Transcript This chapter will deal specifically the first row transition elements

CH# 17
Coordination Chemistry
Transition Metals
 Transition metals show similarities within a period
and a group, different than representative elements
 Differences can be attributed to the fact that
when electrons are added across a period the
valence electrons are not effected.
 Therefore group designations are not important
here
 Behave as metals, strong metallic character
Transition Metals
• Some differences
– Melting point, Tungsten melts 3400°, while mercury 39°C
– Some soft, like sodium that can be cut with a butter
knife
– Reactivity
• Some spontaneously react with oxygen like iron,
which flakes off
• Others react with oxygen to make a colorless tight
fitting oxide, such as chromium, thus protecting the
surface
• Some metals are inert to oxygen such as gold, silver
and platinum
Transition Metals
Ionic compound formation
 More than one oxidation state is often
observed
 Cations, often are complexes, which we
will discuss later in this chapter
 Most compounds are colored, since
complexes absorb visible light
 Many compounds are paramagnetic
This chapter will deal specifically the first
row transition elements
Transition Elements
Electron Configurations
Exceptions to the AUFBAU principle
 Cr prefers a half full d as opposed to a full 4s, thus
4s13d5
 Copper prefers a full 3d as opposed to a full 4s,
thus 4s13d10
 This half filled, or filled d orbital, is used most of
the time to explain this, but other transition
metals do not follow this trend.
Electron Configurations
Many texts explain AUFBAU exceptions of chromium
and copper as a half full sublevel are more stable
than a full 4 s sublevel, or for copper that a full dsublevel is more stable than a half full 4s
 Why is this not the case in periods below?
 The 4 s and the 3 d orbitals are of about the
same energy or nearly degenerate. Perhaps
there is a larger repulsive force in the 4s than
in 3d orbitals.
 I do not think any one knows, but it is good to
think and create right?
Electron Configurations
4d and 5d Transition Series
– See the size relation on next slide
• Decrease in size as we go from left to right,
stopping when the d is half full
• Significant drop in size going from 3d to 4d,
but 4d and 5d remain about the same size
–Called Lanthanide contraction
–Adding f electrons below the d and the
valence shell shel electrons (shielding)
–Thus the effect of the increasing size by
adding another shell of electrons, which is
normally in transition and representative
elements, is offset by the shielding of the
added f electrons
Transition Element Sizes
Oxidation States and IE
 See common oxidation states on Next slide
 The maximum oxidation state for each
transition element going across the row is what
we would get by losing both 4s and 3d
electrons, toward the end only 2+ is observed,
the explanation is that as the effective charge
increases thus holding the d electrons tighter.
 Reducing ability, decreases from left to right
Transition Metal Oxidations #’s
Sc
V
3
2
3
4
5
Ti Cr Mn Fe Co Ni Cu Zn
2
3
4
4
2
3
5
6
2
3
4
2
3
6
7
6
2
3
2
3
1
2
3
2
Ionization Energies
Red dot- First
ionization energy
(removing 4s e)
Blue dot-third
ionization energy
removing 3d
electron, closer to
nucleus, thus
more tightly held
First-row Transition Metals
Scandium
 Rare element most always +3 oxidation state, ie
ScCl3, Sc2O3
 Chemistry of scandium resembles the lanthanides
 Colorless compounds
 Diamagnetic
 Color and magnetic properties are due to d
electron, Sc has no d electrons
First-row Transition Metals
Titanium
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
Found in the earths crust (0.6%)
Low density and high strength
Fairly inert, and is used in pipes
TiO2 is a very common white pigment
Common oxidation state is +4
First-row Transition Metals
Vanadium
 Found in the earth’s crust about 0.02%
 Common oxidation state is +5
 Since vanadium contains d electrons solutions are
colored
 VO2+ is yellow with V in the +5 oxidation state
 VO2+ is blue with V in the +4 oxidation state
 V3+ is blue-green with V in +3 oxidation state
 V2+ is violet with V in +2 oxidation state
First-row Transition Metals
Chromium
– Rare, but important industrial chemical
– Chromium oxide is colorless, tuff, and holds to the
metal strongly, almost invisible
– Chromium compounds in solution are also colored
since they contain d electrons
– Common oxidation states are +2, +3 and +6
– Chromium VI is an excellent oxidizing agent! Why?
• Strength increases as acidity increases
• Chromerge very good glassware cleaning agent
– What would we predict for Cr metal?
– Cr6+ in the form of dichromate ion usually reduces to
the +3 state
First-row Transition Metals
• Iron
– Is the most abundant heavy metal (4.7%) in earth’s
crust, Why?
– Common oxidation states +2 and +3
– Iron solutions are colored since they contain d
electrons
• Cobalt
– Relatively rare
– Hard bluish-white metal
– Common oxidation states are +2 and +3
– Oxidation states +1 and +4 are also known
• Typical color is rose color
First-row Transition Metals
 Nickel
 Most always the +2 oxidation state
 Sometimes +3 oxidation state
 Emerald green colored solutions
First-row Transition Metals
• Copper
– Quite common, as sulfides, arsenides, chlorides
and carbonates
– Great electrical conductor second only to silver
– Widely used in plumbing
– Found in bronze and brass
– Not highly reactive will not reduce H+
– Slowly oxides in air, producing a green oxide
– Common oxidation state +2, +1 is also known
– Aqueous solution are bright Royal blue
– Quite toxic, used to kill bacteria
– Paint often contains copper so algae do not grow
on the paint
First-row Transition Metals
Zinc
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Quite common in earths crust, usually as ZnS
Great reducing agent, quite reactive
Oxidation state of +2
Used to galvanize steel
Coordination compounds
 Transition metals form coordination compounds
 Transition metals contain a complex ion attached to
ligands via coordinate covalent bonds
 Coordination compounds are usually colored and
paramagnetic
Coordination compounds
Complex ions, usually inside [
]
 Transition coordinately bonded to Lewis
bases, the metal is acting as a Lewis
acid
 Example [CoCl(NH3)5]2+ this cation can
combine with anions to balance the
charge, thus forming a salt
 Ligands are the groups of atoms bonded
with a coordinate covalent bond to a
transition metal, or a transition metal ion.

