Field Stabilization Energy (CFSE) of d5 and d10 ions:
energy
For M(II) ions with the same set of ligands, the variation of Δ is not large. One can therefore use the equation for CFSE to calculate CFSE in terms of Δ for d0 through d10 M(II) ions (all metal ions high-spin):
Ca(II) Sc(II) Ti(II) V(II) Cr(II) Mn(II) Fe(II) Co(II) Ni(II) Cu(II) Zn(II) d0 d1 d2 d3 d4 d5 d6 d7 d8 d9 d10
CFSE: 0 0.4Δ 0.8Δ 1.2Δ 0.6Δ 0 0.4Δ 0.8Δ 1.2Δ 0.6Δ 0
This pattern of variation CFSE leads to greater stabilization in the complexes of metal ions with high CFSE, such as Ni(II), and lower stabilization for the complexes of M(II) ions with no CFSE, e.g. Ca(II), Mn(II), and Zn(II). Thevariation in CFSE can be compared with the log K1 values for EDTAcomplexes on the next slide:
Crystal Field Stabilization Energy (CFSE) of d0 to d10 M(II) ions:
CFSE as a function of no of d-electrons
00.20.40.60.8
11.21.4
0 1 2 3 4 5 6 7 8 9 10 11
no of d-electrons
CFSE
in m
ultip
les
of Δ
.
Crystal Field Stabilization Energy (CFSE) of d0 to d10 M(II) ions:
Ca2+ Mn2+ Zn2+
double-humpedcurve
Ni2+
18 Electron "Rule"Organic compounds follow the 8 electron rule: there can only be a maximum of 8 valence electrons around a carbon center. The vast majority of stable diamagnetic organometallic compounds have 16 or 18 valence electrons due to the presence of the five d-orbitals which can hold 10 more electrons relative to C, O, N, etc. Electron counting is the process of determining the number of valence electrons about a metal center in a given transition metal complex. To determine the electron count for a metal complex:1) Determine the oxidation state of the transition metal center(s) and the metal centers resulting d-electron count. To do this one must:
a) note any overall charge on the metal complexb) know the charges of the ligands bound to the metal
center (ionic ligand method)c) know the number of electrons being donated to the metal
center from each ligand (ionic ligand method)2) Add up the electron counts for the metal center and ligandsComplexes with 18 e- counts are referred to as saturated, because there are no empty low-lying orbitals to which another incoming ligand can coordinate. Complexes with counts lower than 18e- are called unsaturated and can electronically bind additional ligands.
Carbonyl complexes
The carbonyl ligand forms a huge number of complexes with metal ions, most commonly in low oxidation states, where it binds to the metal through its C-donor, as in the complexes below, where all the metal ions are zero-valent:
[Ni(CO)4] [Fe(CO)5] [Cr(CO)6] Td TBP (D3h) Oh
Metal Carbonyl compounds
Metal carbonyls form one of the oldest (and important) classes of organometallic complexes. Most metal carbonyls are toxic!
One might wonder why in the above complexes Ni(0) has four C≡O groups attached to it, Fe(0) five C≡O, and Cr(0) six C≡O. A very simple rule allows us to predict the numbers of donor groups attached to metal ions in organometallic complexes, called the eighteen electron rule. The latter rule states that the sum of the d-electrons possessed by the metal plus those donated by the ligands (2 per C≡O) must total eighteen:
[Ni(CO)4] [Fe(CO)5] [Cr(CO)6]
Ni(0) = d10 Fe(0) = d8 Cr(0) = d6
4 x CO = 8 5 x CO 10 6 x CO = 12
18 e 18e 18e
Carbonyl complexes and the 18-electron rule
Formal oxidation states are all zero.
To obey the 18-electron rule, many carbonyl complexes are anions or cations, as in:
[V(CO)6]- [Mn(CO)6]
+ [Fe(CO)4]2-
V(0) = d5 Mn(0) = d7 Fe(0) = d8
6 CO = 12e 6 CO = 12e 4 CO = 8e 1- = +1e 1+ = -1e 2- = 2e
= 18e = 18 e = 18e
Carbonyl complexes and the 18-electron rule
Formal oxidation Formal oxidation Formal oxidationstate = V(-I) state = Mn(I) state = Fe(-II)
[NOTE: In applying the 18-electron rule, metal ions are always considered to be zero-valent, not the formal oxidn. state]
Metal ions in biological system
Periodic Distribution of Biologically Important Elements
The role of the metal center in biomolecules
Metal ions can have structural roles, catalytic roles, or both. Metals that have catalytic roles will be present at the active
site of the biomolecule which will likely be a metalloprotein (a metalloenzyme).
The reactivity of a metalloprotein is defined by the nature of the metal, particularly its electronic structure and oxidation state.
The electronic structure and spin state of a metal center defines its chemical reactivity as a redox center (i.e. it controls its efficiency at accepting or donating electrons)
The electronic structure of a metal center defines its chemical reactivity as a Lewis acid (electron-pair acceptor) which enables it to bind ligands (O2, N2, CO ..) for transport, activation and reaction.
