# Octahedral, Tetrahedral or Square planar? [closed]

In cases where a metal ion like, Fe(III), Cr(III), Pt(II) is made to react with a ligand species like oxalate, CN-, CO etc and it forms a complex then, How do we determine whether 6 ligands will be coordinated to the central atom(i.e. in case of octahedral) or 4 ligands will be coordinated (as in tetrahedral or square planar arrangement) ?

• – Mithoron Mar 24 at 14:15

• the nature of the ligands ($$\pi$$-basic, $$\pi$$-acid, $$\sigma$$-donor and so on).
• Fe(III) has a d$^5$ electronic configuration and oxalate is a $\sigma$-donor ligand, then it will be an high spin complex, no net gain in energy. Anyway, remember for other cases that the octahedral field is always stronger than the tetrahedral one. In this specific case it is more convenient to form six ligands because there is not so much steric hinderance. But, for other metals other factors can play a role like the Jahn-Teller distorsions in Cu(II) ammonia complexes – TheOldJonny Mar 24 at 12:41