Coordinate Covalent Bonding
Coordinate Covalent Bonding
Coordinate Covalent Bonding
Coordinate Covalent Bonding
Coordination compounds
Alfred Werner was the father of
coordination chemistry
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
Alfred Werner called the salt formation the
primary valence
The secondary valence is the formation of
the complex ion itself
The compound above has a secondary
valence of 6, since it combines with 6
ligands


The primary valence is +2 since that is what
needs to be neutralized with anions.
Now days the secondary valence is called the
coordination number and the primary valence
is called the oxidation state
Aqueous Solutions of Metal Ions
Coordination Compounds
 The number of coordinate covalent bonds
formed by the metal ion and the ligands
 Variance of 2-8, with 6 being most common.
 Geometrical Shape
 Ligands = 2, then linear
 Rare for most metals
 Common for d-10 systems (Cu+, Ag+, Au+, Hg2+)
 Ligands = 3, Trigonal planar
 Rare for most metals
 Is known for d-10 systems (example HgI3-)
Coordination Compounds
•
Geometrical Shape
 Ligands = 4, then tetrahedral, or square planar
 Tetrahedral structure is observed for nontransition
metals, BeF42- and d-10 inons such as ZnCl42-,
FeCl4-, FeCl42 Square planar is found with second and third row
transition metals with d-8 Rh+, Pd2+
-Ligands = 5
 trigonal bipyramid
 square pyramidal
Coordination compounds
• Geometrical Shape
− Ligands = 6, then octahedral and prismatic
(rare)
− Ligands = 7 Relatively uncommon, pentagonal
Second and third row transition metals,
lanthanides , and actinides
− Lignads = 8, relatively common for larger
metal ions, common geometry antiprism and
dodecahedron
− Lignads = 9 larger metal ions, geometry
tricapped trigonal prism [Nd(H2O)9]3+
The Ligand
Arrangements for
Coordination
Numbers 2, 4,
and 6
Ligands
 Atoms attached to a transition metal via
coordinate covalent bonds
 They are Lewis bases, since they donate a pair
of electrons to the transition metal.
 Ligands are classified relative to how many
attachments to the metal
 Monodentate forms one bond to a transition metal
 Lignads forming more than bond are called
chelating ligands, or chelates
Ligands
 Ligands are classified relative to how many
attachments to the metal
 Bidentate, a chelating agent, forms two
bonds, examples:
 Oxalate
 Ethylenediamine
 Polydentate forms more than two bonds.
 Diethylenetriamine
 Ethylenediaminetetraacetic acid
Ligands
 EDTA is used to remove lead from animals
 More complicated ligands are found in
biological compounds
 EDTA is used as a preservative to tie up
substances that could catalyze
decomposition of food products
Ethylenediamine
Ethylenediamminetetracidic acid
Coordination
of EDTA with
+
a 2 Metal Ion
Nomenclature
 Cationic species named before anionic species
 Within a complex, the ligands are named first in
alphabetical order followed by the metal atom
 the names of anionic lignads end in the suffix -o chloride ----->chloro
 cyanide ----->cyano
 oxide
----->oxo
 Hydroxide -->hydroxo
 Oxalate------>oxalato
 Sulfate ------>Sulfato
 Nitrate ------>Nitrato
Nomenclature
 lignads whose names end in -ite or ate become
-ito and ato respectively
 carbonate ----> carbonato
 oxalate-----> oxalato
 thiosulfate ----> thiosulfato
 Sulfite
-----> sulfito
 neutral lignads are given the same names as
the neutral molecule
 exceptions, ammonia (ammine), water
(aqua), carbon monoxide (Carbonyl), and NO
(nitrosyl)
Nomenclature
 When there is more than one of a particular ligand,
number is specified by di, tri, tetra, penta, hexa, and
so forth. when confusion might result, the prefixes
bis, tris and tetrakis are employed e.g.
bis(ethylenediaminne)
 negative (anionic) complex ions always end in the
