Preparing to optimise a Molecule of [NiCl₄]^2-

• In the quantum mechanics of the Schrödinger equation we need to know the atom types, but we also need to know the exact number of electrons and the spin state of those electrons. We need to set up the charge and the multiplicity for the calculation.
• Ni(0) has an electronic configuration of [Ar]3d⁸4s² but in our complex it is in a +2 oxidation state Ni(II) ([Ar]3d⁸4s⁰) and it is combined with four -1 ligands (halogens) and so the complex has an overall charge of -2, [NiCl₄]^2-
• Now d8 complexes are unstable in an octahedral environment and distort to either a square planar or tetrahedral geometry. Typically this is in part determined by the ligand size and electronic character. This gives rise to two different formal crystal field splitting patterns,

  

• It is important to recognise that this is just a starting point. The contribution of the ligand orbitals has been completely ignored and as you will see in this lab, the ligand orbitals are critically important. Ligand orbitals can act to reorder the energy levels and you can also have purely ligand based energy levels inter-dispersed among these formal "dAO" levels. Our halide ligands have pi orbitals and so significant MO contributions. You might want to revise some of the MOs in inorganic chemistry course at this point.
• At this stage we are not sure which is the more stable configuration, square planar or tetrahedral. So we need to evaluate the energy of both conformers. So far we have set up a calculation only for the tetrahedral option, we will need to come back and set up a separate calculation for the square planar option.
• For tetrahedral [NiCl₄]^2- the formal splitting pattern for the dAOs, is t above e (the reverse of the octahedral e_g above t_2g). Thus, the 8 d electrons will fill the e level completely and we will have 2 paired and 2 unpaired electrons in the t orbital set (since it costs energy to pair electrons, they would rather remain un-paired).

  

• In setting up a calculation we have to identify the spin state of the system, this is given by the multiplicity=(2S+1) where S is the total spin. Thus for no unpaired electrons the total spin S=0 and the multiplicity is (2.0+1)=1 a singlet state. For two unpaired electrons the total spin state is S=1/2+1/2=1 and the multiplicity is (2.1+1)=3 a triplet state.
• Therefore, for tetrahedral [NiCl₄]^2- the multiplicity is 3 and for square planar [NiCl₄]^2- the multiplicity is 1
• Now we are ready to move on to actually setting up the calculation and optimising your molecule.