Question

How would the 3p orbitals split in an octahedral field? Briefly justify your answer.

Answer 1

Consider an octahedral complex in which each ligand has a single valence orbital directed toward the central metal atom. Each of these orbitals has local σ symmetry with respect to the M-L axis. Such ligands are termed “sigma donor ligands” and can be neutral, such as ammonia, NH3, or charged, such as fluoride, F- . In a strictly octahedral environment, the metal orbitals divide by symmetry into four sets, the orbitals or groups of metal orbitals that are labelled by a particular Mulliken symbol can only interact with a ligand set of orbitals that is described by the same label.

Metal Orbital Symmetry Label Degeneracy s a1g 1 px, py, pz t1u 3 dxy, dyx, dxz t2g 3 dx2-y2, dz2 eg 2

Molecular orbitals for the resulting complex are formed by combining SALCs and metal orbitals of the same symmetry type. Since we have six metal orbitals that can be broken down into three groups according to their symmetry (a1g, tiu and eg) and six ligand SALCs, these combine to form 6 bonding and 6 antibonding MOs. The metal t2g set remains non-bonding The resulting MO diagram can be populated with electrons according to the Aufbau process, Pauli principle and Hund’s rule. For the resulting energy levels, the greatest contribution to the molecular orbital of lowest energy is from atomic orbitals of lowest energy.the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs

The diagram below shows how the SALCs of the ligand orbital sets combine with the metal orbitals to produce molecular orbital energy levels (the MO diagram) of an octahedral ML6 complex. The t2g and eg* (which is now of antibonding symmetry with respect to the M-L bond) are empty suggesting the metal ion, M, has a d0 electron configuration. LFT explains very well the problem that Werner faced regarding the ability of a metal to bind numerous ligands (secondary valence). With several d orbitals (as well as s and p orbitals) available in the valence shell, enough delocalized MOs can be constructed to accommodate all the electrons needed to bind the ligands. Also, since lanthanides have f orbitals as well, they can accommodate even more ligand electrons and, therefore, can easily have coordination numbers higher than 6.

The antibonding π* orbitals of the ligand are vacant and are low enough in energy to undergo orbital overlap with the populated metal t2g set. Therefore, the metal “back-donates” this electron density into the π* orbitals.

Note: you may satify with this justification as well as a deep study also being require to understand concept.