suffix -ate
 aluminum -----> aluminate
 chromium -----> chromate
 manganese ------> manganate
 coblat ------> cobaltate
 For some metals the -ate is appended to the Latin
stem always appears with
Nomenclature
 the common English name for the element
 iron ----> ferr ------> ferrate
 copper ---> cupra -----> cuprate
 lead ----> plumb -----> plumbate
 silver ---> argent ----> argentate
 gold ---> aur ----> aurate
 tin ----> stann -----> stannate
 the oxidation number of the metal in the
complex is written in roman numerals within
parentheses following the name of the metal
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
 tetramminedichlorocobalt(III) ion
 sodium hexanitratochromate(III)
 diamminesilver(I) ion
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
 tetramminedichlorocobalt(III) ion
 sodium hexanitratochromate(III)
 diamminesilver(I) ion
[Ni(CN)4]2-
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
 tetramminedichlorocobalt(III) ion
 sodium hexanitratochromate(III)
 diamminesilver(I) ion
[Ni(CN)4]2[CoCl2(NH3)4]+
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
 tetramminedichlorocobalt(III) ion
 sodium hexanitratochromate(III)
 diamminesilver(I) ion
[Ni(CN)4]2[Co(NH3)4Cl2]+
Na3[Cr(NO3)6]
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
[Ni(CN)4]2 tetramminedichlorocobalt(III) ion
[CoCl2 NH3)4]+
 sodium hexanitratochromate(III)
Na3[Cr(NO3)6]
 diamminesilver(I) ion
[Ag(NH3)2]+
Nomenclature
 Formula writing
 Metal is first, followed by anions, then neutral
molecules
 If two or more anions or neutral molecules are
present, then use alphabetical order.
 Nomenclature Examples
 tetracyanonickelate(II) ion
[Ni(CN)4]2 tetramminedichlorocobalt(III) ion
[CoCl2(NH3)4]+
 sodium hexanitratochromate(III)
Na3[Cr(NO3)6]
 diamminesilver(I) ion
[Ag(NH3)2]+
Nomenclature
Name the following:
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[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3
K4[Mn(CN)6]
K[PtCl5 (NH3)]
[Cu(en)(NH3)2][Co(en)Cl4]
[Pt(en)2Br2](ClO4)2
Nomenclature
Name the following:


[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3
tris(ethylenediamene)chromium(III) chlorate
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K4[Mn(CN)6]
K[PtCl5(NH3)]
[Cu(en)(NH3)2][Co(en)Cl4]
[Pt(en)2Br2](ClO4)2
Nomenclature
Name the following:
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[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3
trisethylenediamenechromium(III) chlorate

K4[Mn(CN)6] Potassium hexacyanomanganate(IV)
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K[PtCl5(NH3)]
[Cu(en)(NH3)2][Co(en)Cl4]
[Pt(en)2Br2](ClO4)2
Nomenclature
Name the following:
[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3 trisethylenediamenechromium(III)
chlorate
K4[Mn(CN)6] Potassium hexacyanomanganate(IV)
K[PtCl5(NH3)] Potassium
monoaminepentachloroplatinate(IV)
[Cu(en)(NH3)2][Co(en)Cl4]
[PtBr2(en)2](ClO4)2
Nomenclature
Name the following:
[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3 trisethylenediamenechromium(III)
chlorate
K4[Mn(CN)6] Potassium hexacyanomanganate(IV)
K[PtCl5(NH3)] Potassium
triaminpentachloroplatinate(IV)
[Cu(en)(NH3)2][Co(en)Cl4]
ethylenediaminediaminecopper(II)
tetrachloroethylenediaminecobaltate(II)
[Pt(Br2en)2](ClO4)2
Nomenclature
Name the following:
[Ni(H2O)6]Cl2
hexaaquanickel(II) chloride
[Cr(en)3](ClO3)3 tris(ethylenediamene)chromium(III)
chlorate
K4[Mn(CN)6] Potassium hexacyanomanganate(IV)
K[PtCl5 (NH3)] Potassium
monoaminpentachloroplatinate(IV)
[Cu(en)(NH3)2][Co(en)Cl4]
ethylenediaminediamincopper(II)
tetrachloroethylenediaminecobaltate(II)
[Pt(en)2Br2](ClO4)2
bis(ethylenediamine)dibromoplatinum(IV)
perchlorate
Some Classes of Isomers
Coordination Isomers
Structural (constitutional) isomers

