One Point Quiz

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One Point Quiz

• One quiz per table, list everyones name
• Agree on an answer
• You have two minutes

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Unknown Dues Dates

• Unknown A Friday, Week 5
• Unknown B Friday, Week 7
• Unknown C Friday, Week 9
• Unknown D Friday, Week 10

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Valence Bond Theory

• Edward A. Mottel
• Department of Chemistry
• Rose-Hulman Institute of Technology

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Bonding Theories

• Ionic Model
• Skeleton Diagrams
• Lewis Dot Diagrams
• Molecular Orbital Theory
• Orbital Hybridization
• Valence Bond Theory

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Valence Bond Theory

• Metal ion-ligand bonding is somewhat different
  than the orbital hybridization of covalent
  nonmetallic compounds.

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Tetrachloroaluminate Ion, AlCl4
Using the orbital hybridization approach the
hybrid description would be
Al 1s2 2s2 2p6 3s2 3p1
Al
1s2 2s2 2p6 (sp3)4
sp3 hybridization
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Tetrachloroaluminate Ion, AlCl4Orbital
Hybridization Approach
Four neutral chlorine atoms each donate one
electron to form a sigma bond to aluminum.
tetrahedral shape
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Problems with the Orbital Hybridization Approach
The description is unreasonable for two reasons
The stable oxidation state of aluminum is 3, not
-1
The negative formal charge should be associated
with the more electronegative chlorine atoms.
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Tetrachloroaluminate Ion, AlCl4Valence Bond
Approach
Al
Al3
1s2 2s2 2p6 3s2 3p1
1s2 2s2 2p6 3s0 3p0
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Tetrachloroaluminate Ion, AlCl4Valence Bond
Approach
Cl
Cl
1s2 2s2 2p6 3s2 3p5
1s2 2s2 2p6 3s2 3p6
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Tetrachloroaluminate Ion, AlCl4Valence Bond
Approach
Each chloride ion donates a pair of electrons to
form a bond with aluminum.
Al3
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Tetrachloroaluminate Ion, AlCl4Valence Bond
Approach
Al3 accepts four pairs of electrons into empty
(not half-filled) orbitals.
An atom, ion or molecule which donates a pair of
electrons to an empty metal orbital is called a
ligand.
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Tetrachloroaluminate Ion, AlCl4Valence Bond
Approach
In this example, each chloride ion is a ligand.
The valence bond approach still allows for
hybridization.
Conceptually, the source of the electrons is
different.
An atom, ion or molecule which donates a pair of
electrons to an empty metal orbital is called a
ligand.
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Hexacyanoferrate(II) Ion,Fe(CN)64

• Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6
• Fe2 1s2 2s2 2p6 3s2 3p6 3d6

Hexacyanoferrate(II) ion contains six cyanide
groups (ligands) bound to the central metal ion
and is diamagnetic.
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Hexacyanoferrate(II) Ion,Fe(CN)64
Fe2 1s2 2s2 2p6 3s2 3p6 3d6
Diamagnetic species have no unpaired electrons.
Which of these patterns represents a diamagnetic
species?
This is a low spin ion because the maximum
number of metal electrons are spin paired.
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Hexacyanoferrate(II) Ion,Fe(CN)64
Fe2 1s2 2s2 2p6 3s2 3p6 3d6
Whats the name of this orbital set?
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A New Hybrid Orbital Set d2sp3




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A New Hybrid Orbital Set d2sp3
octahedral shape all orbitals equal in size
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d2sp3 hybridization
4-
Each cyanide ion donates an unshared pair of
electrons to an empty d2sp3 orbital of iron(II).
Coordination number of iron is 6.
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Hexacyanoferrate(II) Ion,Fe(CN)64
Fe2 1s2 2s2 2p6 3s2 3p6 3d6
The carbon of each cyanide ion is sp hybridized
and donates its unshared pair of electrons to
iron to form an iron-carbon sigma bond.
There are six cyanide ligands.
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Square and Trigonal Symmetry
A wire frame diagram can be used to emphasize the
square or triangular nature of the octahedron.
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Hexafluoroferrate(III) IonFeF63
Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe3 1s2 2s2
2p6 3s2 3p6 3d5

