Which of the d orbitals of metals have possibility to make sigma bond with p orbitals of ligand ? Does only dz2 has this possibility?
I am asking this question for the metal ion in octahedral ligand field. Although we can think about some interactions between dxy,dyz,dzx of metal ion and P orbital of ligands that may be called sigma bond but I have not got any practical example. Can any body give example of such compounds?
Yes dz2 has the possibility and majour combinations d and p orbitals overlap lateraaly
Since the question does not specify the geometry, it is important to keep geometrical considerations in mind.
In an octahedral geometry, both the dz2 and dx2-y2 orbitals participate in sigma bonding with the ligands.
In a tetrahedral geometry, the dxy, dxz and dyz orbitals participate in sigma bonding with the ligands.
So, the answer to the question is that, under the right circumstances, all of the d orbitals can participate in sigma bonding with a ligand.
WITH OUT MENTIONING THE GEOMETRY, ONE CAN NOT GIVE ANSWER TO THE ABOVE QUESTION.
Dear d. Agustin,
this is the correct answer. All d-orbitals can participate in sigma bonding, depending on geometry.
Dear Soumen, d(x2-y2) orbital can also able to participate in making sigma bond with the ligand. As for example you can consult the M.O diagram for octahedral complexes where this orbital take part in bonding with the ligands in the square plane.
Slater-Koster (1954) answers this question thoroughly, given any geometry, I believe.
Most ans. are correct WITHIN the LIGAND FIELD theory, a bonding theory. t2g form sigma bonds M-L type, dx2-y2 and all its linear combinations.
In the crystal field formalisn you cannot speak of formal bond, it is purely charge repulsion between t2g and ligand piont charges oriented along t2g axis. Though both theories use the same geometry terms.
The question presupposes that we are not considering the crystal field formalism since it asks about sigma bonding.
In octahedral symmetry, the t~2g orbitals do not form sigma bonds with the ligands since the lobes of the dxy, dxz and dyz orbitals are off axis with respect to the ligand sigma symmetry orbitals.
In tetrahedral symmetry, there is no center of symmetry and the dxy, dxz and dyz orbitals have t~2 not t~2g symmetry. The t~2 orbitals participate in sigma bonding with the ligands.
Pardon?
>In octahedral symmetry, the t~2g orbitals do not form sigma bonds with the ligands since the lobes of the dxy, dxz and dyz orbitals are off axis with respect to the ligand sigma symmetry orbitals.
The original question was about sigma bonding, not bonding in general. Also, my comments have addressed sigma bonding (since that is what the original question was about).
Most certainly the t~2g levels participate in pi interactions with ligands capable of pi interactions. However, your comment stated "t2g form sigma bonds.." They don't.
If you have an octahedral complex, than two orbitals can form sigma bonds - dz2 and d(x2-y2). In a tetrahedral complex the situation is more difficult, since there is no d orbitals are directed to the ligands, however some hybrid orbitals formed by s and p orbitals of the ligand can be taken into account
According to Valence Bond theory, a sigma bond is formed by one lobe-one lobe overlap. Hence any orbital can form a sigma bond depending on the orientation of the other orbital participating in the sigma bond. Mo theory also allows any of the d orbitals to participate in sigma bond formation provided the symmetry of the combing orbitals match.
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