Definition- Different compounds of the same
formula.
 Types of structural (constitutional) isomers
 Ionization isomerism

 [Cr(NH3)SO4]Cl ppts AgCl when silver nitrate is
added [Cr(NH3Cl)]SO4 ppts barium sulfate when
barium is added
 Hydrate isomers differ in the placement of
water
 Coordination isomers differ in the placement
of ligands between the metal atoms
[CuBr4][PtCl4] or [CuCl4][PtBr4]
Coordination Isomers
Types of structural (constitutional) isomers
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Linkage isomers differ by the atom that is
coordinated to the metal O-N-O-M O2N-M, bond
to oxygen, or bond to nitrogen
Stereoisomerism (octahedral use models)
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
Definition, Same formula, same attachment of
atoms, but atoms are in different volumes of
space
 geometrical isomers (cis and trans)
 cis-[Co(NH3)4Cl2]+ chlorides on same side
 trans- chlorides on the opposite side
As a Ligand,
NO2- can Bond
to a Metal Ion
(a) Through a
Lone Pair on
the Nitrogen
Atom or (b)
Through a Lone
Pair on One of
the Oxygen
Atoms
(a) The cis
Isomer of
Pt(NH3)2Cl2
(b) The
trans
Isomer of
Pt(NH3)2Cl2
The Compound [Co(NH3)4Cl2]Cl has cis
and trans Isomers
Optical Isomers
Optical isomers(nonsuperimposable mirror
images)
 Chiral molecule-nonsuperimposable on it’s mirror
image
 Enantiomers-a pair of nonsuperimposable mirror
images
 Ploarimeter-an instrument that measures the
rotation of polarized light by an optically active
compound
 Dextrorotatory-rotation of polarized light clockwise
 Levorotatory-rotation of polarized light
counterclockwise
 Racemates- a 50/50 mixture of enantiomers (no
rotation of polarized light)
Optical Isomers
A human hand exhibits a nonsuperimposable mirror
image. Note that the mirror image of the right hand
(while identical to the left hand) cannot be turned in
any way to make it identical to (superimposable on) the
actual right hand.
Optical Isomers
Isomers I and II of CO(en)33+ are Mirror Images that
Cannot be Superimposed
Optical Isomers
(a) Superimposable. (b) Not Superimposable
Polarized Light
Polarized Light
Rotating the Plane of Polarization of Light
Valence Bond Model

Valence bond (VB) approach (Localized
Model) relative to octahedral systems


Overlay of the atomic orbitals of the metal and
the ligands
Since the ligands normally do not possess single
electrons, then a pair of electrons from the
ligand, must overlap with the empty orbitals of
the metal
Valence Bond Model
3d
4s
4p
Consider for example the blue-violet [Cr(H2O)6]3+
complex
 This is a 3d3 system, thus the 6 electron pairs
from the water will occupy the d2sp3 hybrid
(note the 4s and 4p orbitals are used here)
 The 4d orbitals are not used in this case, since
the 3d orbitals are lower in energy, and the
bonds formed are stronger
 The orbital notation shows three single
electrons, verified by Gouy Balance
measurements.
Gouy Balance
Valence Bond Model
Consider next the [Ni(H2O)6]2+
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
4s

This is a d8 system, there are no empty 3d
orbitals
The 6 ligands, then will occupy the 4th
energy shell sp3d2
3d
4p
4d
The orbital notation shows two unpaired
electrons, verified by experiment
Valence Bond Model
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
When 3d orbitals are employed, the
system is referred to as an inner
orbital complex; where if the 4d
orbitals are used, then the system is
referred to as an outer orbital
complex.
In the two previous examples there
was no choice where the electrons
are placed.
Coordination Compound Bonding
Co3+ is an ion that can show either
inner orbital, or an outer orbital
complexes
 Co3+ is d 6 system
 Pairing up the 6 electrons, will produce
an inner orbital d2sp3, which is
diamagnetic
 If the d electrons are not paired, then
an outer orbital sp3d2 complex is
formed, with 4 unpaired electrons and
paramagnetic
Valence Bond Model
To pair, or not to pair, that is the question
 This question arises for d 4,5,6 systems
 Failure to pair produces outer orbital systems
 Two factors to consider
 Stronger bonds are formed from 3d
orbitals, than 4d orbitals
 Pairing means putting two electrons in the
same orbital. A higher energy system for
sure (electron repulsion), but the outer
orbital system does not require pairing
electrons
Valence Bond Model
If the formation of bonds releases more energy
than the pairing energy, then the inner orbital
complex is preferred; if not, then the outer
orbital complex is favored
Most first row transition elements when combined
with ligands tend to favor inner orbital
complexes for d4 or d6 systems, with the
exception of the ligands H2O and F-, which usually
prefer outer orbital complexes
Valence Bond Model
The d-5 system produced the half filled
system (chromium), which is weird.