• Hexafluoroferrate(III) ion
• has six fluoride ions bound to iron.
• is experimentally found to have five unpaired
  electrons. (paramagnetic)

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Hexafluoroferrate(III) IonFeF63
Fe3 1s2 2s2 2p6 3s2 3p6 3d5
(Low Spin)
(High Spin)
Which of these patterns represents a paramagnetic
species?
Trick Question!
They BOTH are paramagnetic!
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Hexafluoroferrate(III) IonFeF63
Fe3 1s2 2s2 2p6 3s2 3p6 3d5
This is a high spin ion because the maximum
number of metal d-electrons are spin unpaired.
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A New Hybrid Orbital Set sp3d2

• Composed of
• s px py pz dx2-y2 dz2
• New hybrid set of orbitals
• different sequence of orbitals
• Octahedral coordination

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A New Hybrid Orbital Set sp3d2
octahedral shape all orbitals equal in size
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Hexafluoroferrate(III) IonFeF63
Coordination number is 6. Six ligands bound to
the iron(III) ion.
Each fluoride ion contributes a pair of electrons
to the empty sp3d2 hybrid orbitals of iron(III).
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Hexafluoroferrate(III) IonFeF63
Fe3 1s2 2s2 2p6 3s2 3p6 3d5
There are six fluoride ligands.
An iron-fluorine sigma bond forms.
Why are empty orbitals needed for the hybrid?
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Tetrachloroferrate(III) Ion FeCl4
Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe3 1s2 2s2
2p6 3s2 3p6 3d5

• Tetrachloroferrate(III) ion
• has four chloride ions bound to iron.
• is high spin. Experimentally it is found to have
  five unpaired electrons.

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Tetrachloroferrate(III) Ion FeCl4
Fe3 1s2 2s2 2p6 3s2 3p6 3d5
the next four orbitals are used
How many empty metal orbitals are needed?
Whats the name of this orbital set?
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Tetrachloroferrate(III) Ion FeCl4
Tetrahedral shape Coordination number of iron is
4.
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Tetrachloroferrate(III) Ion FeCl4
Fe3 1s2 2s2 2p6 3s2 3p6 3d5
sp3
There are four chloride ligands.
Each chloride ion donates an unshared pair of
electrons to iron(III) to form an iron-chlorine
sigma bond.
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Prediction of Colored Complexes

• Energy transitions
• are due to electrons being promoted from one
  energy level (orbital) to another.
• result in the absorption of certain wavelengths
  of light.

• Transitions involving d-orbitals occur in the
  visible spectrum.
• If the absorption is in the visible region, then
  the complementary color is observed.

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colorless
color
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d-orbital Transitions
In a complex, the d-orbitals are split and don't
have the same energy.
Octahedral d-orbital splitting pattern
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d Orbital Electron Filling
Low Spin filling sequence
High Spin filling sequence
Electrons have a preferred spin direction (e.g.,
spin down) and fill that way if possible.
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Rules for Electronic Transitions

• For absorption to occur, an electron moves from a
  lower level to a higher level.
• Electrons cannot "spin flip" when moving from one
  level to the other.
• If no transition is allowed, the complex is
  predicted to be colorless.