Biological Roles of Metallic Elements.StructuralSkeletal roles via biomineralizationCa2+, Mg2+, P, O, C, Si, S, F as anions, e.g. PO4
3-, CO32-.
Charge neutralization.Mg2+, Ca2+ to offset charge on DNA - phosphate anions Charge carriers: Na+, K+, Ca2+ Transmembrane concentration gradients ("ion-pumps and channels")Trigger mechanisms in muscle contraction (Ca). Electrical impulses in nerves (Na, K)Heart rhythm (K).Hydrolytic Catalysts: Zn2+ , Mg2+ Lewis acid/Lewis base Catalytic roles. Small labile metals.Redox Catalysts: Fe(II)/Fe(III)/Fe(IV), Cu(I)/Cu(II), Mn(II)/Mn(III)/(Mn(IV),Mo(IV)/Mo(V)/Mo(VI), Co(I)/Co(II)/Co(III)
Transition metals with multiple oxidation states facilitate electron transfer - energy transfer. Biological ligands can stabilize metals in unusual oxidation states and fine tune redox potentials. Activators of small molecules. Transport and storage of O2 (Fe, Cu), Fixation of nitrogen (Mo, Fe, V)Reduction of CO2 (Ni, Fe)
Transition Metals in Biomolecules Iron. Most abundant metal in biology, used by all plants and animals including bacteria. Some roles duplicated by other metals, while others are unique to Fe. Iron use has survived the evolution of the O2 atmosphere on earth and the instability of Fe(II) with respect to oxidation to Fe(III).Zinc.Relatively abundant metal. Major concentration in metallothionein (which also serves as a reservoir for other metals, e.g. Cd, Cu, Hg). Many well characterized Zn proteins, including redox proteins, hydrolases and nucleic acid binding proteins. CopperOften participatse together with Fe in proteins or has equivalent redox roles in same biological reactions. Reversible O2 binding, O2 activation, electron transfer, O2
- dismutation (SOD). Cobalt.
Unique biological role in cobalamin (B12-coenzymes) isomerization reactions.ManganeseCritical role in photosynthetic reaction centers, and SOD enzymes.MolybdenumCentral role in nitrogenase enzymes catalyzing N2 NH3, NO3
- NH3
Chromium, Vanadium and NickelSmall quantities, uncertain biological roles. Sugar metabolism (Cr); Ni only in plants and bacteria (role in CH4 production) and SOD enzymes.
Mg
• Photosynthesis,
• ATP hydrolysis, • Phosphate group transfer reactions (i.e kinase reactions),
• Structure formation, stabilizing DNA and RNA, construction of cell membranes,• DNA polymerase enzyme catalyzing the transcription of DNA.
• Enzymes like cytochrome c oxidase (also Fe), amine oxidase, ascorbic acid oxidase, tyrosinase • Electron transport proteins like plastocyanin, azurin, stellacyanin • Oxygen transport protein hemocyanin (in lower forms of life)• Storage protein ceruloplasmin.
Cu
Fe
• O2 uptake proteins (i.e hemoglobin, myoglobin, hemerythrin);
• Oxygenase enzymes; • Catalase, peroxidase, cytochrome P-450; • Aconitase, in cytochrome c oxidase (also cu); • Nitrogenase (also Mo), in hydrogenase; • Electron transport proteins like Fe-s protein, cytochromes; • In storage protein ferritin; about 70 Fe-proteins are well known
• Lighter elements are more abundant in general and therefore utilized more. 3d metals, rather than 4d, are used as catalytic centers in metalloenzymes.
• Why has Mo (4d) rather than Cr (3d) been utilized more biologically?Although Mo is rare in the earth’s crust, Mo is the most abundant transition metal in sea water as MoO4
- has fairly high solubility in water. Better correlation exists between the abundance of elements in in human body and in sea water than between the human body and the earth's crust. Taken as evidence for the oceans as the site of evolution of life.
• Despite the high abundance of Si, Al and Ti (the 2nd, 3rd and 10th most abundant elements on earth). Why are they are not utilized biologically?
• Because of the insolubility of their naturally occurring oxides (SiO2, Al2O3, TiO2) under physiological conditions. A lower oxidation state is unavailable for Si and Al and unstable for Ti in an aerobic environment and is readily oxidized to Ti(IV)
• Why has iron been used so widely in biology although Fe3+, its most stable
oxidation state, is highly insoluble at pH 7Complex biological mechanisms have been developed to accommodate the low solubility of Fe(OH)3 (Ksp = 1 x 10-38) ~ pH 7, and take advantage of its high "availability".
• Co2+ and Zn2+ have very similar coordination chemistry and ionic size and can be interchanged in many Zn enzymes without loss of activity. Why is Co not utilized more biologically?Zn is much more abundant and therefore has been utilized more.
• Why has cobalt been given an essential role in cobalamins despite its very low availability?
• The unique properties of cobalt (e.g. its oxidation states, redox potentials and coordination chemistry) is needed to achieve essential functions of B12
coenzymes.