It is hard to disturb this stability, thus paring
usually does not happen, thus these systems
prefer outer orbital complexes with most
complexes, but CN- is an exception
Valence Bond Model
Valence bond approach and other geometries

Consider [Ni(CN)4]2- which is square planar and
diamagnetic
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Here Ni2+ is a d-8 system
Here pairing will create an empty d orbital, thus
allowing the dsp2 hybrid orbital system to form
Cyanide, is a strong ligand, as it was in the octahedral
system
The [CoCl4]2- complex forms the tetrahedral
geometry


This is a d-7 system
Tetrahedral here and sp3
Valence Bond Model
One of the most striking physical properties of
the coordination compounds is their color, and
the valence bond theory does little to explain
color.
Crystal Field Splitting Theory
Crystal field splitting theory (CF)



Developed by physics to explain impurities in
crystal lattices
Electrostatic bonds no coordinate covalent
bonds
Ligands are anions or polar particles
Crystal Field Splitting Theory
Another modified version of molecular
orbital theory





Organizes the d orbitals in order of increasing
energy
Organization of energy depends on the geometry
of the complex ion
Consider the geometry of the d orbitals page 967
As a ligand approaches in an octahedral complex,
the nonbonding electrons will repel electrons
found in the dz2 and dx2-y2, thus splitting the
potential energy of the 5 d orbitals, this is where
the name Crystal Field Splitting theory comes
from
Crystal Field Splitting
Theory
Crystal Field Splitting Theory
The average energy of the d orbitals is
not altered
The energy difference is called Δo
where the o means octahedral




The two d orbitals of higher energy are
called eg while the three lower energy
orbitals are called t2g
The eg increases in energy by 0.6, and the
t2g decreases in energy by 0.4, thus total
energy change is zero
Crystal Field Splitting
CF Energy Diagram
Octahedral and Tetrahedral Splits
Some d Systems
Some d Systems
Some d Systems
Some d Systems
Crystal Field Splitting
The energy difference is called Δo where the o
means octahedral



Referred to as the crystal field splitting
Absorption of light corresponds to delta, the
greater the difference the more blue in color
light is absorbed, all other colors are reflected
The energy absorbed is related to the wave
length
Crystal Field Splitting
Placement of electrons into these 5
psudo-molecular orbitals depends on
the magnitude of Δo and the pairing
energy P



If Δo < P, then the next electron goes
into the eg orbital, creating a high spin
case
If Δo > P, then the next electron goes
into t2g creating low spin case
Examples With Pairing Energy
Crystal Field Splitting
Factors affecting magnitude of Δo
 Charge on metal ion



Increasing charge causes radius to decrease,
thus ligands are more strongly attached, thus
increasing Δo
The Δo for a tripositive ion is about 50% when
compared to a dipositive ion
Principle quantum number



With the same charge and same ligands, then as
we travel down a group then Δo increases
This effect is due to the larger radius of the
metal ion
Repulsive ligand forces are important in smaller
metal ions
Crystal Field Splitting
Nature of the ligands



For ligands of the same group, the Δo
decreases as the size of the ligand
increases
Smaller more localized charges interact
more strongly with the d orbitals of the
metal ion.
Small neutral ligands with a localized pair
of electrons, i.e. NH3, gives larger than
expected Δo, when compared to a spherical
ligand such as F-, that has unlocalized
electrons
Crystal Field Splitting


If the ligands cause a large splitting
(large value of Δ) then the electrons
will fill the lower t2g orbital first, thus
minimizing the single electrons (strong
field case) (low spin)
If the ligands cause a small splitting
(low value of Δ) then the electrons will
fill the t2g one at a time and then fill
the eg orbitals one at a time (Weak
field case) (high spin)
Low Spin d-4 System
High Spin d-4 System
Crystal Field Splitting



Only d 4,5,6,7 have a choice of high spin,
or low spin
This model explains magnetic and color
properties of complexes.
Color chart or color wheel
Color Wheel
Visible Spectrum
Crystal Field Splitting
Magnitude of splitting


