YEAH I GUESS IT IS D X2-Y2 AND DZ2( LIKE SANDEE P SAID)... THIS IS BECAUSE WHEN A STRONG LIGAND APPROACHES(IF YOU HAVE NOT YET LEARNT THIS THEN PLEASE ASK) THE ORBITALS ALONG THE AXIS FACE MORE REPULSION AS A RESULT OF WHICH THE DEGENERACY IS DISTURBED AND THE THE ENERGIES OF THESE ORBITALS IS SLIGHTLY HIGHER .

for that use valence bond theory ie first write the configguration of the central ion or metal and the according to crystal field splitting and spectrochemical series check whether pairing will take place if so then pair up the electrons and  use the empty d orbitals and the next or previous s and p for hybridisation

 

 complexs are formed due to synergic effect!!!!

any further doubt, i would be pleased to answer!!!

Consider the simple example of TiCl₆^3- in which six chloride ions octahedrally surround the Ti³⁺ cation. There is only one d-electron to be allocated to one of the five d- orbitals. If it were to occupy the dz2 or dx2-y2 orbital, both of which meet the face of the cube and thus point directly towards the chloride ligands, it would be strongly repelled. The geometry of these orbitals and their nodes would require the electron to stay near the negatively charged ligands causing even more repulsion than a spherically distributed electron would experience. On the other hand, if the electron were to occupy the dxy, dyz or dxz orbital, it would spend less time near the ligands than would a spherically distributed electron and would be repelled less.

This difference between for example the dx2-y2 orbital, the dxy orbital and a spherical distribution can be graphically represented by