Download 4.Valance Bond Theory

Bonding in Metal complexes
Valence Bond Theory
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Postulates
1.The central metal ion provides adequate number of vacant orbitals
for the formation of bond with ligands.
2. The suitable ( s,p & d) pure atomic orbitals hybridise to give an equal number
of hybridised orbitals of equivalent energy .
3. The d- orbitals involved may be inner, (n-1)d or outer, nd orbitals.
The complexes thus formed are Inner , low spin complexes or outer, high spin
complexes. The hybridisation may be d2sp3 or sp3 d2.
4. Each ligand provides at least one orbital containing lone pair of electrons.
5. The empty hybrid orbital overlaps with filled ligand orbital forming a sigma
coordinate bond.
6. The compleses will be paramagnetic if it contains unpaired electron and
diamagnetic if there are no unpaired electrons.
7. Sometimes under the influence of strong ligands electrons in metal orbitals are
forced to pair up against Hund’s rule of maximum multiplicity.
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Examples
Coordination no. - 4
1. Tetrahedral Complexes
a) [Ni(CN)4]2- Tetracyanonickelate (II) ion
Ni 0 –[Ar] 3d8 4s2
Ni 2+ ion – [Ar] 3d8 4s0
1. Ni(II) ion
3d8
4s0
4p0
sp3 hybridisation
2. [Ni(Cl)4]2- ion
X
4 ligands give 4 lone pairs of electrons
X
X
sp3 hybridisation
Paramagnetic, 2 unpaired electrons, Tetrahedral
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X
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Examples
Coordination no. - 4
Tetrahedral Complexes
b) [Zn(NH3)4] 2+
Zn 0 –[Ar] 3d10 4s2
Tetraamminezinc (II) ion
Zn 2+ ion – [Ar] 3d10 4s0
Zn 2+ ion
3d10
4s0
4p0
sp3 hybridisation
[Zn(NH3)4] 2+
X
X
X
X
4 ligands give 4 lone pairs of electrons
Diamagnetic, Tetrahedral
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sp3 hybridisation
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Examples
Coordination no. - 4
2. Square Planar Complexes
a) [Ni(CN) 4 ] 2- Tetracyanonickelate (II) ion
Ni 0 –[Ar] 3d8 4s2
Ni 2+ ion – [Ar] 3d8 4s0
1. Ni(II) ion 3d8
4s0
2. Pairing
against Hund’s
rule1
4p0
dsp2 hybridisation
X
3. [Ni(CN) 4 ] 2-
X
X
X
4 ligands give 4 lone pairs of electrons
dsp2 hybridisation
Diamagnetic, Square planar
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Examples
Coordination no. - 4
Square Planar Complexes
b) [Cu(NH3)4] 2+ Tetraamminecopper (II) ion
Cu 0 –[Ar] 3d10 4s1
Cu 2+ ion – [Ar] 3d9 4s0
1. Cu(II) ion 3d9
4s0
Option 1
X
4p0
X
X
X
sp3 hybridisation
Option 2
X
X
X
X
4 ligands give 4 lone pairs of electrons
dsp2 hybridisation
X- Ray studies confirm Square planar shape
Paramagnetic
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Examples
Coordination no. - 6
1. Octahedral – Low spin - Inner orbital complex
[Fe(CN)6]4- Hexacyanoferrate(II) ion
Fe 0 –[Ar] 3d6 4s2
Fe 3+ ion – [Ar] 3d6s0
1. Fe(II) ion
2. Electronic
rearrangement
against Hund’s
rule
3d6
4s0
X
X
X
4p0
X
X
X
3. [Fe(CN)6] 4- ion
6 ligands give 6 lone pairs of electrons d2sp3 hybridisation
Octahedral
Diamagnetic, Inner, Low spin complex
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Examples
Coordination no. - 6
Octahedral – Low spin - Inner orbital complex
[Cr(NH3)6]3+ Hexaamminechromium(III) ion
Cr 0 –[Ar] 3d5 4s1
Cr 3+ ion – [Ar] 3d3s0
1. Cr 3+ ion
3d3
2. [Cr(NH3)6]3+
4s0
X
X
X
4p0
X
X
X
6 ligands give 6 lone pairs of electrons
d2sp3 hybridisation
Octahedral
Paramagnetic, 3 unpaired electrons, Inner orbital
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Examples
Coordination no. - 6
2. Octahedral – High spin – Outer orbital complex
[Fe(H2O)6]3+ Hexaaquoiron(III) ion
Fe 0 –[Ar] 3d6 4s2
Fe 3+ ion – [Ar] 3d5s0
1. Fe(III) ion
3d5
4s0
4p0
4d0
sp3d2 hybridisation
2. [Fe(H2O)6]3+
X
X
X
X
X
X
6 ligands give 6 lone pairs of electrons
sp3d2 hybridisation
Octahedral
Paramagnetic, 5 unpaired electrons, High spin, Outer complex
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Examples
Coordination no. - 6
Octahedral – High spin – Outer complex
[Co(F)6]3+ Hexafluoridecobolt(III) ion
Co 0 –[Ar] 3d7 4s2
Co 3+ ion – [Ar] 3d6 4s0
Co 3+ ion
3d5
4s0
4p0
4d0
sp3d2 hybridisation
[Co(F)6]3+
X
6 ligands give 6 lone pairs of electrons
X
X
X
X
sp3d2 hybridisation
X
Octahedral
Paramagnetic, 4 unpaired electrons, High spin, Outer complex
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Practice problem
The magnetic moment of [Mn(Cl)6]3- is 2.8 B.M. while that of [Mn(Br)4]2- is 5.9 B.M.
Predict the geometries of the complex ions.
Solution –
[Mn(Cl)6]3Mn 0 –[Ar] 3d5 4s2
Mn 3+ ion – [Ar] 3d4 4s0
U = [n(n+2)],
2.8 = [n(n+2)],
so, n = 2
Complex should have only 2 unpaired electrons. The possible electronic
arrangement would be Mn 3+ ion
3d4
4s0
4p0
d2sp3 hybridisation
[Mn(Cl)6]3-
X
X
X
6 ligands give 6 lone pairs of electrons
X
X
X
d2sp3 hybridisation
Octahedral
Paramagnetic, 2 unpaired electrons, Inner complex
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Practice problem
[Mn(Br)4
]2Mn 0 –[Ar] 3d5 4s2
Mn 2+ ion – [Ar] 3d5 4s0
Solution –
U = [n(n+2)],
5.9 = [n(n+2)],
so, n = 5
Complex has 5 unpaired electrons. The possible electronic arrangement would
be Mn 2+ ion
3d5
4s0
4p0
sp3 hybridisation
[Mn(Br)4]2-
X
X
X
X
4 ligands give 4 lone pairs of electrons sp3 hybridisation, Tetrahedra
Paramagnetic, 5 unpaired electrons, Tetrahedra
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