Magnitude of ΔO depends on the charge of the
central metal, the higher the greater ΔO


For example NH3 is weak field with Co 2+ and strong
field with Co 3+
As charge increases the ligands are drawn closer to
the metal, thus the closer the greater the splitting
Crystal Field Splitting



Depends on polarity, size, etc.
I-<Br-<Cl-<acetate<F-<OH-<oxalate<H2O<SCN<NH3<en<NO2-<CN-≈CO
Iodine is the smallest splitting
Crystal Field Splitting
Cobalt here can have two possibilities





Depends what the ligands are
High spin same as outer
Low spin same as inner
If iodine is used we have high spin

Because delta is not large
Crystal Field Splitting
Bond formation overcomes small delta


If aqua is used we have the low spin case




All paired
Diamagnetic
Different colors because delta is different
The possibility of high spin or low spin
exists when


D=4,5,6 or 7, Choice between High and Low
spin
d 0,1,2,3,8,9,10 High spin only
Crystal Field Splitting
Crystal Field Stabilization Energies




If the t2g populated, then stability is increased, since it is
lower potential energy
Stabilization can be calculated by multiplying the number
of electrons in the t2g orbital by 0.4Δ
If a combination occupied by t2g and eg then subtract 0.4
t2g from 0.6 eg for stabilization energy
Wave numbers