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Hexacyanoferrate(II) IonFe(CN)64
6 d-electrons, low spin
energy absorption
Octahedral d-orbital splitting pattern
electron transition is allowed complex is
expected to have color
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Tetrachloroferrate(III) Ion FeCl4
5 d-electrons, high spin
energy absorption
Tetrahedral d-orbital splitting pattern
NOT ALLOWED an orbital cant have two electrons
with the same spin
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Tetrachloroferrate(III) Ion FeCl4
5 d-electrons, high spin
energy absorption
Tetrahedral d-orbital splitting pattern
NOT ALLOWED an electron cant be promoted
and change spin at the same time
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Tetrachloroferrate(III) Ion FeCl4
5 d-electrons, high spin
energy absorption
No allowed transitions
Tetrahedral d-orbital splitting pattern
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Predictions

• There are 10 different metal ions studied in the
  laboratory.
• Ag, Ba2, Cd2, Co2, Cr3, Cu2, Fe3, Mn2,
  Ni2, Pb2
• Predict whether each ion is expected to be
  colored or colorless.
• Note freshly prepared aqueous iron(III) ion is
  colorless.

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One Point Quiz

• One quiz per table, list everyones name
• Agree on an answer
• You have two minutes

48
Unknown Dues Dates

• Unknown A Friday, Week 5
• Unknown B Friday, Week 7
• Unknown C Friday, Week 9
• Unknown D Friday, Week 10

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Tetrachloronickelate(II) Ion NiCl42

• Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8
• Ni2 1s2 2s2 2p6 3s2 3p6 3d8
• Tetrachloronickelate(II) ion
• has four chloride ions bound to nickel.
• is a blue, paramagnetic ion.

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Tetrachloronickelate(II) Ion NiCl42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
Which of these patterns would predict a colored
complex?
Trick Question!
They BOTH predict a colored complex!
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Tetrachloronickelate(II) Ion NiCl42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
Whats the name of this orbital set?
How many empty metal orbitals are needed?
the next four orbitals are used
sp3
Which of these patterns represents a paramagnetic
species?
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Tetrachloronickelate(II) Ion NiCl42
Tetrahedral shape Coordination number of nickel
is 4.
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Tetrachloronickelate(II) Ion NiCl42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
sp3
There are four chloride ligands.
Each chloride ion donates an unshared pair of
electrons to nickel(II) to form an
nickel-chlorine sigma bond.
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Tetrachloronickelate(II) Ion NiCl42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
Why cant chloride ion donate an unshared pair of
electrons to the half-filled nickel(II)
d-orbitals?
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Tetracyanonickelate(II) Ion Ni(CN)42

• Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8
• Ni2 1s2 2s2 2p6 3s2 3p6 3d8
• Tetracyanonickelate(II) ion
• has four cyanide ligands bound to nickel.
• is an orange, diamagnetic ion.

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Tetracyanonickelate(II) Ion Ni(CN)42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
Which of these patterns represents an orange,
diamagnetic species?
How many empty metal orbitals are needed?
Whats the name of this orbital set?
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A New Hybrid Orbital Set dsp2



square planar shape all lobes are equal in size
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dsp2 hybridization
Each cyanide ion donates an unshared pair of
electrons to an empty dsp2 orbital of nickel(II).
2-
Square planar
Coordination number of nickel is 4.
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Tetracyanonickelate(II) Ion Ni(CN)42
Ni2 1s2 2s2 2p6 3s2 3p6 3d8
There are four cyanide ligands.
The carbon of each cyanide ion is sp hybridized
and donates its unshared pair of electrons to
nickel to form a nickel-carbon sigma bond.
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Square Planar Symmetry
Nickel and the carbon atoms all lie in the same
plane. Coordination number is 4.
Cyanide ligands usually form low spin (spin
paired) complexes.
Are the nitrogen atoms also coplanar with nickel?
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Tetracyanonickelate(II) IonNi(CN)42
Why is this ion diamagnetic?
Why is the ion colored?
Why is its color significantly different
than tetrachloronickelate(II) ion?
Do you think Ni(dmg)2, which is used as
a confirmatory test in the laboratory,
is diamagnetic or paramagnetic?
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Diamminesilver IonAg(NH3)2
Ag 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1
4d10 Ag 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10
Two metal orbitals must be hybridized to bond to
the two ammonia ligands.
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Diamminesilver IonAg(NH3)2
Ag 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10

