Cordination Compounds Essay

* FLEXIDENTATE CHARACTER OF POLYDENTATE LIGANDS* Poly dentate ligands present flexi dentate character in the smell out that all donors specks may not stress coordinate bonds with the central admixture atom or ion. For example EDTA which is the hexa dentate ligand alike acts as tetra dentate or penta dentate in certain complexes. Similarly sulfate ion, which is a bi dentate ligand, in addition acts as a mono dentate ligand in certain complexes, e.g. in Co(SO4)(NH3)5Cl. * labile COMPLEXES* A complex in which the ligands deal be easily replaced by former(a) ligands is called a labile complex. * GEOMETRY OF CuX42- IONS* The halide complex of Cu (2) shows two different stereo chemistries. In (NH4)2CuCl4, CuCl42- ion is squ atomic offspring 18 planar, but Cs2CuCl4 and Cs2CuBr4, the CuX42- ions have a slightly squashed tetrahedral shape. Tetrahedral CuCl42- ions atomic list 18 orange tree whereas, square planar CuCl42- ions are xanthous in colour. * GREATER STABILITY OF Co(+3) C OMPLEXES THAN Co(+2) COMPLEXES * Co+2 ions are very stalls & are difficult to oxidise. Co+3 ions are less s tabulate and are right away reduced by water to Co2+. In contrast Co(+2) are less stable and are readily oxidised to Co(+3) complexes, i.e. Co(3) are very stable. This is because CFSE of Co(+3) with d6 mannikin is higher than Co(+2) with d7 configuration. * SIDGWICK THEORY OR in force(p) ATOMIC NUMBER (EAN) RULE* Sidgwick put forward a rule to explain the stability of complexes on the basis of hard-hitting atomic number as follows- EAN of metal in complex= atomic number oxidation evoke+2 co-ordination number A stable complex is organize if the EAN is equal to the atomic number of the adjacent noble gas.Thus, from the calculation given in the table below, Fe(CN)64- is more stable than Fe(CN)63- though this rule is found to be applicable in many cases, yet it fails in no. Of cases as illustrated by final stage two examples given in the table belowcomplex Oxidation st ate Atomic number Co-ordination number EAN Co(NH3)63+ +3 27 6 27-3+26=36, i.e. KrFe(CN)64- +2 26 6 26-2+26=36, i.e. KrPtCl62- +4 78 6 78-4+26=86, i.e. RnFe(CN)63- +3 26 6 26-3+26=35Ag(NH3)21+ +1 47 2 47-1+22=50