Energies obtained by spectroscopic measurements are
oftern given in units of wave numbers (cm-1)
Wave number is the reciprocal of the wavelength of the
corresponding electromagnetic radiation expressed in cm
cm-1 = 11.96 j
Crystal Field Splitting
Color and the Colors of Complexes
Two types of mixing colors
 Additive, occurs when colored lights are
superimposed on each other
 Subtractive, occurs when colored paints
are mixed with each other
Crystal Field Splitting
Additive Mixing (light beams)
 For additive mixing a primary color is
defined as any three colors that
produce white light
 Examples:
R+G+B=W
 Secondary colors are those that are
produced by combining two primary
colors
 Examples:
R+G=Y
R+B=M
Crystal Field Splitting
Additive and Subtractive Mixing
Complex Solutions
Subtractive Mixing
 Some wave lengths of white light are
removed from absorption (promoting
electrons to higher levels)
 The reflected light, does not contained
the absorbed colors, thus has a color due
to the absence of another color.
 Here primary colors are M,Y, and BG;
while secondary colors are G, B, and R.
 If a material absorbs all three primary
colors, then there in no light left to be
reflected, thus black.
Subtractive Mixing
 If a material absorbs one color,
primary or secondary, the reflected or
transmitted color is the complimentary
color.
 Thus a magenta shirt has that color
because the dye it contains strongly
absorbs green light and reflects the
magenta, the compliment of green
(see color wheel)
Color Wheel
Colored Solutions
 Colored solutions absorb photons of
white light to promote electrons to
higher energy levels.
 White light, minus the absorbed color,
is no longer white, but appears as the
compliment of the color that was
absorbed
 Ions having noble gas configurations do
not have energy absorptions in the
visible range, thus they appear
Colored Solutions
 Crystal field splitting deals with d-electrons
and not Noble gas structures.
 The do absorb in the visible spectrum and we
see the color of the light that is
complimentary to the color absorbed
 A solution containing [Cu(H2O)4]2+ absorbs
most strongly in the yellow region of the
spectrum (about 580 nm)
 The wavelength of the transmitted light is
violet
Acidity of Coordination Compounds
Water molecules coordinately bonded to a metal ion
can lose a proton, thus causing the hydrate to act as an
acid
Acidic Hydrates
Water molecules coordinately bonded to a metal
ion can lose a proton, thus causing the hydrate
to act as an acid
Cr(H2O)63+
Cr(H2O)5(OH)2+ + H+
ka = 1 X 10-4
Acidic Hydrates
Water molecules coordinately bonded to a metal
ion can lose a proton, thus causing the hydrate
to act as an acid
Cr(H2O)63+
Cr(H2O)5(OH)2+ + H+
Ka =
ka = 1 X 10-4
[Cr(H2O)5(OH)2+][H+]
[Cr(H2O)63+]
The hydroxide ion is bonded to the transition
metal cation. The greater the charge of the
metal ion and the smaller it is produces a
stronger attraction to the hydroxide thus making
the complex more stable and more acidic.
Acidic Hydrates
Water molecules coordinately bonded to a metal
ion can lose a proton, thus causing the hydrate
to act as an acid
Cr(H2O)63+
Cr(H2O)5(OH)2+ + H+
Ka =
ka = 1 X 10-4
[Cr(H2O)5(OH)2+][H+]
[Cr(H2O)63+]
Calculate the pH of a 1.00 M solution of Cr3+
Ka =
(x)(x)
1.0-x
= 1.00 X 10-4
Acidic Hydrates
Water molecules coordinately bonded to a metal
ion can lose a proton, thus causing the hydrate
to act as an acid
Cr(H2O)63+
Cr(H2O)5(OH)2+ + H+
Ka =
ka = 1 X 10-4
[Cr(H2O)5(OH)2+][H+]
[Cr(H2O)63+]
Calculate the pH of a 1.00 M solution of Cr3+
Ka =
(x)(x)
1.0-x
= 1.00 X 10-4
[H+] = 10-4
pH = log10-4
pH = 4
Solubility of Ionic Compounds
When a precipitate forms the solution is said
to be saturated
Saturated solutions can also be formed by
adding to much solute
An equilibrium exists between ions forming
solid and the solid dissolving to form ions.
From the equilibrium constant we can
determine the molar solubility
Silver Chloride Solubility
Silver chloride is known to be insoluble
according to our solubility rules, but some
does dissolve.
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp= 1.8X10 -10
Silver Chloride Solubility
Silver chloride is known to be insoluble
according to our solubility rules, but some
does dissolve.
AgCl(s)
Ksp =
Ag+(aq) + Cl-(aq)
[Products]
[Reactants]
=
?
Ksp= 1.8X10 -10
Silver Chloride Solubility
Silver chloride is known to be insoluble
according to our solubility rules, but some
does dissolve.
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
Ksp= 1.8X10 -10
Silver Chloride Solubility
Silver chloride is known to be insoluble
according to our solubility rules, but some
does dissolve.
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = [X][X]
Ksp= 1.8X10 -10
X2 = 1.8 X 10-10
X = 1.34 X 10-5 M
Silver Chloride Solubility
Silver chloride is known to be insoluble
according to our solubility rules, but some
does dissolve.
AgCl(s)
Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = [X][X]
Ksp= 1.8X10 -10
X2 = 1.8 X 10-10
X = 1.34 X 10-5 M
1.34 X 10-5 mole Ag+ mole AgCl 142 g AgCl
mole Ag+ mole AgCl
L
1.91 X 10-3 g AgCl will dissolve in a liter of water
Sample Problem
Determine the molar solubility of magnesium
hydroxide.
Sample Problem