Diamagnetic, colorless coordination number 2
Ð N-Ag-N is linear
bond angle 180º
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Diamminesilver IonAg(NH3)2
Ag 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10
The nitrogen of each ammonia is sp3 hybridized
and donates its unshared pair of electrons to
silver to form a silver-nitrogen sigma bond.
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Diamminesilver IonAg(NH3)2
Why is this ion diamagnetic?
Why is the ion colorless?
This ion is not planar. Why not?
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Pentafluoroarsenic(V)AsF5

• As 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
• As5 1s2 2s2 2p6 3s2 3p6 3d10
• Pentafluoroarsenic(V)
• contains five fluoride ligands bound to
  arsenic(V) ion.
• is diamagnetic.

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Pentafluoroarsenic(V)AsF5
As5 1s2 2s2 2p6 3s2 3p6 3d10
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A New Hybrid Orbital Set sp3d



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A New Hybrid Orbital Set sp3d
What does this new hybrid set look like?
Break it into two smaller sets of orbitals and
analyze each individually.
Combine the results to give the overall orbital
set.
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What hybrid atomic orbital set would be generated
from these three orbitals?
s

3 equal lobes sitting in the xy-plane (like sp2)
px
py
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What hybrid atomic orbital set would be generated
from these two orbitals?
dp hybridization 2 equal lobes lying primarily
along the z-axis. (similar to sp hybridization)
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A New Hybrid Orbital Set sp3d
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A New Hybrid Orbital Set sp3d
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A New Hybrid Orbital Set sp3d
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A New Hybrid Orbital Set sp3d
the equatorial orbitals in the xy-plane are equal
in size
the axial orbitals along the z-axis are equal in
size
trigonal bipyramidal shape
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Pentafluoroarsenic(V)Trigonal Bipyramidal
Geometry
Each fluoride ion donates an unshared pair of
electrons to an empty sp3d hybrid orbital of
arsenic(V).
Coordination number is 5.
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Pentafluoroarsenic(V)Trigonal Bipyramidal
Geometry
Ð Fax-As- Fax 180º
Axial and equatorial positions are non-equivalent.
Ð Fax-As- Feq 90º
Ð Feq-As- Feq 120º
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Pentafluoroarsenic(V)AsF5
As5 1s2 2s2 2p6 3s2 3p6 3d10
There are five fluoride ligands.
Each fluoride ion donates an unshared pair of
electrons to arsenic(V) to form an
arsenic-fluorine sigma bond.
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Hybrid Orbitals
2
sp
linear
180º
3
trigonal
120º
sp2
4
tetrahedral
109.5º
sp3, p3s
4
dsp2
90º
dsp3,sp3d, dp3s
5
90º and 120º
6
d2sp3, sp3d2
octahedral
90º
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Hexaaquacobalt(II) NitrateValence Bond Theory
Predictions

• The hexaaquacobalt(II) nitrate is a high spin
  complex.
• Explain in detail the bonding of both the metal
  and the ligand in this complex.
• Explain in detail the bonding in the anion of
  this compound.
• Predict appropriate physical properties.

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Tetraamminepalladium(II) IonValence Bond Theory
Predictions

• The tetraamminepalladium(II) ion is a low spin
  complex ion.
• Explain in detail the bonding of both the metal
  and the ligand in this complex.
• Predict appropriate physical properties.

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End of Exam IV Material
Be sure to review the Exam IV Objective Sheet
You may use on the exam Concept Maps (web
diagrams) to help relate descriptive chemistry
material from the text reading. Fine, Beall
Stuehr Pg 361-364 and Pg 867-928
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Unknown Dues Dates

• Unknown A Monday, Week 5
• Unknown B Monday, Week 7
• Unknown C Monday, Week 9
• Unknown D Monday, Week 10