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2015/02/16
Tetrahedral and Square-Planar Complexes
• When a complex has four ligands, the
geometry is often tetrahedral
• except for the special case of d8 metal ions
which is square planar.
Tetrahedral
Squareplanar
trans - PtCl2(NH3)2
Tetracarbonylnickel
Tetrahedral Complexes
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Octahedral vs tetrahedral complexes
CFSE = -2/5∆o(t2g) +3/5∆o(eg) + P
CFSE = +2/5∆t(t2g) -3/5∆t(eg) + P
Tetrahedral Complexes
Crystal-field splitting energy is much smaller for
tetrahedral complexes than octahedral complexes
•None of the orbitals are pointing directly at the ligands in
the tetrahedral geometry
•Only four ligands in a tetrahedral complex instead of six,
as in the octahedral case
•The crystal field splitting energy is almost always smaller
than the spin pairing energy P.
•Therefore most tetrahedral complexes are high spin; the
crystal-field splitting energy is about never large enough
to overcome the spin-pairing energies.
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Why are tetrahedral complexes typically high
spin which is not the case for octahedral
complexes?
• Crystal-field splitting energy is much smaller for tetrahedral
complexes than octahedral complexes
• None of the orbitals are pointing directly at the ligands in the
tetrahedral geometry
• Only four ligands in a tetrahedral complex instead of six, as in the
octahedral case
• The crystal field splitting energy is almost always smaller than
the spin pairing energy P.
• Therefore most tetrahedral complexes are high spin; the
crystal-field splitting energy is about never large enough to
overcome the spin-pairing energies.
Are tetrahedral complexes typically
high or low spin complexes? Why?
• High spin
• Crystal-field splitting energy is small for tetrahedral complexes
since:
• None of the orbitals are pointing directly at the ligands in the
tetrahedral geometry
• Only four ligands in a tetrahedral complex instead of six, as in the
octahedral case
• The crystal field splitting energy is almost always smaller than
the spin pairing energy P.
• Therefore most tetrahedral complexes are high spin; the
crystal-field splitting energy is about never large enough to
overcome the spin-pairing energies.
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The “odd” complex
Do not memorise: just for interest. It is the only low spin tetrahedral complex I
have found so far.
Tetrakis(1-norbornyl)cobalt, a low spin tetrahedral complex of a
first row transition metal
Erin K. Byrne, Darrin S. Richeson and Klaus H. Theopold
J. Chem. Soc., Chem. Commun., 1986, 1491-1492
The crystal structure of tetrakis(1-norbornyl)cobalt (1) was determined and, together
with magnetic susceptibility measurements, establishes (1) as the first example of a
low spin tetrahedral complex of a first row transition metal and as the only structurally
characterized cobalt(IV)–alkyl complex.
Square-Planar Complexes
Jahn-Teller distortion
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Octahedral complex vs Square-planar complex
Octahedral complex
Octahedral complex
with Jahn-Teller distortion
Square planar complex
Square-planar complex
• Square planar complexes are similar to
octahedral complexes.
• The difference is that square planar
complexes have two ligands missing in the zaxis.
• There is a very large energy gap between the
x2-y2 orbital and the lower four orbitals.
• Square planar complexes are favoured by
metal ions with d8 electron configurations
• since this configuration favours low-spin
complexes in which the four lower-energy
orbitals are filled and the high energy x2-y2
orbital is empty.
∆t
dx2-y2
dxy
dz2
dxz
dyz
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Why is d8 electron configurations typically squareplanar? Use a drawing of an energy diagram in your
explanation. Don’t only draw a diagram, discuss it as
well.
Square planar complexes are
favoured by metal ions with d8
electron configurations since this
configuration favours low-spin
complexes in which the four lowerenergy orbitals are filled and the
high energy x2-y2 orbital is empty.
∆t
dx2-y2
dxy
dz2
dxz
dyz
Tetrahedral complex vs Square-planar complex
Explain why a d9 electron configurations will have a tetrahedral
geometry and not an square planar geometry. Use a drawing of the
energy diagram of the two geometries in your explanation.
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Self test
• Compare the splitting diagrams of a tetrahedral
complex with that of a square planar complex.
• Compare the splitting diagrams of a tetrahedral
complex with that of an octahedral complex.
• Compare the splitting diagrams of a square planar
complex with that of an octahedral complex.
• The d8 complexes [Ni(CN)4]2- and [NiCl4]2- are square
planar and tetrahedral respectively. Will these
complexes be paramagnetic or diamagnetic? Draw
slitting diagrams and refer to it in your explanation.
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