Determine the molar solubility of magnesium
hydroxide.
Mg(OH)2 (s)
Mg2+ + 2 OH-
ksp= 1.8 X 10-11
Sample Problem
Determine the molar solubility of magnesium
hydroxide.
Mg(OH)2 (s)
Ksp = [Mg2+][OH-]2
Mg2+ + 2 OHx
2x
ksp= 1.8 X 10-11
Sample Problem
Determine the molar solubility of magnesium
hydroxide.
Mg(OH)2 (s)
Mg2+ + 2 OHx
2x
ksp= 1.8 X 10-11
Ksp = [Mg2+][OH-]2 = x(2x)2 = 1.8 X 10-11
x = 1.65 X 10-4 M
Sample Problem
Determine the molar solubility of magnesium
hydroxide.
Mg(OH)2 (s)
Mg2+ + 2 OHx
2x
ksp= 1.8 X 10-11
Ksp = [Mg2+][OH-]2 = x(2x)2 = 1.8 X 10-11
x = 1.65 X 10-4 M
[OH-] = 2(1.65x10-4) = 3.3x10--4
Sample Problem
Determine the additional mass of magnesium
hydroxide dissolved, after the addition of
125 mL of 1.00 X 10-4 M HCl.
Mg(OH)2 (s)
Mg2+ + 2 OH-
Which way will the equilibrium shift?
ksp= 1.8 X 10-11
Sample Problem
Determine the additional mass of magnesium
hydroxide dissolved, after the addition of
125 mL of 1.00 X 10-4 M HCl.
Mg(OH)2 (s)
Mg2+ + 2 OH-
ksp= 1.8 X 10-11
Which way will the equilibrium shift? Right
Sample Problem
Determine the additional mass of magnesium
hydroxide dissolved, after the addition of
125 mL of 1.00 X 10-4 M HCl.
Mg(OH)2 (s)
Mg2+ + 2 OH-
ksp= 1.8 X 10-11
Which way will the equilibrium shift? Right
What is the hydroxide ion concentration after
addition of the HCl?
Sample Problem
Determine the additional mass of magnesium
hydroxide dissolved, after the addition of
125 mL of 1.00 X 10-4 M HCl.
Mg(OH)2 (s)
Mg2+ + 2 OH- ksp= 1.8 X 10-11
Which way will the equilibrium shift? Right
What is the hydroxide ion concentration after
addition of the HCl?
1.00 X 10-4 mole HCl 0.125 L
= 1.25 X10-5 mole HCl
L
3.3x10-4 – 1.25x10-5 = 3.176 x 10-4 mole OH3.176x10-4 mole OH= 2.822X104 M OH1.125 L
Sample Problem
What additional mass of Mg is dissolved
Ksp = [Mg2+][OH-]2 = 1.8 X 10-11
[Mg2+][OH-]2 = 1.8 X 10-11
[Mg2+][2.2822X10-4]2 = 1.8 X 10-11
[Mg2+] = 2.26X10-4M Mg2+
2.26X10-4M Mg2+ - 1.65X10 -4 Mg2+ (original) = 6.1 X10-5 moles Mg2+
6.1 X 10-5 moles Mg2+ 24.3 g Mg2+
= 1.5 X 10-3 g Mg2+
moles Mg2+
About Dissolving Precipitates
Precipitates form when the product of the ion
concentration Q>Ksp. Since precipitates are in
equilibrium with the ions that form the LeChatelier’s
Principle will control the solution process. The previous
sample problem demonstrated how solubility is controlled
by Ph. Clearly pH is dependent upon solubility and this is a
very important consideration in qualitative analysis, which
is always emphasized in prelab lectures. If the procedure
states acidic or basic, use litmus paper. If the procedure
states just basic, then use pH paper and adjust the pH to
8. If the procedure requires a specific pH, then use pH
paper.
Solubility Controlling Factors
Since precipitate formation is related to solubility,
then the following factors control precipitation also.
These factors control the process, by adding or
removing ions from the equilibrium mixture according
to leChateliers principle.
 The pH of the solution
removing H+ or OH- by adding acid or base
 Formation of a complex
Adding a substance to form a coordination compound
 Formation of a gas
Adding a substance to convert one of the ions to a gas.
Dissolving AgCl (s)
One of the most common applications of
precipitate control by complex formation involves
the insoluble silver chloride precipitate.
AgCl(s)
Ag+(aq) + Cl-(aq)
Ammonia will complex with silver ion to from the
diamminosilver(I) complex, thus removing silver
ion from solution. According to LeChatelier’s
principle, silver chloride produces more silver ions
which are then complexed with ammonia until all
of the silver chloride solid is dissolved. The
following slide illustrates this process
quantitatively.
Complex Formation
Consider silver complex formation with ammonia
Ag+ + NH3 ⇄ Ag(NH3)+ K1 = 2.1x103
Ag(NH3)+ + NH3 ⇄ Ag(NH3)2 K2 = 8.2x103
Ag+ + 2NH3 →
Ag(NH3)2+ Kf = K1 x K2
Kf =1.72x107
When using a large excess of NH3 and since the
formation constants are large, the reaction can be
considered to be complete, since the overall
constant is 1.72x107.
Complex Formation
Ag+ + 2NH3 → Ag(NH3)2+
Given: [Ag+]i = 5x10-4 and [NH3]I = 1.0
Then [Ag+]f = 0 and [NH3]f = 1.0- 2(5x10-4), and
[Ag(NH3)2+]=5x10-4
Yes, but there is a small amount of Ag+ so then how
can we calculate this concentration? Next slide!
Complex Formation
How much [Ag(NH3)+] is present can be calculated
using the K2
K2= [Ag(NH3)2+]/[NH3][ Ag(NH3)+]
Solving for [Ag(NH3)+] = [Ag(NH3)2+]/[NH3][ K2]
[Ag(NH3)+]=6.1x10-8, using 1.0 for ammonia
In a similar way, using K1, [Ag+] can be determined
to be 2.9x10-11
Complex Formation
How do you dissolve and insoluble salt? For example AgCl
The equilibrium is AgCl ⇄ Ag+ + ClNeed to shift the equilibrium to the right to dissolve the
solid AgCl
Consider the following process
AgCl ⇄ Ag+ + ClAg+ + NH3 ⇄ Ag(NH3)
Ksp=1.6x10-10
K1 = 2.1x103
Ag(NH3)+ + NH3 ⇄ Ag(NH3)2+
K2 = 8.2x103
AgCl (s) + 2NH3 ⇄ Ag(NH3)2+ + Cl- K1xK2x Ksp
Complex Formation
10.0
0
0
initial
AgCl (s) + 2NH3 ⇄ Ag(NH3)2+ + Cl- K =2.8x103
10-2x
x
x
Use the equilibria expression above to calculate the
solubility of AgCl in a 10.0 M ammonia solution
K = [Ag(NH3)2+][Cl-]/[NH3]2 → 2.8x10-3 = x2/ (10-2x)2
X=0.48 M
The End